Linear Algebra

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Linear Algebra 

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Jim Hefferon 

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Notation 

R real numbers 

N natural numbers: {0, 1, 2,...} 

C complex numbers 

{... � � ...} set of ... such that ... 

〈...〉 sequence; like a set but order matters 

V,W,U vector spaces 

�v, �w vectors 

�0, �0V zero vector, zero vector of V 

B,D bases 

En = 〈�e1, ... , �en〉 standard basis for R n 

�β, � δ basis vectors 

Rep B(�v) matrix representing the vector 

Pn set of n-th degree polynomials 

Mn×m set of n×m matrices 

[S] span of the set S 

M ⊕ N direct sum of subspaces 

V ∼ = W isomorphic spaces 

h, g homomorphisms 

H, G matrices 

t, s transformations; maps from a space to itself 

T,S square matrices 

Rep B,D(h) matrix representing the map h 

hi,j matrix entry from row i, column j 

|T | determinant of the matrix T 

R(h), N (h) rangespace and nullspace of the map h 

R∞(h), N∞(h) generalized rangespace and nullspace 

Lower case Greek alphabet 

name symbol name symbol name symbol 

alpha α iota ι rho ρ 

beta β kappa κ sigma σ 

gamma γ lambda λ tau τ 

delta δ mu µ upsilon υ 

epsilon ɛ nu ν phi φ 

zeta ζ xi ξ chi χ 

eta η omicron o psi ψ 

theta θ pi π omega ω 

Cover. This is Cramer’s Rule applied to the system x +2y =6,3x + y =8. Thearea 

of the first box is the determinant shown. The area of the second box is x times that, 

and equals the area of the final box. Hence, x is the final determinant divided by the 

first determinant.

Preface 

In most mathematics programs linear algebra is taken in the first or second 

year, following or along with at least one course in calculus. While the location 

of this course is stable, lately the content has been under discussion. Some instructors 

have experimented with varying the traditional topics, trying courses 

focused on applications, or on the computer. Despite this (entirely healthy) 

debate, most instructors are still convinced, I think, that the right core material 

is vector spaces, linear maps, determinants, and eigenvalues and eigenvectors. 

Applications and computations certainly can have a part to play but most mathematicians 

agree that the themes of the course should remain unchanged. 

Not that all is fine with the traditional course. Most of us do think that 

the standard text type for this course needs to be reexamined. Elementary 

texts have traditionally started with extensive computations of linear reduction, 

matrix multiplication, and determinants. These take up half of the course. 

Finally, when vector spaces and linear maps appear, and definitions and proofs 

start, the nature of the course takes a sudden turn. In the past, the computation 

drill was there because, as future practitioners, students needed to be fast and 

accurate with these. But that has changed. Being a whiz at 5×5 determinants 

just isn’t important anymore. Instead, the availability of computers gives us an 

opportunity to move toward a focus on concepts. 

This is an opportunity that we should seize. The courses at the start of 

most mathematics programs work at having students correctly apply formulas 

and algorithms, and imitate examples. Later courses require some mathematical 

maturity: reasoning skills that are developed enough to follow different types 

of proofs, a familiarity with the themes that underly many mathematical investigations 

like elementary set and function facts, and an ability to do some 

independent reading and thinking, Where do we work on the transition? 

Linear algebra is an ideal spot. It comes early in a program so that progress 

made here pays off later. The material is straightforward, elegant, and accessible. 

The students are serious about mathematics, often majors and minors. 

There are a variety of argument styles—proofs by contradiction, if and only if 

statements, and proofs by induction, for instance—and examples are plentiful. 

The goal of this text is, along with the development of undergraduate linear 

algebra, to help an instructor raise the students’ level of mathematical sophistication. 

Most of the differences between this book and others follow straight 

from that goal. 

One consequence of this goal of development is that, unlike in many computational 

texts, all of the results here are proved. On the other hand, in contrast 

with more abstract texts, many examples are given, and they are often quite 

detailed. 

Another consequence of the goal is that while we start with a computational 

topic, linear reduction, from the first we do more than just compute. The 

solution of linear systems is done quickly but it is also done completely, proving 

i

everything (really these proofs are just verifications), all the way through the 

uniqueness of reduced echelon form. In particular, in this first chapter, the 

opportunity is taken to present a few induction proofs, where the arguments 

just go over bookkeeping details, so that when induction is needed later (e.g., to 

prove that all bases of a finite dimensional vector space have the same number 

of members), it will be familiar. 

Still another consequence is that the second chapter immediately uses this 

background as motivation for the definition of a real vector space. This typically 

occurs by the end of the third week. We do not stop to introduce matrix 

multiplication and determinants as rote computations. Instead, those topics 

appear naturally in the development, after the definition of linear maps. 

To help students make the transition from earlier courses, the presentation 

here stresses motivation and naturalness. An example is the third chapter, 

on linear maps. It does not start with the definition of homomorphism, as 

is the case in other books, but with the definition of isomorphism. That’s 

because this definition is easily motivated by the observation that some spaces 

are just like each other. After that, the next section takes the reasonable step of 

defining homomorphisms by isolating the operation-preservation idea. A little 

mathematical slickness is lost, but it is in return for a large gain in sensibility 

to students. 

Having extensive motivation in the text helps with time pressures. I ask 

students to, before each class, look ahead in the book, and they follow the 

classwork better because they have some prior exposure to the material. For 

example, I can start the linear independence class with the definition because I 

know students have some idea of what it is about. No book can take the place 

of an instructor, but a helpful book gives the instructor more class time for 

examples and questions. 

Much of a student’s progress takes place while doing the exercises; the exercises 

here work with the rest of the text. Besides computations, there are many 

proofs. These are spread over an approachability range, from simple checks 

to some much more involved arguments. There are even a few exercises that 

are reasonably challenging puzzles taken, with citation, from various journals, 

competitions, or problems collections (as part of the fun of these, the original 

wording has been retained as much as possible). In total, the questions are 

aimed to both build an ability at, and help students experience the pleasure of, 

doing mathematics. 

Applications, and Computers. The point of view taken here, that linear 

algebra is about vector spaces and linear maps, is not taken to the exclusion 

of all other ideas. Applications, and the emerging role of the computer, are 

interesting, important, and vital aspects of the subject. Consequently, every 

chapter closes with a few application or computer-related topics. Some of the 

topics are: network flows, the speed and accuracy of computer linear reductions, 

Leontief Input/Output analysis, dimensional analysis, Markov chains, voting 

paradoxes, analytic projective geometry, and solving difference equations. 

These are brief enough to be done in a day’s class or to be given as indepen- 

ii

dent projects for individuals or small groups. Most simply give a reader a feel 

for the subject, discuss how linear algebra comes in, point to some accessible 

further reading, and give a few exercises. I have kept the exposition lively and 

given an overall sense of breadth of application. In short, these topics invite 

readers to see for themselves that linear algebra is a tool that a professional 

must have. 

For people reading this book on their own. The emphasis on motivation 

and development make this book a good choice for self-study. While a professional 

mathematician knows what pace and topics suit a class, perhaps an 

independent student would find some advice helpful. Here are two timetables 

for a semester. The first focuses on core material. 

week Mon. Wed. Fri. 

1 1.I.1 1.I.1, 2 1.I.2, 3 

2 1.I.3 1.II.1 1.II.2 

3 1.III.1, 2 1.III.2 2.I.1 

4 2.I.2 2.II 2.III.1 

5 2.III.1, 2 2.III.2 exam 

6 2.III.2, 3 2.III.3 3.I.1 

7 3.I.2 3.II.1 3.II.2 

8 3.II.2 3.II.2 3.III.1 

9 3.III.1 3.III.2 3.IV.1, 2 

10 3.IV.2, 3, 4 3.IV.4 exam 

11 3.IV.4, 3.V.1 3.V.1, 2 4.I.1, 2 

12 4.I.3 4.II 4.II 

13 4.III.1 5.I 5.II.1 

14 5.II.2 5.II.3 review 

The second timetable is more ambitious (it presupposes 1.II, the elements of 

vectors, usually covered in third semester calculus). 

week Mon. Wed. Fri. 

1 1.I.1 1.I.2 1.I.3 

2 1.I.3 1.III.1, 2 1.III.2 

3 2.I.1 2.I.2 2.II 

4 2.III.1 2.III.2 2.III.3 

5 2.III.4 3.I.1 exam 

6 3.I.2 3.II.1 3.II.2 

7 3.III.1 3.III.2 3.IV.1, 2 

8 3.IV.2 3.IV.3 3.IV.4 

9 3.V.1 3.V.2 3.VI.1 

10 3.VI.2 4.I.1 exam 

11 4.I.2 4.I.3 4.I.4 

12 4.II 4.II, 4.III.1 4.III.2, 3 

13 5.II.1, 2 5.II.3 5.III.1 

14 5.III.2 5.IV.1, 2 5.IV.2 

See the table of contents for the titles of these subsections. 

iii

For guidance, in the table of contents I have marked some subsections as 

optional if, in my opinion, some instructors will pass over them in favor of 

spending more time elsewhere. These subsections can be dropped or added, as 

desired. You might also adjust the length of your study by picking one or two 

Topics that appeal to you from the end of each chapter. You’ll probably get 

more out of these if you have access to computer software that can do the big 

calculations. 

Do many exercises. (The answers are available.) I have marked a good sample 

with �’s. Be warned about the exercises, however, that few inexperienced 

people can write correct proofs. Try to find a knowledgeable person to work 

with you on this aspect of the material. 

Finally, if I may, a caution: I cannot overemphasize how much the statement 

(which I sometimes hear), “I understand the material, but it’s only that I can’t 

do any of the problems.” reveals a lack of understanding of what we are up 

to. Being able to do particular things with the ideas is the entire point. The 

quote below expresses this sentiment admirably, and captures the essence of 

this book’s approach. It states what I believe is the key to both the beauty and 

the power of mathematics and the sciences in general, and of linear algebra in 

particular. 

I know of no better tactic 

than the illustration of exciting principles 

by well-chosen particulars. 

–Stephen Jay Gould 

Jim Hefferon 

Saint Michael’s College 

Colchester, Vermont USA 

jim@joshua.smcvt.edu 

April 20, 2000 

Author’s Note. Inventing a good exercise, one that enlightens as well as tests, 

is a creative act, and hard work (at least half of the the effort on this text 

has gone into exercises and solutions). The inventor deserves recognition. But, 

somehow, the tradition in texts has been to not give attributions for questions. 

I have changed that here where I was sure of the source. I would greatly appreciate 

hearing from anyone who can help me to correctly attribute others of the 

questions. They will be incorporated into later versions of this book. 

iv

Contents 

1 Linear Systems 1 

1.I Solving Linear Systems ........................ 1 

1.I.1 Gauss’ Method ........................... 2 

1.I.2 Describing the Solution Set .................... 11 

1.I.3 General = Particular + Homogeneous .............. 20 

1.II Linear Geometry of n-Space ...................... 32 

1.II.1 Vectors in Space .......................... 32 

1.II.2 Length and Angle Measures ∗ ................... 38 

1.III Reduced Echelon Form ........................ 45 

1.III.1 Gauss-Jordan Reduction ...................... 45 

1.III.2 Row Equivalence .......................... 51 

Topic: Computer Algebra Systems ..................... 61 

Topic: Input-Output Analysis ....................... 63 

Topic: Accuracy of Computations ..................... 67 

Topic: Analyzing Networks ......................... 72 

2 Vector Spaces 79 

2.I Definition of Vector Space ....................... 80 

2.I.1 Definition and Examples ...................... 80 

2.I.2 Subspaces and Spanning Sets ................... 91 

2.II Linear Independence ..........................102 

2.II.1 Definition and Examples ......................102 

2.III Basis and Dimension ..........................113 

2.III.1 Basis .................................113 

2.III.2 Dimension ..............................119 

2.III.3 Vector Spaces and Linear Systems ................124 

2.III.4 Combining Subspaces ∗ .......................131 

Topic: Fields .................................141 

Topic: Crystals ................................143 

Topic: Voting Paradoxes ..........................147 

Topic: Dimensional Analysis ........................152 

v

3 Maps Between Spaces 159 

3.I Isomorphisms ..............................159 

3.I.1 Definition and Examples ......................159 

3.I.2 Dimension Characterizes Isomorphism ..............169 

3.II Homomorphisms ............................176 

3.II.1 Definition ..............................176 

3.II.2 Rangespace and Nullspace .....................184 

3.III Computing Linear Maps ........................194 

3.III.1 Representing Linear Maps with Matrices ............194 

3.III.2 Any Matrix Represents a Linear Map ∗ ..............204 

3.IV Matrix Operations ...........................211 

3.IV.1 Sums and Scalar Products .....................211 

3.IV.2 Matrix Multiplication .......................214 

3.IV.3 Mechanics of Matrix Multiplication ................221 

3.IV.4 Inverses ...............................230 

3.V Change of Basis ............................238 

3.V.1 Changing Representations of Vectors ...............238 

3.V.2 Changing Map Representations ..................242 

3.VI Projection ................................250 

3.VI.1 Orthogonal Projection Into a Line ∗ ................250 

3.VI.2 Gram-Schmidt Orthogonalization ∗ ................255 

3.VI.3 Projection Into a Subspace ∗ ....................260 

Topic: Line of Best Fit ...........................269 

Topic: Geometry of Linear Maps ......................274 

Topic: Markov Chains ............................280 

Topic: Orthonormal Matrices ........................286 

4 Determinants 293 

4.I Definition ................................294 

4.I.1 Exploration ∗ ............................294 

4.I.2 Properties of Determinants ....................299 

4.I.3 The Permutation Expansion ....................303 

4.I.4 Determinants Exist ∗ ........................312 

4.II Geometry of Determinants ......................319 

4.II.1 Determinants as Size Functions ..................319 

4.III Other Formulas .............................326 

4.III.1 Laplace’s Expansion ∗ .......................326 

Topic: Cramer’s Rule ............................331 

Topic: Speed of Calculating Determinants .................334 

Topic: Projective Geometry .........................337 

5 Similarity 347 

5.I Complex Vector Spaces ........................347 

5.I.1 Factoring and Complex Numbers; A Review ∗ ..........348 

5.I.2 Complex Representations .....................350 

5.II Similarity ................................351 

vi

5.II.1 Definition and Examples ......................351 

5.II.2 Diagonalizability ..........................353 

5.II.3 Eigenvalues and Eigenvectors ...................357 

5.III Nilpotence ...............................365 

5.III.1 Self-Composition ∗ .........................365 

5.III.2 Strings ∗ ...............................368 

5.IV Jordan Form ..............................379 

5.IV.1 Polynomials of Maps and Matrices ∗ ...............379 

5.IV.2 Jordan Canonical Form ∗ ......................386 

Topic: Computing Eigenvalues—the Method of Powers .........399 

Topic: Stable Populations ..........................403 

Topic: Linear Recurrences .........................405 

Appendix A-1 

Introduction .................................A-1 

Propositions .................................A-1 

Quantifiers .................................A-3 

Techniques of Proof ............................A-5 

Sets, Functions, and Relations .......................A-6 

∗ Note: starred subsections are optional. 

vii

Chapter 1 

Linear Systems 

1.I Solving Linear Systems 

Systems of linear equations are common in science and mathematics. These two 

examples from high school science [Onan] give a sense of how they arise. 

The first example is from Physics. Suppose that we are given three objects, 

one with a mass of 2 kg, and are asked to find the unknown masses. Suppose 

further that experimentation with a meter stick produces these two balances. 

h 

40 50 

c 

15 

2 

c 

25 50 

Now, since the sum of moments on the left of each balance equals the sum of 

moments on the right (the moment of an object is its mass times its distance 

from the balance point), the two balances give this system of two equations. 

40h +15c = 100 

25c =50+50h 

The second example of a linear system is from Chemistry. We can mix, 

under controlled conditions, toluene C7H8 and nitric acid HNO3 to produce 

trinitrotoluene C7H5O6N3 along with the byproduct water (conditions have to 

be controlled very well, indeed — trinitrotoluene is better known as TNT). In 

what proportion should those components be mixed? The number of atoms of 

each element present before the reaction 

x C7H8 + y HNO3 −→ z C7H5O6N3 + w H2O 

must equal the number present afterward. Applying that principle to the elements 

C, H, N, and O in turn gives this system. 

7x =7z 

8x +1y =5z +2w 

1y =3z 

3y =6z +1w 

1 

25 

2 

h

2 Chapter 1. Linear Systems 

To finish each of these examples requires solving a system of equations. In 

each, the equations involve only the first power of the variables. This chapter 

shows how to solve any such system. 

1.I.1 Gauss’ Method 

1.1 Definition A linear equation in variables x1,x2,... ,xn has the form 

a1x1 + a2x2 + a3x3 + ···+ anxn = d 

where the numbers a1,... ,an ∈ R are the equation’s coefficients and d ∈ R is 

the constant. Ann-tuple (s1,s2,... ,sn) ∈ R n is a solution of, or satisfies, that 

equation if substituting the numbers s1, ... , sn for the variables gives a true 

statement: a1s1 + a2s2 + ...+ ansn = d. 

A system of linear equations 

a1,1x1 + a1,2x2 + ···+ a1,nxn = d1 

a2,1x1 + a2,2x2 + ···+ a2,nxn = d2 

. 

am,1x1 + am,2x2 + ···+ am,nxn = dm 

has the solution (s1,s2,... ,sn) if that n-tuple is a solution of all of the equations 

in the system. 

1.2 Example The ordered pair (−1, 5) is a solution of this system. 

In contrast, (5, −1) is not a solution. 

3x1 +2x2 =7 

−x1 + x2 =6 

Finding the set of all solutions is solving the system. No guesswork or good 

fortune is needed to solve a linear system. There is an algorithm that always 

works. The next example introduces that algorithm, called Gauss’ method. It 

transforms the system, step by step, into one with a form that is easily solved. 

1.3 Example To solve this system 

3x3 =9 

x1 +5x2 − 2x3 =2 

1 

3 x1 +2x2 

=3

Section I. Solving Linear Systems 3 

we repeatedly transform it until it is in a form that is easy to solve. 

swap row 1 with row 3 

−→ 

multiply row 1 by 3 

−→ 

add −1 times row 1 to row 2 

−→ 

1 

3x1 +2x2 

=3 

x1 +5x2 − 2x3 =2 

3x3 =9 

x1 +6x2 =9 

x1 +5x2 − 2x3 =2 

3x3 =9 

x1 + 6x2 = 9 

−x2 − 2x3 = −7 

3x3 = 9 

The third step is the only nontrivial one. We’ve mentally multiplied both sides 

of the first row by −1, mentally added that to the old second row, and written 

the result in as the new second row. 

Now we can find the value of each variable. The bottom equation shows 

that x3 = 3. Substituting 3 for x3 in the middle equation shows that x2 =1. 

Substituting those two into the top equation gives that x1 = 3 and so the system 

has a unique solution: the solution set is { (3, 1, 3) }. 

Most of this subsection and the next one consists of examples of solving 

linear systems by Gauss’ method. We will use it throughout this book. It is 

fast and easy. But, before we get to those examples, we will first show that 

this method is also safe in that it never loses solutions or picks up extraneous 

solutions. 

1.4 Theorem (Gauss’ method) If a linear system is changed to another by 

one of these operations 

(1) an equation is swapped with another 

(2) an equation has both sides multiplied by a nonzero constant 

(3) an equation is replaced by the sum of itself and a multiple of another 

then the two systems have the same set of solutions. 

Each of those three operations has a restriction. Multiplying a row by 0 is 

not allowed because obviously that can change the solution set of the system. 

Similarly, adding a multiple of a row to itself is not allowed because adding −1 

times the row to itself has the effect of multiplying the row by 0. Finally, swapping 

a row with itself is disallowed to make some results in the fourth chapter 

easier to state and remember (and besides, self-swapping doesn’t accomplish 

anything). 

Proof. We will cover the equation swap operation here and save the other two 

cases for Exercise 29.


Consider this swap of row i with row j. 

a1,1x1 + a1,2x2 + ··· a1,nxn = d1 a1,1x1 + a1,2x2 + ··· a1,nxn = d1 

. 

. 

. 

. 

ai,1x1 + ai,2x2 + ··· ai,nxn = di aj,1x1 + aj,2x2 + ··· aj,nxn = dj 

. −→ 

. 

aj,1x1 + aj,2x2 + ··· aj,nxn = dj ai,1x1 + ai,2x2 + ··· ai,nxn = di 

. 

. 

. 

. 

am,1x1 + am,2x2 + ··· am,nxn = dm am,1x1 + am,2x2 + ··· am,nxn = dm 

The n-tuple (s1,... ,sn) satisfies the system before the swap if and only if 

substituting the values, the s’s, for the variables, the x’s, gives true statements: 

a1,1s1+a1,2s2+···+a1,nsn = d1 and ... ai,1s1+ai,2s2+···+ai,nsn = di and ... 

aj,1s1 + aj,2s2 + ···+ aj,nsn = dj and ... am,1s1 + am,2s2 + ···+ am,nsn = dm. 

In a requirement consisting of statements and-ed together we can rearrange 

the order of the statements, so that this requirement is met if and only if a1,1s1+ 

a1,2s2 + ···+ a1,nsn = d1 and ... aj,1s1 + aj,2s2 + ···+ aj,nsn = dj and ... 

ai,1s1 + ai,2s2 + ···+ ai,nsn = di and ... am,1s1 + am,2s2 + ···+ am,nsn = dm. 

This is exactly the requirement that (s1,... ,sn) solves the system after the row 

swap. QED 

1.5 Definition The three operations from Theorem 1.4 are the elementary reduction 

operations, orrow operations, orGaussian operations. They are swapping, 

multiplying by a scalar or rescaling, andpivoting. 

When writing out the calculations, we will abbreviate ‘row i’ by‘ρi’. For 

instance, we will denote a pivot operation by kρi + ρj, with the row that is 

changed written second. We will also, to save writing, often list pivot steps 

together when they use the same ρi. 

1.6 Example A typical use of Gauss’ method is to solve this system. 

x + y =0 

2x − y +3z =3 

x − 2y − z =3 

The first transformation of the system involves using the first row to eliminate 

the x in the second row and the x in the third. To get rid of the second row’s 

2x, we multiply the entire first row by −2, add that to the second row, and 

write the result in as the new second row. To get rid of the third row’s x, we 

multiply the first row by −1, add that to the third row, and write the result in 

as the new third row. 

−ρ1+ρ3 

−→ 

−2ρ1+ρ2 

x + y =0 

−3y +3z =3 

−3y − z =3 

(Note that the two ρ1 steps −2ρ1 + ρ2 and −ρ1 + ρ3 are written as one operation.) 

In this second system, the last two equations involve only two unknowns.


To finish we transform the second system into a third system, where the last 

equation involves only one unknown. This transformation uses the second row 

to eliminate y from the third row. 

−ρ2+ρ3 

−→ 

x + y =0 

−3y + 3z =3 

−4z =0 

Now we are set up for the solution. The third row shows that z = 0. Substitute 

that back into the second row to get y = −1, and then substitute back into the 

firstrowtogetx =1. 

1.7 Example For the Physics problem from the start of this chapter, Gauss’ 

method gives this. 

40h +15c = 100 

−50h +25c = 50 

5/4ρ1+ρ2 

−→ 

40h + 15c = 100 

(175/4)c = 175 

So c = 4, and back-substitution gives that h = 1. (The Chemistry problem is 

solved later.) 

1.8 Example The reduction 

x + y + z =9 

2x +4y − 3z =1 

3x +6y − 5z =0 

shows that z =3,y = −1, and x =7. 

−2ρ1+ρ2 

−→ 

−3ρ1+ρ3 

−(3/2)ρ2+ρ3 

−→ 

x + y + z = 9 

2y − 5z = −17 

3y − 8z = −27 

x + y + z = 9 

2y − 5z = −17 

− 1 3 

2z = − 2 

As these examples illustrate, Gauss’ method uses the elementary reduction 

operations to set up back-substitution. 

1.9 Definition In each row, the first variable with a nonzero coefficient is the 

row’s leading variable. A system is in echelon form if each leading variable is 

to the right of the leading variable in the row above it (except for the leading 

variable in the first row). 

1.10 Example The only operation needed in the examples above is pivoting. 

Here is a linear system that requires the operation of swapping equations. After 

the first pivot 

x − y =0 

2x − 2y + z +2w =4 

y + w =0 

2z + w =5 

−2ρ1+ρ2 

−→ 

x − y =0 

z +2w =4 

y + w =0 

2z + w =5


the second equation has no leading y. To get one, we look lower down in the 

system for a row that has a leading y and swap it in. 

ρ2↔ρ3 

−→ 

x − y =0 

y + w =0 

z +2w =4 

2z + w =5 

(Had there been more than one row below the second with a leading y then we 

could have swapped in any one.) The rest of Gauss’ method goes as before. 

−2ρ3+ρ4 

−→ 

x − y = 0 

y + w = 0 

z + 2w = 4 

−3w = −3 

Back-substitution gives w =1,z =2,y = −1, and x = −1. 

Strictly speaking, the operation of rescaling rows is not needed to solve linear 

systems. We have included it because we will use it later in this chapter as part 

of a variation on Gauss’ method, the Gauss-Jordan method. 

All of the systems seen so far have the same number of equations as unknowns. 

All of them have a solution, and for all of them there is only one 

solution. We finish this subsection by seeing for contrast some other things that 

can happen. 

1.11 Example Linear systems need not have the same number of equations 

as unknowns. This system 

x +3y = 1 

2x + y = −3 

2x +2y = −2 

has more equations than variables. Gauss’ method helps us understand this 

system also, since this 

−2ρ1+ρ2 

−→ 

−2ρ1+ρ3 

x + 3y = 1 

−5y = −5 

−4y = −4 

shows that one of the equations is redundant. Echelon form 

−(4/5)ρ2+ρ3 

−→ 

x + 3y = 1 

−5y = −5 

0= 0 

gives y =1andx = −2. The ‘0 = 0’ is derived from the redundancy.


That example’s system has more equations than variables. Gauss’ method 

is also useful on systems with more variables than equations. Many examples 

are in the next subsection. 

Another way that linear systems can differ from the examples shown earlier 

is that some linear systems do not have a unique solution. This can happen in 

two ways. 

The first is that it can fail to have any solution at all. 

1.12 Example Contrast the system in the last example with this one. 

x +3y = 1 

2x + y = −3 

2x +2y = 0 

−2ρ1+ρ2 

−→ 

−2ρ1+ρ3 

x + 3y = 1 

−5y = −5 

−4y = −2 

Here the system is inconsistent: no pair of numbers satisfies all of the equations 

simultaneously. Echelon form makes this inconsistency obvious. 

The solution set is empty. 

−(4/5)ρ2+ρ3 

−→ 

x + 3y = 1 

−5y = −5 

0= 2 

1.13 Example The prior system has more equations than unknowns, but that 

is not what causes the inconsistency — Example 1.11 has more equations than 

unknowns and yet is consistent. Nor is having more equations than unknowns 

necessary for inconsistency, as is illustrated by this inconsistent system with the 

same number of equations as unknowns. 

x +2y =8 

2x +4y =8 

−2ρ1+ρ2 

−→ 

x +2y = 8 

0=−8 

The other way that a linear system can fail to have a unique solution is to 

have many solutions. 

1.14 Example In this system 

x + y =4 

2x +2y =8 

any pair of numbers satisfying the first equation automatically satisfies the second. 

The solution set {(x, y) � � x + y =4} is infinite — some of its members 

are (0, 4), (−1, 5), and (2.5, 1.5). The result of applying Gauss’ method here 

contrasts with the prior example because we do not get a contradictory equation. 

−2ρ1+ρ2 

−→ 

x + y =4 

0=0


Don’t be fooled by the ‘0 = 0’ equation in that example. It is not the signal 

that a system has many solutions. 

1.15 Example The absence of a ‘0 = 0’ does not keep a system from having 

many different solutions. This system is in echelon form 

x + y + z =0 

y + z =0 

has no ‘0 = 0’, and yet has infinitely many solutions. (For instance, each of 

these is a solution: (0, 1, −1), (0, 1/2, −1/2), (0, 0, 0), and (0, −π, π). There are 

infinitely many solutions because any triple whose first component is 0 and 

whose second component is the negative of the third is a solution.) 

Nor does the presence of a ‘0 = 0’ mean that the system must have many 

solutions. Example 1.11 shows that. So does this system, which does not have 

many solutions — in fact it has none — despite that when it is brought to 

echelon form it has a ‘0 = 0’ row. 

2x − 2z =6 

y + z =1 

2x + y − z =7 

3y +3z =0 

−ρ1+ρ3 

−→ 

−ρ2+ρ3 

−→ 

−3ρ2+ρ4 

2x − 2z =6 

y + z =1 

y + z =1 

3y +3z =0 

2x − 2z = 6 

y + z = 1 

0= 0 

0=−3 

We will finish this subsection with a summary of what we’ve seen so far 

about Gauss’ method. 

Gauss’ method uses the three row operations to set a system up for back 

substitution. If any step shows a contradictory equation then we can stop 

with the conclusion that the system has no solutions. If we reach echelon form 

without a contradictory equation, and each variable is a leading variable in its 

row, then the system has a unique solution and we find it by back substitution. 

Finally, if we reach echelon form without a contradictory equation, and there is 

not a unique solution (at least one variable is not a leading variable) then the 

system has many solutions. 

The next subsection deals with the third case — we will see how to describe 

the solution set of a system with many solutions. 

Exercises 

� 1.16 Use Gauss’ method to find the unique solution for each system. 

x − z =0 

2x +3y = 13 

(a) (b) 3x + y =1 

x − y = −1 

−x + y + z =4 

� 1.17 Use Gauss’ method to solve each system or conclude ‘many solutions’ or ‘no 

solutions’.


(a) 2x +2y =5 (b) −x + y =1 (c) x − 3y + z = 1 

x − 4y =0 x + y =2 x + y +2z =14 

(d) −x − y =1 (e) 4y + z =20 (f) 2x + z + w = 5 

−3x − 3y =2 2x − 2y + z = 0 

y − w = −1 

x + z = 5 3x − z − w = 0 

x + y − z =10 4x + y +2z + w = 9 

� 1.18 There are methods for solving linear systems other than Gauss’ method. One 

often taught in high school is to solve one of the equations for a variable, then 

substitute the resulting expression into other equations. That step is repeated 

until there is an equation with only one variable. From that, the first number in 

the solution is derived, and then back-substitution can be done. This method both 

takes longer than Gauss’ method, since it involves more arithmetic operations and 

is more likely to lead to errors. To illustrate how it can lead to wrong conclusions, 

we will use the system 

x +3y = 1 

2x + y = −3 

2x +2y = 0 

from Example 1.12. 

(a) Solve the first equation for x and substitute that expression into the second 

equation. Find the resulting y. 

(b) Again solve the first equation for x, but this time substitute that expression 

into the third equation. Find this y. 

What extra step must a user of this method take to avoid erroneously concluding 

a system has a solution? 

� 1.19 For which values of k are there no solutions, many solutions, or a unique 

solution to this system? 

x − y =1 

3x − 3y = k 

� 1.20 This system is not linear: 

2sinα− cos β +3tanγ = 3 

4sinα +2cosβ−2tanγ =10 

6sinα− 3cosβ + tanγ = 9 

but we can nonetheless apply Gauss’ method. 

solution? 

Do so. Does the system have a 

� 1.21 What conditions must the constants, the b’s, satisfy so that each of these 

systems has a solution? Hint. Apply Gauss’ method and see what happens to the 

right side. 

(a) x − 3y = b1 (b) x1 +2x2 +3x3 = b1 

3x + y = b2 2x1 +5x2 +3x3 = b2 

x +7y = b3 

2x +4y = b4 

x1 +8x3 = b3 

1.22 True or false: a system with more unknowns than equations has at least one 

solution. (As always, to say ‘true’ you must prove it, while to say ‘false’ you must 

produce a counterexample.) 

1.23 Must any Chemistry problem like the one that starts this subsection — a 

balance the reaction problem — have infinitely many solutions? 

� 1.24 Find the coefficients a, b, andcso that the graph of f(x) =ax 2 +bx+c passes 

through the points (1, 2), (−1, 6), and (2, 3).


1.25 Gauss’ method works by combining the equations in a system to make new 

equations. 

(a) Can the equation 3x−2y = 5 be derived, by a sequence of Gaussian reduction 

steps, from the equations in this system? 

x + y =1 

4x − y =6 

(b) Can the equation 5x−3y = 2 be derived, by a sequence of Gaussian reduction 

steps, from the equations in this system? 

2x +2y =5 

3x + y =4 

(c) Can the equation 6x − 9y +5z = −2 be derived, by a sequence of Gaussian 

reduction steps, from the equations in the system? 

2x + y − z =4 

6x − 3y + z =5 

1.26 Prove that, where a,b,... ,e are real numbers and a �= 0,if 

ax + by = c 

has the same solution set as 

ax + dy = e 

then they are the same equation. What if a =0? 

� 1.27 Show that if ad − bc �= 0then 

ax + by = j 

cx + dy = k 

has a unique solution. 

� 1.28 In the system 

ax + by = c 

dx + ey = f 

each of the equations describes a line in the xy-plane. By geometrical reasoning, 

show that there are three possibilities: there is a unique solution, there is no 

solution, and there are infinitely many solutions. 

1.29 Finish the proof of Theorem 1.4. 

1.30 Is there a two-unknowns linear system whose solution set is all of R 2 ? 

� 1.31 Are any of the operations used in Gauss’ method redundant? That is, can 

any of the operations be synthesized from the others? 

1.32 Prove that each operation of Gauss’ method is reversible. That is, show that if 

two systems are related by a row operation S1 ↔ S2 then there is a row operation 

to go back S2 ↔ S1. 

1.33 A box holding pennies, nickels and dimes contains thirteen coins with a total 

value of 83 cents. How many coins of each type are in the box? 

1.34 [Con. Prob. 1955] Four positive integers are given. Select any three of the 

integers, find their arithmetic average, and add this result to the fourth integer. 

Thus the numbers 29, 23, 21, and 17 are obtained. One of the original integers 

is:


(a) 19 (b) 21 (c) 23 (d) 29 (e) 17 

� 1.35 [Am. Math. Mon., Jan. 1935] Laugh at this: AHAHA + TEHE = TEHAW. 

It resulted from substituting a code letter for each digit of a simple example in 

addition, and it is required to identify the letters and prove the solution unique. 

1.36 [Wohascum no. 2] The Wohascum County Board of Commissioners, which has 

20 members, recently had to elect a President. There were three candidates (A, B, 

and C); on each ballot the three candidates were to be listed in order of preference, 

with no abstentions. It was found that 11 members, a majority, preferred A over 

B (thus the other 9 preferred B over A). Similarly, it was found that 12 members 

preferred C over A. Given these results, it was suggested that B should withdraw, 

to enable a runoff election between A and C. However, B protested, and it was 

then found that 14 members preferred B over C! The Board has not yet recovered 

from the resulting confusion. Given that every possible order of A, B, C appeared 

on at least one ballot, how many members voted for B as their first choice? 

1.37 [Am. Math. Mon., Jan. 1963] “This system of n linear equations with n unknowns,” 

said the Great Mathematician, “has a curious property.” 

“Good heavens!” said the Poor Nut, “What is it?” 

“Note,” said the Great Mathematician, “that the constants are in arithmetic 

progression.” 

“It’s all so clear when you explain it!” said the Poor Nut. “Do you mean like 

6x +9y =12and15x +18y = 21?” 

“Quite so,” said the Great Mathematician, pulling out his bassoon. “Indeed, 

the system has a unique solution. Can you find it?” 

“Good heavens!” cried the Poor Nut, “I am baffled.” 

Are you? 

1.I.2 Describing the Solution Set 

A linear system with a unique solution has a solution set with one element. 

A linear system with no solution has a solution set that is empty. In these cases 

the solution set is easy to describe. Solution sets are a challenge to describe 

only when they contain many elements. 

2.1 Example This system has many solutions because in echelon form 

2x + z =3 

x − y − z =1 

3x − y =4 

−(1/2)ρ1+ρ2 

−→ 

−(3/2)ρ1+ρ3 

−ρ2+ρ3 

−→ 

2x + z = 3 

−y − (3/2)z = −1/2 

−y − (3/2)z = −1/2 

2x + z = 3 

−y − (3/2)z = −1/2 

0= 0 

not all of the variables are leading variables. The Gauss’ method theorem 

showed that a triple satisfies the first system if and only if it satisfies the third. 

Thus, the solution set {(x, y, z) � � 2x + z =3andx − y − z =1and3x − y =4}


can also be described as {(x, y, z) � � 2x + z =3and−y − 3z/2 =−1/2}. However, 

this second description is not much of an improvement. It has two equations 

instead of three, but it still involves some hard-to-understand interaction 

among the variables. 

To get a description that is free of any such interaction, we take the variable 

that does not lead any equation, z, and use it to describe the variables 

that do lead, x and y. The second equation gives y = (1/2) − (3/2)z and 

the first equation gives x =(3/2) − (1/2)z. Thus, the solution set can be described 

as {(x, y, z) = ((3/2) − (1/2)z,(1/2) − (3/2)z,z) � � z ∈ R}. For instance, 

(1/2, −5/2, 2) is a solution because taking z = 2 gives a first component of 1/2 

and a second component of −5/2. 

The advantage of this description over the ones above is that the only variable 

appearing, z, is unrestricted — it can be any real number. 

2.2 Definition The non-leading variables in an echelon-form linear system are 

free variables. 

In the echelon form system derived in the above example, x and y are leading 

variables and z is free. 

2.3 Example A linear system can end with more than one variable free. This 

row reduction 

x + y + z − w = 1 

y − z + w = −1 

3x +6z − 6w = 6 

−y + z − w = 1 

−3ρ1+ρ3 

−→ 

3ρ2+ρ3 

−→ 

ρ2+ρ4 

x + y + z − w = 1 

y − z + w = −1 

−3y +3z − 3w = 3 

−y + z − w = 1 

x + y + z − w = 1 

y − z + w = −1 

0= 0 

0= 0 

ends with x and y leading, and with both z and w free. To get the description 

that we prefer we will start at the bottom. We first express y in terms of 

the free variables z and w with y = −1 +z − w. Next, moving up to the 

top equation, substituting for y in the first equation x +(−1 +z − w) +z − 

w = 1 and solving for x yields x =2− 2z +2w. Thus, the solution set is 

{2 − 2z +2w, −1+z − w, z, w) � � z,w ∈ R}. 

We prefer this description because the only variables that appear, z and w, 

are unrestricted. This makes the job of deciding which four-tuples are system 

solutions into an easy one. For instance, taking z =1andw = 2 gives the 

solution (4, −2, 1, 2). In contrast, (3, −2, 1, 2) is not a solution, since the first 

component of any solution must be 2 minus twice the third component plus 

twice the fourth.


2.4 Example After this reduction 

2x − 2y =0 

z +3w =2 

3x − 3y =0 

x − y +2z +6w =4 

−(3/2)ρ1+ρ3 

−→ 

−(1/2)ρ1+ρ4 

−2ρ2+ρ4 

−→ 

2x − 2y =0 

z +3w =2 

0=0 

2z +6w =4 

2x − 2y =0 

z +3w =2 

0=0 

0=0 

x and z lead, y and w are free. The solution set is {(y, y, 2 − 3w, w) � � y, w ∈ R}. 

For instance, (1, 1, 2, 0) satisfies the system — take y =1andw =0. The 

four-tuple (1, 0, 5, 4) is not a solution since its first coordinate does not equal its 

second. 

We refer to a variable used to describe a family of solutions as a parameter 

and we say that the set above is paramatrized with y and w. (The terms 

‘parameter’ and ‘free variable’ do not mean the same thing. Above, y and w 

are free because in the echelon form system they do not lead any row. They 

are parameters because they are used in the solution set description. We could 

have instead paramatrized with y and z by rewriting the second equation as 

w =2/3− (1/3)z. In that case, the free variables are still y and w, but the 

parameters are y and z. Notice that we could not have paramatrized with x and 

y, so there is sometimes a restriction on the choice of parameters. The terms 

‘parameter’ and ‘free’ are related because, as we shall show later in this chapter, 

the solution set of a system can always be paramatrized with the free variables. 

Consequenlty, we shall paramatrize all of our descriptions in this way.) 

2.5 Example This is another system with infinitely many solutions. 

x +2y =1 

2x + z =2 

3x +2y + z − w =4 

−2ρ1+ρ2 

−→ 

−3ρ1+ρ3 

−ρ2+ρ3 

−→ 

x + 2y =1 

−4y + z =0 

−4y + z − w =1 

x + 2y =1 

−4y + z =0 

−w =1 

The leading variables are x, y, andw. The variable z is free. (Notice here that, 

although there are infinitely many solutions, the value of one of the variables is 

fixed — w = −1.) Write w in terms of z with w = −1+0z. Then y =(1/4)z. 

To express x in terms of z, substitute for y into the first equation to get x = 

1 − (1/2)z. The solution set is {(1 − (1/2)z,(1/4)z,z,−1) � � z ∈ R}. 

We finish this subsection by developing the notation for linear systems and 

their solution sets that we shall use in the rest of this book. 

2.6 Definition An m×n matrix is a rectangular array of numbers with m rows 

and n columns. Each number in the matrix is an entry,


Matrices are usually named by upper case roman letters, e.g. A. Each entry is 

denoted by the corresponding lower-case letter, e.g. ai,j is the number in row i 

and column j of the array. For instance, 

A = 

� 

1 2.2 

� 

5 

3 4 −7 

has two rows and three columns, and so is a 2×3 matrix. (Read that “twoby-three”; 

the number of rows is always stated first.) The entry in the second 

row and first column is a2,1 = 3. Note that the order of the subscripts matters: 

a1,2 �= a2,1 since a1,2 =2.2. (The parentheses around the array are a typographic 

device so that when two matrices are side by side we can tell where one 

ends and the other starts.) 

2.7 Example We can abbreviate this linear system 

with this matrix. 

x1 +2x2 =4 

x2 − x3 =0 

+2x3 =4 

x1 

⎛ 

1 

⎝0 

2 

1 

0 

−1 

⎞ 

4 

0⎠ 

1 0 2 4 

The vertical bar just reminds a reader of the difference between the coefficients 

on the systems’s left hand side and the constants on the right. When a bar 

is used to divide a matrix into parts, we call it an augmented matrix. In this 

notation, Gauss’ method goes this way. 

⎛ 

⎞ 

1 2 0 4 

⎝0 

1 −1 0⎠ 

1 0 2 4 

−ρ1+ρ3 

⎛ 

⎞ 

1 2 0 4 

−→ ⎝0 

1 −1 0⎠ 

0 −2 2 0 

2ρ2+ρ3 

⎛ 

⎞ 

1 2 0 4 

−→ ⎝0 

1 −1 0⎠ 

0 0 0 0 

The second row stands for y − z = 0 and the first row stands for x +2y =4so 

the solution set is {(4 − 2z,z,z) � � z ∈ R}. One advantage of the new notation is 

that the clerical load of Gauss’ method — the copying of variables, the writing 

of +’s and =’s, etc. — is lighter. 

We will also use the array notation to clarify the descriptions of solution 

sets. A description like {(2 − 2z +2w, −1+z − w, z, w) � � z,w ∈ R} from Example 

2.3 is hard to read. We will rewrite it to group all the constants together, 

all the coefficients of z together, and all the coefficients of w together. We will 

write them vertically, in one-column wide matrices. 

⎛ ⎞ 

2 

⎜ 

{ ⎜−1 

⎟ 

⎝ 0 ⎠ 

0 

+ 

⎛ ⎞ ⎛ ⎞ 

−2 2 

⎜ 1 ⎟ ⎜ 

⎟ 

⎝ 1 ⎠ · z + ⎜−1 

⎟ 

⎝ 0 ⎠ 

0 1 

· w � � z,w ∈ R}


For instance, the top line says that x =2− 2z +2w. The next section gives a 

geometric interpretation that will help us picture the solution sets when they 

are written in this way. 

2.8 Definition A vector (or column vector) is a matrix with a single column. 

A matrix with a single row is a row vector. The entries of a vector are its 

components. 

Vectors are an exception to the convention of representing matrices with 

capital roman letters. We use lower-case roman or greek letters overlined with 

an arrow: �a, �b, ... or �α, � β, ... (boldface is also common: a or α). For 

instance, this is a column vector with a third component of 7. 

⎛ 

�v = ⎝ 1 

⎞ 

3⎠ 

7 

2.9 Definition The linear equation a1x1 + a2x2 + ··· + anxn = d with unknowns 

x1,... ,xn is satisfied by 

⎛ ⎞ 

s1 

⎜ 

�s = ⎝ . 

⎟ 

. ⎠ 

if a1s1 + a2s2 + ··· + ansn = d. A vector satisfies a linear system if it satisfies 

each equation in the system. 

The style of description of solution sets that we use involves adding the 

vectors, and also multiplying them by real numbers, such as the z and w. We 

need to define these operations. 

2.10 Definition The vector sum of �u and �v is this. 

⎛ ⎞ ⎛ ⎞ ⎛ 

u1 v1 

⎜ 

�u + �v = ⎝ . 

⎟ ⎜ 

. ⎠ + ⎝ . 

⎟ ⎜ 

. ⎠ = ⎝ 

un 

sn 

vn 

u1 + v1 

. 

un + vn 

In general, two matrices with the same number of rows and the same number 

of columns add in this way, entry-by-entry. 

2.11 Definition The scalar multiplication of the real number r and the vector 

�v is this. 

⎛ ⎞ ⎛ ⎞ 

v1 rv1 

⎜ 

r · �v = r · 

. 

⎝ . 

⎟ ⎜ 

. ⎠ = 

. 

⎝ . 

⎟ 

. ⎠ 

vn rvn 

In general, any matrix is multiplied by a real number in this entry-by-entry way. 

⎞ 

⎟ 

⎠


Scalar multiplication can be written in either order: r · �v or �v · r, or without 

the ‘·’ symbol: r�v. (Do not refer to scalar multiplication as ‘scalar product’ 

because that name is used for a different operation.) 

2.12 Example 

⎛ 

⎝ 2 

⎞ ⎛ 

3⎠ 

+ ⎝ 

1 

3 

⎞ ⎛ 

−1⎠ 

= ⎝ 

4 

2+3 

⎞ ⎛ 

3 − 1⎠ 

= ⎝ 

1+4 

5 

⎛ ⎞ 

⎞ 1 

⎜ 

2⎠ 

7 · ⎜ 4 ⎟ 

⎝−1⎠ 

5 

−3 

= 

⎛ ⎞ 

7 

⎜ 28 ⎟ 

⎝ −7 ⎠ 

−21 

Notice that the definitions of vector addition and scalar multiplication agree 

where they overlap, for instance, �v + �v =2�v. 

With the notation defined, we can now solve systems in the way that we will 

use throughout this book. 

2.13 Example This system 

2x + y − w =4 

y + w + u =4 

x − z +2w =0 

reduces in this way. 

⎛ 

2 

⎝0 

1 

1 

0 

0 

−1 

1 

0 

1 

⎞ 

4 

4⎠ 

1 0 −1 2 0 0 

−(1/2)ρ1+ρ3 

−→ 

(1/2)ρ2+ρ3 

−→ 

⎛ 

2 

⎝0 

1 

1 

0 

0 

−1 

1 

0 

1 

⎞ 

4 

4 ⎠ 

0 

⎛ 

2 

⎝0 

−1/2 

1 0 

1 0 

−1 5/2 

−1 0 

1 1 

0 −2 

⎞ 

4 

4⎠ 

0 0 −1 3 1/20 The solution set is {(w +(1/2)u, 4 − w − u, 3w +(1/2)u, w, u) � � w, u ∈ R}. We 

write that in vector form. 

⎛ ⎞ 

x 

⎜ 

⎜y 

⎟ 

{ ⎜ 

⎜z 

⎟ 

⎝w⎠ 

u 

= 

⎛ ⎞ 

0 

⎜ 

⎜4 

⎟ 

⎜ 

⎜0 

⎟ 

⎝0⎠ 

0 

+ 

⎛ ⎞ ⎛ ⎞ 

1 1/2 

⎜ 

⎜−1⎟ 

⎜ 

⎟ ⎜−1 

⎟ 

⎜ 3 ⎟ w + ⎜ 

⎜1/2 

⎟ 

⎝ 1 ⎠ ⎝ 0 ⎠ 

0 1 

u � � w, u ∈ R} 

Note again how well vector notation sets off the coefficients of each parameter. 

For instance, the third row of the vector form shows plainly that if u is held 

fixed then z increases three times as fast as w. 

That format also shows plainly that there are infinitely many solutions. For 

example, we can fix u as 0, let w range over the real numbers, and consider the 

first component x. We get infinitely many first components and hence infinitely 

many solutions.


Another thing shown plainly is that setting both w and u to zero gives that 

this 

⎛ ⎞ 

x 

⎜ 

⎜y 

⎟ 

⎜ 

⎜z 

⎟ 

⎝w⎠ 

u 

= 

⎛ ⎞ 

0 

⎜ 

⎜4 

⎟ 

⎜ 

⎜0 

⎟ 

⎝0⎠ 

0 

is a particular solution of the linear system. 

2.14 Example In the same way, this system 

x − y + z =1 

3x + z =3 

5x − 2y +3z =5 

reduces 

⎛ 

1 

⎝3 

5 

−1 

0 

−2 

1 

1 

3 

⎞ 

1 

3⎠ 

5 

−3ρ1+ρ2 

⎛ 

1 

−→ ⎝0 

−5ρ1+ρ3 

0 

−1 

3 

3 

1 

−2 

−2 

⎞ 

1 

0⎠ 

0 

−ρ2+ρ3 

⎛ 

1 

−→ ⎝0 

0 

−1 

3 

0 

1 

−2 

0 

⎞ 

1 

0⎠ 

0 

to a one-parameter solution set. 

⎛ ⎞ ⎛ ⎞ 

1 −1/3 

{ ⎝0⎠ 

+ ⎝ 2/3 ⎠ z 

0 1 

� � z ∈ R} 

Before the exercises, we pause to point out some things that we have yet to 

do. 

The first two subsections have been on the mechanics of Gauss’ method. 

Except for one result, Theorem 1.4 — without which developing the method 

doesn’t make sense since it says that the method gives the right answers — we 

have not stopped to consider any of the interesting questions that arise. 

For example, can we always describe solution sets as above, with a particular 

solution vector added to an unrestricted linear combination of some other vectors? 

The solution sets we described with unrestricted parameters were easily 

seen to have infinitely many solutions so an answer to this question could tell 

us something about the size of solution sets. An answer to that question could 

also help us picture the solution sets — what do they look like in R 2 ,orinR 3 , 

etc? 

Many questions arise from the observation that Gauss’ method can be done 

in more than one way (for instance, when swapping rows, we may have a choice 

of which row to swap with). Theorem 1.4 says that we must get the same 

solution set no matter how we proceed, but if we do Gauss’ method in two 

different ways must we get the same number of free variables both times, so 

that any two solution set descriptions have the same number of parameters?


Must those be the same variables (e.g., is it impossible to solve a problem one 

way and get y and w free or solve it another way and get y and z free)? 

In the rest of this chapter we answer these questions. The answer to each 

is ‘yes’. The first question is answered in the last subsection of this section. In 

the second section we give a geometric description of solution sets. In the final 

section of this chapter we tackle the last set of questions. 

Consequently, by the end of the first chapter we will not only have a solid 

grounding in the practice of Gauss’ method, we will also have a solid grounding 

in the theory. We will be sure of what can and cannot happen in a reduction. 

Exercises 

� 2.15 Find the indicated entry of the matrix, if it is defined. 

� � 

1 3 1 

A = 

2 −1 4 

(a) a2,1 (b) a1,2 (c) a2,2 (d) a3,1 

� 2.16 Give the size of each matrix. 

� � � � 

1 1 

1 0 4 

(a) 

(b) −1 1 

2 1 5 

3 −1 

(c) 

� 

5 

� 

10 

10 5 

� 2.17 Do the indicated vector operation, if it is defined. 

� � � � 

2 3 � � � � � � 

1 3 

4 

(a) 1 + 0 (b) 5 (c) 5 − 1 

−1 

1 4 

1 1 

� � � � � � � � � � 

1 

3 2 1 

1 

(e) + 2 (f) 6 1 − 4 0 +2 1 

2 

3 

1 3 5 

(d) 7 

� � � � 

2 3 

+9 

1 5 

� 2.18 Solve each system using matrix notation. Express the solution using vec- 

tors. 

(a) 3x +6y =18 

x +2y = 6 

(d) 2a + b − c =2 

2a + c =3 

a − b =0 

(b) x + y = 1 

x − y = −1 

(e) x +2y − z =3 

2x + y + w =4 

x − y + z + w =1 

(c) x1 + x3 = 4 

x1 − x2 +2x3 = 5 

4x1 − x2 +5x3 =17 

(f) x + z + w =4 

2x + y − w =2 

3x + y + z =7 

� 2.19 Solve each system using matrix notation. Give each solution set in vector 

notation. 

(a) 2x + y − z =1 

4x − y =3 

(b) x − z =1 

y +2z − w =3 

x +2y +3z − w =7 

(c) x − y + z =0 

y + w =0 

3x − 2y +3z + w =0 

−y − w =0 

(d) a +2b +3c + d − e =1 

3a − b + c + d + e =3 

� 2.20 The vector is in the set. What value of the parameters produces that vector? 

� � � � 

5 1 

(a) , { k 

−5 −1 

� � k ∈ R}


� � � � � � 

−1 −2 3 

(b) 2 , { 1 i + 0 j 

1 0 1 

� � i, j ∈ R} 

� � � � � � 

0 1 2 

(c) −4 , { 1 m + 0 n 

2 0 1 

� � m, n ∈ R} 

2.21 Decide � �if 

the � vector � is in the set. 

3 −6 

(a) , { k 

−1 2 

� � k ∈ R} 

� � � � 

5 5 

(b) , { j 

4 −4 

� � j ∈ R} 

� � � � � � 

2 0 1 

(c) 1 , { 3 + −1 r 

−1 −7 3 

� � r ∈ R} 

� � � � � � 

1 2 −3 

(d) 0 , { 0 j + −1 k 

1 1 1 

� � j, k ∈ R} 

2.22 Paramatrize the solution set of this one-equation system. 

x1 + x2 + ···+ xn =0 

� 2.23 (a) Apply Gauss’ method to the left-hand side to solve 

x +2y − w = a 

2x + z = b 

x + y +2w = c 

for x, y, z, andw, in terms of the constants a, b, andc. 

(b) Use your answer from the prior part to solve this. 

x +2y − w = 3 

2x + z = 1 

x + y +2w = −2 

� 2.24 Why is the comma needed in the notation ‘ai,j’ for matrix entries? 

� 2.25 Give the 4×4 matrix whose i, j-th entry is 

(a) i + j; (b) −1 tothei + j power. 

2.26 For any matrix A, thetranspose of A, written A trans , is the matrix whose 

columns are the rows of A. Findthetransposeofeachofthese. 

� � � � � � � � 

1 

1 2 3 

2 −3 

5 10 

(a) 

(b) 

(c) 

(d) 1 

4 5 6 

1 1 

10 5 

0 

� 2.27 (a) Describe all functions f(x) = ax 2 + bx + c such that f(1) = 2 and 

f(−1) = 6. 

(b) Describe all functions f(x) =ax 2 + bx + c such that f(1) = 2. 

2.28 Show that any set of five points from the plane R 2 lie on a common conic 

section, that is, they all satisfy some equation of the form ax 2 + by 2 + cxy + dx + 

ey + f = 0 where some of a, ... ,f are nonzero. 

2.29 Make up a four equations/four unknowns system having 

(a) a one-parameter solution set; 

(b) a two-parameter solution set; 

(c) a three-parameter solution set.


2.30 [USSR Olympiad no. 174] 

(a) Solve the system of equations. 

ax + y = a 2 

x + ay = 1 

For what values of a does the system fail to have solutions, and for what values 

of a are there infinitely many solutions? 

(b) Answer the above question for the system. 

ax + y = a 3 

x + ay = 1 

2.31 [Math. Mag., Sept. 1952] In air a gold-surfaced sphere weighs 7588 grams. It 

is known that it may contain one or more of the metals aluminum, copper, silver, 

or lead. When weighed successively under standard conditions in water, benzene, 

alcohol, and glycerine its respective weights are 6588, 6688, 6778, and 6328 grams. 

How much, if any, of the forenamed metals does it contain if the specific gravities 

of the designated substances are taken to be as follows? 

Aluminum 2.7 Alcohol 0.81 

Copper 8.9 Benzene 0.90 

Gold 19.3 Glycerine 1.26 

Lead 11.3 Water 1.00 

Silver 10.8 

1.I.3 General = Particular + Homogeneous 

The prior subsection has many descriptions of solution sets. They all fit a 

pattern. They have a vector that is a particular solution of the system added 

to an unrestricted combination of some other vectors. The solution set from 

Example 2.13 illustrates. 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

0 1 1/2 

⎜ 

⎜4⎟ 

⎜ 

⎟ ⎜−1⎟ 

⎜ 

⎟ ⎜−1 

⎟ � 

{ ⎜ 

⎜0 

⎟ + w ⎜ 3 ⎟ + u ⎜ 

⎜1/2⎟ 

� 

⎟ w, u ∈ R} 

⎝0⎠ 

⎝ 1 ⎠ ⎝ 0 ⎠ 

0 0 1 

� �� 

particular 

solution 

� �� 

unrestricted 

combination 

The combination is unrestricted in that w and u can be any real numbers — 

there is no condition like “such that 2w −u = 0” that would restrict which pairs 

w, u can be used to form combinations. 

That example shows an infinite solution set conforming to the pattern. We 

can think of the other two kinds of solution sets as also fitting the same pattern. 

A one-element solution set fits in that it has a particular solution, and 

the unrestricted combination part is a trivial sum (that is, instead of being a 

combination of two vectors, as above, or a combination of one vector, it is a


combination of no vectors). A zero-element solution set fits the pattern since 

there is no particular solution, and so the set of sums of that form is empty. 

We will show that the examples from the prior subsection are representative, 

in that the description pattern discussed above holds for every solution set. 

3.1 Theorem For any linear system there are vectors � β1, ... , � βk such that 

the solution set can be described as 

{�p + c1 � β1 + ··· + ck � � 

βk � c1, ... ,ck ∈ R} 

where �p is any particular solution, and where the system has k free variables. 

This description has two parts, the particular solution �p and also the unrestricted 

linear combination of the � β’s. We shall prove the theorem in two 

corresponding parts, with two lemmas. 

We will focus first on the unrestricted combination part. To do that, we 

consider systems that have the vector of zeroes as one of the particular solutions, 

so that �p + c1 � β1 + ···+ ck � βk can be shortened to c1 � β1 + ···+ ck � βk. 

3.2 Definition A linear equation is homogeneous if it has a constant of zero, 

that is, if it can be put in the form a1x1 + a2x2 + ··· + anxn =0. 

(These are ‘homogeneous’ because all of the terms involve the same power of 

their variable — the first power — including a ‘0x0’ that we can imagine is on 

the right side.) 

3.3 Example With any linear system like 

3x +4y =3 

2x − y =1 

we associate a system of homogeneous equations by setting the right side to 

zeros. 

3x +4y =0 

2x − y =0 

Our interest in the homogeneous system associated with a linear system can be 

understood by comparing the reduction of the system 

3x +4y =3 

2x − y =1 

−(2/3)ρ1+ρ2 

−→ 

3x + 4y =3 

−(11/3)y = −1 

with the reduction of the associated homogeneous system. 

3x +4y =0 

2x − y =0 

−(2/3)ρ1+ρ2 

−→ 

3x + 4y =0 

−(11/3)y =0 

Obviously the two reductions go in the same way. We can study how linear systems 

are reduced by instead studying how the associated homogeneous systems 

are reduced.


Studying the associated homogeneous system has a great advantage over 

studying the original system. Nonhomogeneous systems can be inconsistent. 

But a homogeneous system must be consistent since there is always at least one 

solution, the vector of zeros. 

3.4 Definition A column or row vector of all zeros is a zero vector, denoted �0. 

There are many different zero vectors, e.g., the one-tall zero vector, the two-tall 

zero vector, etc. Nonetheless, people often refer to “the” zero vector, expecting 

that the size of the one being discussed will be clear from the context. 

3.5 Example Some homogeneous systems have the zero vector as their only 

solution. 

3x +2y + z =0 

6x +4y =0 

y + z =0 

−2ρ1+ρ2 

−→ 

3x +2y + z =0 

−2z =0 

y + z =0 

ρ2↔ρ3 

−→ 

3x +2y + z =0 

y + z =0 

−2z =0 

3.6 Example Some homogeneous systems have many solutions. One example 

is the Chemistry problem from the first page of this book. 

The solution set: 

7x − 7j =0 

8x + y − 5j − 2k =0 

y − 3j =0 

3y − 6j − k =0 

−(8/7)ρ1+ρ2 

−→ 

−ρ2+ρ3 

−→ 

−3ρ2+ρ4 

−(5/2)ρ3+ρ4 

−→ 

⎛ ⎞ 

1/3 

⎜ 

{ ⎜ 1 ⎟ 

⎝1/3⎠ 

1 

w � � k ∈ R} 

7x − 7z =0 

y +3z − 2w =0 

y − 3z =0 

3y − 6z − w =0 

7x − 7z =0 

y + 3z − 2w =0 

−6z +2w =0 

−15z +5w =0 

7x − 7z =0 

y + 3z − 2w =0 

−6z +2w =0 

0=0 

has many vectors besides the zero vector (if we interpret w as a number of 

molecules then solutions make sense only when w is a nonnegative multiple of 

3). 

We now have the terminology to prove the two parts of Theorem 3.1. The 

first lemma deals with unrestricted combinations.


3.7 Lemma For any homogeneous linear system there exist vectors � β1, ... , 

�βk such that the solution set of the system is 

{c1 � β1 + ···+ ck � � 

βk � c1,... ,ck ∈ R} 

where k is the number of free variables in an echelon form version of the system. 

Before the proof, we will recall the back substitution calculations that were 

done in the prior subsection. Imagine that we have brought a system to this 

echelon form. 

x + 2y − z +2w =0 

−3y + z =0 

−w =0 

We next perform back-substitution to express each variable in terms of the 

free variable z. Working from the bottom up, we get first that w is 0 · z, 

next that y is (1/3) · z, and then substituting those two into the top equation 

x + 2((1/3)z) − z + 2(0) = 0 gives x =(1/3) · z. So, back substitution gives 

a paramatrization of the solution set by starting at the bottom equation and 

using the free variables as the parameters to work row-by-row to the top. The 

proof below follows this pattern. 

Comment: That is, this proof just does a verification of the bookkeeping in 

back substitution to show that we haven’t overlooked any obscure cases where 

this procedure fails, say, by leading to a division by zero. So this argument, 

while quite detailed, doesn’t give us any new insights. Nevertheless, we have 

written it out for two reasons. The first reason is that we need the result — the 

computational procedure that we employ must be verified to work as promised. 

The second reason is that the row-by-row nature of back substitution leads to a 

proof that uses the technique of mathematical induction. ∗ This is an important, 

and non-obvious, proof technique that we shall use a number of times in this 

book. Doing an induction argument here gives us a chance to see one in a setting 

where the proof material is easy to follow, and so the technique can be studied. 

Readers who are unfamiliar with induction arguments should be sure to master 

this one and the ones later in this chapter before going on to the second chapter. 

Proof. First use Gauss’ method to reduce the homogeneous system to echelon 

form. We will show that each leading variable can be expressed in terms of free 

variables. That will finish the argument because then we can use those free 

variables as the parameters. That is, the � β’s are the vectors of coefficients of 

the free variables (as in Example 3.6, where the solution is x =(1/3)w, y = w, 

z =(1/3)w, andw = w). 

We will proceed by mathematical induction, which has two steps. The base 

step of the argument will be to focus on the bottom-most non-‘0 = 0’ equation 

and write its leading variable in terms of the free variables. The inductive step 

of the argument will be to argue that if we can express the leading variables from 

∗ More information on mathematical induction is in the appendix.


the bottom t rows in terms of free variables, then we can express the leading 

variable of the next row up — the t + 1-th row up from the bottom — in terms 

of free variables. With those two steps, the theorem will be proved because by 

the base step it is true for the bottom equation, and by the inductive step the 

fact that it is true for the bottom equation shows that it is true for the next 

one up, and then another application of the inductive step implies it is true for 

third equation up, etc. 

For the base step, consider the bottom-most non-‘0 = 0’ equation (the case 

where all the equations are ‘0 = 0’ is trivial). We call that the m-th row: 

am,ℓm xℓm + am,ℓm+1xℓm+1 + ···+ am,nxn =0 

where am,ℓm �= 0. (The notation here has ‘ℓ’ stand for ‘leading’, so am,ℓm means 

“the coefficient, from the row m of the variable leading row m”.) Either there 

are variables in this equation other than the leading one xℓm or else there are 

not. If there are other variables xℓm+1, etc., then they must be free variables 

because this is the bottom non-‘0 = 0’ row. Move them to the right and divide 

by am,ℓm 

xℓm =(−am,ℓm+1/am,ℓm )xℓm+1 + ···+(−am,n/am,ℓm )xn 

to expresses this leading variable in terms of free variables. If there are no free 

variables in this equation then xℓm = 0 (see the “tricky point” noted following 

this proof). 

For the inductive step, we assume that for the m-th equation, and for the 

(m − 1)-th equation, ... , and for the (m − t)-th equation, we can express the 

leading variable in terms of free variables (where 0 ≤ t


The next lemma finishes the proof of Theorem 3.1 by considering the particular 

solution part of the solution set’s description. 

3.8 Lemma For a linear system, where �p is any particular solution, the solution 

set equals this set. 

{�p + � h � � � h satisfies the associated homogeneous system} 

Proof. We will show mutual set inclusion, that any solution to the system is 

in the above set and that anything in the set is a solution to the system. ∗ 

For set inclusion the first way, that if a vector solves the system then it is in 

the set described above, assume that �s solves the system. Then �s − �p solves the 

associated homogeneous system since for each equation index i between 1 and 

n, 

ai,1(s1 − p1)+···+ ai,n(sn − pn) =(ai,1s1 + ···+ ai,nsn) 

− (ai,1p1 + ···+ ai,npn) 

= di − di 

=0 

where pj and sj are the j-th components of �p and �s. We can write �s − �p as � h, 

where � h solves the associated homogeneous system, to express �s in the required 

�p + � h form. 

For set inclusion the other way, take a vector of the form �p + � h, where �p 

solves the system and � h solves the associated homogeneous system, and note 

that it solves the given system: for any equation index i, 

ai,1(p1 + h1)+···+ ai,n(pn + hn) =(ai,1p1 + ···+ ai,npn) 

+(ai,1h1 + ···+ ai,nhn) 

= di +0 

= di 

where hj is the j-th component of � h. QED 

The two lemmas above together establish Theorem 3.1. We remember that 

theorem with the slogan “General = Particular + Homogeneous”. 

3.9 Example This system illustrates Theorem 3.1. 

Gauss’ method 

−2ρ1+ρ2 

−→ 

x +2y − z =1 

2x +4y =2 

y − 3z =0 

x +2y − z =1 

2z =0 

y − 3z =0 

ρ2↔ρ3 

−→ 

∗ More information on equality of sets is in the appendix. 

x +2y − z =1 

y − 3z =0 

2z =0


shows that the general solution is a singleton set. 

⎛ 

{ ⎝ 1 

⎞ 

0⎠} 

0 

That single vector is, of course, a particular solution. The associated homogeneous 

system reduces via the same row operations 

x +2y − z =0 

2x +4y =0 

y − 3z =0 

to also give a singleton set. 

−2ρ1+ρ2 

−→ ρ2↔ρ3 

−→ 

⎛ ⎞ 

0 

{ ⎝0⎠} 

0 

x +2y − z =0 

y − 3z =0 

2z =0 

As the theorem states, and as discussed at the start of this subsection, in this 

single-solution case the general solution results from taking the particular solution 

and adding to it the unique solution of the associated homogeneous system. 

3.10 Example Also discussed there is that the case where the general solution 

set is empty fits the ‘General = Particular+Homogeneous’ pattern. This system 

illustrates. Gauss’ method 

x + z + w = −1 

2x − y + w = 3 

x + y +3z +2w = 1 

−2ρ1+ρ2 

−→ 

−ρ1+ρ3 

x + z + w = −1 

−y − 2z − w = 5 

y +2z + w = 2 

shows that it has no solutions. The associated homogeneous system, of course, 

has a solution. 

x + z + w =0 

2x − y + w =0 

x + y +3z +2w =0 

−2ρ1+ρ2 

−→ 

−ρ1+ρ3 

ρ2+ρ3 

−→ 

x + z + w =0 

−y − 2z − w =0 

0=0 

In fact, the solution set of the homogeneous system is infinite. 

⎛ ⎞ ⎛ ⎞ 

−1 −1 

⎜ 

{ ⎜−2⎟ 

⎜ 

⎟ 

⎝ 1 ⎠ z + ⎜−1 

⎟ 

⎝ 0 ⎠ 

0 1 

w � � z,w ∈ R} 

However, because no particular solution of the original system exists, the general 

solution set is empty — there are no vectors of the form �p + � h because there are 

no �p ’s. 

3.11 Corollary Solution sets of linear systems are either empty, have one 

element, or have infinitely many elements.


Proof. We’ve seen examples of all three happening so we need only prove that 

those are the only possibilities. 

First, notice a homogeneous system with at least one non-�0 solution �v has 

infinitely many solutions because the set of multiples s�v is infinite — if s �= 1 

then s�v − �v =(s − 1)�v is easily seen to be non-�0, and so s�v �= �v. 

Now, apply Lemma 3.8 to conclude that a solution set 

{�p + � h � � � h solves the associated homogeneous system} 

is either empty (if there is no particular solution �p), or has one element (if there 

is a �p and the homogeneous system has the unique solution �0), or is infinite (if 

there is a �p and the homogeneous system has a non-�0 solution, and thus by the 

prior paragraph has infinitely many solutions). QED 

This table summarizes the factors affecting the size of a general solution. 

particular 

solution 

exists? 

yes 

no 

number of solutions of the 

associated homogeneous system 

one infinitely many 

unique 

solution 

infinitely many 

solutions 

no 

solutions 

no 

solutions 

The factor on the top of the table is the simpler one. When we perform 

Gauss’ method on a linear system, ignoring the constants on the right side and 

so paying attention only to the coefficients on the left-hand side, we either end 

with every variable leading some row or else we find that some variable does not 

lead a row, that is, that some variable is free. (Of course, “ignoring the constants 

on the right” is formalized by considering the associated homogeneous system. 

We are simply putting aside for the moment the possibility of a contradictory 

equation.) 

A nice insight into the factor on the top of this table at work comes from considering 

the case of a system having the same number of equations as variables. 

This system will have a solution, and the solution will be unique, if and only if it 

reduces to an echelon form system where every variable leads its row, which will 

happen if and only if the associated homogeneous system has a unique solution. 

Thus, the question of uniqueness of solution is especially interesting when the 

system has the same number of equations as variables. 

3.12 Definition A square matrix is nonsingular if it is the matrix of coefficients 

of a homogeneous system with a unique solution. It is singular otherwise, 

that is, if it is the matrix of coefficients of a homogeneous system with infinitely 

many solutions.


3.13 Example The systems from Example 3.3, Example 3.5, and Example 3.9 

each have an associated homogeneous system with a unique solution. Thus these 

matrices are nonsingular. 

� � 

3 4 

2 −1 

⎛ 

3 

⎝6 2 

−4 

⎞ 

1 

0⎠ 

0 1 1 

⎛ 

1 

⎝2 2 

4 

⎞ 

−1 

0 ⎠ 

0 1 −3 

The Chemistry problem from Example 3.6 is a homogeneous system with more 

than one solution so its matrix is singular. 

⎛ 

7 

⎜ 

⎜8 

⎝0 

0 

1 

1 

−7 

−5 

−3 

⎞ 

0 

−2 ⎟ 

0 ⎠ 

0 3 −6 −1 

3.14 Example The first of these matrices is nonsingular while the second is 

singular 

� 

1 

� 

2 

� 

1 

� 

2 

3 4 3 6 

because the first of these homogeneous systems has a unique solution while the 

second has infinitely many solutions. 

x +2y =0 

3x +4y =0 

x +2y =0 

3x +6y =0 

We have made the distinction in the definition because a system (with the same 

number of equations as variables) behaves in one of two ways, depending on 

whether its matrix of coefficients is nonsingular or singular. A system where 

the matrix of coefficients is nonsingular has a unique solution for any constants 

on the right side: for instance, Gauss’ method shows that this system 

x +2y = a 

3x +4y = b 

has the unique solution x = b − 2a and y =(3a − b)/2. On the other hand, a 

system where the matrix of coefficients is singular never has a unique solutions — 

it has either no solutions or else has infinitely many, as with these. 

x +2y =1 

3x +6y =2 

x +2y =1 

3x +6y =3 

Thus, ‘singular’ can be thought of as connoting “troublesome”, or at least “not 

ideal”. 

The above table has two factors. We have already considered the factor 

along the top: we can tell which column a given linear system goes in solely by


considering the system’s left-hand side — the the constants on the right-hand 

side play no role in this factor. The table’s other factor, determining whether a 

particular solution exists, is tougher. Consider these two 

3x +2y =5 

3x +2y =5 

3x +2y =5 

3x +2y =4 

with the same left sides but different right sides. Obviously, the first has a 

solution while the second does not, so here the constants on the right side 

decide if the system has a solution. We could conjecture that the left side of a 

linear system determines the number of solutions while the right side determines 

if solutions exist, but that guess is not correct. Compare these two systems 

3x +2y =5 

4x +2y =4 and 

3x +2y =5 

3x +2y =4 

with the same right sides but different left sides. The first has a solution but 

the second does not. Thus the constants on the right side of a system don’t 

decide alone whether a solution exists; rather, it depends on some interaction 

between the left and right sides. 

For some intuition about that interaction, consider this system with one of 

the coefficients left as the parameter c. 

x +2y +3z =1 

x + y + z =1 

cx +3y +4z =0 

If c = 2 this system has no solution because the left-hand side has the third row 

as a sum of the first two, while the right-hand does not. If c �= 2 this system has 

a unique solution (try it with c = 1). For a system to have a solution, if one row 

of the matrix of coefficients on the left is a linear combination of other rows, 

then on the right the constant from that row must be the same combination of 

constants from the same rows. 

More intuition about the interaction comes from studying linear combinations. 

That will be our focus in the second chapter, after we finish the study of 

Gauss’ method itself in the rest of this chapter. 

Exercises 

� 3.15 Solve each system. Express the solution set using vectors. Identify the par- 

ticular solution and the solution set of the homogeneous system. 

(a) 3x +6y =18 

x +2y = 6 

(d) 2a + b − c =2 

2a + c =3 

a − b =0 

(b) x + y = 1 

x − y = −1 

(e) x +2y − z =3 

2x + y + w =4 

x − y + z + w =1 

(c) x1 + x3 = 4 

x1 − x2 +2x3 = 5 

4x1 − x2 +5x3 =17 

(f) x + z + w =4 

2x + y − w =2 

3x + y + z =7 

3.16 Solve each system, giving the solution set in vector notation. Identify the 

particular solution and the solution of the homogeneous system.


(a) 2x + y − z =1 

4x − y =3 

(b) x − z =1 

y +2z − w =3 

x +2y +3z − w =7 

(c) x − y + z =0 

y + w =0 

3x − 2y +3z + w =0 

−y − w =0 

(d) a +2b +3c + d − e =1 

3a − b + c + d + e =3 

� 3.17 For the system 

2x − y − w = 3 

y + z +2w = 2 

x − 2y − z = −1 

which of these can be used as the particular solution part of some general solution? 

⎛ ⎞ 

0 

⎛ ⎞ 

2 

⎛ ⎞ 

−1 

⎜−3⎟ 

(a) ⎝ 

5 

⎠ 

⎜1⎟ 

(b) ⎝ 

1 

⎠ 

⎜−4⎟ 

(c) ⎝ 

8 

⎠ 

0 

0 

−1 

� 3.18 Lemma 3.8 says that any particular solution may be used for �p. 

possible, a general solution to this system 

Find, if 

x − y + w =4 

2x +3y − z =0 

y + z + w =4 

that uses 

⎛ 

the 

⎞ 

given vector 

⎛ ⎞ 

as its particular 

⎛ ⎞ 

solution. 

0 

−5 

2 

⎜0⎟ 

(a) ⎝ 

0 

⎠ 

⎜ 1 ⎟ 

(b) ⎝ 

−7 

⎠ 

⎜−1⎟ 

(c) ⎝ 

1 

⎠ 

4 

10 

1 

3.19 One of these is nonsingular while the other is singular. Which is which? 

(a) 

� 

1 3 

� 

4 −12 

� 

1 

� 

2 

(b) 

� � 

1 3 

4 12 

� 3.20 Singular or nonsingular? � � 

1 2 

(a) 

(b) 

1 3 

−3 −6 

� � � 

1 2 1 

2 2 

(c) 

� 

1 

� 

1 

1 

2 

3 

� 

1 

(Careful!) 

1 

(d) 1 1 3 (e) 1 0 5 

3 4 7 

−1 1 4 

� 3.21 Is� the � given � �vector 

� �in 

the set generated by the given set? 

2 1 1 

(a) , { , } 

3 4 5 

� � � � � � 

−1 2 1 

(b) 0 , { 1 , 0 } 

1 

� � 

1 

0 

� � 

1 

1 

� � 

2 

� � 

3 

� � 

4 

(c) 3 , { 0 , 1 , 3 , 2 } 

0 

⎛ ⎞ 

1 

4 

⎛ ⎞ 

2 

5 

⎛ ⎞ 

3 

0 1 

⎜0⎟ 

⎜1⎟ 

⎜0⎟ 

(d) ⎝ ⎠ , { ⎝ ⎠ , ⎝ ⎠} 

1 

1 

0 

1 

0 

2


3.22 Prove that any linear system with a nonsingular matrix of coefficients has a 

solution, and that the solution is unique. 

3.23 To tell the whole truth, there is another tricky point to the proof of Lemma 3.7. 

What happens if there are no non-‘0 = 0’ equations? (There aren’t any more tricky 

points after this one.) 

� 3.24 Prove that if �s and �t satisfy a homogeneous system then so do these vectors. 

(a) �s + �t (b) 3�s (c) k�s + m�t for k, m ∈ R 

What’s wrong with: “These three show that if a homogeneous system has one 

solution then it has many solutions — any multiple of a solution is another solution, 

and any sum of solutions is a solution also — so there are no homogeneous systems 

with exactly one solution.”? 

3.25 Prove that if a system with only rational coefficients and constants has a 

solution then it has at least one all-rational solution. Must it have infinitely many?


1.II Linear Geometry of n-Space 

For readers who have seen the elements of vectors before, in calculus or physics, 

this section is an optional review. However, later work in this book will refer to 

this material often, so this section is not optional if it is not a review. 

In the first section, we had to do a bit of work to show that there are only 

three types of solution sets — singleton, empty, and infinite. But for systems 

with two equations and two unknowns, we can just see this. We picture each 

two-unknowns equation as a line in R 2 and then the two lines could have a 

unique intersection, be parallel, or be the same. 

One solution 

3x +2y = 7 

x − y = −1 

No solutions 

3x +2y =7 

3x +2y =4 

Infinitely many 

solutions 

3x +2y = 7 

6x +4y =14 

As this shows, sometimes our results are expressed clearly in a picture. In this 

section we develop the terminology and ideas we need to express our results 

from the prior section, and from some future sections, geometrically. The twodimensional 

case is familiar enough, but to extend to systems with more than 

two unknowns we shall also need some higher-dimensional geometry. 

1.II.1 Vectors in Space 

“Higher-dimensionsional geometry” sounds exotic. It is exotic — interesting 

and eye-opening. But it isn’t distant or unreachable. 

As a start, we define one-dimensional space to be the set R 1 . To see that 

definition is reasonable, draw a one-dimensional space 

and make the usual correspondence with R: pick a point to label 0 and another 

to label 1. 

0 1 

Now, armed with a scale and a direction, finding the point corresponding to, 

say +2.17, is easy — start at 0, head in the direction of 1 (i.e., the positive 

direction), but don’t stop there, go 2.17 times as far. 

The basic idea here, combining magnitude with direction, is the key to extending 

to higher dimensions.

Section II. Linear Geometry of n-Space 33 

An object comprised of a magnitude and a direction is a vector (we will use 

the same word as in the previous section because we shall show below how to 

describe such an object with a column vector). We can draw a vector as having 

some length, and pointing somewhere. 

There is a subtlety here — these 

are equal, even though they start in different places, because they have equal 

lengths and equal directions. Again: those vectors are not just alike, they are 

equal. 

How can things that are in different places be equal? Think of a vector as 

representing a displacement (‘vector’ is Latin for “carrier” or “traveler”). These 

squares undergo the same displacement, despite that those displacements start 

in different places. 

Sometimes, to emphasize this property vectors have of not being anchored, they 

are referred to as free vectors. 

These two, as free vectors, are equal; 

we can think of each as a displacement of one over and two up. More generally, 

two vectors in the plane are the same if and only if they have the same change 

in first components and the same change in second components: the vector 

extending from (a1,a2) to(b1,b2) equals the vector from (c1,c2) to(d1,d2) if 

and only if b1 − a1 = d1 − c1 and b2 − a2 = d2 − c2. 

An expression like ‘the vector that, were it to start at (a1,a2), would stretch 

to (b1,b2)’ is awkward. Instead of that terminology, from among all of these 

we single out the one starting at the origin as being in canonical (or natural) 

position and we describe a vector by stating its endpoint when it is in canonical


position, as a column. For instance, the ‘one over and two up’ vectors above are 

denoted in this way. 

� � 

1 

2 

More generally, the plane vector starting at (a1,a2) and stretching to (b1,b2) is 

denoted 

� � 

b1 − a1 

b2 − a2 

since the prior paragraph shows that when the vector starts at the origin, it 

ends at this location. 

We often just say “the point 

� � 

1 

” 

2 

rather than “the endpoint of the canonical position of” that vector. That is, we 

shall find it convienent to blur the distinction between a point in space and the 

vector that, if it starts at the origin, ends at that point. Thus, we will refer to 

both of these as Rn . 

{(x1,x2) � � � 

x1 �� 

� x1,x2 ∈ R} { x1,x2 ∈ R} 

In the prior section we defined vectors and vector operations with an algebraic 

motivation; 

� � � � � � � � � � 

v1 rv1 v1 w1 v1 + w1 

r · = 

+ = 

v2 rv2 v2 w2 v2 + w2 

we can now interpret those operations geometrically. For instance, if �v represents 

a displacement then 3�v represents a displacement in the same direction 

but three times as far, and −1�v represents a displacement of the same distance 

as �v but in the opposite direction. 

−�v 

�v 

And, where �v and �w represent displacements, �v + �w represents those displacements 

combined. 

�v + �w 

�v 

3�v 

�w 

x2


The long arrow is the combined displacement in this sense: if, in one minute, a 

ship’s motion gives it the displacement relative to the earth of �v and a passenger’s 

motion gives a displacement relative to the ship’s deck of �w, then �v + �w is 

the displacement of the passenger relative to the earth. 

Another way to understand the vector sum is with the parallelogram rule. 

Draw the parallelogram formed by the vectors �v1,�v2 and then the sum �v1 + �v2 

extends along the diagonal to the far corner. 

� � 

� � 

x1 + x2 

x2 

y2 

� � 

x1 

y1 

y1 + y2 

The above drawings show how vectors and vector operations behave in R 2 . 

We can extend to R 3 , or to even higher-dimensional spaces where we have no 

pictures, with the obvious generalization: the free vector that, if it starts at 

(a1,... ,an), ends at (b1,... ,bn), is represented by this column 

⎛ 

⎜ 

⎝ 

b1 − a1 

. 

bn − an 

(vectors are equal if they have the same representation), we aren’t too careful 

to distinguish between a point and the vector whose canonical representation 

ends at that point, 

⎛ ⎞ 

v1 

⎞ 

⎟ 

⎠ 

R n ⎜ 

= { 

. ⎟ 

⎝ . ⎠ � � v1,... ,vn ∈ R} 

vn 

and addition and scalar multiplication are component-wise. 

Having considered points, we now turn to the lines. In R2 , the line through 

(1, 2) and (3, 1) is comprised of (the endpoints of) the vectors in this set 

� � � � 

1 2 �� 

{ + t · t ∈ R} 

2 −1 

That description expresses this picture. 

� � 

2 

= 

−1 

� � 

3 

− 

1 

� � 

1 

2 

The vector associated with the parameter t has its whole body in the line — it 

is a direction vector for the line. Note that points on the line to the left of x =1 

are described using negative values of t.


In R3 , the line through (1, 2, 3) and (5, 5, 5) is the set of (endpoints of) 

vectors of this form 

⎛ 

{ ⎝ 1 

⎞ ⎛ 

2⎠ 

+ t · ⎝ 

3 

4 

⎞ 

3⎠ 

2 

� � t ∈ R} 

and lines in even higher-dimensional spaces work in the same way. 

If a line uses one parameter, so that there is freedom to move back and 

forth in one dimension, then a plane must involve two. For example, the plane 

through the points (1, 0, 5), (2, 1, −3), and (−2, 4, 0.5) consists of (endpoints of) 

the vectors in 

⎛ 

{ ⎝ 1 

⎞ ⎛ 

0⎠ 

+ t · ⎝ 

5 

1 

⎞ ⎛ 

1 ⎠ + s · ⎝ 

−8 

−3 

⎞ 

4 ⎠ 

−4.5 

� � t, s ∈ R} 

(the column vectors associated with the parameters 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

1 2 1 

⎝ 1 ⎠ = ⎝ 1 ⎠ − ⎝0⎠ 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

−3 −2 1 

⎝ 4 ⎠ = ⎝ 4 ⎠ − ⎝0⎠ 

−8 −3 5 −4.5 0.5 5 

are two vectors whose whole bodies lie in the plane). As with the line, note that 

some points in this plane are described with negative t’s or negative s’s or both. 

A description of planes that is often encountered in algebra and calculus uses 

a single equation 

⎛ ⎞ 

x 

P = { ⎝y⎠ 

z 

� � 2x +3y − z =4} 

as the condition that describes the relationship among the first, second, and 

third coordinates of points in a plane. The translation from such a description 

to the vector description that we favor in this book is to think of the condition 

as a one-equation linear system and paramatrize x =(1/2)(4 − 3y + z). 

⎛ 

P = { ⎝ 2 

⎞ ⎛ 

0⎠ 

+ ⎝ 

0 

−3/2 

⎞ ⎛ 

1 ⎠ y + ⎝ 

0 

1/2 

⎞ 

0 ⎠ z 

1 

� � y, z ∈ R} 

Generalizing from lines and planes, we define a k-dimensional linear surface 

(or k-flat) inRn � 

to be {�p + t1�v1 + t2�v2 + ···+ tk�vk 

� t1,... ,tk ∈ R} where 

�v1,... ,�vk ∈ R n . For example, in R 4 , 

⎛ 

⎜ 

{ ⎜ 

⎝ 

2 

π 

3 

⎞ ⎛ ⎞ 

1 

⎟ ⎜ 

⎟ 

⎠ + t ⎜0⎟ 

� 

⎟ � 

⎝0⎠ 

t ∈ R} 

−0.5 0


is a line, 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

0 1 2 

⎜ 

{ ⎜0⎟ 

⎜ 

⎟ 

⎝0⎠ 

+ t ⎜ 1 ⎟ ⎜ 

⎟ 

⎝ 0 ⎠ + s ⎜0⎟ 

� 

⎟ � 

⎝1⎠ 

t, s ∈ R} 

0 −1 0 

is a plane, and 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

3 0 1 2 

⎜ 

{ ⎜ 1 ⎟ ⎜ 

⎟ 

⎝−2⎠ 

+ r ⎜ 0 ⎟ ⎜ 

⎟ 

⎝ 0 ⎠ + s ⎜0⎟ 

⎜ 

⎟ 

⎝1⎠ 

+ t ⎜0⎟ 

� 

⎟ � 

⎝1⎠ 

r, s, t ∈ R} 

0.5 −1 0 0 

is a three-dimensional linear surface. Again, the intuition is that a line permits 

motion in one direction, a plane permits motion in combinations of two 

directions, etc. 

A linear surface description can be misleading about the dimension — this 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

1 1 2 

⎜ 

L = { ⎜ 0 ⎟ ⎜ 

⎟ 

⎝−1⎠ 

+ t ⎜ 1 ⎟ ⎜ 

⎟ 

⎝ 0 ⎠ + s ⎜ 2 ⎟ � 

⎟ � 

⎝ 0 ⎠ t, s ∈ R} 

−2 −1 −2 

is a degenerate plane because it is actually a line. 

⎛ ⎞ ⎛ ⎞ 

1 1 

⎜ 

L = { ⎜ 0 ⎟ ⎜ 

⎟ 

⎝−1⎠ 

+ r ⎜ 1 ⎟ � 

⎟ � 

⎝ 0 ⎠ r ∈ R} 

−2 −1 

We shall see in the Linear Independence section of Chapter Two what relationships 

among vectors causes the linear surface they generate to be degenerate. 

We finish this subsection by restating our conclusions from the first section 

in geometric terms. First, the solution set of a linear system with n unknowns 

is a linear surface in R n . Specifically, it is an k-dimensional linear surface, 

where k is the number of free variables in an echelon form version of the system. 

Second, the solution set of a homogeneous linear system is a linear surface 

passing through the origin. Finally, we can view the general solution set of any 

linear system as being the solution set of its associated homogeneous system 

offset from the origin by a vector, namely by any particular solution. 

Exercises 

� 1.1 Find the canonical name for each vector. 

(a) the vector from (2, 1) to (4, 2) in R 2 

(b) the vector from (3, 3) to (2, 5) in R 2 

(c) the vector from (1, 0, 6) to (5, 0, 3) in R 3 

(d) the vector from (6, 8, 8) to (6, 8, 8) in R 3


� 1.2 Decide if the two vectors are equal. 

(a) the vector from (5, 3) to (6, 2) and the vector from (1, −2) to (1, 1) 

(b) the vector from (2, 1, 1) to (3, 0, 4) and the vector from (5, 1, 4) to (6, 0, 7) 

� 1.3 Does (1, 0, 2, 1) lie on the line through (−2, 1, 1, 0) and (5, 10, −1, 4)? 

� 1.4 (a) Describe the plane through (1, 1, 5, −1), (2, 2, 2, 0), and (3, 1, 0, 4). 

(b) Is the origin in that plane? 

1.5 Describe the plane that contains this point and line. 

� � � � � � 

2 −1 1 

0 { 0 + 1 t 

3 −4 2 

� � t ∈ R} 

� 1.6 Intersect these planes. 

� � � � 

1 0 

{ 1 t + 1 s 

1 3 

� � t, s ∈ R} 

� � � � � � 

1 0 2 

{ 1 + 3 k + 0 m 

0 0 4 

� � k, m ∈ R} 

� 1.7 Intersect each pair, if possible. 

� � � � � � � � 

1 0 �� 1 0 �� 

(a) { 1 + t 1 t ∈ R}, { 3 + s 1 s ∈ R} 

2 1 

−2 2 

� � � � � � � � 

2 1 �� 0 0 �� 

(b) { 0 + t 1 t ∈ R}, {s 1 + w 4 s, w ∈ R} 

1 −1 

2 1 

1.8 Show that the line segments (a1,a2)(b1,b2) and(c1,c2)(d1,d2) have the same 

lengths and slopes if b1 − a1 = d1 − c1 and b2 − a2 = d2 − c2. Is that only if? 

1.9 How should R 0 be defined? 

� 1.10 [Math. Mag., Jan. 1957] A person traveling eastward at a rate of 3 miles per 

hour finds that the wind appears to blow directly from the north. On doubling his 

speed it appears to come from the north east. What was the wind’s velocity? 

1.11 Euclid describes a plane as “a surface which lies evenly with the straight lines 

on itself”. Commentators (e.g., Heron) have interpreted this to mean “(A plane 

surface is) such that, if a straight line pass through two points on it, the line 

coincides wholly with it at every spot, all ways”. (Translations from [Heath], pp. 

171-172.) Do planes, as described in this section, have that property? Does this 

description adequately define planes? 

1.II.2 Length and Angle Measures 

We’ve translated the first section’s results about solution sets into geometric 

terms for insight into how those sets look. But we must watch out not to be 

mislead by our own terms; labeling subsets of R k of the forms {�p + t�v � � t ∈ R} 

and {�p + t�v + s�w � � t, s ∈ R} as “lines” and “planes” doesn’t make them act like 

the lines and planes of our prior experience. Rather, we must ensure that the 

names suit the sets. While we can’t prove that the sets satisfy our intuition — 

we can’t prove anything about intuition — in this subsection we’ll observe that


a result familiar from R 2 and R 3 , when generalized to arbitrary R k , supports 

the idea that a line is straight and a plane is flat. Specifically, we’ll see how to 

do Euclidean geometry in a “plane” by giving a definition of the angle between 

two R n vectors in the plane that they generate. 

2.1 Definition The length of a vector �v ∈ R n is this. 

��v � = 

� 

v 2 1 + ···+ v2 n 

2.2 Remark This is a natural generalization of the Pythagorean Theorem. A 

classic discussion is in [Polya]. 

We can use that definition to derive a formula for the angle between two 

vectors. For a model of what to do, consider two vectors in R3 . 

�v 

Put them in canonical position and, in the plane that they determine, consider 

the triangle formed by �u, �v, and�u − �v. 

To that triangle, apply the Law of Cosines, 

��u − �v � 2 = ��u � 2 + ��v � 2 − 2 ��u ��v � cos θ 

where θ is the angle between �u and �v. Expand both sides 

(u1 − v1) 2 +(u2 − v2) 2 +(u3 − v3) 2 

and simplify. 

�u 

=(u 2 1 + u 2 2 + u 2 3)+(v 2 1 + v 2 2 + v 2 3) − 2 ��u ��v � cos θ 

θ = arccos( u1v1 + u2v2 + u3v3 

) 

��u ��v � 

In higher dimensions no picture suffices but we can make the same argument 

analytically. First, the form of the numerator is clear — it comes from the middle 

terms of the squares (u1 − v1) 2 ,(u2 − v2) 2 ,etc. 

2.3 Definition The dot product (or inner product, orscalar product) of two 

n-component real vectors is the linear combination of their components. 

�u �v = u1v1 + u2v2 + ···+ unvn


Notice that the dot product of two vectors is a real number, not a vector, and 

that the dot product of a vector from R n with a vector from R m is defined 

only when n equals m. Notice also this relationship between dot product and 

length: dotting a vector with itself gives its length squared �u �u = u1u1 + ···+ 

unun = ��u � 2 . 

2.4 Remark The wording in that definition allows one or both of the two to 

be a row vector instead of a column vector. Some books require that the first 

vector be a row vector and that the second vector be a column vector. We shall 

not be that strict. 

Still reasoning with letters, but guided by the pictures, we use the next 

theorem to argue that the triangle formed by �u, �v, and�u − �v in R n lies in the 

planar subset of R n generated by �u and �v. 

2.5 Theorem (Triangle Inequality) For any �u, �v ∈ R n , 

��u + �v �≤��u � + ��v � 

with equality if and only if one of the vectors is a nonnegative scalar multiple 

of the other one. 

This inequality is the source of the familiar saying, “The shortest distance 

between two points is in a straight line.” 

start . 

�u + �v 

�u 

. finish 

Proof. We’ll use some algebraic properties of dot product that we have not 

shown, for instance that �u·(�a+ � b)=�u·�a+�u· � b and that �u·�v = �v ·�u. Verification 

of those properties is Exercise 17. The desired inequality holds if and only if its 

square holds. 

��u + �v � 2 ≤ (��u� + ��v�) 2 

�v 

(�u + �v) (�u + �v) ≤��u � 2 +2��u ��v � + ��v � 2 

�u �u + �u �v + �v �u + �v �v ≤ �u �u +2��u ��v � + �v �v 

2 �u �v ≤ 2 ��u ��v � 

That, in turn, holds if and only if the relationship obtained by multiplying both 

sides by the nonnegative numbers ��u � and ��v � 

and rewriting 

2(��v ��u) (��u ��v) ≤ 2 ��u � 2 ��v � 2 

0 ≤��u � 2 ��v � 2 − 2(��v ��u) (��u ��v)+��u � 2 ��v � 2


is true. But factoring 

0 ≤ (��u ��v −��v ��u) (��u ��v −��v ��u) 

shows that this certainly is true since it only says that the square of the length 

of the vector ��u ��v −��v ��u is not negative. 

As for equality, it holds when, and only when, ��u ��v −��v ��u is �0. The check 

that ��u ��v = ��v ��u if and only if one vector is a nonnegative real scalar multiple 

of the other is easy. QED 

This result supports the intuition that even in higher-dimensional spaces, 

lines are straight and planes are flat. For any two points in a linear surface, the 

line segment connecting them is contained in that surface (this is easily checked 

from the definition). But if the surface has a bend then that would allow for a 

shortcut (shown here dotted, while the line segment from P to Q, contained in 

the linear surface, is solid). 

. P 

. Q 

Because the Triangle Inequality says that in any R n , the shortest cut between 

two endpoints is simply the line segment connecting them, linear surfaces have 

no such bends. 

Back to the definition of angle measure. The heart of the Triangle Inequality’s 

proof is the ‘�u · �v ≤��u ��v �’ line. At first glance, a reader might wonder 

if some pairs of vectors satisfy the inequality in this way: while �u · �v is a large 

number, with absolute value bigger than the right-hand side, it is a negative 

large number. The next result says that no such pair of vectors exists. 

2.6 Corollary (Cauchy-Schwartz Inequality) For any �u, �v ∈ R n , 

|�u · �v |≤��u ��v � 

with equality if and only if one vector is a scalar multiple of the other. 

Proof. The Triangle Inequality’s proof shows that �u �v ≤��u ��v � so if �u �v is 

positive or zero then we are done. If �u �v is negative then this holds. 

|�u �v| = −(�u �v) =(−�u) �v ≤�−�u ��v � = ��u ��v � 

The equality condition is Exercise 18. QED 

The Cauchy-Schwartz inequality assures us that the next definition makes 

sense because the fraction has absolute value less than or equal to one.


2.7 Definition The angle between two nonzero vectors �u, �v ∈ R n is 

�u �v 

θ = arccos( 

��u ��v � ) 

(the angle between the zero vector and any other vector is defined to be a right 

angle). 

Thus vectors from R n are orthogonal if and only if their dot product is zero. 

2.8 Example These vectors are orthogonal. 

� � � � 

1 1 

−1 1 

Although they are shown away from canonical position so that they don’t appear 

to touch, nonetheless they are orthogonal. 

2.9 Example The R 3 angle formula given at the start of this subsection is a 

special case of the definition. Between these two 

the angle is 

arccos( 

� � 

1 

1 

0 

(1)(0) + (1)(3) + (0)(2) 

� � 

0 

3 

2 

√ 1 2 +1 2 +0 2 √ 0 2 +3 2 +2 

=0 

3 

) = arccos( √ √ ) 

2 2 13 

approximately 0.94 radians. Notice that these vectors are not orthogonal. Although 

the yz-plane may appear to be perpendicular to the xy-plane, in fact 

the two planes are that way only in the weak sense that there are vectors in each 

orthogonal to all vectors in the other. Not every vector in each is orthogonal to 

all vectors in the other. 

Exercises 

� 2.10 Find the length of each vector. 

� � � � � � 

4 

3 

−1 

(a) (b) 

(c) 1 

1 

2 

1 

(d) 

� � 

0 

0 

0 

� 2.11 Find the angle between each two, if it is defined. 

(e) 

⎛ ⎞ 

1 

⎜−1⎟ 

⎝ 

1 

⎠ 

0


(a) 

� � � � 

1 1 

, 

2 4 

(b) 

� � 

1 

2 

0 

, 

� � 

0 

4 

1 

(c) 

� � � � 

1 

1 

, 4 

2 

−1 

� 2.12 During maneuvers preceding the Battle of Jutland, the British battle cruiser 

Lion moved as follows (in nautical miles): 1.2 miles north, 6.1 miles 38 degrees 

east of south, 4.0 miles at 89 degrees east of north, and 6.5 miles at 31 degrees 

east of north. Find the distance between starting and ending positions. 

2.13 Find k so that these two vectors are perpendicular. 

� � � � 

k 4 

1 3 

2.14 Describe the set of vectors in R 3 orthogonal to this one. 

� � 

1 

3 

−1 

� 2.15 (a) Find the angle between the diagonal of the unit square in R 2 and one of 

the axes. 

(b) Find the angle between the diagonal of the unit cube in R 3 and one of the 

axes. 

(c) Find the angle between the diagonal of the unit cube in R n and one of the 

axes. 

(d) What is the limit, as n goes to ∞, of the angle between the diagonal of the 

unit cube in R n and one of the axes? 

2.16 Is there any vector that is perpendicular to itself? 

� 2.17 Describe the algebraic properties of dot product. 

(a) Is it right-distributive over addition: (�u + �v) �w = �u �w + �v �w? 

(b) Is is left-distributive (over addition)? 

(c) Does it commute? 

(d) Associate? 

(e) How does it interact with scalar multiplication? 

As always, any assertion must be backed by either a proof or an example. 

2.18 Verify the equality condition in Corollary 2.6, the Cauchy-Schwartz Inequality. 

(a) Show that if �u is a negative scalar multiple of �v then �u �v and �v �u are less 

than or equal to zero. 

(b) Show that |�u �v| = ��u ��v � if and only if one vector is a scalar multiple of 

the other. 

2.19 Suppose that �u �v = �u �w and �u �= �0. Must �v = �w? 

� 2.20 Does any vector have length zero except a zero vector? (If “yes”, produce an 

example. If “no”, prove it.) 

� 2.21 Find the midpoint of the line segment connecting (x1,y1) with(x2,y2) inR 2 . 

Generalize to R n . 

2.22 Show that if �v �= �0 then�v/��v � has length one. What if �v = �0? 

2.23 Show that if r ≥ 0thenr�v is r times as long as �v. Whatifr


2.25 Prove that ��u + �v � 2 + ��u − �v � 2 =2��u � 2 +2��v � 2 . 

2.26 Show that if �x �y =0forevery�ythen �x = �0. 

2.27 Is ��u1 + ···+ �un� ≤��u1� + ···+ ��un�? If it is true then it would generalize 

the Triangle Inequality. 

2.28 What is the ratio between the sides in the Cauchy-Schwartz inequality? 

2.29 Why is the zero vector defined to be perpendicular to every vector? 

2.30 Describe the angle between two vectors in R 1 . 

2.31 Give a simple necessary and sufficient condition to determine whether the 

angle between two vectors is acute, right, or obtuse. 

� 2.32 Generalize to R n the converse of the Pythagorean Theorem, that if �u and �v 

are perpendicular then ��u + �v � 2 = ��u � 2 + ��v � 2 . 

2.33 Show that ��u � = ��v � if and only if �u + �v and �u − �v are perpendicular. Give 

an example in R 2 . 

2.34 Show that if a vector is perpendicular to each of two others then it is perpendicular 

to each vector in the plane they generate. (Remark. They could generate 

a degenerate plane — a line or a point — but the statement remains true.) 

2.35 Prove that, where �u, �v ∈ R n are nonzero vectors, the vector 

�u �v 

+ 

��u � ��v � 

bisects the angle between them. Illustrate in R 2 . 

2.36 Verify that the definition of angle is dimensionally correct: (1) if k>0then 

the cosine of the angle between k�u and �v equals the cosine of the angle between 

�u and �v, and(2)ifk

Section III. Reduced Echelon Form 45 

1.III Reduced Echelon Form 

After developing the mechanics of Gauss’ method, we observed that it can be 

done in more than one way. One example is that we sometimes have to swap 

rows and there can be more than one row to choose from. Another example is 

that from this matrix 

� � 

2 2 

4 3 

Gauss’ method could derive any of these echelon form matrices. 

� 

2 

� 

2 

� 

1 

� 

1 

� 

2 

� 

0 

0 −1 0 −1 0 −1 

The first results from −2ρ1 + ρ2. The second comes from following (1/2)ρ1 with 

−4ρ1 + ρ2. The third comes from −2ρ1 + ρ2 followed by 2ρ2 + ρ1 (after the first 

pivot the matrix is already in echelon form so the second one is extra work but 

it is nonetheless a legal row operation). 

The fact that the echelon form outcome of Gauss’ method is not unique 

leaves us with some questions. Will any two echelon form versions of a system 

have the same number of free variables? Will they in fact have exactly the same 

variables free? In this section we will answer both questions “yes”. We will 

do more than answer the questions. We will give a way to decide if one linear 

system can be derived from another by row operations. The answers to the two 

questions will follow from this larger result. 

1.III.1 Gauss-Jordan Reduction 

Gaussian elimination coupled with back-substitution solves linear systems, 

but it’s not the only method possible. Here is an extension of Gauss’ method 

that has some advantages. 

1.1 Example To solve 

x + y − 2z = −2 

y +3z = 7 

x − z = −1 

we can start by going to echelon form as usual. 

⎛ 

1 

−ρ1+ρ3 

−→ ⎝0 1 

1 

−2 

3 

⎞ 

−2 

7 ⎠ 

0 −1 1 1 

ρ2+ρ3 

⎛ 

1 

−→ ⎝0 1 

1 

−2 

3 

⎞ 

−2 

7 ⎠ 

0 0 4 8


We can keep going to a second stage by making the leading entries into ones 

⎛ 

1 

(1/4)ρ3 

−→ ⎝0 1 

1 

−2 

3 

⎞ 

−2 

7 ⎠ 

0 0 1 2 

and then to a third stage that uses the leading entries to eliminate all of the 

other entries in each column by pivoting upwards. 

⎛ 

⎞ 

1 1 0 2 

−3ρ3+ρ2 

−→ ⎝0 1 0 1⎠ 

2ρ3+ρ1 

0 0 1 2 

−ρ2+ρ1 

⎛ 

⎞ 

1 0 0 1 

−→ ⎝0 1 0 1⎠ 

0 0 1 2 

The answer is x =1,y =1,andz =2. 

Note that the pivot operations in the first stage proceed from column one to 

column three while the pivot operations in the third stage proceed from column 

three to column one. 

1.2 Example We often combine the operations of the middle stage into a 

single step, even though they are operations on different rows. 

� � 

� � 

2 1 7 −2ρ1+ρ2 2 1 7 

−→ 

4 −2 6 

0 −4 −8 

� � 

(1/2)ρ1 1 1/2 7/2 

−→ 

(−1/4)ρ2 0 1 2 

� � 

−(1/2)ρ2+ρ1 1 0 5/2 

−→ 

0 1 2 

The answer is x =5/2 andy =2. 

This extension of Gauss’ method is Gauss-Jordan reduction. It goes past 

echelon form to a more refined, more specialized, matrix form. 

1.3 Definition A matrix is in reduced echelon form if, in addition to being in 

echelon form, each leading entry is a one and is the only nonzero entry in its 

column. 

The disadvantage of using Gauss-Jordan reduction to solve a system is that the 

additional row operations mean additional arithmetic. The advantage is that 

the solution set can just be read off. 

In any echelon form, plain or reduced, we can read off when a system has 

an empty solution set because there is a contradictory equation, we can read off 

when a system has a one-element solution set because there is no contradiction 

and every variable is the leading variable in some row, and we can read off when 

a system has an infinite solution set because there is no contradiction and at 

least one variable is free. 

In reduced echelon form we can read off not just what kind of solution set 

the system has, but also its description. Whether or not the echelon form


is reduced, we have no trouble describing the solution set when it is empty, 

of course. The two examples above show that when the system has a single 

solution then the solution can be read off from the right-hand column. In the 

case when the solution set is infinite, its parametrization can also be read off 

of the reduced echelon form. Consider, for example, this system that is shown 

brought to echelon form and then to reduced echelon form. 

⎛ 

⎞ 

2 6 1 2 5 

⎝0 3 1 4 1⎠ 

0 3 1 2 5 

−ρ2+ρ3 

⎛ 

⎞ 

2 6 1 2 5 

−→ ⎝0 3 1 4 1⎠ 

0 0 0 −2 4 

⎛ 

⎞ 

1 0 −1/2 0 −9/2 

−3ρ2+ρ1 

−→ ⎝ ⎠ 

(1/2)ρ1 

−→ 

(1/3)ρ2 

−(1/2)ρ3 

x4 

(4/3)ρ3+ρ2 

−→ 

−ρ3+ρ1 

0 1 1/3 0 3 

0 0 0 1 −2 

Starting with the middle matrix, the echelon form version, back substitution 

produces −2x4 = 4 so that x4 = −2, then another back substitution gives 

3x2 + x3 +4(−2) = 1 implying that x2 = 3 − (1/3)x3, and then the final 

back substitution gives 2x1 + 6(3 − (1/3)x3) +x3 +2(−2) = 5 implying that 

x1 = −(9/2) + (1/2)x3. Thus the solution set is this. 

⎛ ⎞ 

x1 

⎜x2⎟ 

S = { ⎜ ⎟ 

⎝x3⎠ 

= 

⎛ ⎞ 

−9/2 

⎜ 3 ⎟ 

⎝ 0 ⎠ 

−2 

+ 

⎛ ⎞ 

1/2 

⎜ 

⎜−1/3 

⎟ 

⎝ 1 ⎠ 

0 

x3 

� 

� x3 ∈ R} 

Now, considering the final matrix, the reduced echelon form version, note that 

adjusting the parametrization by moving the x3 terms to the other side does 

indeed give the description of this infinite solution set. 

Part of the reason that this works is straightforward. While a set can have 

many parametrizations that describe it, e.g., both of these also describe the 

above set S (take t to be x3/6 andsto be x3 − 1) 

⎛ ⎞ 

−9/2 

⎜ 

{ ⎜ 3 ⎟ 

⎝ 0 ⎠ 

−2 

+ 

⎛ ⎞ 

3 

⎜ 

⎜−2 

⎟ 

⎝ 6 ⎠ 

0 

t � ⎛ ⎞ 

−4 

⎜ 

� t ∈ R} { ⎜8/3 

⎟ 

⎝ 1 ⎠ 

−2 

+ 

⎛ ⎞ 

1/2 

⎜ 

⎜−1/3 

⎟ 

⎝ 1 ⎠ 

0 

s � � s ∈ R} 

nonetheless we have in this book stuck to a convention of parametrizing using 

the unmodified free variables (that is, x3 = x3 instead of x3 =6t). We can 

easily see that a reduced echelon form version of a system is equivalent to a 

parametrization in terms of unmodified free variables. For instance, 

x1 =4− 2x3 

x2 =3− x3 

⇐⇒ 

⎛ 

1 

⎝0 0 

0 

1 

0 

2 

1 

0 

⎞ 

4 

3⎠ 

0 

(to move from left to right we also need to know how many equations are in the 

system). So, the convention of parametrizing with the free variables by solving


each equation for its leading variable and then eliminating that leading variable 

from every other equation is exactly equivalent to the reduced echelon form 

conditions that each leading entry must be a one and must be the only nonzero 

entry in its column. 

Not as straightforward is the other part of the reason that the reduced 

echelon form version allows us to read off the parametrization that we would 

have gotten had we stopped at echelon form and then done back substitution. 

The prior paragraph shows that reduced echelon form corresponds to some 

parametrization, but why the same parametrization? A solution set can be 

parametrized in many ways, and Gauss’ method or the Gauss-Jordan method 

can be done in many ways, so a first guess might be that we could derive many 

different reduced echelon form versions of the same starting system and many 

different parametrizations. But we never do. Experience shows that starting 

with the same system and proceeding with row operations in many different 

ways always yields the same reduced echelon form and the same parametrization 

(using the unmodified free variables). 

In the rest of this section we will show that the reduced echelon form version 

of a matrix is unique. It follows that the parametrization of a linear system in 

terms of its unmodified free variables is unique because two different ones would 

give two different reduced echelon forms. 

We shall use this result, and the ones that lead up to it, in the rest of the 

book but perhaps a restatement in a way that makes it seem more immediately 

useful may be encouraging. Imagine that we solve a linear system, parametrize, 

and check in the back of the book for the answer. But the parametrization there 

appears different. Have we made a mistake, or could these be different-looking 

descriptions of the same set, as with the three descriptions above of S? The prior 

paragraph notes that we will show here that different-looking parametrizations 

(using the unmodified free variables) describe genuinely different sets. 

Here is an informal argument that the reduced echelon form version of a 

matrix is unique. Consider again the example that started this section of a 

matrix that reduces to three different echelon form matrices. The first matrix 

of the three is the natural echelon form version. The second matrix is the same 

as the first except that a row has been halved. The third matrix, too, is just a 

cosmetic variant of the first. The definition of reduced echelon form outlaws this 

kind of fooling around. In reduced echelon form, halving a row is not possible 

because that would change the row’s leading entry away from one, and neither 

is combining rows possible, because then a leading entry would no longer be 

alone in its column. 

This informal justification is not a proof; we have argued that no two different 

reduced echelon form matrices are related by a single row operation step, but 

we have not ruled out the possibility that multiple steps might do. Before we go 

to that proof, we finish this subsection by rephrasing our work in a terminology 

that will be enlightening. 

Many different matrices yield the same reduced echelon form matrix. The 

three echelon form matrices from the start of this section, and the matrix they


were derived from, all give this reduced echelon form matrix. 

� � 

1 0 

0 1 

We think of these matrices as related to each other. The next result speaks to 

this relationship. 

1.4 Lemma Elementary row operations are reversible. 

Proof. For any matrix A, the effect of swapping rows is reversed by swapping 

them back, multiplying a row by a nonzero k is undone by multiplying by 1/k, 

and adding a multiple of row i to row j (with i �= j) is undone by subtracting 

the same multiple of row i from row j. 

A ρi↔ρj 

−→ ρj↔ρi 

−→ A A kρi 

−→ (1/k)ρi 

−→ A A kρi+ρj 

−→ −kρi+ρj 

−→ A 

(The i �= j conditions is needed. See Exercise 13.) QED 

This lemma suggests that ‘reduces to’ is misleading — where A −→ B, we 

shouldn’t think of B as “after” A or “simpler than” A. Instead we should think 

of them as interreducible or interrelated. Below is a picture of the idea. The 

matrices from the start of this section and their reduced echelon form version 

are shown in a cluster. They are all related; some of the interrelationships are 

shown also. 

� � 

2 2 

4 3 

↔↔ 

↔↔↔ 

� � 

2 0 

0 −1 

↔↔ 

� � 

1 0 

0 1 

↔↔ 

↔↔↔ 

↔↔↔ 

� � 

1 1 

0 −1 

� � 

2 2 

0 −1 

The technical phrase in this situation is that matrices that reduce to each other 

are ‘equivalent with respect to the relationship of row reducibility’. The next 

result verifies this statement using the definition of an equivalence. ∗ 

1.5 Lemma Between matrices, ‘reduces to’ is an equivalence relation. 

Proof. We must check the conditions (i) reflexivity, that any matrix reduces to 

itself, (ii) symmetry, that if A reduces to B then B reduces to A, and (iii) transitivity, 

that if A reduces to B and B reduces to C then A reduces to C. 

Reflexivity is easy; any matrix reduces to itself in zero row operations. 

That the relationship is symmetric is Lemma 1.4 —ifA reduces to B by 

some row operations then also B reduces to A by reversing those operations. 

For transitivity, suppose that A reduces to B and that B reduces to C. 

Linking the reduction steps from A → ··· → B with those from B → ··· → C 

gives a reduction from A to C. QED 

∗ More information on equivalence relations is in the appendix. 

↔


1.6 Definition Two matrices that are interreducible by the elementary row 

operations are row equivalent. 

The diagram below has the collection of all matrices as a box. Inside that 

box, each matrix lies in some class. Matrices are in the same class if and only if 

they are interreducible. The classes are disjoint — no matrix is in two distinct 

classes. The collection of matrices has been partitioned into row equivalence 

classes. One of the reasons that showing the row equivalence relation is an 

equivalence is useful is that any equivalence relation gives rise to a partition. ∗ 

All matrices: 

.A 

✥ 

✪ 

✩ 

✦ ✜ 

✢ 

... 

B. 

A row equivalent 

to B. 

One of the classes in this partition is the cluster of matrices shown above, 

expanded to include all of the nonsingular 2×2 matrices. 

The next subsection proves that the reduced echelon form of a matrix is 

unique; that every matrix reduces to one and only one reduced echelon form 

matrix. Rephrased in the relation language, we shall prove that every matrix is 

row equivalent to one and only one reduced echelon form matrix. In terms of the 

partition in the picture what we shall prove is: every equivalence class contains 

one and only one reduced echelon form matrix. So each reduced echelon form 

matrix serves as a representative of its class. 

After that proof we shall, as mentioned in the introduction to this section, 

have a way to decide if one matrix can be derived from another by row reduction. 

We can just apply the Gauss-Jordan procedure to both and see whether or not 

they come to the same reduced echelon form. 

Exercises 

� 1.7 Use Gauss-Jordan reduction to solve each system. 

(a) x + y =2 (b) x − z =4 (c) 3x − 2y = 1 

x − y =0 2x +2y =1 6x + y =1/2 

(d) 2x − y = −1 

x +3y − z = 5 

y +2z = 5 

� 1.8 Find the reduced echelon form of each matrix. 

� � � � � 

1 3 1 

1 

2 1 

(a) 

(b) 2 0 4 (c) 1 

1 3 

−1 −3 −3 

3 

� � 

0 1 3 2 

0 

4 

4 

3 

2 

8 

1 

1 

1 

� 

2 

5 

2 

(d) 0 0 5 6 

1 5 1 5 

� 1.9 Find each solution set by using Gauss-Jordan reduction, then reading off the 

parametrization. 

∗ More information on partitions and class representatives is in the appendix.


(a) 2x + y − z =1 

4x − y =3 

(b) x − z =1 

y +2z − w =3 

x +2y +3z − w =7 

(d) a +2b +3c + d − e =1 

3a − b + c + d + e =3 

1.10 Give two distinct echelon form versions of this matrix. 

� � 

2 1 1 3 

6 4 1 2 

1 5 1 5 

(c) x − y + z =0 

y + w =0 

3x − 2y +3z + w =0 

−y − w =0 

� 1.11 List the reduced echelon forms possible for each size. 

(a) 2×2 (b) 2×3 (c) 3×2 (d) 3×3 

� 1.12 What results from applying Gauss-Jordan reduction to a nonsingular matrix? 

1.13 The proof of Lemma 1.4 contains a reference to the i �= j condition on the 

row pivoting operation. 

(a) The definition of row operations has an i �= j condition on the swap operation 

ρi ↔ ρj. Show that in A ρi↔ρj −→ ρi↔ρj −→ A this condition is not needed. 

(b) Write down a 2×2 matrix with nonzero entries, and show that the −1·ρ1 +ρ1 

operation is not reversed by 1 · ρ1 + ρ1. 

(c) Expand the proof of that lemma to make explicit exactly where the i �= j 

condition on pivoting is used. 

1.III.2 Row Equivalence 

We will close this section and this chapter by proving that every matrix is 

row equivalent to one and only one reduced echelon form matrix. The ideas 

that appear here will reappear, and be further developed, in the next chapter. 

The underlying theme here is that one way to understand a mathematical 

situation is by being able to classify the cases that can happen. We have met this 

theme several times already. We have classified solution sets of linear systems 

into the no-elements, one-element, and infinitely-many elements cases. We have 

also classified linear systems with the same number of equations as unknowns 

into the nonsingular and singular cases. We adopted these classifications because 

they give us a way to understand the situations that we were investigating. Here, 

where we are investigating row equivalence, we know that the set of all matrices 

breaks into the row equivalence classes. When we finish the proof here, we will 

have a way to understand each of those classes — its matrices can be thought 

of as derived by row operations from the unique reduced echelon form matrix 

in that class. 

To understand how row operations act to transform one matrix into another, 

we consider the effect that they have on the parts of a matrix. The crucial 

observation is that row operations combine the rows linearly.


2.1 Definition A linear combination of x1,... ,xm is an expression of the form 

c1x1 + c2x2 + ··· + cmxm where the c’s are scalars. 

(We have already used the phrase ‘linear combination’ in this book. The meaning 

is unchanged, but the next result’s statement makes a more formal definition 

in order.) 

2.2 Lemma (Linear Combination Lemma) A linear combination of linear 

combinations is a linear combination. 

Proof. Given the linear combinations c1,1x1 + ···+ c1,nxn through cm,1x1 + 

···+ cm,nxn, consider a combination of those 

d1(c1,1x1 + ···+ c1,nxn) +···+ dm(cm,1x1 + ···+ cm,nxn) 

where the d’s are scalars along with the c’s. Distributing those d’s and regrouping 

gives 

= d1c1,1x1 + ···+ d1c1,nxn + d2c2,1x1 + ···+ dmc1,1x1 + ···+ dmc1,nxn 

=(d1c1,1 + ···+ dmcm,1)x1 + ···+(d1c1,n + ···+ dmcm,n)xn 

which is indeed a linear combination of the x’s. QED 

In this subsection we will use the convention that, where a matrix is named 

with an upper case roman letter, the matching lower-case greek letter names 

the rows. 

⎛ 

⎞ ⎛ 

⎞ 

α1 

β1 

⎜ 

⎟ ⎜ 

⎟ 

⎜ 

⎟ 

⎜ 

A = ⎜ 

⎝ 

α2 

. 

αm 

⎟ 

⎠ 

⎜ 

B = ⎜ 

⎝ 

2.3 Corollary Where one matrix row reduces to another, each row of the 

second is a linear combination of the rows of the first. 

The proof below uses induction on the number of row operations used to 

reduce one matrix to the other. Before we proceed, here is an outline of the argument 

(readers unfamiliar with induction may want to compare this argument 

with the one used in the ‘General = Particular + Homogeneous’ proof). ∗ First, 

for the base step of the argument, we will verify that the proposition is true 

when reduction can be done in zero row operations. Second, for the inductive 

step, we will argue that if being able to reduce the first matrix to the second 

in some number t ≥ 0 of operations implies that each row of the second is a 

linear combination of the rows of the first, then being able to reduce the first to 

the second in t + 1 operations implies the same thing. Together, this base step 

and induction step prove this result because by the base step the proposition 

β2 

. 

βm 

∗ More information on mathematical induction is in the appendix. 

⎟ 

⎠


is true in the zero operations case, and by the inductive step the fact that it is 

true in the zero operations case implies that it is true in the one operation case, 

and the inductive step applied again gives that it is therefore true in the two 

operations case, etc. 

Proof. We proceed by induction on the minimum number of row operations 

that take a first matrix A to a second one B. 

In the base step, that zero reduction operations suffice, the two matrices 

are equal and each row of B is obviously a combination of A’s rows: � βi = 

0 · �α1 + ···+1· �αi + ···+0· �αm. 

For the inductive step, assume the inductive hypothesis: with t ≥ 0, if a 

matrix can be derived from A in t or fewer operations then its rows are linear 

combinations of the A’s rows. Consider a B that takes t+1 operations. Because 

there are more than zero operations, there must be a next-to-last matrix G so 

that A −→ · · · −→ G −→ B. This G is only t operations away from A and so the 

inductive hypothesis applies to it, that is, each row of G is a linear combination 

of the rows of A. 

If the last operation, the one from G to B, isarowswapthentherows 

of B are just the rows of G reordered and thus each row of B is also a linear 

combination of the rows of A. The other two possibilities for this last operation, 

that it multiplies a row by a scalar and that it adds a multiple of one row to 

another, both result in the rows of B being linear combinations of the rows of 

G. But therefore, by the Linear Combination Lemma, each row of B is a linear 

combination of the rows of A. 

With that, we have both the base step and the inductive step, and so the 

proposition follows. QED 

2.4 Example In the reduction 

� 

0 

1 

� � 

2 ρ1↔ρ2 1 

−→ 

1 0 

� 

1 (1/2)ρ2 

−→ 

2 

� � 

1 1 −ρ2+ρ1 

−→ 

0 1 

� � 

1 0 

, 

0 1 

call the matrices A, D, G, andB. The methods of the proof show that there 

are three sets of linear relationships. 

δ1 =0· α1 +1· α2 

δ2 =1· α1 +0· α2 

γ1 =0· α1 +1· α2 

γ2 =(1/2)α1 +0· α2 

β1 =(−1/2)α1 +1· α2 

β2 =(1/2)α1 +0· α2 

The prior result gives us the insight that Gauss’ method works by taking 

linear combinations of the rows. But to what end; why do we go to echelon 

form as a particularly simple, or basic, version of a linear system? The answer, 

of course, is that echelon form is suitable for back substitution, because we have 

isolated the variables. For instance, in this matrix 

⎛ 

⎞ 

2 3 7 8 0 0 

⎜ 

R = ⎜0 

0 1 5 1 1 ⎟ 

⎝0 

0 0 3 3 0⎠ 

0 0 0 0 2 1


x1 has been removed from x5’s equation. That is, Gauss’ method has made x5’s 

row independent of x1’s row. 

Independence of a collection of row vectors, or of any kind of vectors, will 

be precisely defined and explored in the next chapter. But a first take on it is 

that we can show that, say, the third row above is not comprised of the other 

rows, that ρ3 �= c1ρ1 + c2ρ2 + c4ρ4. For, suppose that there are scalars c1, c2, 

and c4 such that this relationship holds. 

� 

0 0 0 3 3 

� � 

0 = c1 2 3 7 8 0 

� 

0 

� � 

+ c2 0 0 1 5 1 1 

� � 

+ c4 0 0 0 0 2 1 

The first row’s leading entry is in the first column and narrowing our consideration 

of the above relationship to consideration only of the entries from the first 

column 0 = 2c1+0c2+0c4 gives that c1 = 0. The second row’s leading entry is in 

the third column and the equation of entries in that column 0 = 7c1 +1c2 +0c4, 

along with the knowledge that c1 = 0, gives that c2 = 0. Now, to finish, the 

third row’s leading entry is in the fourth column and the equation of entries 

in that column 3 = 8c1 +5c2 +0c4, along with c1 =0andc2 = 0, gives an 

impossibility. 

The following result shows that this effect always holds. It shows that what 

Gauss’ linear elimination method eliminates is linear relationships among the 

rows. 

2.5 Lemma In an echelon form matrix, no nonzero row is a linear combination 

of the other rows. 

Proof. Let R be in echelon form. Suppose, to obtain a contradiction, that 

some nonzero row is a linear combination of the others. 

ρi = c1ρ1 + ...+ ci−1ρi−1 + ci+1ρi+1 + ...+ cmρm 

We will first use induction to show that the coefficients c1, ... , ci−1 associated 

with rows above ρi are all zero. The contradiction will come from consideration 

of ρi and the rows below it. 

The base step of the induction argument is to show that the first coefficient 

c1 is zero. Let the first row’s leading entry be in column number ℓ1 be the 

column number of the leading entry of the first row and consider the equation 

of entries in that column. 

ρi,ℓ1 

= c1ρ1,ℓ1 + ...+ ci−1ρi−1,ℓ1 + ci+1ρi+1,ℓ1 + ...+ cmρm,ℓ1 

The matrix is in echelon form so the entries ρ2,ℓ1 , ... , ρm,ℓ1 , including ρi,ℓ1 ,are 

all zero. 

0=c1ρ1,ℓ1 + ···+ ci−1 · 0+ci+1 · 0+···+ cm · 0 

Because the entry ρ1,ℓ1 is nonzero as it leads its row, the coefficient c1 must be 

zero.


The inductive step is to show that for each row index k between 1 and i − 2, 

if the coefficient c1 and the coefficients c2, ... , ck are all zero then ck+1 is also 

zero. That argument, and the contradiction that finishes this proof, is saved for 

Exercise 21. QED 

We can now prove that each matrix is row equivalent to one and only one 

reduced echelon form matrix. We will find it convenient to break the first half 

of the argument off as a preliminary lemma. For one thing, it holds for any 

echelon form whatever, not just reduced echelon form. 

2.6 Lemma If two echelon form matrices are row equivalent then the leading 

entries in their first rows lie in the same column. The same is true of all the 

nonzero rows — the leading entries in their second rows lie in the same column, 

etc. 

For the proof we rephrase the result in more technical terms. Define the form 

of an m×n matrix to be the sequence 〈ℓ1,ℓ2,... ,ℓm〉 where ℓi is the column 

number of the leading entry in row i and ℓi = ∞ if there is no leading entry 

in that column. The lemma says that if two echelon form matrices are row 

equivalent then their forms are equal sequences. 

Proof. Let B and D be echelon form matrices that are row equivalent. Because 

they are row equivalent they must be the same size, say n×m. Let the column 

number of the leading entry in row i of B be ℓi and let the column number of 

the leading entry in row j of D be kj. We will show that ℓ1 = k1, that ℓ2 = k2, 

etc., by induction. 

This induction argument relies on the fact that the matrices are row equivalent, 

because the Linear Combination Lemma and its corollary therefore give 

that each row of B is a linear combination of the rows of D and vice versa: 

βi = si,1δ1 + si,2δ2 + ···+ si,mδm and δj = tj,1β1 + tj,2β2 + ···+ tj,mβm 

where the s’s and t’s are scalars. 

The base step of the induction is to verify the lemma for the first rows of 

the matrices, that is, to verify that ℓ1 = k1. If either row is a zero row then 

the entire matrix is a zero matrix since it is in echelon form, and hterefore both 

matrices are zero matrices (by Corollary 2.3), and so both ℓ1 and k1 are ∞. For 

the case where neither β1 nor δ1 is a zero row, consider the i = 1 instance of 

the linear relationship above. 

β1 = s1,1δ1 + s1,2δ2 + ···+ s1,mδm 

� � � � 

0 ··· b1,ℓ1 ··· = s1,1 0 ··· d1,k1 ··· 

� � 

+ s1,2 0 ··· 0 ··· 

. 

� � 

+ s1,m 0 ··· 0 ···


First, note that ℓ1


gives this set of equations for i =1uptoi = r. 

b1,ℓj 

bj,ℓj 

br,ℓj 

= c1,1d1,ℓj + ···+ c1,jdj,ℓj + ···+ c1,rdr,ℓj 

. 

= cj,1d1,ℓj + ···+ cj,jdj,ℓj + ···+ cj,rdr,ℓj 

. 

= cr,1d1,ℓj + ···+ cr,jdj,ℓj + ···+ cr,rdr,ℓj 

Since D is in reduced echelon form, all of the d’s in column ℓj are zero except for 

dj,ℓj , which is 1. Thus each equation above simplifies to bi,ℓj = ci,jdj,ℓj = ci,j ·1. 

But B is also in reduced echelon form and so all of the b’s in column ℓj are zero 

except for bj,ℓj , which is 1. Therefore, each ci,j is zero, except that c1,1 =1, 

and c2,2 =1,... ,andcr,r =1. 

We have shown that the only nonzero coefficient in the linear combination 

labelled (∗) iscj,j, which is 1. Therefore βj = δj. Because this holds for all 

nonzero rows, B = D. QED 

We end with a recap. In Gauss’ method we start with a matrix and then 

derive a sequence of other matrices. We defined two matrices to be related if one 

can be derived from the other. That relation is an equivalence relation, called 

row equivalence, and so partitions the set of all matrices into row equivalence 

classes. 

All matrices: 

✥ 

✪ 

✩ 

✦ ✜ 

. ( ✢ 

... 

13 

27 ) 

. ( 13 

01 ) 

each class 

consists of 

row equivalent 

matrices 

(There are infinitely many matrices in the pictured class, but we’ve only got 

room to show two.) We have proved there is one and only one reduced echelon 

form matrix in each row equivalence class. So the reduced echelon form is a 

canonical form ∗ for row equivalence: the reduced echelon form matrices are 

representatives of the classes. 

All matrices: 

⋆ 

⋆ 

✥ 

✪ 

✩ 

✦ ✜ 

⋆ 

✢ 

⋆ 

... 

( 10 

01 ) 

⋆ 

one reduced 

echelon form matrix 

from each class 

We can answer questions about the classes by translating them into questions 

about the representatives. 

∗ More information on canonical representatives is in the appendix.


2.8 Example We can decide if matrices are interreducible by seeing if Gauss- 

Jordan reduction produces the same reduced echelon form result. Thus, these 

are not row equivalent � 1 −3 

−2 6 

� � 1 −3 

−2 5 

because their reduced echelon forms are not equal. 

� 

1 

� 

−3 

� 

1 

� 

0 

0 0 0 1 

2.9 Example Any nonsingular 3×3 matrix Gauss-Jordan reduces to this. 

⎛ 

1 

⎝0 0 

1 

⎞ 

0 

0⎠ 

0 0 1 

2.10 Example We can describe the classes by listing all possible reduced echelon 

form matrices. Any 2×2 matrix lies in one of these: the class of matrices 

row equivalent to this, 

� � 

0 0 

0 0 

the infinitely many classes of matrices row equivalent to one of this type 

� � 

1 a 

0 0 

where a ∈ R (including a = 0), the class of matrices row equivalent to this, 

� � 

0 1 

0 0 

and the class of matrices row equivalent to this 

� � 

1 0 

0 1 

(this the class of nonsingular 2×2 matrices). 

Exercises 

� 2.11 Decide if the matrices are row equivalent. 

� � � � � � � 

1 0 2 1 

1 2 0 1 

(a) , 

(b) 3 −1 1 , 0 

4 8 1 2 

5 −1 5 2 

� � 

2 1 −1 � � � 

1 0 2 

1 1 

(c) 1 1 0 , 

(d) 

0 2 10 

−1 2 

4 3 −1 

� � � � 

1 1 1 0 1 2 

(e) 

, 

0 0 3 1 −1 1 

� 

0 2 

2 10 

0 4 

� � 

1 0 

, 

2 2 

3 

2 

� 

−1 

5 

2.12 Describe the matrices in each of the classes represented in Example 2.10. 

2.13 Describe all matrices in the row equivalence class of these. 

�


(a) 

� � 

1 0 

0 0 

(b) 

� � 

1 2 

2 4 

(c) 

� � 

1 1 

1 3 

2.14 How many row equivalence classes are there? 

2.15 Can row equivalence classes contain different-sized matrices? 

2.16 How big are the row equivalence classes? 

(a) Show that the class of any zero matrix is finite. 

(b) Do any other classes contain only finitely many members? 

� 2.17 Give two reduced echelon form matrices that have their leading entries in the 

same columns, but that are not row equivalent. 

� 2.18 Show that any two n×n nonsingular matrices are row equivalent. Are any 

two singular matrices row equivalent? 

� 2.19 Describe all of the row equivalence classes containing these. 

(a) 2 × 2matrices (b) 2 × 3matrices (c) 3 × 2 matrices 

(d) 3×3 matrices 

2.20 (a) Show that a vector � β0 is a linear combination of members of the set 

{ � β1,... , � βn} if and only there is a linear relationship �0 =c0� β0 + ··· + cn� βn 

where c0 is not zero. (Watch out for the � β0 = �0 case.) 

(b) Derive Lemma 2.5. 

� 2.21 Finish the proof of Lemma 2.5. 

(a) First illustrate the inductive step by showing that ℓ2 = k2. 

(b) Do the full inductive step: assume that ck is zero for 1 ≤ k


(2) Can any equation be derived from an inconsistent system? 

2.27 Extend the definition of row equivalence to linear systems. Under your definition, 

do equivalent systems have the same solution set? 

� 2.28 In this matrix 

� � 

1 2 3 

3 0 3 

1 4 5 

the first and second columns add to the third. 

(a) Show that remains true under any row operation. 

(b) Make a conjecture. 

(c) Prove that it holds.

Topic: Computer Algebra Systems 61 

Topic: Computer Algebra Systems 

The linear systems in this chapter are small enough that their solution by hand 

is easy. But large systems are easiest, and safest, to do on a computer. There 

are special purpose programs such as LINPACK for this job. Another popular 

tool is a general purpose computer algebra system, including both commercial 

packages such as Maple, Mathematica, or MATLAB, or free packages such as 

SciLab, or Octave. 

For example, in the Topic on Networks, we need to solve this. 

i0 − i1 − i2 

= 0 

i1 − i3 − i5 = 0 

i2 − i4 + i5 = 0 

i3 + i4 − i6 = 0 

5i1 +10i3 =10 

2i2 +4i4 =10 

5i1 − 2i2 +50i5 = 0 

It can be done by hand, but it would take a while and be error-prone. Using a 

computer is better. 

We illustrate by solving that system under Maple (for another system, a 

user’s manual would obviously detail the exact syntax needed). The array of 

coefficients can be entered in this way 

> A:=array( [[1,-1,-1,0,0,0,0], 

[0,1,0,-1,0,-1,0], 

[0,0,1,0,-1,1,0], 

[0,0,0,1,1,0,-1], 

[0,5,0,10,0,0,0], 

[0,0,2,0,4,0,0], 

[0,5,-1,0,0,10,0]] ); 

(putting the rows on separate lines is not necessary, but is done for clarity). 

The vector of constants is entered similarly. 

> u:=array( [0,0,0,0,10,10,0] ); 

Then the system is solved, like magic. 

> linsolve(A,u); 

7 2 5 2 5 7 

[ -, -, -, -, -, 0, - ] 

3 3 3 3 3 3 

Systems with infinitely many solutions are solved in the same way — the computer 

simply returns a parametrization. 

Exercises 

1 Use the computer to solve the two problems that opened this chapter. 

(a) This is the Statics problem. 

40h +15c =100 

25c =50+50h


(b) This is the Chemistry problem. 

7h =7j 

8h +1i =5j +2k 

1i =3j 

3i =6j +1k 

2 Use the computer to solve these systems from the first subsection, or conclude 

‘many solutions’ or ‘no solutions’. 

(a) 2x +2y =5 (b) −x + y =1 (c) x − 3y + z = 1 

x − 4y =0 x + y =2 x + y +2z =14 

(d) −x − y =1 (e) 4y + z =20 (f) 2x + z + w = 5 

−3x − 3y =2 2x − 2y + z = 0 

y − w = −1 

x + z = 5 3x − z − w = 0 

x + y − z =10 4x + y +2z + w = 9 

3 Use the computer to solve these systems from the second subsection. 

(a) 3x +6y =18 (b) x + y = 1 (c) x1 + x3 = 4 

x +2y = 6 x − y = −1 x1 − x2 +2x3 = 5 

4x1 − x2 +5x3 =17 

(d) 2a + b − c =2 (e) x +2y − z =3 (f) x + z + w =4 

2a + c =3 2x + y + w =4 2x + y − w =2 

a − b =0 x − y + z + w =1 3x + y + z =7 

4 What does the computer give for the solution of the general 2×2 system? 

ax + cy = p 

bx + dy = q

Topic: Input-Output Analysis 63 

Topic: Input-Output Analysis 

An economy is an immensely complicated network of interdependences. Changes 

in one part can ripple out to affect other parts. Economists have struggled to 

be able to describe, and to make predictions about, such a complicated object. 

Mathematical models using systems of linear equations have emerged as a key 

tool. One is Input-Output Analysis, pioneered by W. Leontief, who won the 

1973 Nobel Prize in Economics. 

Consider an economy with many parts, two of which are the steel industry 

and the auto industry. As they work to meet the demand for their product from 

other parts of the economy, that is, from users external to the steel and auto 

sectors, these two interact tightly. For instance, should the external demand 

for autos go up, that would lead to an increase in the auto industry’s usage of 

steel. Or, should the external demand for steel fall, then it would lead to a fall 

in steel’s purchase of autos. The type of Input-Output model we will consider 

takes in the external demands and then predicts how the two interact to meet 

those demands. 

We start with a listing of production and consumption statistics. (These 

numbers, giving dollar values in millions, are excerpted from [Leontief 1965], 

describing the 1958 U.S. economy. Today’s statistics would be quite different, 

both because of inflation and because of technical changes in the industries.) 

value of 

steel 

value of 

auto 

used by 

steel 

used by 

auto 

used by 

others total 

5 395 2 664 25 448 

48 9 030 30 346 

For instance, the dollar value of steel used by the auto industry in this year is 

2, 664 million. Note that industries may consume some of their own output. 

We can fill in the blanks for the external demand. This year’s value of the 

steel used by others this year is 17, 389 and this year’s value of the auto used 

by others is 21, 268. With that, we have a complete description of the external 

demands and of how auto and steel interact, this year, to meet them. 

Now, imagine that the external demand for steel has recently been going up 

by 200 per year and so we estimate that next year it will be 17, 589. Imagine 

also that for similar reasons we estimate that next year’s external demand for 

autos will be down 25 to 21, 243. We wish to predict next year’s total outputs. 

That prediction isn’t as simple as adding 200 to this year’s steel total and 

subtracting 25 from this year’s auto total. For one thing, a rise in steel will 

cause that industry to have an increased demand for autos, which will mitigate, 

to some extent, the loss in external demand for autos. On the other hand, the 

drop in external demand for autos will cause the auto industry to use less steel, 

and so lessen somewhat the upswing in steel’s business. In short, these two 

industries form a system, and we need to predict the totals at which the system 

as a whole will settle.


For that prediction, let s be next years total production of steel and let a be 

next year’s total output of autos. We form these equations. 

next year’s production of steel = next year’s use of steel by steel 

+ next year’s use of steel by auto 

+ next year’s use of steel by others 

next year’s production of autos = next year’s use of autos by steel 

+ next year’s use of autos by auto 

+ next year’s use of autos by others 

On the left side of those equations go the unknowns s and a. At the ends of the 

right sides go our external demand estimates for next year 17, 589 and 21, 243. 

For the remaining four terms, we look to the table of this year’s information 

about how the industries interact. 

For instance, for next year’s use of steel by steel, we note that this year the 

steel industry used 5395 units of steel input to produce 25, 448 units of steel 

output. So next year, when the steel industry will produce s units out, we 

expect that doing so will take s · (5395)/(25 448) units of steel input — this is 

simply the assumption that input is proportional to output. (We are assuming 

that the ratio of input to output remains constant over time; in practice, models 

may try to take account of trends of change in the ratios.) 

Next year’s use of steel by the auto industry is similar. This year, the auto 

industry uses 2664 units of steel input to produce 30346 units of auto output. So 

next year, when the auto industry’s total output is a, we expect it to consume 

a · (2664)/(30346) units of steel. 

Filling in the other equation in the same way, we get this system of linear 

equation. 

5 395 2 664 

· s + · a + 17 589 = s 

25 448 30 346 

48 9 030 

· s + · a + 21 243 = a 

25 448 30 346 

Rounding to four decimal places and putting it into the form for Gauss’ method 

gives this. 

0.7880s − 0.0879a = 17 589 

−0.0019s +0.7024a = 21 268 

The solution is s = 25 708 and a = 30 350. 

Looking back, recall that above we described why the prediction of next 

year’s totals isn’t as simple as adding 200 to last year’s steel total and subtracting 

25 from last year’s auto total. In fact, comparing these totals for next year 

to the ones given at the start for the current year shows that, despite the drop 

in external demand, the total production of the auto industry is predicted to 

rise. The increase in internal demand for autos caused by steel’s sharp rise in 

business more than makes up for the loss in external demand for autos.

Topic: Input-Output Analysis 65 

One of the advantages of having a mathematical model is that we can ask 

“What if ... ?” questions. For instance, we can ask “What if the estimates for 

next year’s external demands are somewhat off?” To try to understand how 

much the model’s predictions change in reaction to changes in our estimates, we 

can try revising our estimate of next year’s external steel demand from 17, 589 

down to 17, 489, while keeping the assumption of next year’s external demand 

for autos fixed at 21, 243. The resulting system 

0.7880s − 0.0879a = 17 489 

−0.0019s +0.7024a = 21 243 

when solved gives s = 25 577 and a = 30 314. This kind of exploration of the 

model is sensitivity analysis. We are seeing how sensitive the predictions of our 

model are to the accuracy of the assumptions. 

Obviously, we can consider larger models that detail the interactions among 

more sectors of an economy. These models are typically solved on a computer, 

using the techniques of matrix algebra that we will develop in Chapter Three. 

Some examples are given in the exercises. Obviously also, a single model does 

not suit every case; expert judgment is needed to see if the assumptions underlying 

the model can are reasonable ones to apply to a particular case. With 

those caveats, however, this model has proven in practice to be a useful and accurate 

tool for economic analysis. For further reading, try [Leontief 1951] and 

[Leontief 1965]. 

Exercises 

Hint: these systems are easiest to solve on a computer. 

1 With the steel-auto system given above, estimate next year’s total productions 

in these cases. 

(a) Next year’s external demands are: up 200 from this year for steel, and unchanged 

for autos. 

(b) Next year’s external demands are: up 100 for steel, and up 200 for autos. 

(c) Next year’s external demands are: up 200 for steel, and up 200 for autos. 

2 Imagine a new process for making autos is pioneered. The ratio for use of steel 

by the auto industry falls to .0500 (that is, the new process is more efficient in its 

use of steel). 

(a) How will the predictions for next year’s total productions change compared 

to the first example discussed above (i.e., taking next year’s external demands 

to be 17, 589 for steel and 21, 243 for autos)? 

(b) Predict next year’s totals if, in addition, the external demand for autos rises 

to be 21, 500 because the new cars are cheaper. 

3 This table gives the numbers for the auto-steel system from a different year, 1947 

(see [Leontief 1951]). The units here are billions of 1947 dollars. 

value of 

steel 

value of 

autos 

used by 

steel 

used by 

auto 

used by 

others total 

6.90 1.28 18.69 

0 4.40 14.27


(a) Fill in the missing external demands, and compute the ratios. 

(b) Solve for total output if next year’s external demands are: steel’s demand 

up 10% and auto’s demand up 15%. 

(c) How do the ratios compare to those given above in the discussion for the 

1958 economy? 

(d) Solve these equations with the 1958 external demands (note the difference 

in units; a 1947 dollar buys about what $1.30 in 1958 dollars buys). How far off 

are the predictions for total output? 

4 Predict next year’s total productions of each of the three sectors of the hypothet- 

ical economy shown below 

used by 

used by 

rail 

used by 

shipping 

used by 

others total 

value of 

farm 

farm 

25 50 100 800 

value of 

rail 

25 50 50 300 

value of 

shipping 

15 10 0 500 

if next year’s external demands are as stated. 

(a) 625 for farm, 200 for rail, 475 for shipping 

(b) 650 for farm, 150 for rail, 450 for shipping 

5 This table gives the interrelationships among three segments of an economy (see 

[Clark & Coupe]). 

used by 

food 

used by 

wholesale 

used by 

retail 

used by 

others 

total 

value of 

food 

value of 

0 2 318 4 679 11 869 

wholesale 

value of 

393 1 089 22 459 122 242 

retail 3 53 75 116 041 

We will do an Input-Output analysis on this system. 

(a) Fill in the numbers for this year’s external demands. 

(b) Set up the linear system, leaving next year’s external demands blank. 

(c) Solve the system where next year’s external demands are calculated by taking 

this year’s external demands and inflating them 10%. Do all three sectors 

increase their total business by 10%? 

rate? 

Do they all even increase at the same 

(d) Solve the system where next year’s external demands are calculated by taking 

this year’s external demands and reducing them 7%. (The study from which 

these numbers are taken concluded that because of the closing of a local military 

facility, overall personal income in the area would fall 7%, so this might be a 

first guess at what would actually happen.)

Topic: Accuracy of Computations 67 

Topic: Accuracy of Computations 

Gauss’ method lends itself nicely to computerization. The code below illustrates. 

It operates on an n×n matrix a, pivoting with the first row, then with 

the second row, etc. (This code is in the C language. For readers unfamiliar 

with this concise language, here is a brief translation. The loop construct 

for(pivot row=1;pivot row


4 

3 

2 

1 

0 

-1 

(1,1) 

-1 0 1 2 3 4 

At the scale of this graph, the two lines are hard to resolve apart. This system 

is nearly singular in the sense that the two lines are nearly the same line. Nearsingularity 

gives this system the property that a small change in the system 

can cause a large change in its solution; for instance, changing the 3.000 000 01 

to 3.000 000 03 changes the intersection point from (1, 1) to (3, 0). This system 

changes radically depending on a ninth digit, which explains why the eightplace 

computer is stumped. A problem that is very sensitive to inaccuracy or 

uncertainties in the input values is ill-conditioned. 

The above example gives one way in which a system can be difficult to solve 

on a computer. It has the advantage that the picture of nearly-equal lines 

gives a memorable insight into one way that numerical difficulties can arise. 

Unfortunately, though, this insight isn’t very useful when we wish to solve some 

large system. We cannot, typically, hope to understand the geometry of an 

arbitrary large system. And, in addition, the reasons that the computer’s results 

may be unreliable are more complicated than only that the angle between some 

of the linear surfaces is quite small. 

For an example, consider the system below, from [Hamming]. 

0.001x + y =1 

x − y =0 

The second equation gives x = y, sox = y =1/1.001 and thus both variables 

have values that are just less than 1. A computer using two digits represents 

the system internally in this way (we will do this example in two-digit floating 

point arithmetic, but a similar one with eight digits is easy to invent). 

(1.0 × 10 −2 )x +(1.0 × 10 0 )y =1.0 × 10 0 

(1.0 × 10 0 )x − (1.0 × 10 0 )y =0.0 × 10 0 

The computer’s row reduction step −1000ρ1 + ρ2 produces a second equation 

−1001y = −999, which the computer rounds to two places as (−1.0 × 10 3 )y = 

−1.0 × 10 3 . Then the computer decides from the second equation that y =1 

and from the first equation that x = 0. This y value is fairly good, but the x 

is way off. Thus, another cause of unreliable output is the mixture of floating 

point arithmetic and a reliance on pivots that are small.


An experienced programmer may respond that we should go to double precision 

where, usually, sixteen significant digits are retained. It is true, this will 

solve many problems. However, there are some difficulties with it as a general 

approach. For one thing, double precision takes longer than single precision (on 

a ’486 chip, multiplication takes eleven ticks in single precision but fourteen in 

double precision [Programmer’s Ref.]) and has twice the memory requirements. 

So attempting to do all calculations in double precision is just not practical. And 

besides, the above systems can obviously be tweaked to give the same trouble in 

the seventeenth digit, so double precision won’t fix all problems. What we need 

is a strategy to minimize the numerical trouble arising from solving systems 

on a computer, and some guidance as to how far the reported solutions can be 

trusted. 

Mathematicians have made a careful study of how to get the most reliable 

results. A basic improvement on the naive code above is to not simply take 

the entry in the pivot row , pivot row position for the pivot, but rather to look 

at all of the entries in the pivot row column below the pivot row row, and take 

the one that is most likely to give reliable results (e.g., take one that is not too 

small). This strategy is partial pivoting. For example, to solve the troublesome 

system (∗) above, we start by looking at both equations for a best first pivot, 

and taking the 1 in the second equation as more likely to give good results. 

Then, the pivot step of −.001ρ2 + ρ1 gives a first equation of 1.001y =1,which 

the computer will represent as (1.0×100 )y =1.0×100 , leading to the conclusion 

that y = 1 and, after back-substitution, x = 1, both of which are close to right. 

The code from above can be adapted to this purpose. 

for(pivot_row=1;pivot_row


experts is a variation on the code above that first finds the best pivot among 

the candidates, and then scales it to a number that is less likely to give trouble. 

This is scaled partial pivoting. 

In addition to returning a result that is likely to be reliable, most welldone 

code will return a number, called the conditioning number of the matrix, 

that describes the factor by which uncertainties in the input numbers could be 

magnified to become possible inaccuracies in the results returned (see [Rice]). 

The lesson of this discussion is that just because Gauss’ method always works 

in theory, and just because computer code correctly implements that method, 

and just because the answer appears on green-bar paper, doesn’t mean that the 

answer is reliable. In practice, always use a package where experts have worked 

hard to counter what can go wrong. 

Exercises 

1 Using two decimal places, add 253 and 2/3. 

2 This intersect-the-lines problem contrasts with the example discussed above. 

4 

3 

2 

1 

0 

-1 

(1,1) 

x +2y =3 

3x − 2y =1 

-1 0 1 2 3 4 

Illustrate that, in the resulting system, some small change in the numbers will 

produce only a small change in the solution by changing the constant in the bottom 

equation to 1.008 and solving. Compare it to the solution of the unchanged 

system. 

3 Solve this system by hand ([Rice]). 

0.000 3x +1.556y =1.569 

0.345 4x − 2.346y =1.018 

(a) Solve it accurately, by hand. (b) Solve it by rounding at each step to 

four significant digits. 

4 Rounding inside the computer often has an effect on the result. Assume that 

your machine has eight significant digits. 

(a) Show that the machine will compute (2/3) + ((2/3) − (1/3)) as unequal to 

((2/3) + (2/3)) − (1/3). Thus, computer arithmetic is not associative. 

(b) Compare the computer’s version of (1/3)x + y = 0 and (2/3)x +2y =0. Is 

twice the first equation the same as the second? 

5 Ill-conditioning is not only dependent on the matrix of coefficients. This example 

[Hamming] shows that it can arise from an interaction between the left and right 

sides of the system. Let ε be a small real. 

3x + 2y + z = 6 

2x +2εy +2εz =2+4ε 

x +2εy − εz = 1+ε


(a) Solve the system by hand. Notice that the ε’s divide out only because there 

is an exact cancelation of the integer parts on the right side as well as on the 

left. 

(b) Solve the system by hand, rounding to two decimal places, and with ε = 

0.001.


Topic: Analyzing Networks 

This is the diagram of an electrical circuit. It happens to describe some of the 

connections between a car’s battery and lights, but it is typical of such diagrams. 

To read it, we can think of the electricity as coming out of one end of the battery 

(labeled 6V OR 12V), flowing through the wires (drawn as straight lines to make 

the diagram more readable), and back into the other end of the battery. If, in 

making its way from one end of the battery to the other through the network of 

wires, some electricity flows through a light bulb (drawn as a circle enclosing a 

loop of wire), then that light lights. For instance, when the driver steps on the 

brake at point A then the switch makes contact and electricity flows through 

the brake lights at point B. 

This network of connections and components is complicated enough that to 

analyze it — for instance, to find out how much electricity is used when both 

the headlights and the brake lights are on — then we need systematic tools. 

One such tool is linear systems. To illustrate this application, we first need a 

few facts about electricity and networks. 

The two facts that we need about electricity concern how the electrical components 

act. First, the battery is like a pump for electricity; it provides a force 

or push so that the electricity will flow, if there is at least one available path for 

it. The second fact about the components is the observation that (in the materials 

commonly used in components) the amount of current flow is proportional 

to the force pushing it. For each electrical component there is a constant of 

proportionality, called its resistance, satisfying that potential = flow·resistance. 

(The units are: potential to flow is described in volts, the rate of flow itself is 

given in amperes, and resistance to the flow is in ohms. These units are set up 

so that volts = amperes · ohms.) 

For example, suppose a bulb has a resistance of 25 ohms. Wiring its ends 

to a battery with 12 volts results in a flow of electrical current of 12/25 = 

0.48 amperes. Conversely, with that same bulb, if we have flow of electrical 

current of 2 amperes through it, then the potential difference between one end

Topic: Analyzing Networks 73 

of the bulb and the other end will be 2 · 25 = 50 volts. This is the voltage drop 

across this bulb. One way to think of the above circuit is that the battery is a 

voltage source, or rise, and the other components are voltage sinks, or drops, 

that use up the force provided by the battery. 

The two facts that we need about networks are Kirchhoff’s Laws. 

First Law. The flow into any spot equals the flow out. 

Second Law. Around a circuit the total drop equals the total rise. 

(In the above circuit the only voltage rise is at the one battery, but some circuits 

have more than one rise.) 

We can use these facts for a simple analysis of the circuit shown below. 

There are three components; they might be bulbs, or they might be some other 

component that resists the flow of electricity (resistors are drawn as zig-zags ). 

When components are wired one after another, as these are, they are said to be 

in series. 

20 volt 

potential 

3ohm 

resistance 

2ohm 

resistance 

5ohm 

resistance 

By Kirchhoff’s Second Law, because the voltage rise in this circuit is 20 volts, so 

too, the total voltage drop around this circuit is 20 volts. Since the resistance in 

total, from start to finish, in this circuit is 10 ohms (we can take the resistance 

of a wire to be negligible), we get that the current is (20/10) = 2 amperes. Now, 

Kirchhoff’s First Law says that there are 2 amperes through each resistor, and 

so the voltage drops are 4 volts, 10 volts, and 6 volts. 

Linear systems appear in the analysis of the next network. In this one, the 

resistors are not in series. They are instead in parallel. This network is more 

like the car’s lighting diagram. 

20 volts 12 ohm 8ohm


We begin by labeling the branches of the network. Call the flow of current 

coming out of the top of the battery and through the top wire i0, call the 

current through the left branch of the parallel portion i1, that through the right 

branch i2, and call the current flowing through the bottom wire and into the 

bottom of the battery i3. (Remark: in labeling, we don’t have to know the 

actual direction of flow. We arbitrarily choose a direction to establish a sign 

convention for the equations.) 

i0 

The fact that i0 splits into i1 and i2, on application of Kirchhoff’s First Law, 

gives that i1 + i2 = i0. Similarly, we have that i1 + i2 = i3. In the circuit that 

loops out of the top of the battery, down the left branch of the parallel portion, 

and back into the bottom of the battery, the voltage rise is 20 and the voltage 

drop is i1 ·12, so Kirchoff’s Second Law gives that 12i1 = 20. In the circuit from 

the battery to the right branch and back to the battery there is a voltage rise of 

20 and a voltage drop of i2 ·8, so Kirchoff’s Second law gives that 8i2 = 20. And 

finally, in the circuit that just loops around in the left and right branches of the 

parallel portion (taken clockwise), there is a voltage rise of 0 and a voltage drop 

of 8i2 − 12i1 so Kirchoff’s Second Law gives 8i2 − 12i1 =0. 

All of these equations taken together make this system. 

i3 

i1 

i0 − i1 − i2 = 0 

− i1 − i2 + i3 = 0 

12i1 =20 

8i2 =20 

−12i1 +8i2 = 0 

The solution is i0 =25/6, i1 =5/3, i2 =5/2, and i3 =25/6 (all in amperes). 

(Incidentally, this illustrates that redundant equations do arise in practice, since 

the fifth equation here is redundant.) 

Kirchhoff’s laws can be used to establish the electrical properties of networks 

of great complexity. The next circuit has five resistors, wired in a combination 

of series and parallel. It is said to be a series-parallel circuit. 

i2


10 volts 

5ohm 

10 ohm 

50 ohm 

2ohm 

4ohm 

This circuit is a Wheatstone bridge. It is used to measure the resistance of an 

component placed at, say, the location labeled 5 ohms, against known resistances 

placed in the other positions (see Exercise 7). To analyze it, we can establish 

the arrows in this way. 

i0 

Kirchoff’s First Law, applied to the top node, the left node, the right node, and 

the bottom node gives these equations. 

i0 = i1 + i2 

i1 = i3 + i5 

i2 + i5 = i4 

i3 + i4 = i6 

Kirchhoff’s Second Law, applied to the inside loop (i0-i1-i3-i6), the outside loop, 

and the upper loop not involving the battery, gives these equations. 

i6 

5i1 +10i3 =10 

2i2 +4i4 =10 

5i1 +50i5− 2i2 =0 

We could get more equations, but these are enough to produce a solution: i0 = 

7/3, i1 =2/3, i2 =5/3, i3 =2/3, i4 =5/3, i5 =0,andi6 =7/3. 

Networks of other kinds, not just electrical ones, can also be analyzed in this 

way. For instance, a network of streets in given in the exercises. 

Exercises 

Hint: Most of the linear systems are large enough that they are best solved on a 

computer. 

i1 

i3 

i5 

i2 

i4


1 Calculate the amperages in each part of each network. 

(a) This is a relatively simple network. 

9volt 

3ohm 

2ohm 

2ohm 

(b) Compare this one with the parallel case discussed above. 

9volt 

3ohm 

2ohm 2ohm 

2ohm 

(c) This is a reasonably complicated network. 

9volt 

3ohm 

3ohm 2ohm 

2ohm 

2ohm 

3ohm 

4ohm 

2 Kirchhoff’s laws can apply to a network of streets, as here. On Cape Cod, in 

Massachusetts, there are many intersections that involve traffic circles like this 

one. 

Main St 

North Ave 

Pier Bvd 

Assume the traffic is as below. 

North Pier Main 

into 100 150 25 

out of 75 150 50


We can use Kirchhoff’s Law, that the flow into any intersection equals the flow 

out, to establish some equations modeling how traffic flows work here. 

(a) Label each of the three arcs of road in the circle with a variable. For each of 

the three in-out intersections, get an equation describing the traffic flow at that 

node. 

(b) Solve that system. 

3 This is a map of a network of streets. Below we will describe the flow of cars 

into, and out of, this network. 

Willow 

Shelburne St 

Jay Ln 

west Winooski Ave 

The hourly flow of cars into this network’s entrances, and out of its exits can be 

observed. 

east Winooski west Winooski Willow Jay Shelburne 

into 100 150 25 – 200 

out of 125 150 50 25 125 

(The total in must approximately equal the total out over a long period of time.) 

Once inside the network, the traffic may proceed in different ways, perhaps 

filling Willow and leaving Jay mostly empty, or perhaps flowing in some other 

way. We can use Kirchhoff’s Law that the flow into any intersection equals the 

flow out. 

(a) Determine the restrictions on the flow inside this network of streets by setting 

up a variable for each block, establishing the equations, and solving them. Notice 

that some streets are one-way only. (Hint: this will not yield a unique solution, 

since traffic can flow through this network in various ways. You should get at 

least one free variable.) 

(b) Suppose some construction is proposed for Winooski Avenue East between 

Willow and Jay, so traffic on that block will be reduced. What is the least 

amount of traffic flow that can be allowed on that block without disrupting the 

hourly flow into and out of the network? 

4 Calculate the amperages in this network with more than one voltage rise. 

1.5 volt 

5ohm 

2ohm 3volt 

10 ohm 

east 

3ohm 

6ohm 

5 In the circuit with the 8 ohm and 12 ohm resistors in parallel, the electric current 

away from and back to the battery was found to be 25/6 amperes. Thus, the


parallel pair can be said to be equivalent to a single resistor having a value of 

20/(25/6) = 24/5 =4.8 ohms. 

(a) What is the equivalent resistance if the two resistors in parallel are 8 ohms 

and 5 ohms? Has the equivalent resistance risen or fallen? 

(b) What is the equivalent resistance if the two are both 8 ohms? 

(c) Find the formula for the equivalent resistance R if the two resistors in parallel 

are R1 ohms and R2 ohms. 

(d) What is the formula for more than two resistors in parallel? 

6 In the car dashboard example that begins the discussion, solve for these amperages. 

Assume all resistances are 15 ohms. 

(a) If the driver is stepping on the brakes, so the brake lights are on, and no 

other circuit is closed. 

(b) If all the switches are closed (suppose both the high beams and the low beams 

rate 15 ohms). 

7 Show that, in the Wheatstone Bridge, if r2r6 = r3r5 then i4 =0. (Thewaythis 

device is used in practice is that an unknown resistance, say at r1, iscompared 

to three known resistances. At r3 is placed a meter that shows the current. The 

known resistances are varied until the current is read as 0, and then from the above 

equation the value of the resistor at r1 can be calculated.)

Chapter 2 

Vector Spaces 

The first chapter began by introducing Gauss’ method and finished with a fair 

understanding, keyed on the Linear Combination Lemma, of how it finds the 

solution set of a linear system. Gauss’ method systematically takes linear combinations 

of the rows. With that insight, we now move to a general study of 

linear combinations. 

We need a setting for this study. At times in the first chapter, we’ve combined 

vectors from R 2 , at other times vectors from R 3 , and at other times vectors 

from even higher-dimensional spaces. Thus, our first impulse might be to work 

in R n , leaving n unspecified. This would have the advantage that any of the 

results would hold for R 2 and for R 3 and for many other spaces, simultaneously. 

But, if having the results apply to many spaces at once is advantageous then 

sticking only to R n ’s is overly restrictive. We’d like the results to also apply to 

combinations of row vectors, as in the final section of the first chapter. We’ve 

even seen some spaces that are not just a collection of all of the same-sized 

column vectors or row vectors. For instance, we’ve seen a solution set of a 

homogeneous system that is a plane, inside of R 3 . This solution set is a closed 

system in the sense that a linear combination of these solutions is also a solution. 

But it is not just a collection of all of the three-tall column vectors; only some 

of them are in this solution set. 

We want the results about linear combinations to apply anywhere that linear 

combinations are sensible. We shall call any such set a vector space. Our results, 

instead of being phrased as “Whenever we have a collection in which we can 

sensibly take linear combinations ... ”, will be stated as “In any vector space 

... ”. 

Such a statement describes at once what happens in many spaces. The step 

up in abstraction from studying a single space at a time to studying a class 

of spaces can be hard to make. To understand its advantages, consider this 

analogy. Imagine that the government made laws one person at a time: “Leslie 

Jones can’t jay walk.” That would be a bad idea; statements have the virtue of 

economy when they apply to many cases at once. Or, suppose that they ruled, 

“Kim Ke must stop when passing the scene of an accident.” Contrast that with, 

“Any doctor must stop when passing the scene of an accident.” More general 

statements, in some ways, are clearer. 

79

80 Chapter 2. Vector Spaces 

2.I Definition of Vector Space 

We shall study structures with two operations, an addition and a scalar multiplication, 

that are subject to some simple conditions. We will reflect more on 

the conditions later, but on first reading notice how reasonable they are. For 

instance, surely any operation that can be called an addition (e.g., column vector 

addition, row vector addition, or real number addition) will satisfy all the 

conditions in (1) below. 

2.I.1 Definition and Examples 

1.1 Definition A vector space (over R) consists of a set V along with two 

operations ‘+’ and ‘·’ such that 

(1) if �v, �w ∈ V then their vector sum �v + �w is in V and 

• �v + �w = �w + �v 

• (�v + �w)+�u = �v +(�w + �u) (where �u ∈ V ) 

• there is a zero vector �0 ∈ V such that �v + �0 =�v for all �v ∈ V 

• each �v ∈ V has an additive inverse �w ∈ V such that �w + �v = �0 

(2) if r, s are scalars (members of R) and�v, �w ∈ V then each scalar multiple 

r · �v is in V and 

• (r + s) · �v = r · �v + s · �v 

• r · (�v + �w) =r · �v + r · �w 

• (rs) · �v = r · (s · �v) 

• 1 · �v = �v. 

1.2 Remark Because it involves two kinds of addition and two kinds of multiplication, 

that definition may seem confused. For instance, in ‘(r + s) · �v = 

r · �v + s · �v ’, the first ‘+’ is the real number addition operator while the ‘+’ to 

the right of the equals sign represents vector addition in the structure V . These 

expressions aren’t ambiguous because, e.g., r and s are real numbers so ‘r + s’ 

can only mean real number addition. 

The best way to go through the examples below is to check all of the conditions 

in the definition. That check is written out in the first example. Use 

it as a model for the others. Especially important are the two: ‘�v + �w is in 

V ’ and ‘r · �v is in V ’. These are the closure conditions. They specify that the 

addition and scalar multiplication operations are always sensible — they must 

be defined for every pair of vectors, and every scalar and vector, and the result 

of the operation must be a member of the set (see Example 1.4).

Section I. Definition of Vector Space 81 

1.3 Example The set R2 is a vector space if the operations ‘+’ and ‘·’ have 

their usual meaning. 

� � � � � � � � � � 

x1 y1 x1 + y1 x1 rx1 

+ = 

r · = 

x2 y2 x2 + y2 x2 rx2 

We shall check all of the conditions in the definition. 

There are five conditions in item (1). First, for closure of addition, note that 

for any v1,v2,w1,w2 ∈ R the result of the sum 

� � � � � � 

v1 w1 v1 + w1 

+ = 

v2 

w2 

v2 + w2 

is a column array with two real entries, and so is in R2 . Second, to show that 

addition of vectors commutes, take all entries to be real numbers and compute 

� � � � � � � � � � � � 

v1 w1 v1 + w1 w1 + v1 w1 v1 

+ = 

= 

= + 

v2 

w2 

v2 + w2 

w2 + v2 

(the second equality follows from the fact that the components of the vectors 

are real numbers, and the addition of real numbers is commutative). The third 

condition, associativity of vector addition, is similar. 

� � � � � � � � 

v1 w1 u1 (v1 + w1)+u1 

( + )+ = 

v2 w2 u2 (v2 + w2)+u2 

� � 

v1 +(w1 + u1) 

= 

v2 +(w2 + u2) 

� � � � � � 

v1 w1 u1 

= +( + ) 

For the fourth we must produce a zero element — the vector of zeroes is it. 

� � � � � � 

v1 0 v1 

+ = 

0 

v2 

Fifth, to produce an additive inverse, note that for any v1,v2 ∈ R we have 

� � � � � � 

−v1 v1 0 

+ = 

−v2 v2 0 

so the first vector is the desired additive inverse of the second. 

The checks for the five conditions in item (2) are just as routine. First, for 

closure under scalar multiplication, where r, v1,v2 ∈ R, 

� � � � 

v1 rv1 

r · = 

v2 rv2 

is a column array with two real entries, and so is in R2 . This checks the second 

condition. 

� � � � � � � � � � 

v1 (r + s)v1 rv1 + sv1 v1 v1 

(r + s) · = 

= 

= r · + s · 

v2 (r + s)v2 rv2 + sv2 v2 v2 

v2 

v2 

w2 

w2 

u2 

v2


For the third condition, that scalar multiplication distributes from the left over 

vector addition, the check is also straightforward. 

� � � � � � � � � � � � 

v1 w1 r(v1 + w1) rv1 + rw1 v1 w1 

r · ( + )= 

= 

= r · + r · 

v2 w2 r(v2 + w2) rv2 + rw2 v2 w2 

The fourth 

� � � � � � � � 

v1 (rs)v1 r(sv1) 

v1 

(rs) · = = = r · (s · ) 

v2 (rs)v2 r(sv2) 

v2 

and fifth conditions are also easy. 

� � � � � � 

v1 1v1 v1 

1 · = = 

v2 1v2 v2 

In a similar way, each R n is a vector space with the usual operations of 

vector addition and scalar multiplication. (In R 1 , we usually do not write the 

members as column vectors, i.e., we usually do not write ‘(π)’. Instead we just 

write ‘π’.) 

1.4 Example This subset of R3 that is a plane through the origin 

⎛ ⎞ 

x 

P = { ⎝y⎠ 

z 

� � x + y + z =0} 

is a vector space if ‘+’ and ‘·’ are interpreted in this way. 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

x1 x2 x1 + x2 

x rx 

⎝y1⎠ 

+ ⎝y2⎠ 

= ⎝y1 

+ y2 ⎠ r · ⎝y⎠ 

= ⎝ry⎠ 

z rz 

z1 

z2 

z1 + z2 

The addition and scalar multiplication operations here are just the ones of R3 , 

reused on its subset P .WesayPinherits these operations from R3 .Hereisa 

typical addition in P . 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

1 −1 0 

⎝ 1 ⎠ + ⎝ 0 ⎠ = ⎝ 1 ⎠ 

−2 1 −1 

This illustrates that P is closed under addition. We’ve added two vectors from 

P — that is, with the property that the sum of their three entries is zero — 

and we’ve gotten a vector also in P . Of course, this example of closure is not 

a proof of closure. To prove that P is closed under addition, take two elements 

of P 

⎛ 

⎝ x1 

⎞ ⎛ 

⎠ , ⎝ x2 

⎞ 

⎠ 

y1 

z1 

y2 

z2


(membership in P means that x1 + y1 + z1 =0andx2 + y2 + z2 = 0), and 

observe that their sum 

⎛ ⎞ 

⎠ 

⎝ x1 + x2 

y1 + y2 

z1 + z2 

is also in P since (x1+x2)+(y1+y2)+(z1+z2) =(x1+y1+z1)+(x2+y2+z2) =0. 

To show that P is closed under scalar multiplication, start with a vector from 

P 

⎛ 

⎝ x 

⎞ 

y⎠ 

z 

(so that x + y + z = 0), and then for r ∈ R observe that the scalar multiple 

⎛ ⎞ ⎛ ⎞ 

x rx 

r · ⎝y⎠ 

= ⎝ry⎠ 

z rz 

satisfies that rx + ry + rz = r(x + y + z) = 0. Thus the two closure conditions 

are satisfied. The checks for the other conditions in the definition of a vector 

space are just as easy. 

1.5 Example Example 1.3 shows that the set of all two-tall vectors with real 

entries is a vector space. Example 1.4 gives a subset of an Rn that is also a 

vector space. In contrast with those two, consider the set of two-tall columns 

with entries that are integers (under the obvious operations). This is a subset 

of a vector space, but it is not itself a vector space. The reason is that this set is 

not closed under scalar multiplication, that is, it does not satisfy requirement (2) 

in the definition. Here is a column with integer entries, and a scalar, such that 

the outcome of the operation 

0.5 · 

� � � � 

4 2 

= 

3 1.5 

is not a member of the set, since its entries are not all integers. 

1.6 Example The singleton set 

⎛ ⎞ 

0 

⎜ 

{ ⎜0 

⎟ 

⎝0⎠ 

0 

} 

is a vector space under the operations 

⎛ ⎞ 

0 

⎜ 

⎜0 

⎟ 

⎝0⎠ 

0 

+ 

⎛ ⎞ 

0 

⎜ 

⎜0 

⎟ 

⎝0⎠ 

0 

= 

⎛ ⎞ 

0 

⎜ 

⎜0 

⎟ 

⎝0⎠ 

0 

that it inherits from R 4 . 

⎛ ⎞ 

0 

⎜ 

r · ⎜0 

⎟ 

⎝0⎠ 

0 

= 

⎛ ⎞ 

0 

⎜ 

⎜0 

⎟ 

⎝0⎠ 

0


A vector space must have at least one element, its zero vector. Thus a 

one-element vector space is the smallest one possible. 

1.7 Definition A one-element vector space is a trivial space. 

Warning! The examples so far involve sets of column vectors with the usual 

operations. But vector spaces need not be collections of column vectors, or even 

of row vectors. Below are some other types of vector spaces. The term ‘vector 

space’ does not mean ‘collection of columns of reals’. It means something more 

like ‘collection in which any linear combination is sensible’. 

1.8 Example Consider P3 = {a0 + a1x + a2x 2 + a3x 3 � � a0,... ,a3 ∈ R}, the 

set of polynomials of degree three or less (in this book, we’ll take constant 

polynomials, including the zero polynomial, to be of degree zero). It is a vector 

space under the operations 

(a0 + a1x + a2x 2 + a3x 3 )+(b0 + b1x + b2x 2 + b3x 3 ) 

=(a0 + b0)+(a1 + b1)x +(a2 + b2)x 2 +(a3 + b3)x 3 

and 

r · (a0 + a1x + a2x 2 + a3x 3 )=(ra0)+(ra1)x +(ra2)x 2 +(ra3)x 3 

(the verification is easy). This vector space is worthy of attention because these 

are the polynomial operations familiar from high school algebra. For instance, 

3 · (1 − 2x +3x 2 − 4x 3 ) − 2 · (2 − 3x + x 2 − (1/2)x 3 )=−1+7x 2 − 11x 3 . 

Although this space is not a subset of any R n , there is a sense in which we 

can think of P3 as “the same” as R 4 . If we identify these two spaces’s elements 

in this way 

a0 + a1x + a2x 2 + a3x 3 

corresponds to 

then the operations also correspond. Here is an example of corresponding additions. 

1 − 2x +0x2 +1x3 + 2+3x +7x2 − 4x3 3+1x +7x2 − 3x3 ⎛ ⎞ 

1 

⎜ 

corresponds to ⎜−2 

⎟ 

⎝ 0 ⎠ 

1 

+ 

⎛ ⎞ 

2 

⎜ 3 ⎟ 

⎝ 7 ⎠ 

−4 

= 

⎛ ⎞ 

3 

⎜ 1 ⎟ 

⎝ 7 ⎠ 

−3 

Things we are thinking of as “the same” add to “the same” sum. Chapter Three 

makes precise this idea of vector space correspondence. For now we shall just 

leave it as an intuition. 

⎛ 

⎜ 

⎝ 

a0 

a1 

a2 

a3 

⎞ 

⎟ 

⎠


1.9 Example The set {f � � f : N → R} of all real-valued functions of one natural 

number variable is a vector space under the operations 

(f1 + f2)(n) =f1(n)+f2(n) (r · f)(n) =rf(n) 

so that if, for example, f1(n) =n 2 +2sin(n) andf2(n) =− sin(n) +0.5 then 

(f1 +2f2)(n) =n 2 +1. 

We can view this space as a generalization of Example 1.3 by thinking of 

these functions as “the same” as infinitely-tall vectors: 

n f(n) =n 2 +1 

0 1 

1 2 

2 5 

3 10 

. 

. 

corresponds to 

⎛ ⎞ 

1 

⎜ 2 ⎟ 

⎜ 5 ⎟ 

⎜ 

⎝ 

10⎟ 

⎠ 

. 

with addition and scalar multiplication are component-wise, as before. (The 

“infinitely-tall” vector can be formalized as an infinite sequence, or just as a 

function from N to R, in which case the above correspondence is an equality.) 

1.10 Example The set of polynomials with real coefficients 

{a0 + a1x + ···+ anx n � � n ∈ N and a0,... ,an ∈ R} 

makes a vector space when given the natural ‘+’ 

(a0 + a1x + ···+ anx n )+(b0 + b1x + ···+ bnx n ) 

and ‘·’. 

=(a0 + b0)+(a1 + b1)x + ···+(an + bn)x n 

r · (a0 + a1x + ...anx n )=(ra0)+(ra1)x + ...(ran)x n 

This space differs from the space P3 of Example 1.8. This space contains not just 

degree three polynomials, but degree thirty polynomials and degree three hundred 

polynomials, too. Each individual polynomial of course is of a finite degree, 

but the set has no single bound on the degree of all of its members. 

This example, like the prior one, can be thought of in terms of infinite-tuples. 

For instance, we can think of 1 + 3x +5x 2 as corresponding to (1, 3, 5, 0, 0,...). 

However, don’t confuse this space with the one from Example 1.9. Each member 

of this set has a bounded degree, so under our correspondence there are no 

elements from this space matching (1, 2, 5, 10, ...). The vectors in this space 

correspond to infinite-tuples that end in zeroes. 

1.11 Example The set {f � � f : R → R} of all real-valued functions of one real 

variable is a vector space under these. 

(f1 + f2)(x) =f1(x)+f2(x) (r · f)(x) =rf(x) 

The difference between this and Example 1.9 is the domain of the functions.


1.12 Example The set F = {a cos θ+b sin θ � � a, b ∈ R} of real-valued functions 

of the real variable θ is a vector space under the operations 

and 

(a1 cos θ + b1 sin θ)+(a2 cos θ + b2 sin θ) =(a1 + a2)cosθ +(b1 + b2)sinθ 

r · (a cos θ + b sin θ) =(ra)cosθ +(rb)sinθ 

inherited from the space in the prior example. (We can think of F as “the same” 

as R 2 in that a cos θ + b sin θ corresponds to the vector with components a and 

b.) 

1.13 Example The set 

{f : R → R � � d2f + f =0} 

dx2 is a vector space under the, by now natural, interpretation. 

(f + g)(x) =f(x)+g(x) (r · f)(x) =rf(x) 

In particular, notice that closure is a consequence: 

and 

d 2 (f + g) 

dx 2 

+(f + g) =( d2f dx2 + f)+(d2 g 

+ g) 

dx2 d2 (rf) 

dx2 +(rf) =r(d2 f 

+ f) 

dx2 of basic Calculus. This turns out to equal the space from the prior example — 

functions satisfying this differential equation have the form a cos θ + b sin θ — 

but this description suggests an extension to solutions sets of other differential 

equations. 

1.14 Example The set of solutions of a homogeneous linear system in n variables 

is a vector space under the operations inherited from Rn . 

under addition, if 

For closure 

⎛ ⎞ ⎛ ⎞ 

�v = 

⎜ 

⎝ 

v1 

. 

vn 

⎟ 

⎠ �w = 

both satisfy the condition that their entries add to zero then �v + �w also satisfies 

that condition: c1(v1 + w1)+···+ cn(vn + wn) =(c1v1 + ···+ cnvn)+(c1w1 + 

···+ cnwn) = 0. The checks of the other conditions are just as routine. 

⎜ 

⎝ 

w1 

. 

wn 

⎟ 

⎠


As we’ve done in those equations, we often omit the multiplication symbol ‘·’. 

We can distinguish the multiplication in ‘c1v1’ fromthatin‘r�v ’ since if both 

multiplicands are real numbers then real-real multiplication must be meant, 

while if one is a vector then scalar-vector multiplication must be meant. 

The prior example has brought us full circle since it is one of our motivating 

examples. 

1.15 Remark Now, with some feel for the kinds of structures that satisfy the 

definition of a vector space, we can reflect on that definition. For example, why 

specify in the definition the condition that 1 · �v = �v but not a condition that 

0 · �v = �0? 

One answer is that this is just a definition — it gives the rules of the game 

from here on, and if you don’t like it, put the book down and walk away. 

Another answer is perhaps more satisfying. People in this area have worked 

hard to develop the right balance of power and generality. This definition has 

been shaped so that it contains the conditions needed to prove all of the interesting 

and important properties of spaces of linear combinations, and so that it 

does not contain extra conditions that only bar as examples spaces where those 

properties occur. As we proceed, we shall derive all of the properties natural to 

collections of linear combinations from the conditions given in the definition. 

The next result is an example. We do not need to include these properties 

in the definition of vector space because they follow from the properties already 

listed there. 

1.16 Lemma In any vector space V , 

(1) 0 · �v = �0 

(2) (−1 · �v)+�v = �0 

(3) r · �0 =�0 

for any �v ∈ V and r ∈ R. 

Proof. For the first item, note that �v =(1+0)· �v = �v +(0· �v). Add to both 

sides the additive inverse of �v, the vector �w such that �w + �v = �0. 

�w + �v = �w + �v +0· �v 

�0 =�0+0· �v 

�0 =0· �v 

The second item is easy: (−1 · �v)+�v =(−1+1)· �v =0· �v = �0 shows that 

we can write ‘−�v ’ for the additive inverse of �v without worrying about possible 

confusion with (−1) · �v. 

For the third one, this r · �0 =r · (0 · �0) = (r · 0) · �0 =�0 will do. QED 

We finish this subsection with an recap, and a comment.


Chapter One studied Gaussian reduction. That lead us here to the study of 

collections of linear combinations. We have named any such structure a ‘vector 

space’. In a phrase, the point of this material is that vector spaces are the right 

context in which to study linearity. 

Finally, a comment. From the fact that it forms a whole chapter, and especially 

because that chapter is the first one, a reader could come to think that 

the study of linear systems is our purpose. The truth is, we will not so much 

use vector spaces in the study of linear systems as we will instead have linear 

systems lead us into the study of vector spaces. The wide variety of examples 

from this subsection shows that the study of vector spaces is interesting and important 

in its own right, aside from how it helps us understand linear systems. 

Linear systems won’t go away. But from now on our primary objects of study 

will be vector spaces. 

Exercises 

1.17 Give the zero vector from each of these vector spaces. 

(a) The space of degree three polynomials under the natural operations 

(b) The space of 2×4 matrices 

(c) The space {f :[0..1] → R � � f is continuous} 

(d) The space of real-valued functions of one natural number variable 

� 1.18 Find the additive inverse, in the vector space, of the vector. 

(a) In P3, thevector−3 − 2x + x 2 

(b) In the space of 2×2 matrices with real number entries under the usual matrix 

addition and scalar multiplication, 

�1 � 

−1 

0 3 

(c) In {ae x + be −x � � a, b ∈ R}, a space of functions of the real variable x under 

the natural operations, the vector 3e x − 2e −x . 

� 1.19 Show that each of these is a vector space. 

(a) The set of linear polynomials P1 = {a0 + a1x � � a0,a1 ∈ R} under the usual 

polynomial addition and scalar multiplication operations 

(b) The set of 2×2 matrices with real entries under the usual matrix operations 

(c) The set of three-component row vectors with their usual operations 

(d) The set 

⎛ ⎞ 

x 

⎜y 

⎟ 

L = { ⎝ 

z 

⎠ ∈ R 

w 

4 � � x + y − z + w =0} 

under the operations inherited from R 4 

� 1.20 Show that each set is not a vector space. (Hint. Start by listing two members 

of each set.) 

(a) Under the operations inherited from R 3 ,thisset 

� � 

x 

{ y ∈ R 

z 

3 � � x + y + z =1}


(b) Under the operations inherited from R 3 ,thisset 

� � 

x 

{ y 

z 

∈ R 3 � � x 2 + y 2 + z 2 =1} 

(c) Under the usual matrix operations, 

� � 

a 1 �� 

{ a, b, c ∈ R} 

b c 

(d) Under the usual polynomial operations, 

{a0 + a1x + a2x 2 � � a0,a1,a2 ∈ R + } 

where R + is the set of reals greater than zero 

(e) Under the inherited operations, 

� � 

x 

{ ∈ R 

y 

2 � � x +3y =4and2x−y =3and6x +4y =10} 

1.21 Define addition and scalar multiplication operations to make the complex 

numbers a vector space over R. 

� 1.22 Is the set of rational numbers a vector space over R under the usual addition 

and scalar multiplication operations? 

1.23 Show that the set of linear combinations of the variables x, y, z is a vector 

space under the natural addition and scalar multiplication operations. 

1.24 Prove that this is not a vector space: the set of two-tall column vectors with 

real entries subject to these operations. 

� � � � � � � � � � 

x1 x2 x1 − x2 

x rx 

+ = 

r · = 

y ry 

y1 

y2 

y1 − y2 

1.25 Prove or disprove that R 3 is a vector space under these operations. 

� � � � � � � � � � 

x1 x2 0 

x rx 

(a) y1 + y2 = 0 and r y = ry 

z1 

� � 

x1 

z2 

� � 

x2 

0 

� � 

0 

z 

� � 

x 

rz 

� � 

0 

(b) y1 + y2 = 0 and r y = 0 

z1 z2 0 

z 0 

� 1.26 For each, decide if it is a vector space; the intended operations are the natural 

ones. 

(a) The diagonal 2×2 matrices 

� 

a 

{ 

0 

� 

0 �� 

a, b ∈ R} 

b 

(b) This set of 2×2 matrices 

� � 

x x+ y �� 

{ 

x, y ∈ R} 

x + y y 

(c) This set 

⎛ ⎞ 

x 

⎜y 

⎟ 

{ ⎝ 

z 

⎠ ∈ R 

w 

4 � � x + y + w =1}


(d) The set of functions {f : R → R � � df/dx +2f =0} 

(e) The set of functions {f : R → R � � df /dx +2f =1} 

� 1.27 Prove or disprove that this is a vector space: the real-valued functions f of 

one real variable such that f(7) = 0. 

� 1.28 Show that the set R + of positive reals is a vector space when ‘x + y’ isinterpreted 

to mean the product of x and y (so that 2 + 3 is 6), and ‘r · x’ is interpreted 

as the r-th power of x. 

1.29 Is {(x, y) � � x, y ∈ R} a vector space under these operations? 

(a) (x1,y1)+(x2,y2) =(x1 + x2,y1 + y2) andr(x, y) =(rx, y) 

(b) (x1,y1)+(x2,y2) =(x1 + x2,y1 + y2) andr · (x, y) =(rx, 0) 

1.30 Prove or disprove that this is a vector space: the set of polynomials of degree 

greater than or equal to two, along with the zero polynomial. 

1.31 At this point “the same” is only an intuitive notion, but nonetheless for each 

vector space identify the k for which the space is “the same” as R k . 

(a) The 2×3 matrices under the usual operations 

(b) The n×m matrices (under their usual operations) 

(c) This set of 2×2 matrices 

� � 

a 0 �� 

{ a, b, c ∈ R} 

b c 

(d) This set of 2×2 matrices 

� � 

a 0 �� 

{ a + b + c =0} 

b c 

� 1.32 Using �+ to represent vector addition and �· for scalar multiplication, restate 

the definition of vector space. 

� 1.33 Prove these. 

(a) Any vector is the additive inverse of the additive inverse of itself. 

(b) Vector addition left-cancels: if �v,�s,�t ∈ V then �v + �s = �v + �t implies that 

�s = �t. 

1.34 The definition of vector spaces does not explicitly say that �0+�v = �v (check 

the order in which the summands appear). Show that it must nonetheless hold in 

any vector space. 

� 1.35 Prove or disprove that this is a vector space: the set of all matrices, under 

the usual operations. 

1.36 In a vector space every element has an additive inverse. Can some elements 

have two or more? 

1.37 (a) Prove that every point, line, or plane thru the origin in R 3 is a vector 

space under the inherited operations. 

(b) What if it doesn’t contain the origin? 

� 1.38 Using the idea of a vector space we can easily reprove that the solution set of 

a homogeneous linear system has either one element or infinitely many elements. 

Assume that �v ∈ V is not �0. 

(a) Prove that r · �v = �0 if and only if r =0. 

(b) Prove that r1 · �v = r2 · �v if and only if r1 = r2. 

(c) Prove that any nontrivial vector space is infinite. 

(d) Use the fact that a nonempty solution set of a homogeneous linear system is 

a vector space to draw the conclusion.


1.39 Is this a vector space under the natural operations: the real-valued functions 

of one real variable that are differentiable? 

1.40 A vector space over the complex numbers C has the same definition as a vector 

space over the reals except that scalars are drawn from C insteadoffromR. Show 

that each of these is a vector space over the complex numbers. (Recall how complex 

numbers add and multiply: (a0 + a1i) +(b0 + b1i) =(a0 + b0) +(a1 + b1)i and 

(a0 + a1i)(b0 + b1i) =(a0b0 − a1b1)+(a0b1 + a1b0)i.) 

(a) The set of degree two polynomials with complex coefficients 

(b) This set 

� � 

0 a �� 

{ a, b ∈ C and a + b =0+0i} 

b 0 

1.41 Find a property shared by all of the R n ’s not listed as a requirement for a 

vector space. 

� 1.42 (a) Prove that a sum of four vectors �v1,... ,�v4 ∈ V can be associated in 

any way without changing the result. 

((�v1 + �v2)+�v3)+�v4 =(�v1 +(�v2 + �v3)) + �v4 

=(�v1 + �v2)+(�v3 + �v4) 

= �v1 +((�v2 + �v3)+�v4) 

= �v1 +(�v2 +(�v3 + �v4)) 

This allows us to simply write ‘�v1 + �v2 + �v3 + �v4’ without ambiguity. 

(b) Prove that any two ways of associating a sum of any number of vectors give 

the same sum. (Hint. Use induction on the number of vectors.) 

1.43 For any vector space, a subset that is itself a vector space under the inherited 

operations (e.g., a plane through the origin inside of R 3 )isasubspace. 

(a) Show that {a0 + a1x + a2x 2 � � a0 + a1 + a2 =0} is a subspace of the vector 

space of degree two polynomials. 

(b) Show that this is a subspace of the 2×2 matrices. 

� � 

a b �� 

{ a + b =0} 

c 0 

(c) Show that a nonempty subset S of a real vector space is a subspace if and only 

if it is closed under linear combinations of pairs of vectors: whenever c1,c2 ∈ R 

and �s1,�s2 ∈ S then the combination c1�v1 + c2�v2 is in S. 

2.I.2 Subspaces and Spanning Sets 

One of the examples that led us to introduce the idea of a vector space 

was the solution set of a homogeneous system. For instance, we’ve seen in 

Example 1.4 such a space that is a planar subset of R 3 . There, the vector space 

R 3 contains inside it another vector space, the plane. 

2.1 Definition For any vector space, a subspace is a subset that is itself a 

vector space, under the inherited operations.


2.2 Example The plane from the prior subsection, 

⎛ 

P = { ⎝ x 

⎞ 

y⎠ 

z 

� � x + y + z =0} 

is a subspace of R3 . As specified in the definition, the operations are the ones 

that are inherited from the larger space, that is, vectors add in P3 as they add 

in R3 ⎛ 

⎝ x1 

⎞ ⎛ 

⎠ + ⎝ x2 

⎞ ⎛ 

⎠ = ⎝ x1 

⎞ 

+ x2 

⎠ 

y1 

z1 

y2 

z2 

y1 + y2 

z1 + z2 

and scalar multiplication is also the same as it is in R 3 . To show that P is a 

subspace, we need only note that it is a subset and then verify that it is a space. 

Checking that P satisfies the conditions in the definition of a vector space is 

routine. For instance, for closure under addition, just note that if the summands 

satisfy that x1 + y1 + z1 =0andx2 + y2 + z2 = 0 then the sum satisfies that 

(x1 + x2)+(y1 + y2)+(z1 + z2) =(x1 + y1 + z1)+(x2 + y2 + z2) =0. 

2.3 Example The x-axis in R2 is a subspace where the addition and scalar 

multiplication operations are the inherited ones. 

� � � � � � � � � � 

x1 x2 x1 + x2 x rx 

+ = 

r · = 

0 0 0 

0 0 

As above, to verify that this is a subspace, we simply note that it is a subset 

and then check that it satisfies the conditions in definition of a vector space. 

For instance, the two closure conditions are satisfied: (1) adding two vectors 

with a second component of zero results in a vector with a second component 

of zero, and (2) multiplying a scalar times a vector with a second component of 

zero results in a vector with a second component of zero. 

2.4 Example Another subspace of R2 is 

� � 

0 

{ } 

0 

its trivial subspace. 

Any vector space has a trivial subspace {�0 }. At the opposite extreme, any 

vector space has itself for a subspace. These two are the improper subspaces. 

Other subspaces are proper. 

2.5 Example The condition in the definition requiring that the addition and 

scalar multiplication operations must be the ones inherited from the larger space 

is important. Consider the subset {1} of the vector space R 1 . Under the operations 

1+1 = 1 and r·1 = 1 that set is a vector space, specifically, a trivial space. 

But it is not a subspace of R 1 because those aren’t the inherited operations, since 

of course R 1 has 1 + 1 = 2.


2.6 Example All kinds of vector spaces, not just R n ’s, have subspaces. The 

vector space of cubic polynomials {a + bx + cx 2 + dx 3 � � a, b, c, d ∈ R} has a subspace 

comprised of all linear polynomials {m + nx � � m, n ∈ R}. 

2.7 Example Another example of a subspace not taken from an R n is one 

from the examples following the definition of a vector space. The space of all 

real-valued functions of one real variable f : R → R has a subspace of functions 

satisfying the restriction (d 2 f/dx 2 )+f =0. 

2.8 Example Being vector spaces themselves, subspaces must satisfy the closure 

conditions. The set R + is not a subspace of the vector space R 1 because 

with the inherited operations it is not closed under scalar multiplication: if 

�v = 1 then −1 · �v �∈ R + . 

The next result says that Example 2.8 is prototypical. The only way that a 

subset can fail to be a subspace (if it is nonempty and the inherited operations 

are used) is if it isn’t closed. 

2.9 Lemma For a nonempty subset S of a vector space, under the inherited 

operations, the following are equivalent statements. ∗ 

(1) S is a subspace of that vector space 

(2) S is closed under linear combinations of pairs of vectors: for any vectors 

�s1,�s2 ∈ S and scalars r1,r2 the vector r1�s1 + r2�s2 is in S 

(3) S is closed under linear combinations of any number of vectors: for any 

vectors �s1,... ,�sn ∈ S and scalars r1,... ,rn the vector r1�s1 + ···+ rn�sn is 

in S. 

Briefly, the way that a subset gets to be a subspace is by being closed under 

linear combinations. 

Proof. ‘The following are equivalent’ means that each pair of statements are 

equivalent. 

(1) ⇐⇒ (2) (2) ⇐⇒ (3) (3) ⇐⇒ (1) 

We will show this equivalence by establishing that (1) =⇒ (3) =⇒ (2) =⇒ 

(1). This strategy is suggested by noticing that (1) =⇒ (3) and (3) =⇒ (2) 

are easy and so we need only argue the single implication (2) =⇒ (1). 

For that argument, assume that S is a nonempty subset of a vector space V 

and that S is closed under combinations of pairs of vectors. We will show that 

S is a vector space by checking the conditions. 

The first item in the vector space definition has five conditions. First, for 

closure under addition, if �s1,�s2 ∈ S then �s1 + �s2 ∈ S, as�s1 + �s2 =1· �s1 +1· �s2. 

Second, for any �s1,�s2 ∈ S, because addition is inherited from V , the sum �s1 +�s2 

in S equals the sum �s1 +�s2 in V , and that equals the sum �s2 +�s1 in V (because 

V is a vector space, its addition is commutative), and that in turn equals the 

sum �s2 +�s1 in S. The argument for the third condition is similar to that for the 

∗ More information on equivalence of statements is in the appendix.


second. For the fourth, consider the zero vector of V and note that closure of S 

under linear combinations of pairs of vectors gives that (where �s is any member 

of the nonempty set S) 0· �s +0· �s = �0 isinS; showing that �0 acts under the 

inherited operations as the additive identity of S is easy. The fifth condition is 

satisfied because for any �s ∈ S, closure under linear combinations shows that 

the vector 0 · �0 +(−1) · �s is in S; showing that it is the additive inverse of �s 

under the inherited operations is routine. 

The checks for item (2) are similar and are saved for Exercise 32. QED 

We usually show that a subset is a subspace with (2) =⇒ (1). 

2.10 Remark At the start of this chapter we introduced vector spaces as collections 

in which linear combinations are “sensible”. The above result speaks 

to this. 

The vector space definition has ten conditions but eight of them, the ones 

stated there with the ‘•’ bullets, simply ensure that referring to the operations 

as an ‘addition’ and a ‘scalar multiplication’ is sensible. The proof above checks 

that if the nonempty set S satisfies statement (2) then inheritance of the operations 

from the surrounding vector space brings with it the inheritance of these 

eight properties also (i.e., commutativity of addition in S follows right from 

commutativity of addition in V ). So, in this context, this meaning of “sensible” 

is automatically satisfied. 

In assuring us that this first meaning of the word is met, the result draws 

our attention to the second meaning. It has to do with the two remaining 

conditions, the closure conditions. Above, the two separate closure conditions 

inherent in statement (1) are combined in statement (2) into the single condition 

of closure under all linear combinations of two vectors, which is then extended 

in statement (3) to closure under combinations of any number of vectors. The 

latter two statements say that we can always make sense of an expression like 

r1�s1 + r2�s2, without restrictions on the r’s — such expressions are “sensible” in 

that the vector described is defined and is in the set S. 

This second meaning suggests that a good way to think of a vector space 

is as a collection of unrestricted linear combinations. The next two examples 

take some spaces and describe them in this way. That is, in these examples we 

paramatrize, just as we did in Chapter One to describe the solution set of a 

homogeneous linear system. 

2.11 Example This subset of R 3 

⎛ 

S = { ⎝ x 

⎞ 

y⎠ 

z 

� � x − 2y + z =0} 

is a subspace under the usual addition and scalar multiplication operations of 

column vectors (the check that it is nonempty and closed under linear combinations 

of two vectors is just like the one in Example 2.2). To paramatrize, we


can take x − 2y + z = 0 to be a one-equation linear system and expressing the 

leading variable in terms of the free variables x =2y− z. 

⎛ ⎞ 

2y − z 

S = { ⎝ y ⎠ 

z 

� ⎛ 

� y, z ∈ R} = {y ⎝ 2 

⎞ ⎛ 

1⎠ 

+ z ⎝ 

0 

−1 

⎞ 

0 ⎠ 

1 

� � y, z ∈ R} 

Now the subspace is described as the collection of unrestricted linear combinations 

of those two vectors. Of course, in either description, this is a plane 

through the origin. 

2.12 Example This is a subspace of the 2×2 matrices 

� � 

a 0 �� 

L = { a + b + c =0} 

b c 

(checking that it is nonempty and closed under linear combinations is easy). To 

paramatrize, express the condition as a = −b − c. 

� � � � � � 

−b − c 0 �� −1 0 −1 0 �� 

L = { 

b, c ∈ R} = {b + c 

b, c ∈ R} 

b c 

1 0 0 1 

As above, we’ve described the subspace as a collection of unrestricted linear 

combinations (by coincidence, also of two elements). 

Paramatrization is an easy technique, but it is important. We shall use it 

often. 

2.13 Definition The span (or linear closure) of a nonempty subset S of a 

vector space is the set of all linear combinations of vectors from S. 

� 

[S] ={c1�s1 + ···+ cn�sn � c1,... ,cn ∈ R and �s1,... ,�sn ∈ S} 

The span of the empty subset of a vector space is the trivial subspace. 

No notation for the span is completely standard. The square brackets used here 

are common, but so are ‘span(S)’ and ‘sp(S)’. 

2.14 Remark In Chapter One, after we showed that the solution set of a 

homogeneous linear system can written as {c1 � β1 + ···+ ck � � 

βk 

� c1,... ,ck ∈ R}, 

we described that as the set ‘generated’ by the � β’s. We now have the technical 

term; we call that the ‘span’ of the set { � β1,... , � βk}. 

Recall also the discussion of the “tricky point” in that proof. The span of 

the empty set is defined to be the set {�0} because we follow the convention that 

a linear combination of no vectors sums to �0. Besides, defining the empty set’s 

span to be the trivial subspace is a convienence in that it keeps results like the 

next one from having annoying exceptional cases. 

2.15 Lemma In a vector space, the span of any subset is a subspace.


Proof. Call the subset S. IfS is empty then by definition its span is the trivial 

subspace. If S is not empty then by Lemma 2.9 we need only check that the 

span [S] is closed under linear combinations. For a pair of vectors from that 

span, �v = c1�s1 +···+cn�sn and �w = cn+1�sn+1 +···+cm�sm, a linear combination 

p · (c1�s1 + ···+ cn�sn)+r · (cn+1�sn+1 + ···+ cm�sm) 

= pc1�s1 + ···+ pcn�sn + rcn+1�sn+1 + ···+ rcm�sm 

(p, r scalars) is a linear combination of elements of S and so is in [S] (possibly 

some of the �si’s forming �v equal some of the �sj’s from �w, but it does not 

matter). QED 

The converse of the lemma holds: any subspace is the span of some set, 

because a subspace is obviously the span of the set of its members. Thus a 

subset of a vector space is a subspace if and only if it is a span. This fits the 

intuition that a good way to think of a vector space is as a collection in which 

linear combinations are sensible. 

Taken together, Lemma 2.9 and Lemma 2.15 show that the span of a subset 

S of a vector space is the smallest subspace containing all the members of S. 

2.16 Example In any vector space V , for any vector �v, the set {r · �v � � r ∈ R} 

is a subspace of V . For instance, for any vector �v ∈ R 3 , the line through the 

origin containing that vector, {k�v � � k ∈ R} is a subspace of R 3 . This is true even 

when �v is the zero vector, in which case the subspace is the degenerate line, the 

trivial subspace. 

2.17 Example The span of this set is all of R2 . 

� � � � 

1 1 

{ , } 

1 −1 

Tocheck this we must show that any member of R 2 is a linear combination of 

these two vectors. So we ask: for which vectors (with real components x and y) 

are there scalars c1 and c2 such that this holds? 


c1 

c1 + c2 = x 

c1 − c2 = y 

� � 

1 

1 

+ c2 

� � 

1 

−1 

= 

� � 

x 

y 

−ρ1+ρ2 

−→ c1 + c2 = x 

−2c2 = −x + y 

with back substitution gives c2 =(x − y)/2 andc1 =(x + y)/2. These two 

equations show that for any x and y that we start with, there are appropriate 

coefficients c1 and c2 making the above vector equation true. For instance, for 

x =1andy = 2 the coefficients c2 = −1/2 andc1 =3/2 will do. That is, any 

vector in R 2 can be written as a linear combination of the two given vectors.


Since spans are subspaces, and we know that a good way to understand a 

subspace is to paramatrize its description, we can try to understand a set’s span 

in that way. 

2.18 Example Consider, in P2, the span of the set {3x − x 2 , 2x}. By the 

definition of span, it is the subspace of unrestricted linear combinations of the 

two {c1(3x − x 2 )+c2(2x) � � c1,c2 ∈ R}. Clearly polynomials in this span must 

have a constant term of zero. Is that necessary condition also sufficient? 

We are asking: for which members a2x 2 + a1x + a0 of P2 are there c1 and c2 

such that a2x 2 + a1x + a0 = c1(3x − x 2 )+c2(2x)? Since polynomials are equal 

if and only if their coefficients are equal, we are looking for conditions on a2, 

a1, anda0 satisfying these. 

−c1 = a2 

3c1 +2c2 = a1 

0=a0 

Gauss’ method gives that c1 = −a2, c2 =(3/2)a2 +(1/2)a1, and0=a0. Thus 

the only condition on polynomials in the span is the condition that we knew 

of — as long as a0 = 0, we can give appropriate coefficients c1 and c2 to describe 

the polynomial a0 + a1x + a2x 2 as in the span. For instance, for the polynomial 

0 − 4x +3x 2 , the coefficients c1 = −3 andc2 =5/2 will do. So the span of the 

given set is {a1x + a2x 2 � � a1,a2 ∈ R}. 

This shows, incidentally, that the set {x, x 2 } also spans this subspace. A 

space can have more than one spanning set. Two other sets spanning this subspace 

are {x, x 2 , −x +2x 2 } and {x, x + x 2 ,x+2x 2 ,...}. (Naturally, we usually 

prefer to work with spanning sets that have only a few members.) 

2.19 Example These are the subspaces of R 3 that we now know of, the trivial 

subspace, the lines through the origin, the planes through the origin, and the 

whole space (of course, the picture shows only a few of the infinitely many 

subspaces). In the next section we will prove that R 3 has no other type of 

subspaces, so in fact this picture shows them all. 

{x 

{x 

� � � � 1 

0 

0 + y 1 } 

0 

0 

� 1 

0 

0 

� } 

✄ 

✄ 

{x 

� � 1 

0 + y 

0 

✘✘✘ 

✏ 

✘✘ 

✏✏ 

✘✘✘ 

✏✏ 

� 

✘✘ 

✏ 

� 

� � 

{x 

} {x 

} 

� 1 

0 

0 

� + z 

� 0 

0 

1 

� � � 1 

0 

1 + z 0 

0 

1 

❍ ✏✏ 

❆✏✏❍❍❍{y �� 

✏❆ 

� � � � � � 0 

2 

1 

{y 1 } 

1 } {y 1 } ... 

0 

0 

1 

❳❳ � 

❳❳❳❳ �� ❍ 

❍❍❍❍ 

❳❳❳❳ 

❳❳ 

❅ ❅ � � 0 

{ } 

0 

0 

� � 0 

1 + z 

0 

... 

� � 0 

0 } 

1


The subsets are described as spans of sets, using a minimal number of members, 

and are shown connected to their supersets. Note that these subspaces fall 

naturally into levels — planes on one level, lines on another, etc. — according 

to how many vectors are in a minimal-sized spanning set. 

So far in this chapter we have seen that to study the properties of linear 

combinations, the right setting is a collection that is closed under these combinations. 

In the first subsection we introduced such collections, vector spaces, 

and we saw a great variety of examples. In this subsection we saw still more 

spaces, ones that happen to be subspaces of others. In all of the variety we’ve 

seen a commonality. Example 2.19 above brings it out: vector spaces and subspaces 

are best understood as a span, and especially as a span of a small number 

of vectors. The next section studies spanning sets that are minimal. 

Exercises 

� 2.20 Which of these subsets of the vector space of 2 × 2 matrices are subspaces 

under the inherited operations? For each one that is a subspace, paramatrize its 

description. � For � each that is not, give a condition that fails. 

a 0 �� 

(a) { a, b ∈ R} 

0 b 

� � 

a 0 �� 

(b) { a + b =0} 

0 b 

� � 

a 0 �� 

(c) { a + b =5} 

0 b 

� � 

a c �� 

(d) { a + b =0,c∈ R} 

0 b 

� 2.21 Is this a subspace of P2: {a0 + a1x + a2x 2 � � a0 +2a1 + a2 =4}? If so, paramatrize 

its description. 

� 2.22 Decide if the vector lies in the span of the set, inside of the space. 

� � � � � � 

2 1 0 

(a) 0 , { 0 , 0 }, inR 

1 0 1 

3 

(b) x − x 3 , {x 2 , 2x + x 2 ,x+ x 3 },inP3 

� � � � � � 

0 1 1 0 2 0 

(c) , { , }, inM2×2 

4 2 1 1 2 3 

2.23 Which of these are members of the span [{cos 2 x, sin 2 x}] in the vector space 

of real-valued functions of one real variable? 

(a) f(x) =1 (b) f(x) =3+x 2 

(c) f(x) =sinx (d) f(x) = cos(2x) 

� 2.24 Which of these sets spans R 3 ? That is, which of these sets has the property 

that any three-tall vector can be expressed as a suitable linear combination of the 

set’s elements? � � � � 

1 0 

� � 

0 

� � 

2 

� � 

1 

� � 

0 

� � 

1 

� � 

3 

(a) { 0 , 2 , 0 } (b) { 0 , 1 , 0 } (c) { 1 , 0 } 

0 

� � 

1 

0 

� � 

3 

3 

� � 

−1 

� � 

2 

1 0 

� � 

2 

1 

� � 

3 

� � 

5 

0 

� � 

6 

0 

(d) { 0 , 1 , 0 , 1 } (e) { 1 , 0 , 1 , 0 } 

1 0 0 5 

1 1 2 2


� 2.25 Paramatrize each subspace’s description. Then express each subspace as a 

span. 

(a) The subset {a + bx + cx 3 � � a − 2b + c =0} of P3 

(b) The subset { � a b c � � � a − c =0} of the three-wide row vectors 

(c) This subset of M2×2 

� 

a 

{ 

c 

� 

b �� 

a + d =0} 

d 

(d) This subset of M2×2 

� � 

a b �� 

{ 2a − c − d =0anda +3b =0} 

c d 

(e) The subset of P2 of quadratic polynomials p such that p(7) = 0 

� 2.26 Find a set to span the given subspace of the given space. (Hint. Paramatrize 

each.) 

(a) the xz-plane in R 3 

� � 

x �� 

(b) { y 3x +2y + z =0} in R 

z 

3 

⎛ ⎞ 

x 

⎜y 

⎟ 

(c) { ⎝ 

z 

⎠ 

w 

� � 2x + y + w =0andy +2z =0} in R 4 

(d) {a0 + a1x + a2x 2 + a3x 3 � � a0 + a1 =0anda2 − a3 =0} in P3 

(e) The set P4 in the space P4 

(f) M2×2 in M2×2 

2.27 Is R 2 a subspace of R 3 ? 

� 2.28 Decide if each is a subspace of the vector space of real-valued functions of one 

real variable. 

(a) The even functions {f : R → R � � f(−x) =f(x) for all x}. For example, two 

membersofthissetaref1(x) =x 2 and f2(x) =cos(x). 

(b) The odd functions {f : R → R � � f(−x) =−f(x) for all x}. Twomembersare 

f3(x) =x 3 and f4(x) =sin(x). 

2.29 Example 2.16 says that for any vector �v in any vector space V , the set 

{r · �v � � r ∈ R} is a subspace of V . (This is of course, simply the span of the 

singleton set {�v}.) Must any such subspace be a proper subspace, or can it be 

improper? 

2.30 An example following the definition of a vector space shows that the solution 

set of a homogeneous linear system is a vector space. In the terminology of this 

subsection, it is a subspace of R n where the system has n variables. What about 

a non-homogeneous linear system; do its solutions form a subspace (under the 

inherited operations)? 

2.31 Example 2.19 shows that R 3 has infinitely many subspaces. Does every nontrivial 

space have infinitely many subspaces? 

2.32 Finish the proof of Lemma 2.9. 

2.33 Show that each vector space has only one trivial subspace. 

� 2.34 Show that for any subset S of a vector space, the span of the span equals the 

span [[S]]=[S]. (Hint. Members of [S] are linear combinations of members of


S. Membersof[[S]] are linear combinations of linear combinations of members of 

S.) 

2.35 All of the subspaces that we’ve seen use zero in their description in some 

way. For example, the subspace in Example 2.3 consists of all the vectors from R 2 

with a second component of zero. In contrast, the collection of vectors from R 2 

with a second component of one does not form a subspace (it is not closed under 

scalar multiplication). Another example is Example 2.2, where the condition on 

the vectors is that the three components add to zero. If the condition were that the 

three components add to ong then it would not be a subspace (again, it would fail 

to be closed). This exercise shows that a reliance on zero is not strictly necessary. 

Consider the set 

� � 

x �� 

{ y x + y + z =1} 

z 

under these operations. 

� � � � 

x1 x2 

� � 

x1 + x2 − 1 

� � 

x 

� � 

rx − r +1 

+ = y1 + y2 r y = ry 

z rz 

y1 

z1 

y2 

z2 

z1 + z2 

(a) Show that it is not a subspace of R 3 .(Hint. See Example 2.5). 

(b) Show that it is a vector space. Note that by the prior item, Lemma 2.9 can 

not apply. 

(c) Show that any subspace of R 3 must pass thru the origin, and so any subspace 

of R 3 must involve zero in its description. Does the converse hold? Does any 

subset of R 3 that contains the origin become a subspace when given the inherited 

operations? 

2.36 We can give a justification for the convention that the sum of no vectors equals 

the zero vector. Consider this sum of three vectors �v1 + �v2 + �v3. 

(a) What is the difference between this sum of three vectors and the sum of the 

first two of this three? 

(b) What is the difference between the prior sum and the sum of just the first 

one vector? 

(c) What should be the difference between the prior sum of one vector and the 

sum of no vectors? 

(d) So what should be the definition of the sum of no vectors? 

2.37 Is a space determined by its subspaces? That is, if two vector spaces have the 

same subspaces, must the two be equal? 

2.38 (a) Give a set that is closed under scalar multiplication but not addition. 

(b) Give a set closed under addition but not scalar multiplication. 

(c) Give a set closed under neither. 

2.39 Show that the span of a set of vectors does not depend on the order in which 

the vectors are listed in that set. 

2.40 Which trivial subspace is the span of the empty set? Is it 

� � 

0 

{ 0 

0 

}⊆R 3 or some other subspace? 

, or {0+0x} ⊆P1, 

2.41 Show that if a vector is in the span of a set then adding that vector to the set 

won’t make the span any bigger. Is that also ‘only if’?


� 2.42 Subspaces are subsets and so we naturally consider how ‘is a subspace of’ 

interacts with the usual set operations. 

(a) If A, B are subspaces of a vector space, must A ∩ B be a subspace? Always? 

Sometimes? Never? 

(b) Must A ∪ B be a subspace? 

(c) If A is a subspace, must its complement be a subspace? 

(Hint. Try some test subspaces from Example 2.19.) 

� 2.43 Does the span of a set depend on the enclosing space? That is, if W is a 

subspace of V and S is a subset of W (and so also a subset of V ), might the span 

of S in W differ from the span of S in V ? 

2.44 Is the relation ‘is a subspace of’ transitive? That is, if V is a subspace of W 

and W is a subspace of X, mustV be a subspace of X? 

� 2.45 Because ‘span of’ is an operation on sets we naturally consider how it interacts 

with the usual set operations. 

(a) If S ⊆ T are subsets of a vector space, is [S] ⊆ [T ]? Always? Sometimes? 

Never? 

(b) If S, T are subsets of a vector space, is [S ∪ T ]=[S] ∪ [T ]? 

(c) If S, T are subsets of a vector space, is [S ∩ T ]=[S] ∩ [T ]? 

(d) Is the span of the complement equal to the complement of the span? 

2.46 Reprove Lemma 2.15 without doing the empty set separately. 

2.47 Find a structure that is closed under linear combinations, and yet is not a 

vector space. (Remark. This is a bit of a trick question.)


2.II Linear Independence 

The prior section shows that a vector space can be understood as an unrestricted 

linear combination of some of its elements — that is, as a span. For example, 

the space of linear polynomials {a + bx � � a, b ∈ R} is spanned by the set {1,x}. 

The prior section also showed that a space can have many sets that span it. 

The space of linear polynomials is also spanned by {1, 2x} and {1,x,2x}. 

At the end of that section we described some spanning sets as ‘minimal’, 

but we never precisely defined that word. We could take ‘minimal’ to mean one 

of two things. We could mean that a spanning set is minimal if it contains the 

smallest number of members of any set with the same span. With this meaning 

{1,x,2x} is not minimal because it has one member more than the other two. 

Or we could mean that a spanning set is minimal when it has no elements that 

can be removed without changing the span. Under this meaning {1,x,2x} is not 

minimal because removing the 2x and getting {1,x} leaves the span unchanged. 

The first sense of minimality appears to be a global requirement, in that to 

check if a spanning set is minimal we seemingly must look at all the spanning sets 

of a subspace and find one with the least number of elements. The second sense 

of minimality is local in that we need to look only at the set under discussion 

and consider the span with and without various elements. For instance, using 

the second sense, we could compare the span of {1,x,2x} with the span of {1,x} 

and note that the 2x is a “repeat” in that its removal doesn’t shrink the span. 

In this section we will use the second sense of ‘minimal spanning set’ because 

of this technical convenience. However, the most important result of this book 

is that the two senses coincide; we will prove that in the section after this one. 

2.II.1 Definition and Examples 

We first characterize when a vector can be removed from a set without 

changing its span. 

1.1 Lemma Where S is a subset of a vector space, 

for any �v in that space. 

[S] =[S ∪{�v}] if and only if �v ∈ [S] 

Proof. The left to right implication is easy. If [S] = [S ∪{�v}] then, since 

obviously �v ∈ [S ∪{�v}], the equality of the two sets gives that �v ∈ [S]. 

For the right to left implication assume that �v ∈ [S] toshowthat[S] =[S ∪ 

{�v}] by mutual inclusion. The inclusion [S] ⊆ [S ∪{�v}] is obvious. For the other 

inclusion [S] ⊇ [S ∪{�v}], write an element of [S ∪{�v}] asd0�v+d1�s1 +···+dm�sm, 

and substitute �v’s expansion as a linear combination of members of the same set 

d0(c0 �t0 + ···+ ck �tk)+d1�s1 + ···+ dm�sm. This is a linear combination of linear 

combinations, and so after distributing d0 we end with a linear combination of 

vectors from S. Hence each member of [S ∪{�v}] isalsoamemberof[S]. QED

Section II. Linear Independence 103 

1.2 Example In R3 , where 

⎛ 

�v1 = ⎝ 1 

⎞ 

0⎠ 

, 

⎛ 

�v2 = ⎝ 

0 

0 

⎞ 

1⎠ 

, 

⎛ 

�v3 = ⎝ 

0 

2 

⎞ 

1⎠ 

0 

the spans [{�v1,�v2}] and [{�v1,�v2,�v3}] are equal since �v3 is in the span [{�v1,�v2}]. 

The lemma says that if we have a spanning set then we can remove a �v to 

get a new set S with the same span if and only if �v is a linear combination of 

vectors from S. Thus, under the second sense described above, a spanning set 

is minimal if and only if it contains no vectors that are linear combinations of 

the others in that set. We have a term for this important property. 

1.3 Definition A subset of a vector space is linearly independent if none of its 

elements is a linear combination of the others. Otherwise it is linearly dependent. 

Here is a small but useful observation: although this way of writing one 

vector as a combination of the others 

�s0 = c1�s1 + c2�s2 + ···+ cn�sn 

visually sets �s0 off from the other vectors, algebraically there is nothing special 

in that equation about �s0. For any �si with a coefficient ci that is nonzero, we 

can rewrite the relationship to set off �si. 

�si =(1/ci)�s0 +(−c1/ci)�s1 + ···+(−cn/ci)�sn 

When we don’t want to single out any vector by writing it alone on one side of 

the equation we will instead say that �s0,�s1,...,�sn areinalinear relationship and 

write the relationship with all of the vectors on the same side. The next result 

rephrases the linear independence definition in this style. It gives what is usually 

the easiest way to compute whether a finite set is dependent or independent. 

1.4 Lemma A subset S of a vector space is linearly independent if and only if 

for any distinct �s1,...,�sn ∈ S the only linear relationship among those vectors 

is the trivial one: c1 =0,..., cn =0. 

c1�s1 + ···+ cn�sn = �0 c1,...,cn ∈ R 

Proof. This is a direct consequence of the observation above. 

If the set S is linearly independent then no vector �si can be written as a linear 

combination of the other vectors from S so there is no linear relationship where 

some of the �s ’s have nonzero coefficients. If S is not linearly independent then 

some �si is a linear combination �si = c1�s1+···+ci−1�si−1+ci+1�si+1+···+cn�sn of 

other vectors from S, and subtracting �si from both sides of that equation gives 

a linear relationship involving a nonzero coefficient, namely the −1 infrontof 

�si. QED


1.5 Example In the vector space of two-wide row vectors, the two-element set 

{ � 40 15 � , � −50 25 � } is linearly independent. To check this, set 

c1 · � 40 15 � + c2 · � −50 25 � = � 0 0 � 

and solve the resulting system. 

40c1 − 50c2 =0 

15c1 +25c2 =0 

−(15/40)ρ1+ρ2 

−→ 

40c1 − 50c2 =0 

(175/4)c2 =0 

Therefore c1 and c2 both equal zero and so the only linear relationship between 

the two given row vectors is the trivial relationship. 

In the same vector space, { � 40 15 � , � 20 7.5 � } is linearly dependent since 

we can satisfy 

� � 

c1 40 15 + c2 · � 20 7.5 � = � 0 0 � 

with c1 =1andc2 = −2. 

1.6 Remark Recall the Statics example that began this book. We first set the 

unknown-mass objects at 40 cm and 15 cm and got a balance, and then we set 

the objects at −50 cm and 25 cm and got a balance. With those two pieces of 

information we could compute values of the unknown masses. Had we instead 

first set the unknown-mass objects at 40 cm and 15 cm, and then at 20 cm and 

7.5 cm, we would not have been able to compute the values of the unknown 

masses (try it). Intuitively, the problem is that the � 20 7.5 � information is a 

“repeat” of the � 40 15 � information — that is, � 20 7.5 � is in the span of the 

set { � 40 15 � } — and so we would be trying to solve a two-unknowns problem 

with what is essentially one piece of information. 

1.7 Example The set {1+x, 1 − x} is linearly independent in P2, the space 

of quadratic polynomials with real coefficients, because 

gives 

0+0x +0x 2 = c1(1 + x)+c2(1 − x) =(c1 + c2)+(c1 − c2)x +0x 2 

c1 + c2 =0 

c1 − c2 =0 

−ρ1+ρ2 

−→ 

c1 + c2 =0 

2c2 =0 

(since polynomials are equal only if their coefficients are equal). Thus, the only 

linear relationship between these two members of P2 is the trivial one. 

1.8 Example In R3 , where 

⎛ 

�v1 = ⎝ 3 

⎞ 

4⎠ 

, 

⎛ 

�v2 = ⎝ 

5 

2 

⎞ 

9⎠ 

, 

⎛ 

�v3 = ⎝ 

2 

4 

⎞ 

18⎠ 

4


the set S = {�v1,�v2,�v3} is linearly dependent because this is a relationship 

0 · �v1 +2· �v2 − 1 · �v3 = �0 

where not all of the scalars are zero (the fact that some scalars are zero is 

irrelevant). 

1.9 Remark That example shows why, although Definition 1.3 is a clearer 

statement of what independence is, Lemma 1.4 is more useful for computations. 

Working straight from the definition, someone trying to compute whether S is 

linearly independent would start by setting �v1 = c2�v2+c3�v3 and concluding that 

there are no such c2 and c3. But knowing that the first vector is not dependent 

on the other two is not enough. Working straight from the definition, this 

personwouldhavetogoontotry�v2 = c3�v3 in order to find the dependence 

c3 =1/2. Similarly, working straight from the definition, a set with four vectors 

would require checking three vector equations. Lemma 1.4 makes the job easier 

because it allows us to get the conclusion with only one computation. 

1.10 Example The empty subset of a vector space is linearly independent. 

There is no nontrivial linear relationship among its members as it has no members. 

1.11 Example In any vector space, any subset containing the zero vector is 

linearly dependent. For example, in the space P2 of quadratic polynomials, 

consider the subset {1+x, x + x 2 , 0}. 

One way to see that this subset is linearly dependent is to use Lemma 1.4: we 

have 0 ·�v1 +0·�v2 +1·�0 =�0, and this is a nontrivial relationship as not all of the 

coefficients are zero. Another way to see that this subset is linearly dependent 

is to go straight to Definition 1.3: we can express the third member of the subset 

as a linear combination of the first two, namely, c1�v1 + c2�v2 = �0 is satisfied by 

taking c1 =0andc2 = 0 (in contrast to the lemma, the definition allows all of 

the coefficients to be zero). 

(There is still another way to see this that is somewhat trickier. The zero 

vector is equal to the trivial sum, that is, it is the sum of no vectors. So in 

a set containing the zero vector, there is an element that can be written as a 

combination of a collection of other vectors from the set, specifically, the zero 

vector can be written as a combination of the empty collection.) 

Lemma 1.1 suggests how to turn a spanning set into a spanning set that is 

minimal. Given a finite spanning set, we can repeatedly pull out vectors that 

are a linear combination of the others, until there aren’t any more such vectors 

left. 

1.12 Example This set spans R3 . 

⎛ 

S0 = { ⎝ 1 

⎞ ⎛ 

0⎠ 

, ⎝ 

0 

0 

⎞ ⎛ 

2⎠ 

, ⎝ 

0 

1 

⎞ ⎛ 

2⎠ 

, ⎝ 

0 

0 

⎞ ⎛ 

−1⎠ 

, ⎝ 

1 

3 

⎞ 

3⎠} 

0


Looking for a linear relationship 

⎛ 

c1 ⎝ 1 

⎞ ⎛ 

0⎠ 

+ c2 ⎝ 

0 

0 

⎞ ⎛ 

2⎠ 

+ c3 ⎝ 

0 

1 

⎞ ⎛ 

2⎠ 

+ c4 ⎝ 

0 

0 

⎞ ⎛ 

−1⎠ 

+ c5 ⎝ 

1 

3 

⎞ ⎛ 

3⎠ 

= ⎝ 

0 

0 

⎞ 

0⎠ 

0 

gives a three equations/five unknowns linear system whose solution set can be 

paramatrized in this way. 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

c1 −1 −3 

⎜c2⎟ 

⎜ 

⎜ ⎟ ⎜−1⎟ 

⎜ 

⎟ ⎜−3/2 

⎟ � 

{ ⎜c3⎟ 

⎜ ⎟ = c3 

⎜ 1 ⎟ + c5 

⎜ 0 ⎟ � 

⎟ c3,c5 ∈ R} 

⎝c4⎠ 

⎝ 0 ⎠ ⎝ 0 ⎠ 

0 

1 

c5 

Setting, say, c3 =0andc5 = 1 shows that the fifth vector is a linear combination 

of the first two. Thus, Lemma 1.1 gives that this set 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

1 0 1 0 

S1 = { ⎝0⎠ 

, ⎝2⎠ 

, ⎝2⎠ 

, ⎝−1⎠} 

0 0 0 1 

has the same span as S0. Similarly, the third vector of the new set S1 is a linear 

combination of the first two and we get 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

1 0 0 

S2 = { ⎝0⎠ 

, ⎝2⎠ 

, ⎝−1⎠} 

0 0 1 

with the same span as S1 and S0, but with one difference. This last set is 

linearly independent (this is easily checked), and so removal of any of its vectors 

will shrink the span. 

We finish this subsection by recasting that example as a theorem that any 

finite spanning set has a subset with the same span that is linearly independent. 

To prove that result we will first need some facts about how linear independence 

and dependence, which are properties of sets, interact with the subset relation 

between sets. 

1.13 Lemma Any subset of a linearly independent set is also linearly independent. 

Any superset of a linearly dependent set is also linearly dependent. 

Proof. This is clear. QED 

Restated, independence is preserved by subset and dependence is preserved 

by superset. Those are two of the four possible cases of interaction that we 

can consider. The third case, whether linear dependence is preserved by the 

subset operation, is covered by Example 1.12, which gives a linearly dependent 

set S0 with a subset S1 that is linearly dependent and another subset S2 that 

is linearly independent. 

That leaves one case, whether linear independence is preserved by superset. 

The next example shows what can happen.


1.14 Example Here are some linearly independent sets from R3 and their 

supersets. 

⎛ 

(1) If S1 = { ⎝ 1 

⎞ 

0⎠} 

then the span [S1] isthex-axis. 

0 

A linearly dependent superset: 

⎛ 

{ ⎝ 1 

⎞ ⎛ 

0⎠ 

, ⎝ 

0 

−3 

⎞ 

0 ⎠} 

A linearly independent superset: 

⎛ 

0 

{ ⎝ 1 

⎞ ⎛ 

0⎠ 

, ⎝ 

0 

0 

⎞ 

1⎠} 

0 

⎛ ⎞ ⎛ ⎞ 

1 0 

(2) If S2 = { ⎝0⎠ 

, ⎝1⎠} 

then [S2] isthexy-plane. 

0 0 

⎛ ⎞ 

1 

⎛ ⎞ 

0 

⎛ ⎞ 

3 

A linearly dependent superset: { ⎝0⎠ 

, ⎝1⎠ 

, ⎝−2⎠} 

⎛ 

0 

⎞ 

1 

⎛ 

0 

⎞ 

0 

⎛ 

0 

⎞ 

0 

A linearly independent superset: { ⎝0⎠ 

, ⎝1⎠ 

, ⎝0⎠} 

0 0 1 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

1 0 0 

(3) If S3 = { ⎝0⎠ 

, ⎝1⎠ 

, ⎝0⎠} 

then [S3] is all of R 

0 0 1 

3 . 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

1 0 0 2 

A linearly dependent superset: { ⎝0⎠ 

, ⎝1⎠ 

, ⎝0⎠ 

, ⎝−1⎠} 

0 0 1 3 

There are no linearly independent supersets. 

(Checking the dependence or independence of these sets is easy.) 

So in general a linearly independent set can have some supersets that are dependent 

and some supersets that are independent. We can characterize when a 

superset of a independent set is dependent and when it is independent. 

1.15 Lemma Where S is a linearly independent subset of a vector space V , 

S ∪{�v} is linearly dependent if and only if �v ∈ [S] 

for any �v ∈ V with �v �∈ S.


Proof. One implication is clear: if �v ∈ [S] then �v = c1�s1 + c2�s2 + ···+ cn�sn 

where each �si ∈ S and ci ∈ R, andso�0 =c1�s1 + c2�s2 + ···+ cn�sn +(−1)�v is a 

nontrivial linear relationship among elements of S ∪{�v}. 

The other implication requires the assumption that S is linearly independent. 

With S ∪{�v} linearly dependent, there is a nontrivial linear relationship c0�v + 

c1�s1 + c2�s2 + ···+ cn�sn = �0 and independence of S then implies that c0 �= 0,or 

else that would be a nontrivial relationship among members of S. Now rewriting 

this equation as �v = −(c1/c0)�s1 −···−(cn/c0)�sn shows that �v ∈ [S]. QED 

(Compare this result with Lemma 1.1. Note the additional hypothesis here of 

linear independence.) 

1.16 Corollary A subset S = {�s1,...,�sn} of a vector space is linearly dependent 

if and only if some �si is a linear combination of the vectors �s1, ... , �si−1 

listed before it. 

Proof. Consider S0 = {}, S1 = { �s1}, S2 = {�s1,�s2}, etc. Some index i ≥ 1is 

the first one with Si−1 ∪{�si} linearly dependent, and there �si ∈ [Si−1]. QED 

Lemma 1.15 can be restated in terms of independence instead of dependence: 

if S is linearly independent (and �v �∈ S) then the set S ∪{�v} is also linearly 

independent if and only if �v �∈ [S]. Applying Lemma 1.1, we conclude that if 

S is linearly independent and �v �∈ S then S ∪{�v} is also linearly independent 

if and only if [S ∪{�v}] �= [S]. Briefly, to preserve linear independence through 

superset we must expand the span. 

Example 1.14 shows that some linearly independent sets are maximal — 

have as many elements as possible — in that they have no linearly independent 

supersets. By the prior paragraph, linearly independent sets are maximal if and 

only if their span is the entire space, because then no vector exists that is not 

already in the span. 

This table summarizes the interaction between the properties of independence 

and dependence and the relations of subset and superset. 

K ⊂ S K ⊃ S 

S independent K must be independent K may be either 

S dependent K may be either K must be dependent 

In developing this table we’ve uncovered an intimate relationship between linear 

independence and span. Complementing the fact that a spanning set is minimal 

if and only if it is linearly independent, a linearly independent set is maximal if 

and only if it spans the space. 

We close with the result promised earlier that recasts Example 1.12 as a 

theorem. 

1.17 Theorem In a vector space, any finite subset has a linearly independent 

subset with the same span.


Proof. If the finite set S is linearly independent then there is nothing to prove 

so assume that S = {�s1,...,�sn} is linearly dependent. By Corollary 1.16, there 

is a vector �si that is a linear combination of �s1, ... , �si−1. Define S1 to be the 

set S −{�si}. Lemma 1.1 then says that the span does not shrink: [S1] =[S]. 

If S1 is linearly independent then we are finished. Otherwise repeat the 

prior paragraph to derive S2 ⊂ S1 such that [S2] =[S1]. Repeat this process 

until a linearly independent set appears; one must eventually appear because 

S is finite. (Formally, this part of the argument uses mathematical induction. 

Exercise 37 asks for the details.) QED 

In summary, we have introduced the definition of linear independence to 

formalize the idea of the minimality of a spanning set. We have developed some 

elementary properties of this idea. The most important is Lemma 1.15, which, 

complementing that a spanning set is minimal when it linearly independent, 

tells us that a linearly independent set is maximal when it spans the space. 

Exercises 

� 1.18 Decide whether each subset of R 3 is linearly dependent or linearly independent. 

� � 

1 

� � 

2 

� � 

4 

(a) { −3 , 2 , −4 } 

5 

� � 

1 

4 

� � 

2 

14 

� � 

3 

(b) { 7 , 7 , 7 } 

7 

� � 

0 

7 

� � 

1 

7 

(c) { 0 , 0 } 

−1 

� � 

9 

4 

� � 

2 

� � 

3 

� � 

12 

(d) { 9 , 0 , 5 , 12 } 

0 1 −4 −1 

� 1.19 Which of these subsets of P3 are linearly dependent and which are independent? 

(a) {3 − x +9x 2 , 5 − 6x +3x 2 , 1+1x− 5x 2 } 

(b) {−x 2 , 1+4x 2 } 

(c) {2+x +7x 2 , 3 − x +2x 2 , 4 − 3x 2 } 

(d) {8+3x +3x 2 ,x+2x 2 , 2+2x +2x 2 , 8 − 2x +5x 2 } 

� 1.20 Prove that each set {f,g} is linearly independent in the vector space of all 

functions from R + to R. 

(a) f(x) =x and g(x) =1/x 

(b) f(x) = cos(x) andg(x) =sin(x) 

(c) f(x) =e x and g(x) =ln(x) 

� 1.21 Which of these subsets of the space of real-valued functions of one real variable 

is linearly dependent and which is linearly independent? (Note that we have 

abbreviated some constant functions; e.g., in the first item, the ‘2’ stands for the 

constant function f(x) =2.)


(a) {2, 4sin 2 (x), cos 2 (x)} (b) {1, sin(x), sin(2x)} (c) {x, cos(x)} 

(d) {(1 + x) 2 ,x 2 +2x, 3} (e) {cos(2x), sin 2 (x), cos 2 (x)} (f) {0,x,x 2 } 

1.22 Does the equation sin 2 (x)/ cos 2 (x) =tan 2 (x) show that this set of functions 

{sin 2 (x), cos 2 (x), tan 2 (x)} is a linearly dependent subset of the set of all real-valued 

functions with domain (−π/2..π/2)? 

1.23 Why does Lemma 1.4 say “distinct”? 

� 1.24 Show that the nonzero rows of an echelon form matrix form a linearly independent 

set. 

� 1.25 (a) Show that if the set {�u, �v, �w} linearly independent set then so is the set 

{�u, �u + �v, �u + �v + �w}. 

(b) What is the relationship between the linear independence or dependence of 

the set {�u, �v, �w} and the independence or dependence of {�u − �v,�v − �w, �w − �u}? 

1.26 Example 1.10 shows that the empty set is linearly independent. 

(a) When is a one-element set linearly independent? 

(b) How about a set with two elements? 

1.27 In any vector space V , the empty set is linearly independent. What about all 

of V ? 

1.28 Show that if {�x, �y,�z} is linearly independent then so are all of its proper 

subsets: {�x, �y}, {�x, �z}, {�y,�z}, {�x},{�y}, {�z}, and{}. Is that ‘only if’ also? 

1.29 (a) Show that this 

� � � � 

1 −1 

S = { 1 , 2 } 

0 0 

is a linearly independent subset of R 3 . 

(b) Show that 

� � 

3 

2 

0 

is in the span of S by finding c1 and c2 giving a linear relationship. 

� � � � � � 

1 −1 3 

c1 1 + c2 2 = 2 

0 0 0 

Show that the pair c1,c2 is unique. 

(c) Assume that S is a subset of a vector space and that �v is in [S], so that �v is 

a linear combination of vectors from S. Prove that if S is linearly independent 

then a linear combination of vectors from S adding to �v is unique (that is, unique 

up to reordering and adding or taking away terms of the form 0 · �s). Thus S 

as a spanning set is minimal in this strong sense: each vector in [S] is“hit”a 

minimum number of times — only once. 

(d) Prove that it can happen when S is not linearly independent that distinct 

linear combinations sum to the same vector. 

1.30 Prove that a polynomial gives rise to the zero function if and only if it is 

the zero polynomial. (Comment. This question is not a Linear Algebra matter, 

but we often use the result. A polynomial gives rise to a function in the obvious 

way: x ↦→ cnx n + ···+ c1x + c0.) 

1.31 Return to Section 1.2 and redefine point, line, plane, and other linear surfaces 

to avoid degenerate cases.


1.32 (a) Show that any set of four vectors in R 2 is linearly dependent. 

(b) Is this true for any set of five? Any set of three? 

(c) What is the most number of elements that a linearly independent subset of 

R 2 can have? 

� 1.33 Is there a set of four vectors in R 3 , any three of which form a linearly independent 

set? 

1.34 Must every linearly dependent set have a subset that is dependent and a 

subset that is independent? 

1.35 In R 4 , what is the biggest linearly independent set you can find? The smallest? 

The biggest linearly dependent set? The smallest? (‘Biggest’ and ‘smallest’ mean 

that there are no supersets or subsets with the same property.) 

� 1.36 Linear independence and linear dependence are properties of sets. We can 

thus naturally ask how those properties act with respect to the familiar elementary 

set relations and operations. In this body of this subsection we have covered the 

subset and superset relations. We can also consider the operations of intersection, 

complementation, and union. 

(a) How does linear independence relate to intersection: can an intersection of 

linearly independent sets be independent? Must it be? 

(b) How does linear independence relate to complementation? 

(c) Show that the union of two linearly independent sets need not be linearly 

independent. 

(d) Characterize when the union of two linearly independent sets is linearly independent, 

in terms of the intersection of the span of each. 

� 1.37 For Theorem 1.17, 

(a) fill in the induction for the proof; 

(b) give an alternate proof that starts with the empty set and builds a sequence 

of linearly independent subsets of the given finite set until one appears with the 

same span as the given set. 

1.38 With a little calculation we can get formulas to determine whether or not a 

set of vectors is linearly independent. 

(a) Show that this subset of R 2 

� � � � 

a b 

{ , } 

c d 

is linearly independent if and only if ad − bc �= 0. 

(b) Show that this subset of R 3 

� � 

a 

� � 

b 

� � 

c 

{ d , e , f } 

g h i 

is linearly independent iff aei + bfg + cdh − hfa − idb − gec �= 0. 

(c) When is this subset of R 3 

� � 

a 

� � 

b 

{ d , e } 

g h 

linearly independent? 

(d) This is an opinion question: for a set of four vectors from R 4 , must there be 

a formula involving the sixteen entries that determines independence of the set? 

(You needn’t produce such a formula, just decide if one exists.)


� 1.39 (a) Prove that a set of two perpendicular nonzero vectors from R n is linearly 

independent when n>1. 

(b) What if n =1? n =0? 

(c) Generalize to more than two vectors. 

1.40 Consider the set of functions from the open interval (−1..1) to R. 

(a) Show that this set is a vector space under the usual operations. 

(b) Recall the formula for the sum of an infinite geometric series: 1+x+x 2 +···= 

1/(1−x) for all x ∈ (−1..1). Why does this not express a dependence inside of the 

set {g(x) =1/(1 − x),f0(x) =1,f1(x) =x, f2(x) =x 2 ,...} (in the vector space 

that we are considering)? (Hint. Review the definition of linear combination.) 

(c) Show that the set in the prior item is linearly independent. 

This shows that some vector spaces exist with linearly independent subsets that 

are infinite. 

1.41 Show that, where S is a subspace of V , if a subset T of S is linearly independent 

in S then T is also linearly independent in V . Is that ‘only if’?

Section III. Basis and Dimension 113 

2.III Basis and Dimension 

The prior section ends with the statement that a spanning set is minimal when 

it is linearly independent and that a linearly independent set is maximal when 

it spans the space. So the notions of minimal spanning set and maximal independent 

set coincide. In this section we will name this notion and study some 

of its properties. 

2.III.1 Basis 

1.1 Definition A basis for a vector space is a sequence of vectors that form a 

set that is linearly independent and that spans the space. 

We denote a basis with angle brackets 〈 � β1, � β2,...〉 to signify that this collection 

is a sequence ∗ — the order of the elements is significant. (The requirement 

that a basis be ordered will be needed, for instance, in Definition 1.13.) 

1.2 Example This is a basis for R2 . 

� � � � 

2 1 

〈 , 〉 

4 1 

It is linearly independent 

� � � � � � 

2 1 0 

c1 + c2 = 

4 1 0 

and it spans R 2 . 

2c1 +1c2 = x 

4c1 +1c2 = y 

1.3 Example This basis for R 2 

=⇒ 

2c1 +1c2 =0 

4c1 +1c2 =0 =⇒ c1 = c2 =0 

=⇒ c2 =2x − y and c1 =(y − x)/2 

� � � � 

1 2 

〈 , 〉 

1 4 

differs from the prior one because of its different order. The verification that it 

is a basis is just as in the prior example. 

1.4 Example The space R2 has many bases. Another one is this. 

� � � � 

1 0 

〈 , 〉 

0 1 

The verification is easy. 

∗ More information on sequences is in the appendix.


1.5 Definition For any Rn , 

⎛ ⎞ 

1 

⎜ 

⎜0 

⎟ 

En = 〈 ⎜. 

⎝. 

⎟ 

. ⎠ 

0 

, 

⎛ ⎞ 

0 

⎜ 

⎜1 

⎟ 

⎜. 

⎝. 

⎟ 

. ⎠ 

0 

,..., 

⎛ ⎞ 

0 

⎜ 

⎜0 

⎟ 

⎜. 

⎝. 

⎟ 

. ⎠ 

1 

〉 

is the standard (or natural) basis. We denote these vectors by �e1,...,�en. 

Note that the symbol ‘�e1’ means something different in a discussion of R 3 than 

it means in a discussion of R 2 . (Calculus books call R 2 ’s standard basis vectors 

�ı and �j instead of �e1 and �e2, and they call R 3 ’s standard basis vectors �ı, �j, and 

� k instead of �e1, �e2, and�e3.) 

1.6 Example We can give bases for spaces other than just those comprised of 

column vectors. For instance, consider the space {a · cos θ + b · sin θ � � a, b ∈ R} 

of function of the real variable θ. This is a natural basis 

〈1 · cos θ +0· sin θ, 0 · cos θ +1· sin θ〉 = 〈cos θ, sin θ〉 

while, another, more generic, basis is 〈cos θ − sin θ, 2cosθ +3sinθ〉. Verfication 

that these two are bases is Exercise 22. 

1.7 Example A natural basis for the vector space of cubic polynomials P3 is 

〈1,x,x 2 ,x 3 〉. Two other bases for this space are 〈x 3 , 3x 2 , 6x, 6〉 and 〈1, 1+x, 1+ 

x + x 2 , 1+x + x 2 + x 3 〉. Checking that these are linearly independent and span 

the space is easy. 

1.8 Example The trivial space {�0} has only one basis, the empty one 〈〉. 

1.9 Example The space of finite degree polynomials has a basis with infinitely 

many elements 〈1,x,x 2 ,...〉. 

1.10 Example We have seen bases before. For instance, we have described 

the solution set of homogeneous systems such as this one 

by paramatrizing. 

x + y − w =0 

z + w =0 

⎛ ⎞ ⎛ ⎞ 

−1 1 

⎜ 

{ ⎜ 1 ⎟ ⎜ 

⎟ 

⎝ 0 ⎠ y + ⎜ 0 ⎟ 

⎝−1⎠ 

0 1 

w � � y, w ∈ R} 

That is, we have described the vector space of solutions as the span of a twoelement 

set. We can easily check that this two-vector set is also linearly independent. 

Thus the solution set is a subspace of R 4 with a two-element basis.


1.11 Example Parameterization helps find bases for other vector spaces, not 

just for solution sets of homogeneous systems. To find a basis for this subspace 

of M2×2 

� � 

a b �� 

{ a + b − 2c =0} 

c 0 

we rewrite the condition as a = −b +2c to get this. 

� 

−b +2c 

{ 

c 

� � 

b �� −1 

b, c ∈ R} = {b 

0 

0 

� � 

1 2 

+ c 

0 1 

� 

0 �� 

b, c ∈ R} 

0 

Thus, this is a natural candidate for a basis. 

� 

−1 

〈 

0 

� � 

1 2 

, 

0 1 

� 

0 

〉 

0 

The above work shows that it spans the space. 

independent is routine. 

To show that it is linearly 

Consider Example 1.2 again. To show that the basis spans the space we 

looked at a general vector � � x 

2 

y from R . We found a formula for coefficients c1 

and c2 in terms of x and y. Although we did not mention it in the example, 

the formula shows that for each vector there is only one suitable coefficient pair. 

This always happens. 

1.12 Theorem In any vector space, a subset is a basis if and only if each 

vector in the space can be expressed as a linear combination of elements of the 

subset in a unique way. (We consider combinations to be the same if they differ 

only in the order of summands or in the addition or deletion of terms of the 

form ‘0 · � β’.) 

Proof. By definition, a sequence is a basis if and only if its vectors form both 

a spanning set and a linearly independent set. A subset is a spanning set if 

and only if each vector in the space is a linear combination of elements of that 

subset in at least one way. 

Thus, to finish we need only show that a subset is linearly independent if 

and only if every vector in the space is a linear combination of elements from 

the subset in at most one way. Consider two expressions of a vector as a linear 

combination of the members of the basis. We can rearrange the two sums, and 

if necessary add some 0 � βi’s, so that the two combine the same � β’s in the same 

order: �v = c1 � β1 + c2 � β2 + ···+ cn � βn and �v = d1 � β1 + d2 � β2 + ···+ dn � βn. Now, 

equality 

holds if and only if 

c1 � β1 + c2 � β2 + ···+ cn � βn = d1 � β1 + d2 � β2 + ···+ dn � βn 

(c1 − d1) � β1 + ···+(cn − dn) � βn = �0 

holds, and so asserting that each coefficient in the lower equation is zero is the 

same thing as asserting that ci = di for each i. QED


1.13 Definition In a vector space with basis B the representation of �v with 

respect to B is the column vector of the coefficients used to express �v as a linear 

combination of the basis vectors. That is, 

⎛ ⎞ 

c1 

⎜c2⎟ 

⎜ ⎟ 

RepB(�v) = ⎜ . 

⎝ . 

⎟ 

. ⎠ 

where B = 〈 � β1,..., � βn〉 and �v = c1 � β1 + c2 � β2 + ··· + cn � βn. The c’s are the 

coordinates of �v with respect to B. 

1.14 Example In P3, with respect to the basis B = 〈1, 2x, 2x 2 , 2x 3 〉, the rep- 

resentation of x + x 2 is 

cn 

B 

RepB(x + x 2 ⎛ ⎞ 

0 

⎜ 

)= ⎜1/2 

⎟ 

⎝1/2⎠ 

0 

(note that the coordinates are scalars, not vectors). With respect to a different 

basis D = 〈1+x, 1 − x, x + x 2 ,x+ x 3 〉, the representation 

is different. 

RepD(x + x 2 ⎛ ⎞ 

0 

⎜ 

)= ⎜0 

⎟ 

⎝1⎠ 

0 

1.15 Remark This use of column notation and the term ‘coordinates’ has 

both a down side and an up side. 

The down side is that representations look like vectors from R n , and that 

can be confusing when the vector space we are working with is R n , especially 

since we sometimes omit the subscript base. We must then infer the intent from 

the context. For example, the phrase ‘in R2 , where 

� � 

3 

�v = , ... ’ 

2 

refers to the plane vector that, when in canonical position, ends at (3, 2). To 

find the coordinates of that vector with respect to the basis 

� � � � 

1 0 

B = 〈 , 〉 

1 2 

we solve 

� � � � � � 

1 0 3 

c1 + c2 = 

1 2 2 

D 

B


to get that c1 =3andc2 =1/2. Then we have this. 

� � 

3 

RepB(�v) = 

−1/2 

Here, although we’ve ommited the subscript B from the column, the fact that 

the right side it is a representation is clear from the context. 

The up side of the notation and the term ‘coordinates’ is that they generalize 

the use that we are familiar with: in Rn and with respect to the standard 

basis En, the vector starting at the origin and ending at (v1,...,vn) has this 

representation. 

⎛ ⎞ ⎛ ⎞ 

v1 

vn 

v1 

⎜ 

RepEn ( ⎝ . 

⎟ ⎜ 

. ⎠) = ⎝ . 

⎟ 

. ⎠ 

Our main use of representations will come in the third chapter. The definition 

appears here because the fact that every vector is a linear combination of 

basis vectors in a unique way is a crucial property of bases, and also to help make 

two points. First, we put the elements of a basis in a fixed order so that coordinates 

can stated in that order. Second, for calculation of coordinates, among 

other things, we shall want our bases to have only finitely many elements. We 

will see that in the next subsection. 

Exercises 

� 1.16 Decide if each is a basis for R 3 � � � � � � . � � 

1 3 0 

1 

� � 

3 

� � 

0 

� � 

1 

� � 

2 

(a) 〈 2 , 2 , 0 〉 (b) 〈 2 , 2 〉 (c) 〈 2 , 1 , 5 〉 

3 

� � 

0 

1 

� � 

1 

1 

� � 

1 

3 1 

−1 1 0 

(d) 〈 2 , 1 , 3 〉 

−1 1 0 

� 1.17 Represent � � the� vector � � with � respect to the basis. 

1 1 −1 

(a) , B = 〈 , 〉⊆R 

2 1 1 

2 

(b) x 2 + x 3 , D = 〈1, 1+x, 1+x + x 2 , 1+x + x 2 + x 3 〉⊆P3 

⎛ ⎞ 

0 

⎜−1⎟ 

(c) ⎝ 

0 

⎠, E4 ⊆ R 

1 

4 

1.18 Find a basis for P2, the space of all quadratic polynomials. Must any such 

basis contain a polynomial of each degree: degree zero, degree one, and degree 

two? 

1.19 Find a basis for the solution set of this system. 

x1 − 4x2 +3x3 − x4 =0 

2x1 − 8x2 +6x3 − 2x4 =0 

� 1.20 Find a basis for M2×2, the space of 2×2 matrices. 

vn 

En


� 1.21 Find a basis for each. 

(a) The subspace {a2x 2 � 

+ a1x + a0 � a2 − 2a1 = a0} of P2 

(b) The space of three-wide row vectors whose first and second components add 

to zero 

(c) This subspace of the 2×2 matrices 

� � 

a b �� 

{ c − 2b =0} 

0 c 

1.22 Check Example 1.6. 

� 1.23 Find the span of each set and then find a basis for that span. 

(a) {1+x, 1+2x} in P2 (b) {2 − 2x, 3+4x 2 } in P2 

� 1.24 Find a basis for each of these subspaces of the space P3 of cubic polynomials. 

(a) The subspace of cubic polynomials p(x) such that p(7) = 0 

(b) The subspace of polynomials p(x) such that p(7) = 0 and p(5) = 0 

(c) The subspace of polynomials p(x) such that p(7) = 0, p(5) = 0, and p(3) = 0 

(d) The space of polynomials p(x) such that p(7) = 0, p(5) = 0, p(3) = 0, 

and p(1) = 0 

1.25 We’ve seen that it is possible for a basis to remain a basis when it is reordered. 

Must it remain a basis? 

1.26 Can a basis contain a zero vector? 

� 1.27 Let 〈 � β1, � β2, � β3〉 beabasisforavectorspace. 

(a) Show that 〈c1� β1,c2� β2,c3� β3〉 is a basis when c1,c2,c3 �= 0. What happens 

when at least one ci is 0? 

(b) Prove that 〈�α1,�α2,�α3〉 is a basis where �αi = � β1 + � βi. 

1.28 Give one more vector �v that will make each into a basis for the indicated 

space. 

� � 

1 

(a) 〈 ,�v〉 in R 

1 

2 

� � � � 

1 0 

(b) 〈 1 , 1 ,�v〉 in R 

0 0 

3 

(c) 〈x, 1+x 2 ,�v〉 in P2 

� 1.29 Where 〈 � β1,..., � βn〉 is a basis, show that in this equation 

c1 � β1 + ···+ ck � βk = ck+1 � βk+1 + ···+ cn � βn 

each of the ci’s is zero. Generalize. 

1.30 A basis contains some of the vectors from a vector space; can it contain them 

all? 

1.31 Theorem 1.12 shows that, with respect to a basis, every linear combination is 

unique. If a subset is not a basis, can linear combinations be not unique? If so, 

must they be? 

� 1.32 A square matrix is symmetric if for all indices i and j, entryi, j equals entry 

j, i. 

(a) Find a basis for the vector space of symmetric 2×2 matrices. 

(b) Find a basis for the space of symmetric 3×3 matrices. 

(c) Find a basis for the space of symmetric n×n matrices. 

� 1.33 We can show that every basis for R 3 contains the same number of vectors, 

specifically, three of them. 

(a) Show that no linearly independent subset of R 3 contains more than three 

vectors.


(b) Show that no spanning subset of R 3 contains fewer than three vectors. (Hint. 

Recall how to calculate the span of a set and show that this method, when applied 

to two vectors, cannot yield all of R 3 .) 

1.34 One of the exercises in the Subspaces subsection shows that the set 

� � 

x �� 

{ y x + y + z =1} 

z 

is a vector space under these operations. 

� � � � � � 

x1 x2 x1 + x2 − 1 

� � 

x 

� � 

rx − r +1 

y1 + y2 = y1 + y2 r y = ry 

z1 

Find a basis. 

z2 z1 + z2 

z rz 

2.III.2 Dimension 

In the prior subsection we saw that a vector space can have many different 

bases. For example, following the definition of a basis, we saw three different 

bases for R 2 . So we cannot talk about “the” basis for a vector space. 

True, some vector spaces have bases that strike us as more natural than others, 

for instance, R 2 ’s basis E2 or R 3 ’s basis E3 or P2’s basis 〈1,x,x 2 〉. But 

the idea of “natural” is hard to make formal. For example, with the space 

{a2x2 � 

+ a1x + a0 

� 2a2 − a0 = a1}, no particular basis leaps out at us as “the” 

natural one. We cannot, in general, associate with a space any single basis that 

best describes that space. 

We can, however, find something about the bases that is uniquely associated 

with the space. This subsection shows that any two bases for a space have the 

same number of elements. So, with each space we can associate a number, the 

number of vectors in any of its bases. 

This brings us back to when we considered the two things that could be 

meant by the term ‘minimal spanning set’. At that point we defined ‘minimal’ 

as linearly independent, but we noted that another reasonable interpretation of 

the term is that a spanning set is ‘minimal’ when it has the fewest number of 

elements of any set with the same span. At the end of this subsection, after we 

have shown that all bases have the same number of elements, then we will have 

shown that the two senses of ‘minimal’ are equivalent. 

Before we start, we first limit our attention to spaces where at least one basis 

has only finitely many members. 

2.1 Definition A vector space is finite-dimensional if it has a basis with only 

finitely many vectors. 

(One reason for sticking to finite-dimensional spaces is so that the representation 

of a vector with respect to a basis is a finitely-tall vector, and so can be easily 

written out. A further remark is at the end of this subsection.) From now on


we study only finite-dimensional vector spaces. We shall take the term ‘vector 

space’ to mean ‘finite-dimensional vector space’. Infinite-dimensional spaces are 

interesting and important, but they lie outside of our scope. 

To prove the main theorem we shall use a technical result. 

2.2 Lemma (Exchange Lemma) Assume that B = 〈 � β1,..., � βn〉 is a basis 

for a vector space, and that for the vector �v the relationship �v = c1 � β1 + c2 � β2 + 

···+ cn � βn has ci �= 0. Then exchanging � βi for �v yields another basis for the 

space. 

Proof. Call the outcome of the exchange ˆ B = 〈 � β1,..., � βi−1,�v, � βi+1,..., � βn〉. 

We first show that ˆ B is linearly independent. Any relationship d1 � β1 + ···+ 

di�v + ···+ dn � βn = �0 among the members of ˆ B, after substitution for �v, 

d1 � β1 + ···+ di · (c1 � β1 + ···+ ci � βi + ···+ cn � βn)+···+ dn � βn = �0 (∗) 

gives a linear relationship among the members of B. The basis B is linearly 

independent, so the coefficient dici of � βi is zero. Because ci is assumed to be 

nonzero, di = 0. Using this in equation (∗) above gives that all of the other d’s 

are also zero. Therefore ˆ B is linearly independent. 

We finish by showing that ˆ B has the same span as B. Half of this argument, 

that [ ˆ B] ⊆ [B], is easy; any member d1 � β1 + ···+ di�v + ···+ dn � βn of [ ˆ B]can 

be written d1 � β1 + ···+ di · (c1 � β1 + ···+ cn � βn)+···+ dn � βn, which is a linear 

combination of linear combinations of members of B, and hence is in [B]. For 

the [B] ⊆ [ ˆ B] half of the argument, recall that when �v = c1 � β1 + ···+ cn � βn with 

ci �= 0, then the equation can be rearranged to � βi =(−c1/ci) � β1+···+(−1/ci)�v+ 

···+(−cn/ci) � βn. Now, consider any member d1 � β1 + ···+ di � βi + ···+ dn � βn of 

[B], substitute for � βi its expression as a linear combination of the members 

of ˆ B, and recognize (as in the first half of this argument) that the result is a 

linear combination of linear combinations, of members of ˆ B, and hence is in 

[ ˆ B]. QED 

2.3 Theorem In any finite-dimensional vector space, all of the bases have the 

same number of elements. 

Proof. Fix a vector space with at least one finite basis. Choose, from among all 

of this space’s bases, B = 〈 � β1,..., � βn〉 of minimal size. We will show that any 

other basis D = 〈 � δ1, � δ2,...〉 also has the same number of members, n. Because 

B has minimal size, D has no fewer than n vectors. We will argue that it cannot 

have more. 

The basis B spans the space and � δ1 is in the space, so � δ1 is a nontrivial linear 

combination of elements of B. By the Exchange Lemma, � δ1 can be swapped for 

a vector from B, resulting in a basis B1, where one element is � δ and all of the 

n − 1 other elements are � β’s. 

The prior paragraph forms the basis step for an induction argument. The 

inductive step starts with a basis Bk (for 1 ≤ k


� δk+1. Represent it as a linear combination of elements of Bk. The key point: in 

that representation, at least one of the nonzero scalars must be associated with 

a � βi or else that representation would be a nontrivial linear relationship among 

elements of the linearly independent set D. Exchange � δk+1 for � βi to get a new 

basis Bk+1 with one � δ more and one � β fewer than the previous basis Bk. 

Repeat the inductive step until no � β’s remain, so that Bn contains � δ1,..., � δn. 

Now, D cannot have more than these n vectors because any � δn+1 that remains 

would be in the span of Bn (since it is a basis) and hence would be a linear combination 

of the other � δ’s, contradicting that D is linearly independent. QED 

2.4 Definition The dimension of a vector space is the number of vectors in 

any of its bases. 

2.5 Example Any basis for R n has n vectors since the standard basis En has 

n vectors. Thus, this definition generalizes the most familiar use of term, that 

R n is n-dimensional. 

2.6 Example The space Pn of polynomials of degree at most n has dimension 

n + 1. We can show this by exhibiting any basis — 〈1,x,...,xn 〉 comes to 

mind — and counting its members. 

2.7 Example A trivial space is zero-dimensional since its basis is empty. 

Again, although we sometimes say ‘finite-dimensional’ as a reminder, in the 

rest of this book all vector spaces are assumed to be finite-dimensional. An 

instance of this is that in the next result the word ‘space’ should be taken to 

mean ‘finite-dimensional vector space’. 

2.8 Corollary No linearly independent set can have a size greater than the 

dimension of the enclosing space. 

Proof. Inspection of the above proof shows that it never uses that D spans the 

space, only that D is linearly independent. QED 

2.9 Example Recall the subspace diagram from the prior section showing the 

subspaces of R 3 . Each subspaces shown is described with a minimal spanning 

set, for which we now have the term ‘basis’. The whole space has a basis with 

three members, the plane subspaces have bases with two members, the line 

subspaces have bases with one member, and the trivial subspace has a basis 

with zero members. When we saw that diagram we could not show that these 

are the only subspaces that this space has. We can show it now. The prior 

corollary proves the only subspaces of R 3 are either three-, two-, one-, or zerodimensional. 

Therefore, the diagram indicates all of the subspaces. There are 

no subspaces somehow, say, between lines and planes. 

2.10 Corollary Any linearly independent set can be expanded to make a basis. 

Proof. If a linearly independent set is not already a basis then it must not 

span the space. Adding to it a vector that is not in the span preserves linear 

independence. Keep adding, until the resulting set does span the space, which 

the prior corollary shows will happen after only a finite number of steps. QED


2.11 Corollary Any spanning set can be shrunk to a basis. 

Proof. Call the spanning set S. If S is empty then it is already a basis. If 

S = {�0} then it can be shrunk to the empty basis without changing the span. 

Otherwise, S contains a vector �s1 with �s1 �= �0 and we can form a basis 

B1 = 〈�s1〉. If[B1] =[S] then we are done. 

If not then there is a �s2 ∈ [S] such that �s2 �∈ [B1]. Let B2 = 〈�s1, �s2〉; if 

[B2] =[S] then we are done. 

We can repeat this process until the spans are equal, which must happen in 

at most finitely many steps. QED 

2.12 Corollary In an n-dimensional space, a set of n vectors is linearly independent 

if and only if it spans the space. 

Proof. First we will show that a subset with n vectors is linearly independent 

if and only if it is a basis. ‘If’ is trivially true — bases are linearly independent. 

‘Only if’ holds because a linearly independent set can be expanded to a basis, 

but a basis has n elements, so that this expansion is actually the set we began 

with. 

To finish, we will show that any subset with n vectors spans the space if and 

only if it is a basis. Again, ‘if’ is trivial. ‘Only if’ holds because any spanning 

set can be shrunk to a basis, but a basis has n elements and so this shrunken 

set is just the one we started with. QED 

The main result of this subsection, that all of the bases in a finite-dimensional 

vector space have the same number of elements, is the single most important 

result in this book because, as Example 2.9 shows, it describes what vector 

spaces and subspaces there can be. We will see more in the next chapter. 

2.13 Remark The case of infinite-dimensional vector spaces is somewhat controversial. 

The statement ‘any infinite-dimensional vector space has a basis’ 

is known to be equivalent to a statement called the Axiom of Choice (see 

[Blass 1984]). Mathematicians differ philosophically on whether to accept or 

reject this statement as an axiom on which to base mathematics. Consequently 

the question about infinite-dimensional vector spaces is still somewhat up in the 

air. (A discussion of the Axiom of Choice can be found in the Frequently Asked 

Questions list for the Usenet group sci.math. Another accessible reference is 

[Rucker].) 

Exercises 

Assume that all spaces are finite-dimensional unless otherwise stated. 

� 2.14 Find a basis for, and the dimension of, P2. 

2.15 Find a basis for, and the dimension of, the solution set of this system. 

x1 − 4x2 +3x3 − x4 =0 

2x1 − 8x2 +6x3 − 2x4 =0 

� 2.16 Find a basis for, and the dimension of, M2×2, the vector space of 2×2matrices.


2.17 Find the dimension of the vector space of matrices 

� � 

a b 

c d 

subject to each condition. 

(a) a, b, c, d ∈ R 

(b) a − b +2c =0andd ∈ R 

(c) a + b + c =0,a + b − c =0,andd ∈ R 

� 2.18 Find the dimension of each. 

(a) The space of cubic polynomials p(x) such that p(7) = 0 

(b) The space of cubic polynomials p(x) such that p(7) = 0 and p(5) = 0 

(c) The space of cubic polynomials p(x) such that p(7) = 0, p(5) = 0, and p(3) = 

0 

(d) The space of cubic polynomials p(x) such that p(7) = 0, p(5) = 0, p(3) = 0, 

and p(1) = 0 

2.19 What is the dimension of the span of the set {cos 2 θ, sin 2 θ, cos 2θ, sin 2θ}? This 

span is a subspace of the space of all real-valued functions of one real variable. 

2.20 Find the dimension of C 47 , the vector space of 47-tuples of complex numbers. 

2.21 What is the dimension of the vector space M3×5 of 3×5 matrices? 

� 2.22 Show that this is a basis for R 4 . 

⎛ ⎞ ⎛ ⎞ 

1 1 

⎛ ⎞ 

1 

⎛ ⎞ 

1 

⎜0⎟ 

⎜1⎟ 

⎜1⎟ 

⎜1⎟ 

〈 ⎝ 

0 

⎠ , ⎝ 

0 

⎠ , ⎝ 

1 

⎠ , ⎝ 

1 

⎠〉 

0 0 0 1 

(The results of this subsection can be used to simplify this job.) 

2.23 Refer to Example 2.9. 

(a) Sketch a similar subspace diagram for P2. 

(b) Sketch one for M2×2. 

� 2.24 Observe that, where S is a set, the functions f : S → R form a vector space 

under the natural operations: f + g (s) =f(s)+g(s) andr · f (s) =r · f(s). What 

is the dimension of the space resulting for each domain? 

(a) S = {1} (b) S = {1, 2} (c) S = {1,... ,n} 

2.25 (See Exercise 24.) Prove that this is an infinite-dimensional space: the set of 

all functions f : R → R under the natural operations. 

2.26 (See Exercise 24.) What is the dimension of the vector space of functions 

f : S → R, under the natural operations, where the domain S is the empty set? 

2.27 Show that any set of four vectors in R 2 is linearly dependent. 

2.28 Show that the set 〈�α1,�α2,�α3〉 ⊂R 3 is a basis if and only if there is no plane 

through the origin containing all three vectors. 

2.29 (a) Prove that any subspace of a finite dimensional space has a basis. 

(b) Prove that any subspace of a finite dimensional space is finite dimensional. 

2.30 Where is the finiteness of B used in Theorem 2.3? 

� 2.31 Prove that if U and W are both three-dimensional subspaces of R 5 then U ∩W 

is non-trivial. Generalize. 

2.32 Because a basis for a space is a subset of that space, we are naturally led to 

how the property ‘is a basis’ interacts with set operations. 

(a) Consider first how bases might be related by ‘subset’. Assume that U, W are


subspaces of some vector space and that U ⊆ W . Can there exist bases BU for 

U and BW for W such that BU ⊆ BW ? Must such bases exist? 

For any basis BU for U, must there be a basis BW for W such that BU ⊆ BW ? 

For any basis BW for W , must there be a basis BU for U such that BU ⊆ BW ? 

For any bases BU ,BW for U and W ,mustBU be a subset of BW ? 

(b) Is the intersection of bases a basis? For what space? 

(c) Is the union of bases a basis? For what space? 

(d) What about complement? 

(Hint. Test any conjectures against some subspaces of R 3 .) 

� 2.33 Consider how ‘dimension’ interacts with ‘subset’. Assume U and W are both 

subspaces of some vector space, and that U ⊆ W . 

(a) Prove that dim(U) ≤ dim(W ). 

(b) Prove that equality of dimension holds if and only if U = W . 

(c) Show that the prior item does not hold if they are infinite-dimensional. 

2.34 [Wohascum no. 47] Foranyvector�v in R n and any permutation σ of the 

numbers 1, 2, ... , n (that is, σ is a rearrangement of those numbers into a new 

order), define σ(�v) to be the vector whose components are vσ(1), vσ(2), ... ,and 

vσ(n) (where σ(1) is the first number in the rearrangement, etc.). Now fix �v and 

let V be the span of {σ(�v) � � σ permutes 1, ... , n}. What are the possibilities for 

the dimension of V ? 

2.III.3 Vector Spaces and Linear Systems 

We will now reconsider linear systems and Gauss’ method, aided by the tools 

and terms of this chapter. We will make three points. 

For the first point, recall the Linear Combination Lemma and its corollary: if 

two matrices are related by row operations A −→ · · · −→ B then each row of B 

is a linear combination of the rows of A. That is, Gauss’ method works by taking 

linear combinations of rows. Therefore, the right setting in which to study row 

operations in general, and Gauss’ method in particular, is the following vector 

space. 

3.1 Definition The row space of a matrix is the span of the set of its rows. The 

row rank is the dimension of the row space, the number of linearly independent 

rows. 

3.2 Example If 

A = 

� � 

2 3 

4 6 

then Rowspace(A) is this subspace of the space of two-component row vectors. 

{c1 · � 2 3 � + c2 · � 4 6 � � � c1,c2 ∈ R} 

The linear dependence of the second on the first is obvious and so we can simplify 

this description to {c · � 2 3 � � � c ∈ R}.


3.3 Lemma If the matrices A and B are related by a row operation 

A ρi↔ρj 

−→ B or A kρi 

−→ B or A kρi+ρj 

−→ B 

(for i �= j and k �= 0) then their row spaces are equal. Hence, row-equivalent 

matrices have the same row space, and hence also, the same row rank. 

Proof. The row space of A is the set of all linear combinations of the rows 

of A. By the Linear Combination Lemma then, each row of B is in the row 

space of A. Further, Rowspace(B) ⊆ Rowspace(A) because a member of the 

set Rowspace(B) is a linear combination of the rows of B, which means it is a 

combination of a combination of the rows of A, and hence is also a member of 

Rowspace(A). 

For the other containment, recall that row operations are reversible: A −→ B 

if and only if B −→ A. With that, Rowspace(A) ⊆ Rowspace(B) also follows 

from the prior paragraph, and hence the two sets are equal. QED 

So, row operations leave the row space unchanged. But of course, Gauss’ 

method performs the row operations systematically, with a specific goal in mind, 

echelon form. 

3.4 Lemma The nonzero rows of an echelon form matrix make up a linearly 

independent set. 

Proof. A result in the first chapter, Lemma III.2.5, states that in an echelon 

form matrix, no nonzero row is a linear combination of the other rows. This is 

a restatement of that result into new terminology. QED 

Thus, in the language of this chapter, Gaussian reduction works by eliminating 

linear dependences among rows, leaving the span unchanged, until no 

nontrivial linear relationships remain (among the nonzero rows). That is, Gauss’ 

method produces a basis for the row space. 

3.5 Example From any matrix, we can produce a basis for the row space by 

performing Gauss’ method and taking the nonzero rows of the resulting echelon 

form matrix. For instance, 

⎛ ⎞ 

⎛ ⎞ 

1 3 1 

⎝1 4 1⎠ 

2 0 5 

−ρ1+ρ2 

−→ 

−2ρ1+ρ3 

6ρ2+ρ3 

−→ 

1 

⎝0 3 

1 

1 

0⎠ 

0 0 3 

produces the basis 〈 � 1 3 1 � , � 0 1 0 � , � 0 0 3 � 〉 for the row space. This 

is a basis for the row space of both the starting and ending matrices, since the 

two row spaces are equal. 

Using this technique, we can also find bases for spans not directly involving 

row vectors.


3.6 Definition The column space of a matrix is the span of the set of its 

columns. The column rank is the dimension of the column space, the number 

of linearly independent columns. 

Our interest in column spaces stems from our study of linear systems. An 

example is that this system 

c1 +3c2 +7c3 = d1 

2c1 +3c2 +8c3 = d2 

4c1 

c2 +2c3 = d3 

+4c3 = d4 

has a solution if and only if the vector of d’s is a linear combination of the other 

column vectors, 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

1 3 7 

⎜ 

c1 

⎜2⎟ 

⎜ 

⎟ 

⎝0⎠ 

+ c2 

⎜3⎟ 

⎜ 

⎟ 

⎝1⎠ 

+ c3 

⎜8 

⎟ 

⎝2⎠ 

4 0 4 

= 

⎛ ⎞ 

d1 

⎜d2⎟ 

⎜ ⎟ 

⎝d3⎠ 

meaning that the vector of d’s is in the column space of the matrix of coefficients. 

3.7 Example Given this matrix, 

⎛ 

1 

⎜ 

⎜2 

⎝0 

3 

3 

1 

⎞ 

7 

8 ⎟ 

2⎠ 

4 0 4 

to get a basis for the column space, temporarily turn the columns into rows and 

reduce. 

⎛ 

⎞ 

1 2 0 4 

⎝3 

3 1 0⎠ 

7 8 2 4 

−3ρ1+ρ2 

⎛ 

⎞ 

1 2 0 4 

−2ρ2+ρ3 

−→ −→ ⎝0 

−3 1 −12⎠ 

−7ρ1+ρ3 

0 0 0 0 

Now turn the rows back to columns. 

⎛ ⎞ 

1 

⎜ 

〈 ⎜2 

⎟ 

⎝0⎠ 

4 

, 

⎛ ⎞ 

0 

⎜ −3 ⎟ 

⎝ 1 ⎠ 

−12 

〉 

The result is a basis for the column space of the given matrix. 

3.8 Definition The transpose of a matrix is the result of interchanging the 

rows and columns of that matrix. That is, column j of the matrix A is row j of 

A trans , and vice versa. 

d4


So the instructions for the prior example are “transpose, reduce, and transpose 

back”. 

We can even, at the price of tolerating the as-yet-vague idea of vector spaces 

being “the same”, use Gauss’ method to find bases for spans in other types of 

vector spaces. 

3.9 Example To get a basis for the span of {x 2 + x 4 , 2x 2 +3x 4 , −x 2 − 3x 4 } 

in the space P4, think of these three polynomials as “the same” as the row 

vectors � 0 0 1 0 1 � , � 0 0 2 0 3 � , and � 0 0 −1 0 −3 � , apply 


⎛ 

0 

⎝0 0 

0 

0 

0 

1 

2 

−1 

0 

0 

0 

⎞ 

1 

3 ⎠ 

−3 

−2ρ1+ρ2 2ρ2+ρ3 

−→ −→ 

ρ1+ρ3 

⎛ 

0 

⎝0 0 

0 

1 

0 

0 

0 

⎞ 

1 

1⎠ 

0 0 0 0 0 

and translate back to get the basis 〈x 2 + x 4 ,x 4 〉. (As mentioned earlier, we will 

make the phrase “the same” precise at the start of the next chapter.) 

Thus, our first point in this subsection is that the tools of this chapter give 

us a more conceptual understanding of Gaussian reduction. 

For the second point of this subsection, consider the effect on the column 

space of this row reduction. 

� � 

1 2 −2ρ1+ρ2 

−→ 

2 4 

� � 

1 2 

0 0 

The column space of the left-hand matrix contains vectors with a second component 

that is nonzero. But the column space of the right-hand matrix is different 

because it contains only vectors whose second component is zero. It is this 

knowledge that row operations can change the column space that makes next 

result surprising. 

3.10 Lemma Row operations do not change the column rank. 

Proof. Restated, if A reduces to B then the column rank of B equals the 

column rank of A. 

We will be done if we can show that row operations do not affect linear relationships 

among columns (e.g., if the fifth column is twice the second plus the 

fourth before a row operation then that relationship still holds afterwards), because 

the column rank is just the size of the largest set of unrelated columns. But 

this is exactly the first theorem of this book: in a relationship among columns, 

⎛ ⎞ 

a1,1 

⎜ a2,1 ⎟ 

⎜ ⎟ 

c1 · ⎜ . 

⎝ . 

⎟ 

. ⎠ + ···+ cn 

⎛ ⎞ 

a1,n 

⎜ a2,n ⎟ 

⎜ ⎟ 

· ⎜ . 

⎝ . 

⎟ 

. ⎠ = 

⎛ ⎞ 

0 

⎜ 

⎜0 

⎟ 

⎜. 

⎝. 

⎟ 

. ⎠ 

0 

am,1 

am,n 

row operations leave unchanged the set of solutions (c1,... ,cn). QED


Another way, besides the prior result, to state that Gauss’ method has something 

to say about the column space as well as about the row space is to consider 

again Gauss-Jordan reduction. Recall that it ends with the reduced echelon form 

of a matrix, as here. 

⎛ 

⎞ 

⎛ ⎞ 

1 

⎝2 3 

6 

1 

3 

6 

16⎠−→ 

··· −→ 

1 3 1 6 

1 

⎝0 3 

0 

0 

1 

2 

4⎠ 

0 0 0 0 

Consider the row space and the column space of this result. Our first point 

made above says that a basis for the row space is easy to get: simply collect 

together all of the rows with leading entries. However, because this is a reduced 

echelon form matrix, a basis for the column space is just as easy: take the 

columns containing the leading entries, that is, 〈�e1,�e2〉. (Linear independence 

is obvious. The other columns are in the span of this set, since they all have 

a third component of zero.) Thus, for a reduced echelon form matrix, bases 

for the row and column spaces can be found in essentially the same way — 

by taking the parts of the matrix, the rows or columns, containing the leading 

entries. 

3.11 Theorem The row rank and column rank of a matrix are equal. 

Proof. First bring the matrix to reduced echelon form. At that point, the 

row rank equals the number of leading entries since each equals the number 

of nonzero rows. Also at that point, the number of leading entries equals the 

column rank because the set of columns containing leading entries consists of 

some of the �ei’s from a standard basis, and that set is linearly independent and 

spans the set of columns. Hence, in the reduced echelon form matrix, the row 

rank equals the column rank, because each equals the number of leading entries. 

But Lemma 3.3 and Lemma 3.10 show that the row rank and column rank 

are not changed by using row operations to get to reduced echelon form. Thus 

the row rank and the column rank of the original matrix are also equal. QED 

3.12 Definition The rank of a matrix is its row rank or column rank. 

So our second point in this subsection is that the column space and row 

space of a matrix have the same dimension. Our third and final point is that 

the concepts that we’ve seen arising naturally in the study of vector spaces are 

exactly the ones that we have studied with linear systems. 

3.13 Theorem For linear systems with n unknowns and with matrix of coefficients 

A, the statements 

(1) the rank of A is r 

(2) the space of solutions of the associated homogeneous system has dimension 

n − r 

are equivalent.


So if the system has at least one particular solution then for the set of solutions, 

the number of parameters equals n − r, the number of variables minus the rank 

of the matrix of coefficients. 

Proof. The rank of A is r if and only if Gaussian reduction on A ends with r 

nonzero rows. That’s true if and only if echelon form matrices row equivalent 

to A have r-many leading variables. That in turn holds if and only if there are 

n − r free variables. QED 

3.14 Remark [Munkres] Sometimes that result is mistakenly remembered to 

say that the general solution of an n unknown system of m equations uses n−m 

parameters. The number of equations is not the relevant figure, rather, what 

matters is the number of independent equations (the number of equations in 

a maximal independent set). Where there are r independent equations, the 

general solution involves n − r parameters. 

3.15 Corollary Where the matrix A is n×n, the statements 

(1) the rank of A is n 

(2) A is nonsingular 

(3) the rows of A form a linearly independent set 

(4) the columns of A form a linearly independent set 

(5) any linear system whose matrix of coefficients is A has one and only one 

solution 

are equivalent. 

Proof. Clearly (1) ⇐⇒ (2) ⇐⇒ (3) ⇐⇒ (4). The last, (4) ⇐⇒ (5), holds 

because a set of n column vectors is linearly independent if and only if it is a 

basis for Rn , but the system 

⎛ ⎞ ⎛ ⎞ 

a1,1 

a1,n 

⎜ a2,1 ⎟ ⎜ a2,n ⎟ 

⎜ ⎟ ⎜ ⎟ 

c1 ⎜ 

⎝ . 

⎟ + ···+ cn ⎜ 

. ⎠ ⎝ . 

⎟ 

. ⎠ = 

⎛ ⎞ 

d1 

⎜d2⎟ 

⎜ ⎟ 

⎜ 

⎝ . 

⎟ 

. ⎠ 

am,1 

am,n 

has a unique solution for all choices of d1,...,dn ∈ R if and only if the vectors 

of a’s form a basis. QED 

Exercises 

3.16 Transpose each. 

� � 

2 1 

(a) 

(b) 

3 1 

� 

2 

1 

� 

1 

3 

(c) 

� 

1 

6 

4 

7 

� 

3 

8 

(d) 

(e) � −1 −2 � 

� 3.17 Decide if the vector is in the row space of the matrix. 

� � 

2 1 

(a) , 

3 1 

� 1 0 � 

� � 

0 1 3 

(b) −1 0 1 , 

−1 2 7 

� 1 1 1 � 

� 3.18 Decide if the vector is in the column space. 

dn 

� � 

0 

0 

0


(a) 

� 

1 

1 

� � � 

1 1 

, 

1 3 

(b) 

� 

1 

2 

1 

3 

0 

−3 

� � � 

1 1 

4 , 0 

−3 0 

� 3.19 Find a basis for the row space of this matrix. 

⎛ 

⎞ 

2 0 3 4 

⎜0 

⎝ 

3 

1 

1 

1 

0 

−1⎟ 

2 

⎠ 

1 0 −4 −1 

� 3.20 Find � the rank�of each matrix. � 

2 1 3 

1 −1 

� 

2 

(a) 1 −1 2 (b) 3 −3 6 (c) 

1 

� 

0 

0 

0 

3 

� 

0 

−2 2 −4 

(d) 0 0 0 

0 0 0 

� 3.21 Find a basis for the span of each set. 

(a) { � 1 3 � , � −1 3 � , � 1 4 � , � 2 1 � }⊆M1×2 

� � � � � � 

1 3 1 

(b) { 2 , 1 , −3 }⊆R 

1 −1 −3 

3 

(c) {1+x, 1 − x 2 , 3+2x− x 2 }⊆P3 

� � � � � 

1 0 1 1 0 3 −1 

(d) { 

, 

, 

3 1 −1 2 1 4 −1 

0 

−1 

� 

−5 

−9 

� 

1 3 

� 

2 

5 1 1 

6 4 3 

}⊆M2×3 

3.22 Which matrices have rank zero? Rank one? 

� 3.23 Given a, b, c ∈ R, what choice of d will cause this matrix to have the rank of 

one? 

� 

a 

� 

b 

c d 

3.24 Find the column rank of this matrix. 

� 

1 3 −1 5 0 

� 

4 

2 0 1 0 4 1 

3.25 Show that a linear system with at least one solution has at most one solution if 

and only if the matrix of coefficients has rank equal to the number of its columns. 

� 3.26 If a matrix is 5×9, which set must be dependent, its set of rows or its set of 

columns? 

3.27 Give an example to show that, despite that they have the same dimension, 

the row space and column space of a matrix need not be equal. Are they ever 

equal? 

3.28 Show that the set {(1, −1, 2, −3), (1, 1, 2, 0), (3, −1, 6, −6)} does not have the 

same span as {(1, 0, 1, 0), (0, 2, 0, 3)}. What, by the way, is the vector space? 

� 3.29 Show that this set of column vectors 

�� 

d1 �� 3x +2y +4z = d1 

d2 there are x, y, andzsuch that x − z = d2 

d3 

2x +2y +5z = d3 

is a subspace of R 3 . Find a basis.


3.30 Show that the transpose operation is linear: 

(rA + sB) trans = rA trans + sB trans 

for r, s ∈ R and A, B ∈Mm×n, 

� 3.31 In this subsection we have shown that Gaussian reduction finds a basis for 

the row space. 

(a) Show that this basis is not unique — different reductions may yield different 

bases. 

(b) Produce matrices with equal row spaces but unequal numbers of rows. 

(c) Prove that two matrices have equal row spaces if and only if after Gauss- 

Jordan reduction they have the same nonzero rows. 

3.32 Why is there not a problem with Remark 3.14 in the case that r is bigger 

than n? 

3.33 Show that the row rank of an m×n matrix is at most m. Is there a better 

bound? 

� 3.34 Show that the rank of a matrix equals the rank of its transpose. 

3.35 True or false: the column space of a matrix equals the row space of its transpose. 

� 3.36 We have seen that a row operation may change the column space. Must it? 

3.37 Prove that a linear system has a solution if and only if that system’s matrix 

of coefficients has the same rank as its augmented matrix. 

3.38 An m×n matrix has full row rank if its row rank is m, and it has full column 

rank if its column rank is n. 

(a) Show that a matrix can have both full row rank and full column rank only 

if it is square. 

(b) Prove that the linear system with matrix of coefficients A has a solution for 

any d1, ... , dn’s on the right side if and only if A has full row rank. 

(c) Prove that a homogeneous system has a unique solution if and only if its 

matrix of coefficients A has full column rank. 

(d) Prove that the statement “if a system with matrix of coefficients A has any 

solution then it has a unique solution” holds if and only if A has full column 

rank. 

3.39 How would the conclusion of Lemma 3.3 change if Gauss’ method is changed 

to allow multiplying a row by zero? 

� 3.40 What is the relationship between rank(A) and rank(−A)? Between rank(A) 

and rank(kA)? What, if any, is the relationship between rank(A), rank(B), and 

rank(A + B)? 

2.III.4 Combining Subspaces 

This subsection is optional. It is required only for the last sections of Chapter 

Three and Chapter Five and for occasional exercises, and can be passed over 

without loss of continuity. 

This chapter opened with the definition of a vector space, and the middle 

consisted of a first analysis of the idea. This subsection closes the chapter


by finishing the analysis, in the sense that ‘analysis’ means “method of determining 

the ... essential features of something by separating it into parts” 

[Macmillan Dictionary]. 

A common way to understand things is to see how they can be built from 

component parts. For instance, we think of R 3 as put together, in some way, 

from the x-axis, the y-axis, and z-axis. In this subsection we will make this 

precise; we will describe how to decompose a vector space into a combination of 

some of its subspaces. In developing this idea of subspace combination, we will 

keep the R 3 example in mind as a benchmark model. 

Subspaces are subsets and sets combine via union. But taking the combination 

operation for subspaces to be the simple union operation isn’t what we 

want. For one thing, the union of the x-axis, the y-axis, and z-axis is not all of 

R 3 , so the benchmark model would be left out. Besides, union is all wrong for 

this reason: a union of subspaces need not be a subspace (it need not be closed; 

for instance, this R3 vector 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

1 0 0 1 

⎝0⎠ 

+ ⎝1⎠ 

+ ⎝0⎠ 

= ⎝1⎠ 

0 0 1 1 

is in none of the three axes and hence is not in the union). In addition to the 

members of the subspaces, we must at a minimum also include all possible linear 

combinations. 

4.1 Definition Where W1,...,Wk are subspaces of a vector space, their sum 

is the span of their union W1 + W2 + ···+ Wk =[W1 ∪ W2 ∪ ...Wk]. 

(The notation, writing the ‘+’ between sets in addition to using it between 

vectors, fits with the practice of using this symbol for any natural accumulation 

operation.) 

4.2 Example The R 3 model fits with this operation. Any vector �w ∈ R 3 can 

be written as a linear combination c1�v1 + c2�v2 + c3�v3 where �v1 is a member of 

the x-axis, etc., in this way 

⎛ 

⎝ w1 

⎞ ⎛ 

w2⎠ 

=1· ⎝ w1 

⎞ ⎛ 

0 ⎠ +1· ⎝ 

0 

0 

⎞ ⎛ 

w2⎠ 

+1· ⎝ 

0 

0 

0 

w3 

and so R 3 = x-axis + y-axis + z-axis. 

4.3 Example A sum of subspaces can be less than the entire space. Inside of 

P4, letL be the subspace of linear polynomials {a + bx � � a, b ∈ R} and let C be 

the subspace of purely-cubic polynomials {cx 3 � � c ∈ R}. Then L + C is not all 

of P4. Instead, it is the subspace L + C = {a + bx + cx 3 � � a, b, c ∈ R}. 

4.4 Example A space can be described as a combination of subspaces in more 

than one way. Besides the decomposition R 3 = x-axis + y-axis + z-axis, we can 

w3 

⎞ 

⎠


also write R3 = xy-plane + yz-plane. To check this, we simply note that any 

�w ∈ R3 can be written 

⎛ 

⎝ w1 

⎞ ⎛ 

w2⎠ 

=1· ⎝ w1 

⎞ ⎛ 

w2⎠ 

+1· ⎝ 

0 

0 

⎞ 

0 ⎠ 

w3 

as a linear combination of a member of the xy-plane and a member of the 

yz-plane. 

The above definition gives one way in which a space can be thought of as a 

combination of some of its parts. However, the prior example shows that there is 

at least one interesting property of our benchmark model that is not captured by 

the definition of the sum of subspaces. In the familiar decomposition of R3 ,we 

often speak of a vector’s ‘x part’or‘y part’or‘z part’. That is, in this model, 

each vector has a unique decomposition into parts that come from the parts 

making up the whole space. But in the decomposition used in Example 4.4, we 

cannot refer to the “xy part” of a vector — these three sums 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

1 1 0 1 0 1 0 

⎝2⎠ 

= ⎝2⎠ 

+ ⎝0⎠ 

= ⎝0⎠ 

+ ⎝2⎠ 

= ⎝1⎠ 

+ ⎝1⎠ 

3 0 3 0 3 0 3 

all describe the vector as comprised of something from the first plane plus something 

from the second plane, but the “xy part” is different in each. 

That is, when we consider how R 3 is put together from the three axes “in 

some way”, we might mean “in such a way that every vector has at least one 

decomposition”, and that leads to the definition above. But if we take it to 

mean “in such a way that every vector has one and only one decomposition” 

then we need another condition on combinations. To see what this condition 

is, recall that vectors are uniquely represented in terms of a basis. We can use 

this to break a space into a sum of subspaces such that any vector in the space 

breaks uniquely into a sum of members of those subspaces. 

4.5 Example The benchmark is R 3 with its standard basis E3 = 〈�e1,�e2,�e3〉. 

The subspace with the basis B1 = 〈�e1〉 is the x-axis. The subspace with the 

basis B2 = 〈�e2〉 is the y-axis. The subspace with the basis B3 = 〈�e3〉 is the 

z-axis. The fact that any member of R 3 is expressible as a sum of vectors from 

these subspaces 

w3 

⎛ 

⎝ x 

⎞ ⎛ 

y⎠ 

= ⎝ 

z 

x 

⎞ ⎛ 

0⎠ 

+ ⎝ 

0 

0 

⎞ ⎛ 

y⎠ 

+ ⎝ 

0 

0 

⎞ 

0⎠ 

z 

is a reflection of the fact that E3 spans the space — this equation 

⎛ 

⎝ x 

⎞ ⎛ 

y⎠ 

= c1 ⎝ 

z 

1 

⎞ ⎛ 

0⎠ 

+ c2 ⎝ 

0 

0 

⎞ ⎛ 

1⎠ 

+ c3 ⎝ 

0 

0 

⎞ 

0⎠ 

1


has a solution for any x, y, z ∈ R. And, the fact that each such expression is 

unique reflects that fact that E3 is linearly independent — any equation like the 

one above has a unique solution. 

4.6 Example We don’t have to take the basis vectors one at a time, the same 

idea works if we conglomerate them into larger sequences. Consider again the 

space R3 and the vectors from the standard basis E3. The subspace with the 

basis B1 = 〈�e1,�e3〉 is the xz-plane. The subspace with the basis B2 = 〈�e2〉 is 

the y-axis. As in the prior example, the fact that any member of the space is a 

sum of members of the two subspaces in one and only one way 

⎛ 

⎝ x 

⎞ ⎛ 

y⎠ 

= ⎝ 

z 

x 

⎞ ⎛ 

0⎠ 

+ ⎝ 

z 

0 

⎞ 

y⎠ 

0 

is a reflection of the fact that these vectors form a basis — this system 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

x 1 0 0 

⎝y⎠ 

=(c1⎝0⎠ 

+ c3 ⎝0⎠)+c2 

⎝1⎠ 

z 0 1 0 

has one and only one solution for any x, y, z ∈ R. 

These examples illustrate a natural way to decompose a space into a sum 

of subspaces in such a way that each vector decomposes uniquely into a sum of 

vectors from the parts. The next result says that this way is the only way. 

4.7 Definition The concatenation of the sequences B1 = 〈 � β1,1,..., � β1,n1 〉, ... , 

Bk = 〈 � βk,1,..., � βk,nk 〉 is their adjoinment. 

⌢ ⌢ ⌢ 

B1 B2 ··· Bk = 〈 � β1,1,..., � β1,n1 , � β2,1,..., � βk,nk 〉 

4.8 Lemma Let V be a vector space that is the sum of some of its subspaces 

V = W1 + ···+ Wk. Let B1, ... , Bk be any bases for these subspaces. Then 

the following are equivalent. 

(1) For every �v ∈ V , the expression �v = �w1 + ···+ �wk (with �wi ∈ Wi) is 

unique. 

(2) The concatenation B1 

⌢ ··· ⌢ Bk is a basis for V . 

(3) The nonzero members of { �w1,..., �wk} (with �wi ∈ Wi) form a linearly 

independent set — among nonzero vectors from different Wi’s, every linear 

relationship is trivial. 

Proof. We will show that (1) =⇒ (2), that (2) =⇒ (3), and finally that 

(3) =⇒ (1). For these arguments, observe that we can pass from a combination 

of �w’s to a combination of � β’s 

d1 �w1 + ···+ dk �wk 

= d1(c1,1 � �β1,n1 β1,1 + ···+ c1,n1 )+···+ dk(ck,1 � �βk,nk βk,1 + ···+ ck,nk ) 

= d1c1,1 · � β1,1 + ···+ dkck,nk · � βk,nk 

(∗)


and vice versa. 

For (1) =⇒ (2), assume that all decompositions are unique. We will show 

⌢ ⌢ 

that B1 ··· Bk spans the space and is linearly independent. It spans the 

space because the assumption that V = W1 + ··· + Wk means that every �v 

canbeexpressedas�v = �w1 + ···+ �wk, which translates by equation (∗) toan 

expression of �v as a linear combination of the � β’s from the concatenation. For 

linear independence, consider this linear relationship. 

�0 =c1,1 � �βk,nk 

β1,1 + ···+ ck,nk 

Regroup as in (∗) (that is, take d1, ... , dk to be 1 and move from bottom to 

top) to get the decomposition �0 = �w1 + ···+ �wk. Because of the assumption 

that decompositions are unique, and because the zero vector obviously has the 

decomposition �0 =�0+···+�0, we now have that each �wi is the zero vector. This 

means that ci,1 � �βi,ni βi,1 + ···+ ci,ni = �0. Thus, since each Bi is a basis, we have 

the desired conclusion that all of the c’s are zero. 

⌢ ⌢ 

For (2) =⇒ (3), assume that B1 ··· Bk is a basis for the space. Consider 

a linear relationship among nonzero vectors from different Wi’s, 

�0 = ···+ di �wi + ··· 

in order to show that it is trivial. (The relationship is written in this way 

because we are considering a combination of nonzero vectors from only some of 

the Wi’s; for instance, there might not be a �w1 in this combination.) As in (∗), 

�0 = ···+di(ci,1 � �βi,ni βi,1+···+ci,ni )+··· = ···+dici,1· � βi,1+···+dici,ni ·� βi,ni +··· 

⌢ ⌢ 

and the linear independence of B1 ··· Bk gives that each coefficient dici,j is 

zero. Now, �wi is a nonzero vector, so at least one of the ci,j’s is zero, and thus 

di is zero. This holds for each di, and therefore the linear relationship is trivial. 

Finally, for (3) =⇒ (1), assume that, among nonzero vectors from different 

Wi’s, any linear relationship is trivial. Consider two decompositions of a vector 

�v = �w1 + ···+ �wk and �v = �u1 + ···+ �uk in order to show that the two are the 

same. We have 

�0 =(�w1 + ···+ �wk) − (�u1 + ···+ �uk) =(�w1 − �u1)+···+(�wk − �uk) 

which violates the assumption unless each �wi − �ui is the zero vector. Hence, 

decompositions are unique. QED 

4.9 Definition A collection of subspaces {W1,... ,Wk} is independent if no 

nonzero vector from any Wi is a linear combination of vectors from the other 

subspaces W1,...,Wi−1,Wi+1,...,Wk. 

4.10 Definition A vector space V is the direct sum (or internal direct sum) 

of its subspaces W1,...,Wk if V = W1 + W2 + ··· + Wk and the collection 

{W1,...,Wk} is independent. We write V = W1 ⊕ W2 ⊕ ...⊕ Wk. 

4.11 Example The benchmark model fits: R 3 = x-axis ⊕ y-axis ⊕ z-axis.


4.12 Example The space of 2×2 matrices is this direct sum. 

� 

a 

{ 

0 

� � 

0 �� 0 

a, d ∈ R} ⊕{ 

d 

0 

� � 

b �� 0 

b ∈ R} ⊕{ 

0 

c 

� 

0 �� 

c ∈ R} 

0 

It is the direct sum of subspaces in many other ways as well; direct sum decompositions 

are not unique. 

4.13 Corollary The dimension of a direct sum is the sum of the dimensions 

of its summands. 

Proof. In Lemma 4.8, the number of basis vectors in the concatenation equals 

the sum of the number of vectors in the subbases that make up the concatenation. 

QED 

The special case of two subspaces is worth mentioning separately. 

4.14 Definition When a vector space is the direct sum of two of its subspaces, 

then they are said to be complements. 

4.15 Lemma A vector space V is the direct sum of two of its subspaces W1 

and W2 if and only if it is the sum of the two V = W1+W2 and their intersection 

is trivial W1 ∩ W2 = {�0 }. 

Proof. Suppose first that V = W1 ⊕ W2. By definition, V is the sum of the 

two. To show that the two have a trivial intersection, let �v be a vector from 

W1 ∩ W2 and consider the equation �v = �v. On the left side of that equation 

is a member of W1, and on the right side is a linear combination of members 

(actually, of only one member) of W2. But the independence of the spaces then 

implies that �v = �0, as desired. 

For the other direction, suppose that V is the sum of two spaces with a 

trivial intersection. To show that V is a direct sum of the two, we need only 

show that the spaces are independent — no nonzero member of the first is 

expressible as a linear combination of members of the second, and vice versa. 

This is true because any relationship �w1 = c1 �w2,1 + ···+ dk �w2,k (with �w1 ∈ W1 

and �w2,j ∈ W2 for all j) shows that the vector on the left is also in W2, since 

the right side is a combination of members of W2. The intersection of these two 

spaces is trivial, so �w1 = �0. The same argument works for any �w2. QED 

4.16 Example In the space R 2 ,thex-axis and the y-axis are complements, 

that is, R 2 = x-axis ⊕ y-axis. A space can have more than one pair of complementary 

subspaces; another pair here are the subspaces consisting of the lines 

y = x and y =2x. 

4.17 Example In the space F = {a cos θ + b sin θ � � a, b ∈ R}, the subspaces 

W1 = {a cos θ � � a ∈ R} and W2 = {b sin θ � � b ∈ R} are complements. In addition 

to the fact that a space like F can have more than one pair of complementary 

subspaces, inside of the space a single subspace like W1 can have more than one 

complement — another complement of W1 is W3 = {b sin θ + b cos θ � � b ∈ R}.


4.18 Example In R 3 ,thexy-plane and the yz-planes are not complements, 

which is the point of the discussion following Example 4.4. One complement of 

the xy-plane is the z-axis. A complement of the yz-plane is the line through 

(1, 1, 1). 

4.19 Example Following Lemma 4.15, here is a natural question: is the simple 

sum V = W1 + ···+ Wk also a direct sum if and only if the intersection of the 

subspaces is trivial? The answer is that if there are more than two subspaces 

then having a trivial intersection is not enough to guarantee unique decomposition 

(i.e., is not enough to ensure that the spaces are independent). In R3 ,let 

W1 be the x-axis, let W2 be the y-axis, and let W3 be this. 

⎛ 

W3 = { ⎝ q 

⎞ 

q⎠ 

r 

� � q, r ∈ R} 

The check that R3 = W1 + W2 + W3 is easy. The intersection W1 ∩ W2 ∩ W3 is 

trivial, but decompositions aren’t unique. 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

x 0 0 x x − y 0 y 

⎝y⎠ 

= ⎝0⎠ 

+ ⎝y 

− x⎠ 

+ ⎝x⎠ 

= ⎝ 0 ⎠ + ⎝0⎠ 

+ ⎝y⎠ 

z 0 0 z 0 0 z 

(This example also shows that this requirement is also not enough: that all 

pairwise intersections of the subspaces be trivial. See Exercise 30.) 

In this subsection we have seen two ways to regard a space as built up from 

component parts. Both are useful; in particular, in this book the direct sum 

definition is needed to do the Jordan Form construction in the fifth chapter. 

Exercises 

� 4.20 Decide if R 2 � �is the direct sum of � each � W1 and W2. 

x �� x �� 

(a) W1 = { x ∈ R}, W2 = { x ∈ R} 

0 

x 

� � � � 

s �� s �� 

(b) W1 = { s ∈ R}, W2 = { s ∈ R} 

s 

1.1s 

(c) W1 = R 2 , W2 �= {�0} � 

t �� 

(d) W1 = W2 = { t ∈ R} 

t 

� � � � � � � � 

1 x �� −1 0 �� 

(e) W1 = { + x ∈ R}, W2 = { + y ∈ R} 

0 0 

0 y 

� 4.21 Show that R 3 is the direct sum of the xy-plane with each of these. 

(a) the z-axis 

(b) the line 

� � 

z �� 

{ z z ∈ R} 

z


4.22 Is P2 the direct sum of {a + bx 2 � � a, b ∈ R} and {cx � � c ∈ R}? 

� 4.23 In Pn, theeven polynomials are � the members of this set 

E = {p ∈Pn � p(−x) =p(x) for all x} 

and the odd polynomials are the members � of this set. 

O = {p ∈Pn � p(−x) =−p(x) for all x} 

Show that these are complementary subspaces. 

4.24 Which of these subspaces of R 3 

W1: thex-axis, W2: they-axis, W3: thez-axis, 

W4: the plane x + y + z =0, W5: theyz-plane 

can be combined to 

(a) sum to R 3 ? (b) direct sum to R 3 � 

? 

� 4.25 Show that Pn = {a0 � a0 ∈ R}⊕...⊕{anx n � � an ∈ R}. 

4.26 What is W1 + W2 if W1 ⊆ W2? 

4.27 Does Example 4.5 generalize? That is, is this true or false: if a vector space V 

has a basis 〈 � β1,..., � βn〉 then it is the direct sum of the spans of the one-dimensional 

subspaces V =[{ � β1}] ⊕ ...⊕ [{ � βn}]? 

4.28 Can R 4 be decomposed as a direct sum in two different ways? Can R 1 ? 

4.29 This exercise makes the notation of writing ‘+’ between sets more natural. 

Prove that, where W1,...,Wk are subspaces of�a vector space, 

W1 + ···+ Wk = { �w1 + �w2 + ···+ �wk � �w1 ∈ W1,..., �wk ∈ Wk}, 

and so the sum of subspaces is the subspace of all sums. 

4.30 (Refer to Example 4.19. This exercise shows that the requirement that pariwise 

intersections be trivial is genuinely stronger than the requirement only that 

the intersection of all of the subspaces be trivial.) Give a vector space and three 

subspaces W1, W2, andW3such that the space is the sum of the subspaces, the 

intersection of all three subspaces W1 ∩ W2 ∩ W3 is trivial, but the pairwise intersections 

W1 ∩ W2, W1 ∩ W3, andW2∩W3 are nontrivial. 

� 4.31 Prove that if V = W1 ⊕ ...⊕ Wk then Wi ∩ Wj is trivial whenever i �= j. This 

shows that the first half of the proof of Lemma 4.15 extends to the case of more 

than two subspaces. (Example 4.19 shows that this implication does not reverse; 

the other half does not extend.) 

4.32 Recall that no linearly independent set contains the zero vector. Can an 

independent set of subspaces contain the trivial subspace? 

� 4.33 Does every subspace have a complement? 

� 4.34 Let W1,W2 be subspaces of a vector space. 

(a) Assume that the set S1 spans W1, and that the set S2 spans W2. CanS1∪S2 span W1 + W2? Mustit? 

(b) Assume that S1 is a linearly independent subset of W1 and that S2 is a 

linearly independent subset of W2. CanS1∪S2bea linearly independent subset 

of W1 + W2? Mustit? 

4.35 When a vector space is decomposed as a direct sum, the dimensions of the 

subspaces add to the dimension of the space. The situation with a space that is 

given as the sum of its subspaces is not as simple. This exercise considers the 

two-subspace special case. 

(a) For these subspaces of M2×2 find W1 ∩ W2, dim(W1∩W2), W1 + W2, and 

dim(W1 + W2). 

� � � � 

0 0 �� 0 b �� 

W1 = { c, d ∈ R} W2 = { b, c ∈ R} 

c d 

c 0


(b) Suppose that U and W are subspaces of a vector space. Suppose that the 

sequence 〈 � β1,..., � βk〉 is a basis for U ∩ W . Finally, suppose that the prior 

sequence has been expanded to give a sequence 〈�µ1,...,�µj, � β1,..., � βk〉 that is a 

basis for U, and a sequence 〈 � β1,..., � βk,�ω1,...,�ωp〉 that is a basis for W .Prove 

that this sequence 

〈�µ1,...,�µj, � β1,..., � βk,�ω1,...,�ωp〉 

is a basis for for the sum U + W . 

(c) Conclude that dim(U + W )=dim(U)+dim(W ) − dim(U ∩ W ). 

(d) Let W1 and W2 be eight-dimensional subspaces of a ten-dimensional space. 

List all values possible for dim(W1 ∩ W2). 

4.36 Let V = W1 ⊕ ...⊕ Wk and for each index i suppose that Si is a linearly 

independent subset of Wi. Prove that the union of the Si’s is linearly independent. 

4.37 Amatrixissymmetric if for each pair of indices i and j, thei, j entry equals 

the j, i entry. A matrix is antisymmetric if each i, j entry is the negative of the j, i 

entry. 

(a) Give a symmetric 2×2 matrix and an antisymmetric 2×2 matrix. (Remark. 

For the second one, be careful about the entries on the diagional.) 

(b) What is the relationship between a square symmetric matrix and its transpose? 

Between a square antisymmetric matrix and its transpose? 

(c) Show that Mn×n is the direct sum of the space of symmetric matrices and 

the space of antisymmetric matrices. 

4.38 Let W1,W2,W3 be subspaces of a vector space. Prove that (W1 ∩W2)+(W1∩ W3) ⊆ W1 ∩ (W2 + W3). Does the inclusion reverse? 

4.39 The example of the x-axis and the y-axis in R 2 shows that W1 ⊕ W2 = V does 

not imply that W1 ∪ W2 = V .CanW1⊕W2 = V and W1 ∪ W2 = V happen? 

� 4.40 Our model for complementary subspaces, the x-axis and the y-axis in R 2 , 

has one property not used here. Where U is a subspace of R n we define the 

orthocomplement of U to be 

U ⊥ = {�v ∈ R n � � �v �u =0forall�u∈ U} 

(read “U perp”). 

(a) Find the orthocomplement of the x-axis in R 2 . 

(b) Find the orthocomplement of the x-axis in R 3 . 

(c) Find the orthocomplement of the xy-plane in R 3 . 

(d) Show that the orthocomplement of a subspace is a subspace. 

(e) Show that if W is the orthocomplement of U then U is the orthocomplement 

of W . 

(f) Prove that a subspace and its orthocomplement have a trivial intersection. 

(g) Conclude that for any n and subspace U ⊆ R n we have that R n = U ⊕ U ⊥ . 

(h) Show that dim(U)+dim(U ⊥ ) equals the dimension of the enclosing space. 

� 4.41 Consider Corollary 4.13. Does it work both ways — that is, supposing that 

V = W1 + ···+ Wk, isV = W1 ⊕ ...⊕ Wk if and only if dim(V )=dim(W1) + 

···+dim(Wk)? 

4.42 We know that if V = W1 ⊕ W2 then there is a basis for V that splits into a 

basis for W1 and a basis for W2. Can we make the stronger statement that every 

basis for V splits into a basis for W1 and a basis for W2? 

4.43 We can ask about the algebra of the ‘+’ operation. 

(a) Is it commutative; is W1 + W2 = W2 + W1? 

(b) Is it associative; is (W1 + W2)+W3 = W1 +(W2 + W3)?


(c) Let W be a subspace of some vector space. Show that W + W = W . 

(d) Must there be an identity element, a subspace I such that I + W = W + I = 

W for all subspaces W ? 

(e) Does left-cancelation hold: if W1 + W2 = W1 + W3 then W2 = W3? Right 

cancelation? 

4.44 Consider the algebraic properties of the direct sum operation. 

(a) Does direct sum commute: does V = W1 ⊕ W2 imply that V = W2 ⊕ W1? 

(b) Prove that direct sum is associative: (W1 ⊕ W2) ⊕ W3 = W1 ⊕ (W2 ⊕ W3). 

(c) Show that R 3 is the direct sum of the three axes (the relevance here is that by 

the previous item, we needn’t specify which two of the threee axes are combined 

first). 

(d) Does the direct sum operation left-cancel: does W1 ⊕ W2 = W1 ⊕ W3 imply 

W2 = W3? Does it right-cancel? 

(e) There is an identity element with respect to this operation. Find it. 

(f) Do some, or all, subspaces have inverses with respect to this operation: is 

there a subspace W of some vector space such that there is a subspace U with 

the property that U ⊕ W equals the identity element from the prior item?

Topic: Fields 141 

Topic: Fields 

Linear combinations involving only fractions or only integers are much easier 

for computations than combinations involving real numbers, because computing 

with irrational numbers is awkward. Could other number systems, like the 

rationals or the integers, work in the place of R in the definition of a vector 

space? 

Yes and no. If we take “work” to mean that the results of this chapter 

remain true then an analysis of which properties of the reals we have used in 

this chapter gives the following list of conditions an algebraic system needs in 

order to “work” in the place of R. 

Definition. A field is a set F with two operations ‘+’ and ‘·’ such that 

(1) for any a, b ∈F the result of a + b is in F and 

• a + b = b + a 

• if c ∈F then a +(b + c) =(a + b)+c 

(2) for any a, b ∈F the result of a · b is in F and 

• a · b = b · a 

• if c ∈F then a · (b · c) =(a · b) · c 

(3) if a, b, c ∈F then a · (b + c) =a · b + a · c 

(4) there is an element 0 ∈F such that 

• if a ∈F then a +0=a 

• for each a ∈F there is an element −a ∈F such that (−a)+a =0 

(5) there is an element 1 ∈F such that 

• if a ∈F then a · 1=a 

• for each non-0 element a ∈F there is an element a −1 ∈F such that 

a −1 · a =1. 

The number system comsisting of the set of real numbers along with the 

usual addition and multiplication operation is a field, naturally. Another field is 

the set of rational numbers with its usual addition and multiplication operations. 

An example of an algebraic structure that is not a field is the integer number 

system—it fails the final condition. 

Some examples are surprising. The set {0, 1} under these operations: 

isafield(seeExercise4). 

+ 0 1 

0 0 1 

1 1 0 

· 0 1 

0 0 0 

1 0 1


We could develop Linear Algebra as the theory of vector spaces with scalars 

from an arbitrary field, instead of sticking to taking the scalars only from R. In 

that case, almost all of the statements in this book would carry over by replacing 

‘R’ with ‘F’, and thus by taking coefficients, vector entries, and matrix entries 

to be elements of F. (This says “almost all” because statements involving 

distances or angles are exceptions.) Here are some examples; each applies to a 

vector space V over a field F. 

∗ For any �v ∈ V and a ∈F, (i) 0 · �v = �0, and (ii) −1 · �v + �v = �0, and 

(iii) a · �0 =�0. 

∗ The span (the set of linear combinations) of a subset of V is a subspace 

of V . 

∗ Any subset of a linearly independent set is also linearly independent. 

∗ In a finite-dimensional vector space, any two bases have the same number 

of elements. 

(Even statements that don’t explicitly mention F use field properties in their 

proof.) 

We won’t develop vector spaces in this more general setting because the 

additional abstraction can be a distraction. The ideas we want to bring out 

already appear when we stick to the reals. 

The only exception is in Chapter Five. In that chapter we must factor 

polynomials, so we will switch to considering vector spaces over the field of 

complex numbers. We will discuss this more, including a brief review of complex 

arithmetic, when we get there. 

Exercises 

1 Show that the real numbers form a field. 

2 Prove that these are fields: 

(a) the rational numbers (b) the complex numbers. 

3 Give an example that shows that the integer number system is not a field. 

4 Consider the set {0, 1} subject to the operations given above. Show that it is a 

field. 

5 Come up with suitable operations to make the set {0, 1, 2} afield.

Topic: Crystals 143 

Topic: Crystals 

Everyone has noticed that table salt comes in little cubes. 

Remarkably, the explanation for the cubical external shape is the simplest one 

possible: the internal shape, the way the atoms lie, is also cubical. The internal 

structure is pictured below. Salt is sodium cloride, and the small spheres shown 

are sodium while the big ones are cloride. (To simplify the view, only the 

sodiums and clorides on the front, top, and right are shown.) 

The specks of salt that we see when we spread a little out on the table consist of 

many repetitions of this fundamental unit. That is, these cubes of atoms stack 

up to make the larger cubical structure that we see. A solid, such as table salt, 

with a regular internal structure is a crystal. 

We can restrict our attention to the front face. There, we have this pattern 

repeated many times. 

The distance between the corners of this cell is about 3.34 ˚Angstroms (an 

˚Angstrom is 10−10 meters). Obviously that unit is unwieldly for describing 

points in the crystal lattice. Instead, the thing to do is to take as a unit the 

length of each side of the square. That is, we naturally adopt this basis. 

� � � � 

3.34 0 

〈 , 〉 

0 3.34


Then we can describe, say, the corner in the upper right of the picture above as 

3 � β1 +2 � β2. 

Another crystal from everyday experience is pencil lead. It is graphite, 

formed from carbon atoms arranged in this shape. 

This is a single plane of graphite. A piece of graphite consists of many of these 

planes layered in a stack. (The chemical bonds between the planes are much 

weaker than the bonds inside the planes, which explains why graphite writes— 

it can be sheared so that the planes slide off and are left on the paper.) A 

convienent unit of length can be made by decomposing the hexagonal ring into 

three regions that are rotations of this unit cell. 

A natural basis then would consist of the vectors that form the sides of that 

unit cell. The distance along the bottom and slant is 1.42 ˚Angstroms, so this 

� � � � 

1.42 1.23 

〈 , 〉 

0 .71 

is a good basis. 

The selection of convienent bases extends to three dimensions. Another 

familiar crystal formed from carbon is diamond. Like table salt, it is built from 

cubes, but the structure inside each cube is more complicated than salt’s. In 

addition to carbons at each corner, 

there are carbons in the middle of each face.

Topic: Crystals 145 

(To show the added face carbons clearly, the corner carbons have been reduced 

to dots.) There are also four more carbons inside the cube, two that are a 

quarter of the way up from the bottom and two that are a quarter of the way 

down from the top. 

(As before, carbons shown earlier have been reduced here to dots.) The distance 

along any edge of the cube is 2.18 ˚Angstroms. Thus, a natural basis for 

describing the locations of the carbons, and the bonds between them, is this. 

⎛ ⎞ ⎛ ⎞ ⎛ 

2.18 0 

〈 ⎝ 0 ⎠ , ⎝2.18⎠ 

, ⎝ 

0 

0 

⎞ 

⎠〉 

0 0 2.18 

Even the few examples given here show that the structures of crystals is complicated 

enough that some organized system to give the locations of the atoms, 

and how they are chemically bound, is needed. One tool for that organization 

is a convienent basis. This application of bases is simple, but it shows a context 

where the idea arises naturally. The work in this chapter just takes this simple 

idea and develops it. 

Exercises 

1 How many fundamental regions are there in one face of a speck of salt? (With a 

ruler, we can estimate that face is a square that is 0.1 cmonaside.) 

2 In the graphite picture, imagine that we are interested in a point 5.67 ˚Angstroms 

up and 3.14 ˚Angstroms over from the origin. 

(a) Express that point in terms of the basis given for graphite. 

(b) How many hexagonal shapes away is this point from the origin? 

(c) Express that point in terms of a second basis, where the first basis vector is 

the same, but the second is perpendicular to the first (going up the plane) and 

of the same length. 

3 Give the locations of the atoms in the diamond cube both in terms of the basis, 

and in ˚Angstroms. 

4 This illustrates how the dimensions of a unit cell could be computed from the 

shape in which a substance crystalizes ([Ebbing], p. 462). 

(a) Recall that there are 6.022×10 23 atoms in a mole (this is Avagadro’s number). 

From that, and the fact that platinum has a mass of 195.08 grams per mole, 

calculate the mass of each atom. 

(b) Platinum crystalizes in a face-centered cubic lattice with atoms at each lattice 

point, that is, it looks like the middle picture given above for the diamond crystal. 

Find the number of platinums per unit cell (hint: sum the fractions of platinums 

that are inside of a single cell). 

(c) From that, find the mass of a unit cell.


(d) Platinum crystal has a density of 21.45 grams per cubic centimeter. From 

this, and the mass of a unit cell, calculate the volume of a unit cell. 

(e) Find the length of each edge. 

(f) Describe a natural three-dimensional basis.

Topic: Voting Paradoxes 147 

Topic: Voting Paradoxes 

Imagine that a Political Science class studying the American presidential process 

holds a mock election. Members of the class are asked to rank, from most 

preferred to least preferred, the nominees from the Democratic Party, the Republican 

Party, and the Third Party, and this is the result (> means ‘is preferred 

to’). 

preference order 

number with 

that preference 

Democrat > Republican > Third 5 

Democrat > Third > Republican 4 

Republican > Democrat > Third 2 

Republican > Third > Democrat 8 

Third > Democrat > Republican 8 

Third > Republican > Democrat 2 

total 29 

What is the preference of the group as a whole? 

Overall, the group prefers the Democrat to the Republican (by five votes; 

seventeen voters ranked the Democrat above the Republican versus twelve the 

other way). And, overall, the group prefers the Republican to the Third’s 

nominee (by one vote; fifteen to fourteen). But, strangely enough, the group 

also prefers the Third to the Democrat (by seven votes; eighteen to eleven). 

7voters 

Third 

Democrat 

1voter 

5voters 

Republican 

This is an example of a voting paradox, specifically, a majority cycle. 

Voting paradoxes are studied in part because of their implications for practical 

politics. For instance, the instructor can manipulate the class into choosing 

the Democrat as the overall winner by first asking the class to choose between 

the Republican and the Third, and then asking the class to choose between the 

winner of that contest (the Republican) and the Democrat. By similar manipulations, 

any of the other two candidates can be made to come out as the winner. 

(In this Topic we will stick to three-candidate elections, but similar results apply 

to larger elections.) 

Voting paradoxes are also studied simply because they are mathematically 

interesting. One interesting aspect is that the group’s overall majority cycle 

occurs despite that each single voters’s preference list is rational—in a straightline 

order. That is, the majority cycle seems to arise in the aggregate, without 

being present in the elements of that aggregate, the preference lists. Recently,


however, linear algebra has been used [Zwicker] to argue that a tendency toward 

cyclic preference is actually present in each voter’s list, and that it surfaces when 

there is more adding of the tendency than cancelling. 

For this argument, abbreviating the choices as D, R, andT , we can describe 

how a voter with preference order D>R>T contributes to the above cycle 

−1 voterD 

1voter 

T R 

1voter 

(the negative sign is here because the arrow describes T as preferred to D, but 

this voter likes them the other way). The descriptions for the other preference 

lists are in the table on page 150. Now, to conduct the election, we linearly 

combine these descriptions; for instance, the Political Science mock election 

D D D −1 voter 1voter −1 voter 1voter 

1voter −1voter 

5 · T R +4· T R + ··· +2· T R 

1voter 

−1 voter 

−1 voter 

yields the circular group preference shown earlier. 

Of course, taking linear combinations is linear algebra. The above cycle notation 

is suggestive but inconvienent, so we temporarily switch to using column 

vectors by starting at the D and taking the numbers from the cycle in counterclockwise 

order. Thus, the mock election and a single D>R>Tvote are 

represented in this way. 

⎛ ⎞ ⎛ ⎞ 

7 

⎝1⎠ 

and 

5 

−1 

⎝ 1 ⎠ 

1 

We will decompose vote vectors into two parts, one cyclic and the other acyclic. 

For the first part, we say that a vector is purely cyclic if it is in this subspace 

of R3 . 

⎛ ⎞ 

k 

C = { ⎝k⎠ 

k 

� ⎛ ⎞ 

1 

� k ∈ R} = {k · ⎝1⎠ 

1 

� � k ∈ R} 

For the second part, consider the subspace (see Exercise 6) of vectors that are 

perpendicular to all of the vectors in C. 

C ⊥ ⎛ 

= { ⎝ c1 

⎞ 

c2⎠ 

c3 

� ⎛ 

� ⎝ c1 

⎞ ⎛ 

c2⎠ 

⎝ 

c3 

k 

⎞ 

k⎠ 

= 0 for all k ∈ R} 

k 

⎛ 

= { ⎝ c1 

⎞ 

⎠ � � c1 + c2 + c3 =0} 

c2 

c3 

⎛ 

= {c2 ⎝ −1 

⎞ ⎛ 

1 ⎠ + c3 ⎝ 

0 

−1 

⎞ 

0 ⎠ 

1 

� � c2,c3 ∈ R}


(Read that aloud as “C perp”.) Consideration of those two has led to this basis 

of R3 . 

⎛ 

〈 ⎝ 1 

⎞ ⎛ 

1⎠ 

, ⎝ 

1 

−1 

⎞ ⎛ 

1 ⎠ , ⎝ 

0 

−1 

⎞ 

0 ⎠〉 

1 

We can represent votes with respect to this basis, and thereby decompose them 

into a cyclic part and an acyclic part. (Note for readers who have covered the 

optional section: that is, the space is the direct sum of C and C ⊥ .) 

For example, consider the D>R>T voter discussed above. The representation 

in terms of the basis is easily found, 

c1 − c2 − c3 = −1 

c1 + c2 = 1 

c1 + c3 = 1 

−ρ1+ρ2 (−1/2)ρ2+ρ3 

−→ −→ 

−ρ1+ρ3 

so that c1 =1/3, c2 =2/3, and c3 =2/3. Then 

c1 − c2 − c3 = −1 

2c2 + c3 = 2 

(3/2)c3 = 1 

⎛ ⎞ 

−1 

⎝ 1 ⎠ = 

1 

1 

3 · 

⎛ ⎞ 

1 

⎝1⎠ 

+ 

1 

2 

3 · 

⎛ ⎞ 

−1 

⎝ 1 ⎠ + 

0 

2 

3 · 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

−1 1/3 −4/3 

⎝ 0 ⎠ = ⎝1/3⎠ 

+ ⎝ 2/3 ⎠ 

1 1/3 2/3 

gives the desired decomposition into a cyclic part and and an acyclic part. 

−1 

T 

D 1 

1 

R 

= 

1/3 

D 

1/3 

T 

1/3 

R 

+ 

−4/3 

D 

2/3 

T 

2/3 

R 

Thus, this D>R>Tvoter’s rational preference list can indeed be seen to 

have a cyclic part. 

The T>R>Dvoter is opposite to the one just considered in that the ‘>’ 

symbols are reversed. This voter’s decomposition 

1 

D 

−1 

T 

−1 

R 

= 

−1/3 

D 

−1/3 

T R 

−1/3 

+ 

4/3 

D 

−2/3 

T R 

−2/3 

shows that these opposite preferences have decompositions that are opposite. 

We say that the first voter has positive spin since the cycle part is with the 

direction we have chosen for the arrows, while the second voter’s spin is negative. 

The fact that that these opposite voters cancel each other is reflected in the 

fact that their vote vectors add to zero. This suggests an alternate way to tally 

an election. We could first cancel as many opposite preference lists as possible, 

and then determine the outcome by adding the remaining lists. 

The rows of the table below contain the three pairs of opposite preference 

lists, and the columns group those pairs by spin. For instance, the first row 

contains the two voters just considered.


positive spin negative spin 

Democrat > Republican > Third 

−1 D 

1 

T 

1 

R 

= 

1/3 D 1/3 

T R 

1/3 

−4/3 D 2/3 

T R 

2/3 

+ 

Republican > Third > Democrat 

1 

D −1 

T 

1 

R 

= 

1/3 D 1/3 

T R 

1/3 

+ 

2/3 D −4/3 

T R 

2/3 

Third > Democrat > Republican 

1 

D 

1 

T R 

−1 

= 

1/3 D 1/3 

T R 

1/3 

+ 

2/3 D 2/3 

T R 

−4/3 

Third > Republican > Democrat 

1 

D −1 

T R 

−1 

−1/3 D −1/3 

T R 

−1/3 

= 

+ 

4/3 D −2/3 

T R 

−2/3 

Democrat > Third > Republican 

−1 D 

1 

T R 

−1 

−1/3 D −1/3 −2/3 D 4/3 

T R + T R 

−1/3 

−2/3 

= 

Republican > Democrat > Third 

−1 D −1 

T 

1 

R 

−1/3 D −1/3 −2/3 D −2/3 

T R + T R 

−1/3 

4/3 

If we conduct the election as just described then after the cancellation of as many 

opposite pairs of voters as possible, there will be left three sets of preference 

lists, one set from the first row, one set from the second row, and one set from 

the third row. We will finish by proving that a voting paradox can happen 

only if the spins of these three sets are in the same direction. That is, for a 

voting paradox to occur, the three remaining sets must all come from the left 

of the table or all come from the right (see Exercise 3). This shows that there 

is some connection between the majority cycle and the decomposition that we 

are using—a voting paradox can happen only when the tendencies toward cyclic 

preference reinforce each other. 

For the proof, assume that opposite preference orders have been cancelled, 

and we are left with one set of preference lists from each of the three rows. 

Consider the sum of these three (here, a, b, andc could be positive, negative, 

or zero). 

−a 

D 

a 

T 

a 

R 

+ 

b 

T 

D −b 

b 

R 

+ 

c 

D 

c 

T 

−c 

R 

= 

= 

−a + b + c 

D 

a − b + c 

T R 

a + b − c 

A voting paradox occurs when the three numbers on the right, a − b + c and 

a + b − c and −a + b + c, are all nonnegative or all nonpositive. On the left, 

at least two of the three numbers, a and b and c, are both nonnegative or both 

nonpositive. We can assume that they are a and b. That makes four cases: the 

cycle is nonnegative and a and b are nonnegative, the cycle is nonpositive and 

a and b are nonpositive, etc. We will do only the first case, since the second is 

similar and the other two are also easy. 

So assume that the cycle is nonnegative and that a and b are nonnegative. 

The conditions 0 ≤ a − b + c and 0 ≤−a + b + c add to give that 0 ≤ 2c, which 

implies that c is also nonnegative, as desired. That ends the proof. 

This result only says that having all three spin in the same direction is a 

necessary condition for a majority cycle. It is not also a sufficient condition; see 

Exercise 4.


Voting theory and associated topics are the subject of current research. The 

are many surprising and intriguing results, most notably the one produced by 

K. Arrow [Arrow], who won the Nobel Prize in part for this work, showing, essentially, 

that no voting system is entirely fair. For more information, some good 

introductory articles are [Gardner, 1970], [Gardner, 1974], [Gardner, 1980], and 

[Neimi & Riker]. A quite readable recent book is [Taylor]. The material of this 

Topic is largely drawn from [Zwicker]. (Author’s Note: I would like to thank 

Professor Zwicker for his kind and illuminating discussions.) 

Exercises 

1 Here is a reasonable way in which a voter could have a cyclic preference. Suppose 

that this voter ranks each candidate on each of three criteria. 

(a) Draw up a table with the rows labelled ‘Democrat’, ‘Republican’, and ‘Third’, 

and the columns labelled ‘character’, ‘experience’, and ‘policies’. Inside each 

column, rank some candidate as most preferred, rank another as in the middle, 

and rank the remaining one as least preferred. 

(b) In this ranking, is the Democrat preferred to the Republican in (at least) two 

out of three criteria, or vice versa? Is the Republican preferred to the Third? 

(c) Does the table that was just constructed have a cyclic preference order? If 

not, make one that does. 

So it is possible for a voter to have a cyclic preference among candidates. The 

paradox described above, however, is that even if each voter has a straight-line 

preference list, there can still be a cyclic group preference. 

2 Compute the values in the table of decompositions. 

3 Do the cancellations of opposite preference orders for the Political Science class’s 

mock election. Are all the remaining preferences from the left three rows of the 

tableorfromtheright? 

4 The necessary condition that is proved above—a voting paradox can happen only 

if all three preference lists remaining after cancellation have the same spin—is not 

also sufficient. 

(a) Continuing the positive cycle case considered in the proof, use the two inequalities 

0 ≤ a − b + c and 0 ≤−a + b + c to show that |a − b| ≤c. 

(b) Also show that c ≤ a + b, and hence that |a − b| ≤c ≤ a + b. 

(c) Give an example of a vote where there is a majority cycle, and addition of 

one more voter with the same spin causes the cycle to go away. 

(d) Can the opposite happen; can addition of one voter with a “wrong” spin 

cause a cycle to appear? 

(e) Give a condition that is both necessary and sufficient to get a majority cycle. 

5 A one-voter election cannot have a majority cycle because of the requirement 

that we’ve imposed that the voter’s list must be rational. 

(a) Show that a two-voter election may have a majority cycle. (We consider the 

group preference a majority cycle if all three group totals are nonnegative or if 

all three are nonpositive—that is, we allow some zero’s in the group preference.) 

(b) Show that for any number of voters greater than one, there is an election 

involving that many voters that results in a majority cycle. 

6 Let U be a subspace of R 3 . Prove that the set U ⊥ = {�v � � �v �u =0forall�u ∈ U} 

of vectors that are perpendicular to each vector in U is also a subspace of R 3 .


Topic: Dimensional Analysis 

“You can’t add apples and oranges,” the old saying goes. It reflects the common 

experience that in applications the numbers are associated with units, and 

keeping track of the units is worthwhile. Everyone is familiar with calculations 

such as this one that use the units as a check. 

60 sec min hr day 

sec 

· 60 · 24 · 365 = 31 536 000 

min hr day year year 

However, the idea of paying attention to how the quantities are measured can 

be pushed beyond bookkeeping. It can be used to draw conclusions about the 

nature of relationships among physical quantities. 

Consider this equation expressing a relationship: dist = 16 · (time) 2 . If 

distance is taken in feet and time in seconds then this is a true statement about 

the motion of a falling body. But this equation is a correct description only in 

the foot-second unit system. In the yard-second unit system it is not the case 

that d =16t2 . To get a complete equation—one that holds irrespective of the 

size of the units—we will make the 16 a dimensional constant. 

dist = 16 ft 

· (time)2 

sec2 Now, the equation holds in any units system, e.g., in yards and seconds we have 

this. 

dist in yd = 16 

(1/3) yd 

sec 2 

· (time in sec) 2 = 16 

3 

yd 

· (time in sec)2 

sec2 The results below hold for complete equations. 

Dimensional analysis can be applied to many areas, but we shall stick to 

Newtonian dynamics. In the light of the prior paragraph, we shall work outside 

of any particular unit system, and instead say that all quantities are measured 

in combinations of (some units of) length L, mass M, and time T . Thus, for 

instance, the dimensional formula of velocity is L/T and that of density is 

M/L 3 . We shall prefer to write those by including even the dimensions with a 

zero exponent, e.g., as L 1 M 0 T −1 and L −3 M 1 T 0 . 

In this terminology, the saying “You can’t add apples to oranges” becomes 

the advice to have all of the terms in an equation have the same dimensional 

formula. Such an equation is dimensionally homogeneous. An example is this 

version of the falling body equation: d − gt 2 = 0 where the dimensional formula 

of d is L 1 M 0 T 0 , that of g is L 1 M 0 T −2 , and that of t is L 0 M 0 T 1 (g is the 

dimensional constant expressed above in units of ft/sec 2 ). The gt 2 term works 

out as L 1 M 0 T −2 (L 0 M 0 T 1 ) 2 = L 1 M 0 T 0 , and so it has the same dimensional 

formula as the d term. 

Quantities with dimensional formula L 0 M 0 T 0 , are said to be dimensionless. 

An example of such a quantity is the measure of an angle. An angle measured 

in radians is the ratio of the subtended arc to the radius.

Topic: Dimensional Analysis 153 

arc 

θ 

rad 

This is the ratio of a length to a length L 1 M 0 T 0 /L 1 M 0 T 0 and thus angles have 

the dimensional formula L 0 M 0 T 0 . 

Paying attention to the dimensional formulas of the physical quantities will 

help us to see which relationships are possible or impossible among the quantities. 

For instance, suppose that we want to give the period of a pendulum as 

some formula p = ··· involving the other relevant physical quantities, length of 

the string, etc. (see the table on page 154). The period is expressed in units of 

time—it has dimensional formula L 0 M 0 T 1 —and so the quantities on the other 

side of the equation must have their dimensional formulas combine in such a 

way that the L’s and M’s cancel and only a T is left. For instance, in that table, 

the only quantities involving L are the length of the string and the acceleration 

due to gravity. For these L’s to cancel, the quantities must enter the equation 

in ratio, e.g., as (ℓ/g) 2 or as cos(ℓ/g), or as (ℓ/g) −1 . In this way, simply from 

consideration of the dimensional formulas, we know that that the period can be 

written as a function of ℓ/g; the formula cannot possibly involve, say, ℓ 3 and 

g −2 because the dimensional formulas wouldn’t cancel their L’s. 

To do dimensional analysis systematically, we need two results (for proofs, 

see [Bridgman], Chapter II and IV). First, each equation relating physical quantities 

that we shall see involves a sum of terms, where each term has the form 

m p1 

1 mp2 

2 ...mpk 

k 

for numbers m1, ... , mk that measure the quantities. 

Next, observe that an easy way to construct a dimensionally homogeneous 

expression is by taking a product of dimensionless quantities, or by adding 

such dimensionless terms. The second result, Buckingham’s Theorem, is that 

any complete relationship among quantities with dimensional formulas can be 

algebraically manipulated into a form where there is some function f such that 

f(Π1,... ,Πn) =0 

for a complete set {Π1,... ,Πn} of dimensionless products. (We shall see what 

makes a set of dimensionless products ‘complete’ in the examples below.) We 

usually want to express one of the quantities, m1 for instance, in terms of the 

others, and for that we will assume that the above equality can be rewritten 

m1 = m −p2 

2 

...m −pk 

k 

· ˆ f(Π2,...,Πn) 

where Π1 = m1m p2 

2 ...mpk 

k is dimensionless and the products Π2, ... ,Πn don’t 

involve m1 (as with f, here ˆ f is just some function, this time of n−1 arguments). 

Thus, Buckingham’s Theorem says that to investigate the complete relationships 

that are possible, we can look into the dimensionless products that are possible. 

For that we will use the material of this chapter.


The classic example is a pendulum. An investigator trying to determine 

the formula for its period might conjecture that these are the relevant physical 

quantities. 

quantity 

dimensional 

formula 

period p L 0 M 0 T 1 

length of string ℓ L 1 M 0 T 0 

mass of bob m L 0 M 1 T 0 

acceleration due to gravity g L 1 M 0 T −2 

arc of swing θ L 0 M 0 T 0 

To find which combinations of the powers in p p1 ℓ p2 m p3 g p4 θ p5 yield dimensionless 

products, consider this equation. 

(L 0 M 0 T 1 ) p1 (L 1 M 0 T 0 ) p2 (L 0 M 1 T 0 ) p3 (L 1 M 0 T −2 ) p4 (L 0 M 0 T 0 ) p5 = L 0 M 0 T 0 

It gives three conditions on the powers. 

p2 + p4 =0 

p3 =0 

p1 − 2p4 =0 

Note that p3 is 0—the mass of the bob does not affect the period. The system’s 

solution space can be described in this way (p1 is taken as one of the parameters 

in order to express the period in terms of the other quantities). 

⎛ 

⎜ 

{ ⎜ 

⎝ 

p1 

p2 

p3 

p4 

p5 

⎞ 

⎟ 

⎠ = 

⎛ ⎞ 

1 

⎜ 

⎜−1/2 

⎟ 

⎜ 0 ⎟ 

⎝ 1/2 ⎠ 

0 

p1 

⎛ ⎞ 

0 

⎜ 

⎜0 

⎟ 

+ ⎜ 

⎜0 

⎟ 

⎝0⎠ 

1 

p5 

� 

� p1,p5 ∈ R} 

Here is the linear algebra. The set of dimensionless products is the set of 

products p p1 ℓ p2 m p3 a p4 θ p5 subject to the conditions in the above linear system. 

This forms a vector space under the ‘+’ addition operation of multiplying two 

such products and the ‘·’ scalar multiplication operation of raising such a product 

to the power of the scalar (see Exercise 5). The term ‘complete set of 

dimensionless products’ in Buckingham’s Theorem means a basis for this vector 

space. 

We can get a basis by first taking p1 =1andp5 = 0, and then taking p1 =0 

and p5 = 1. The associated dimensionless products are Π1 = pℓ −1/2 g 1/2 and 

Π2 = θ. The set {Π1, Π2} is complete, so we have 

p = ℓ 1/2 g −1/2 · ˆ f(θ) 

= � ℓ/g · ˆ f(θ)


where ˆ f is a function that we cannot determine from this analysis (by other 

means we know that for small angles it is approximately the constant function 

ˆf(θ) =2π). 

Thus, analysis of the relationships that are possible between the quantities 

with the given dimensional formulas has given us a fair amount of information: a 

pendulum’s period does not depend on the mass of the bob, and it rises with 

the square root of the length of the string. 

For the next example we try to determine the period of revolution of two 

bodies in space orbiting each other under mutual gravitational attraction. An 

experienced investigator could expect that these are the relevant quantities. 

quantity 

dimensional 

formula 

period of revolution p L 0 M 0 T 1 

mean radius of separation r L 1 M 0 T 0 

mass of the first m1 L 0 M 1 T 0 

mass of the second m2 L 0 M 1 T 0 

gravitational constant G L 3 M −1 T −2 

To get the complete set of dimensionless products we consider the equation 

(L 0 M 0 T 1 ) p1 (L 1 M 0 T 0 ) p2 (L 0 M 1 T 0 ) p3 (L 0 M 1 T 0 ) p4 (L 3 M −1 T −2 ) p5 = L 0 M 0 T 0 

which gives rise to these relationships among the powers 

p1 

p2 

+3p5 =0 

p3 + p4 − p5 =0 

− 2p5 =0 

with the solution space 

⎛ ⎞ 

1 

⎜ 

⎜−3/2 

⎟ 

{ ⎜ 1/2 ⎟ 

⎝ 0 ⎠ 

1/2 

p1 

⎛ ⎞ 

0 

⎜ 0 ⎟ 

+ ⎜ 

⎜−1 

⎟ 

⎝ 1 ⎠ 

0 

p4 

� 

� p1,p4 ∈ R} 

(p1 is taken as a parameter so that we can state the period as a function of the 

other quantities). As with the pendulum example, the linear algebra here is 

that the set of dimensionless products of these quantities forms a vector space, 

and we want to produce a basis for that space, a ‘complete’ set of dimensionless 

products. One such set, gotten from setting p1 = 1 and p4 = 0, and also 

setting p1 = 0 and p4 = 1 is {Π1 = pr−3/2m 1/2 

1 G1/2 , Π2 = m −1 

1 m2}. With 

that, Buckingham’s Theorem says that any complete relationship among these 

quantities must be stateable this form. 

p = r 3/2 m −1/2 

1 G−1/2 · ˆ f(m −1 

1 m2) 

= r3/2 

√ · 

Gm1 

ˆ f(m2/m1)


Remark. An especially interesting application of the above formula occurs 

when when the two bodies are a planet and the sun. The mass of the sun m1 

is much larger than that of the planet m2. Thus the argument to ˆ f is approximately 

0, and we can wonder if this part of the formula remains approximately 

constant as m2 varies. One way to see that it does is this. The sun’s mass is 

much larger than the planet’s mass and so the mutual rotation is approximately 

about the sun’s center. If we vary the planet’s mass m2 by a factor of x then the 

force of attraction is multiplied by x, andx times the force acting on x times 

the mass results in the same acceleration, about the same center. Hence, the 

orbit will be the same, and so its period will be the same, and thus the right side 

of the above equation also remains unchanged (approximately). Therefore, for 

m2’s much smaller than m1, the value of ˆ f(m2/m1) is approximately constant 

as m2 varies. This result is Kepler’s Third Law: the square of the period of a 

planet is proportional to the cube of the mean radius of its orbit about the sun. 

In the final example, we will see that sometimes dimensional analysis alone 

suffices to essentially determine the entire formula. One of the earliest applications 

of the technique was to give the formula for the speed of a wave in deep 

water. Lord Raleigh put these down as the relevant quantities. 

Considering 

quantity 

dimensional 

formula 

velocity of the wave v L 1 M 0 T −1 

density of the water d L −3 M 1 T 0 


wavelength λ L 1 M 0 T 0 

(L 1 M 0 T −1 ) p1 (L −3 M 1 T 0 ) p2 (L 1 M 0 T −2 ) p3 (L 1 M 0 T 0 ) p4 = L 0 M 0 T 0 

gives this system 

with this solution space 

p1 − 3p2 + p3 + p4 =0 

p2 

=0 

−p1 − 2p3 =0 

⎛ ⎞ 

1 

⎜ 

{ ⎜ 0 ⎟ 

⎝−1/2⎠ 

−1/2 

p1 

� 

� p1 ∈ R} 

(as in the pendulum example, one of the quantities d turns out not to be involved 

in the relationship). There is thus one dimensionless product, Π1 = 

vg −1/2 λ −1/2 , and we have that v is √ λg times a constant ( ˆ f is constant since 

it is a function of no arguments). 

As those three examples show, analysis of the relationships possible among 

quantities of the given dimensional formulas can bring us far toward expressing


the relationship among the quantities. For further reading, the classic reference 

is [Bridgman]—this brief book is a delight to read. Another source is 

[Giordano, Wells, Wilde]. A description of how dimensional analysis fits into 

the process of mathematical modeling is [Giordano, Jaye, Weir]. 

Exercises 

1 [de Mestre] Consider a projectile, launched with initial velocity v0, atanangle 

θ. An investigation of this motion might start with the guess that these are the 

relevant quantities. 

dimensional 

quantity formula 

horizontal position x L 1 M 0 T 0 

vertical position y L 1 M 0 T 0 

initial speed v0 L 1 M 0 T −1 

angle of launch θ L 0 M 0 T 0 


time t L 0 M 0 T 1 

(a) Show that {gt/v0,gx/v 2 0, gy/v 2 0,θ} is a complete set of dimensionless products. 

(Hint. This can be done by finding the appropriate free variables in the 

linear system that arises, but there is a shortcut that uses the properties of a 

basis.) 

(b) These two equations of motion for projectiles are familiar: x = v0 cos(θ)t and 

y = v0 sin(θ)t−(g/2)t 2 . Algebraic manipulate each to rewrite it as a relationship 

among the dimensionless products of the prior item. 

2 [Einstein] conjectured that the infrared characteristic frequencies of a solid may 

be determined by the same forces between atoms as determine the solid’s ordanary 

elastic behavior. The relevant quantities are 

dimensional 

quantity formula 

characteristic frequency ν L 0 M 0 T −1 

compressibility k L 1 M −1 T 2 

number of atoms per cubic cm N L −3 M 0 T 0 

mass of an atom m L 0 M 1 T 0 

Show that there is one dimensionless product. Conclude that, in any complete 

relationship among quantities with these dimensional formulas, k is a constant 

times ν −2 N −1/3 m −1 . This conclusion played an important role in the early study 

of quantum phenomena. 

3 [Giordano, Wells, Wilde] Consider the torque produced by an engine. Torque 

has dimensional formula L 2 M 1 T −2 . We may first guess that it depends on the 

engine’s rotation rate (with dimensional formula L 0 M 0 T −1 ), and the volume of 

air displaced (with dimensional formula L 3 M 0 T 0 ). 

(a) Try to find a complete set of dimensionless products. What goes wrong? 

(b) Adjust the guess by adding the density of the air (with dimensional formula 

L −3 M 1 T 0 ). Now find a complete set of dimensionless products. 

4 [Tilley] Dominoes falling make a wave. We may conjecture that the wave speed v 

depends on the the spacing d between the dominoes, the height h of each domino, 

and the acceleration due to gravity g. 

(a) Find the dimensional formula for each of the four quantities.


(b) Show that {Π1 = h/d, Π2 = dg/v 2 } is a complete set of dimensionless products. 

(c) Show that if h/d is fixed then the propagation speed is proportional to the 

square root of d. 

5 Prove that the dimensionless products form a vector space under the �+ operation 

of multiplying two such products and the �· operation of raising such the product 

to the power of the scalar. (The vector arrows are a precaution against confusion.) 

That is, prove that, for any particular homogeneous system, this set of products 

of powers of m1, ... , mk 

{m p1 

1 ...mp � 

k � 

k p1, ... , pk satisfy the system} 

is a vector space under: 

m p1 

1 ...mp k q1 �+m k 1 ...mq k 

k = mp1+q1 1 ...m pk+qk k 

and 

r�·(m p1 

1 ...mp k 

k )=mrp1 

1 

...m rp k 

k 

(assume that all variables represent real numbers). 

6 The advice about apples and oranges is not right. Consider the familiar equations 

for a circle C =2πr and A = πr 2 . 

(a) Check that C and A have different dimensional formulas. 

(b) Produce an equation that is not dimensionally homogeneous (i.e., it adds 

apples and oranges) but is nonetheless true of any circle. 

(c) The prior item asks for an equation that is complete but not dimensionally 

homogeneous. Produce an equation that is dimensionally homogeneous but not 

complete. 

(Just because the old saying isn’t strictly right, doesn’t keep it from being a useful 

strategy. Dimensional homogeneity is often used as a check on the plausibility 

of equations used in models. For an argument that any complete equation can 

easily be made dimensionally homogeneous, see [Bridgman], Chapter I, especially 

page 15.)

Chapter 3 

Maps Between Spaces 

3.I Isomorphisms 

In the examples following the definition of a vector space we developed the 

intuition that some spaces are “the same” as others. For instance, the space 

of two-tall column vectors and the space of two-wide row vectors are not equal 

because their elements—column vectors and row vectors—are not equal, but we 

have the idea that these spaces differ only in how their elements appear. We 

will now make this idea precise. 

This section illustrates a common aspect of a mathematical investigation. 

With the help of some examples, we’ve gotten an idea. We will next give a formal 

definition, and then we will produce some results backing our contention that 

the definition captures the idea. We’ve seen this happen already, for instance, in 

the first section of the Vector Space chapter. There, the study of linear systems 

led us to consider collections closed under linear combinations. We defined such 

a collection as a vector space, and we followed it with some supporting results. 

Of course, that definition wasn’t an end point, instead it led to new insights 

such as the idea of a basis. Here too, after producing a definition, and supporting 

it, we will get two (pleasant) surprises. First, we will find that the definition 

applies to some unforeseen, and interesting, cases. Second, the study of the 

definition will lead to new ideas. In this way, our investigation will build a 

momentum. 

3.I.1 Definition and Examples 

We start with two examples that suggest the right definition. 

1.1 Example Consider the example mentioned above, the space of two-wide 

row vectors and the space of two-tall column vectors. They are “the same” in 

that if we associate the vectors that have the same components, e.g., 

� � 

� � 1 

1 2 ←→ 

2 

159

160 Chapter 3. Maps Between Spaces 

then this correspondence preserves the operations, for instance this addition 

� 1 2 � + � 3 4 � = � 4 6 � ←→ 

and this scalar multiplication. 

5 · � 1 2 � = � 5 10 � ←→ 5 · 

More generally stated, under the correspondence 

both operations are preserved: 

and 

� � 

a0 a1 ←→ 

� � 

1 

+ 

2 

� � 

a0 

� � � � � � 

a0 a1 + b0 b1 = a0 + b0 a1 + b1 ←→ 

a1 

r · � � � � 

a0 a1 = ra0 ra1 ←→ r · 

(all of the variables are real numbers). 

� � 

1 

= 

2 

� � 

3 

= 

4 

� � 

5 

10 

� � 

4 

6 

� � � � 

a0 b0 

+ = 

a1 

b1 

� � � � 

a0 ra0 

= 

a1 ra1 

� a0 + b0 

a1 + b1 

1.2 Example Another two spaces we can think of as “the same” are P2, the 

space of quadratic polynomials, and R 3 . A natural correspondence is this. 

a0 + a1x + a2x 2 ←→ 

⎛ 

⎝ 

a0 

a1 

a2 

⎞ 

⎠ (e.g., 1 + 2x +3x2 ⎛ ⎞ 

1 

←→ ⎝2⎠) 

3 

The structure is preserved: corresponding elements add in a corresponding way 

a0 + a1x + a2x 2 

+ b0 + b1x + b2x 2 

(a0 + b0)+(a1 + b1)x +(a2 + b2)x 2 

and scalar multiplication corresponds also. 

←→ 

⎛ 

⎝ a0 

⎞ ⎛ 

⎠ + ⎝ b0 

⎞ ⎛ 

⎠ = 

r · (a0 + a1x + a2x 2 )=(ra0)+(ra1)x +(ra2)x 2 ←→ r · 

a1 

a2 

b1 

b2 

� 

⎝ a0 + b0 

a1 + b1 

a2 + b2 

⎞ 

⎠ 

⎛ 

⎝ a0 

⎞ ⎛ 

a1⎠ 

= ⎝ 

a2 

ra0 

⎞ 

ra1⎠ 

ra2

Section I. Isomorphisms 161 

1.3 Definition An isomorphism between two vector spaces V and W is a map 

f : V → W that 

(1) is a correspondence: f is one-to-one and onto; ∗ 

(2) preserves structure: if �v1,�v2 ∈ V then 

and if �v ∈ V and r ∈ R then 

f(�v1 + �v2) =f(�v1)+f(�v2) 

f(r�v) =rf(�v) 

(we write V ∼ = W ,read“V is isomorphic to W ”, when such a map exists). 

(“Morphism” means map, so “isomorphism” means a map expressing sameness.) 

1.4 Example The vector space G = {c1 cos θ + c2 sin θ � � c1,c2 ∈ R} of functions 

of θ is isomorphic to the vector space R2 under this map. 

c1 cos θ + c2 sin θ f 

� � 

c1 

↦−→ 

We will check this by going through the conditions in the definition. 

We will first verify condition (1), that the map is a correspondence between 

the sets underlying the spaces. 

To establish that f is one-to-one, we must prove that f(�a) =f( � b) only when 

�a = � b.If 

then, by the definition of f, 

f(a1 cos θ + a2 sin θ) =f(b1 cos θ + b2 sin θ) 

� � � � 

a1 b1 

= 

a2 

from which we can conclude that a1 = b1 and a2 = b2 because column vectors are 

equal only when they have equal components. We’ve proved that f(�a) =f( �b) implies that �a = �b, which shows that f is one-to-one. 

To check that f is onto we must check that any member of the codomain R2 mapped to. But that’s clear—any 

� � 

x 

∈ R 

y 

2 

is the image, under f, of this member of the domain: x cos θ + y sin θ ∈ G. 

Next we will verify condition (2), that f preserves structure. 

∗ More information on one-to-one and onto maps is in the appendix. 

b2 

c2


This computation shows that f preserves addition. 

f � (a1 cos θ + a2 sin θ)+(b1cos θ + b2 sin θ) � 

= f � (a1 + b1)cosθ +(a2 + b2)sinθ � 

� � 

a1 + b1 

= 

a2 + b2 

� � � � 

a1 b1 

= + 

a2 

b2 

= f(a1 cos θ + a2 sin θ)+f(b1 cos θ + b2 sin θ) 

A similar computation shows that f preserves scalar multiplication. 

f � r · (a1 cos θ + a2 sin θ) � = f( ra1 cos θ + ra2 sin θ ) 

� � 

ra1 

= 

ra2 

� � 

a1 

= r · 

a2 

= r · f(a1 cos θ + a2 sin θ) 

With that, conditions (1) and (2) are verified, so we know that f is an 

isomorphism, and we can say that the spaces are isomorphic G ∼ = R 2 . 

1.5 Example Let V be the space {c1x + c2y + c3z � � c1,c2,c3 ∈ R} of linear 

combinations of three variables x, y, andz, under the natural addition and 

scalar multiplication operations. Then V is isomorphic to P2, the space of 

quadratic polynomials. 

To show this we will produce an isomorphism map. There is more than one 

possibility; for instance, here are four. 

c1x + c2y + c3z 

f1 

↦−→ c1 + c2x + c3x2 f2 

↦−→ c2 + c3x + c1x2 f3 

↦−→ −c1 − c2x − c3x2 f4 

↦−→ c1 +(c1 + c2)x +(c1 + c3)x2 Although the first map is the more natural correspondence, below we shall 

verify that the second one is an isomorphism, to underline that there are many 

isomorphisms other than the obvious one that just carries the coefficients over 

(showing that f1 is an isomorphism is Exercise 12). 

To show that f2 is one-to-one, we will prove that if f2(c1x + c2y + c3z) = 

f2(d1x + d2y + d3z) then c1x + c2y + c3z = d1x + d2y + d3z. The assumption 

that f2(c1x+c2y +c3z) =f2(d1x+d2y +d3z) gives, by the definition of f2, that 

c2 + c3x + c1x 2 = d2 + d3 + d1x 2 . Equal polynomials have equal coefficients, so 

c2 = d2, c3 = d3, andc1 = d1. Thusf2(c1x + c2y + c3z) =f2(d1x + d2y + d3z) 

implies that c1x + c2y + c3z = d1x + d2y + d3z and therefore f2 is one-to-one.


The map f2 is onto because any member a + bx + cx 2 of the codomain is the 

image of some member of the domain, namely it is the image of cx + ay + bz. 

(For instance, 2 + 3x − 4x 2 is f2(−4x +2y +3z).) 

The computations for structure preservation for this map are like those in 

the prior example. This map preserves addition 

� 

f2 (c1x + c2y + c3z)+(d1x + d2y + d3z) � 

� 

= f2 (c1 + d1)x +(c2 + d2)y +(c3 + d3)z � 

=(c2 + d2)+(c3 + d3)x +(c1 + d1)x 2 

=(c2 + c3x + c1x 2 )+(d2 + d3x + d1x 2 ) 

= f2(c1x + c2y + c3z)+f2(d1x + d2y + d3z) 

and scalar multiplication. 

� 

f2 r · (c1x + c2y + c3z) � = f2(rc1x + rc2y + rc3z) 

= rc2 + rc3x + rc1x 2 

Thus f2 is an isomorphism and we write V ∼ = P2. 

= r · (c2 + c3x + c1x 2 ) 

= r · f2(c1x + c2y + c3z) 

We are sometimes interested in an isomorphism of a space with itself, called 

an automorphism. The identity map is easily seen to be an automorphism. The 

next example shows that there are others. 

1.6 Example Consider the space P5 of polynomials of degree 5 or less and the 

map f that sends a polynomial p(x) top(x − 1). For instance, under this map 

x 2 ↦→ (x−1) 2 = x 2 −2x+1 and x 3 +2x ↦→ (x−1) 3 +2(x−1) = x 3 −3x 2 +5x−3. 

This map is an automorphism of this space, the check is Exercise 21. 

This isomorphism of P5 with itself does more than just tell us that the space 

is “the same” as itself. It gives us some insight into the space’s structure. For 

instance, below is shown a family of parabolas, graphs of members of P5. Each 

has a vertex at y = −1, and the left-most one has zeroes at −2.25 and −1.75, 

the next one has zeroes at −1.25 and −0.75, etc. 

p 0 (x) p 1 (x)


Geometrically, the substitution of x − 1forx in any function’s argument shifts 

its graph to the right by one. In the case of the above picture, f(p0) =p1, and 

more generally, f’s action is to shift all of the parabolas to the right by one. 

Observe, though, that the picture before f is applied is the same as the picture 

after f is applied, because while each parabola moves to the right, another one 

comes in from the left to take its place. This also holds true for cubics, etc. 

So the automorphism f gives us the insight that P5 has a certain horizontalhomogeneity—the 

space looks the same near x = 1 as near x =0. 

1.7 Example A dilation map ds : R 2 → R 2 that multiplies all vectors by a 

nonzero scalar s is an automorphism of R 2 . 

�v 

�u 

d1.5 

↦−→ 

d1.5(�u) 

d1.5(�v) 

A rotation or turning map tθ : R 2 → R 2 that rotates all vectors through an angle 

θ is an automorphism. 

�v 

t π/3 

↦−→ 

t π/3(�v) 

A third type of automorphism of R 2 is a map fℓ : R 2 → R 2 that flips or reflects 

all vectors over a line ℓ through the origin. 

See Exercise 29. 

�v 

f ℓ 

↦−→ 

fℓ(�v) 

As described in the preamble to this section, we will next produce some 

results supporting the contention that the definition of isomorphism above captures 

our intuition of vector spaces being the same. 

Of course the definition itself is persuasive: a vector space consists of two 

components, a set and some structure, and the definition simply requires that 

the sets correspond and that the structures correspond also. Also persuasive are 

the examples above. In particular, Example 1.1, which gives an isomorphism 

between the space of two-wide row vectors and the space of two-tall column 

vectors, dramatizes our intuition that isomorphic spaces are the same in all


relevant respects. Sometimes people say, where V ∼ = W ,that“W is just V 

painted green”—any differences are merely cosmetic. 

Further support for the definition, in case it is needed, is provided by the 

following results that, taken together, suggest that all the things of interest 

in a vector space correspond under an isomorphism. Since we studied vector 

spaces to study linear combinations, “of interest” means “pertaining to linear 

combinations”. Not of interest is the way that the vectors are typographically 

laid out (or their color!). 

As an example, although the definition of isomorphism doesn’t explicitly say 

that the zero vectors must correspond, it is a consequence of that definition. 

1.8 Lemma An isomorphism maps a zero vector to a zero vector. 

Proof. Where f : V → W is an isomorphism, fix any �v ∈ V . Then f(�0V )= 

f(0 · �v) =0· f(�v) =�0W . QED 

The definition of isomorphism requires that sums of two vectors correspond 

and that so do scalar multiples. We can extend that to say that all linear 

combinations correspond. 

1.9 Lemma For any map f : V → W between vector spaces the statements 

(1) f preserves structure 

f(�v1 + �v2) =f(�v1)+f(�v2) and f(c�v) =cf(�v) 

(2) f preserves linear combinations of two vectors 

f(c1�v1 + c2�v2) =c1f(�v1)+c2f(�v2) 

(3) f preserves linear combinations of any finite number of vectors 

are equivalent. 

f(c1�v1 + ···+ cn�vn) =c1f(�v1)+···+ cnf(�vn) 

Proof. Since the implications (3) =⇒ (2) and (2) =⇒ (1) are clear, we need 

only show that (1) =⇒ (3). Assume statement (1). We will prove statement (3) 

by induction on the number of summands n. 

The one-summand base case, that f(c�v) = cf(�v), is covered by statement 

(1). 

For the inductive step assume that statement (3) holds whenever there are k 

or fewer summands, that is, whenever n =1,orn =2,... ,orn = k. Consider 

the k + 1-summand case. The first half of statement (1) gives 

f(c1�v1 + ···+ ck�vk + ck+1�vk+1) =f(c1�v1 + ···+ ck�vk)+f(ck+1�vk+1)


by breaking the sum along the final +. Then the inductive hypothesis lets us 

break up the sum of the k things. 

= f(c1�v1)+···+ f(ck�vk)+f(ck+1�vk+1) 

Finally, the second half of statement (1) gives 

= c1 f(�v1)+···+ ck f(�vk)+ck+1 f(�vk+1) 

when applied k + 1 times. QED 

In addition to adding to the intuition that the definition of isomorphism 

does indeed preserve things of interest in a vector space, that lemma’s second 

item is an especially handy way of checking that a map preserves structure. 

We close with a summary. We have defined the isomorphism relation ‘ ∼ =’ 

between vector spaces. We have argued that it is the right way to split the 

collection of vector spaces into cases because it preserves the features of interest 

in a vector space—in particular, it preserves linear combinations. The material 

in this section augments the chapter on Vector Spaces. There, after giving the 

definition of a vector space, we informally looked at what different things can 

happen. We have now said precisely what we mean by ‘different’, and by ‘the 

same’, and so we have precisely classified the vector spaces. 

Exercises 

� 1.10 Verify, using Example 1.4 as a model, that the two correspondences given 

before the definition are isomorphisms. 

(a) Example 1.1 (b) Example 1.2 

� 1.11 For the map f : P1 → R 2 given by 

a + bx f 

� � 

a − b 

↦−→ 

b 

Find the image of each of these elements of the domain. 

(a) 3 − 2x (b) 2+2x (c) x 

Show that this map is an isomorphism. 

1.12 Show that the natural map f1 from Example 1.5 is an isomorphism. 

� 1.13 Decide whether each map is an isomorphism (of course, if it is an isomorphism 

then prove it and if it isn’t then state a condition that it fails to satisfy). 

(a) f : M2×2 → R given by 

� � 

a b 

↦→ ad − bc 

c d 

(b) f : M2×2 → R 4 given by 

⎛ 

⎞ 

� � a + b + c + d 

a b ⎜ a + b + c ⎟ 

↦→ 

c d 

⎝ 

a + b 

⎠ 

a 

(c) f : M2×2 →P3 given by 

� � 

a b 

c d 

↦→ c +(d + c)x +(b + a)x 2 + ax 3


(d) f : M2×2 →P3 given by 

� � 

a b 

c d 

↦→ c +(d + c)x +(b + a +1)x 2 + ax 3 

1.14 Show that the map f : R 1 → R 1 given by f(x) =x 3 is one-to-one and onto. 

Is it an isomorphism? 

� 1.15 Refer to Example 1.1. Produce two more isomorphisms (of course, that they 

satisfy the conditions in the definition of isomorphism must be verified). 

1.16 Refer to Example 1.2. Produce two more isomorphisms (and verify that they 

satisfy the conditions). 

� 1.17 Show that, although R 2 is not itself a subspace of R 3 , it is isomorphic to the 

xy-plane subspace of R 3 . 

1.18 Find two isomorphisms between R 16 and M4×4. 

� 1.19 For what k is Mm×n isomorphic to R k ? 

1.20 For what k is Pk isomorphic to R n ? 

1.21 Prove that the map in Example 1.6, fromP5 to P5 given by p(x) ↦→ p(x − 1), 

is a vector space isomorphism. 

1.22 Why, in Lemma 1.8, must there be a �v ∈ V ? That is, why must V be 

nonempty? 

1.23 Are any two trivial spaces isomorphic? 

1.24 In the proof of Lemma 1.9, what about the zero-summands case (that is, if n 

is zero)? 

1.25 Show that any isomorphism f : P0 → R 1 has the form a ↦→ ka for some nonzero 

real number k. 

� 1.26 These prove that isomorphism is an equivalence relation. 

(a) Show that the identity map id: V → V is an isomorphism. Thus, any vector 

space is isomorphic to itself. 

(b) Show that if f : V → W is an isomorphism then so is its inverse f −1 : W → V . 

Thus, if V is isomorphic to W then also W is isomorphic to V . 

(c) Show that a composition of isomorphisms is an isomorphism: if f : V → W is 

an isomorphism and g : W → U is an isomorphism then so also is g ◦ f : V → U. 

Thus, if V is isomorphic to W and W is isomorphic to U, thenalsoV is isomorphic 

to U. 

1.27 Suppose that f : V → W preserves structure. Show that f is one-to-one if and 

only if the unique member of V mapped by f to �0W is �0V . 

1.28 Suppose that f : V → W is an isomorphism. Prove that the set {�v1,...,�vk} ⊆ 

V is linearly dependent if and only if the set of images {f(�v1),...,f(�vk)} ⊆W is 

linearly dependent. 

� 1.29 Show that each type of map from Example 1.7 is an automorphism. 

(a) Dilation ds by a nonzero scalar s. 

(b) Rotation tθ throughanangleθ. 

(c) Reflection fℓ over a line through the origin. 

Hint. For the second and third items, polar coordinates are useful. 

1.30 Produce an automorphism of P2 other than the identity map, and other than 

ashiftmapp(x) ↦→ p(x − k). 

1.31 (a) Show that a function f : R 1 → R 1 is an automorphism if and only if it 

has the form x ↦→ kx for some k �= 0. 

(b) Let f be an automorphism of R 1 such that f(3) = 7. Find f(−2).


(c) Show that a function f : R 2 → R 2 is an automorphism if and only if it has 

the form 

� � 

x 

↦→ 

y 

� � 

ax + by 

cx + dy 

for some a, b, c, d ∈ R with ad − bc �= 0. Hint. Exercises in prior subsections 

have shown that 

� � 

b 

is not a multiple of 

d 

� � 

a 

c 

if and only if ad − bc �= 0. 

(d) Let f be an automorphism of R 2 with 

� � � � 

1 2 

f( )= and 

3 −1 

� � � � 

1 0 

f( )= . 

4 1 

Find 

� � 

0 

f( ). 

−1 

1.32 Refer to Lemma 1.8 and Lemma 1.9. 

isomorphism. 

Find two more things preserved by 

1.33 We show that isomorphisms can be tailored to fit in that, sometimes, given 

vectors in the domain and in the range we can produce an isomorphism associating 

those vectors. 

(a) Let B = 〈 � β1, � β2, � β3〉 be a basis for P2 so that any �p ∈ P2 has a unique 

representation as �p = c1� β1 + c2� β2 + c3� β3, which we denote in this way. 

� � 

c1 

RepB(�p) = 

Show that the Rep B(·) operation is a function from P2 to R 3 (this entails showing 

that with every domain vector �v ∈P2 there is an associated image vector in R 3 , 

and further, that with every domain vector �v ∈P2 there is at most one associated 

image vector). 

(b) Show that this Rep B(·) function is one-to-one and onto. 

(c) Show that it preserves structure. 

(d) Produce an isomorphism from P2 to R 3 that fits these specifications. 

x + x 2 ↦→ 

� � 

1 

0 

0 

c2 

c3 

and 1 − x ↦→ 

� � 

0 

1 

0 

1.34 Prove that a space is n-dimensional if and only if it is isomorphic to R n . 

Hint. Fix a basis B for the space and consider the map sending a vector over to 

its representation with respect to B. 

1.35 (Requires the subsection on Combining Subspaces, which is optional.) Let U 

and W be vector spaces. Define a new vector space, consisting of the set U × W = 

{(�u, �w) � � �u ∈ U and �w ∈ W } along with these operations. 

(�u1, �w1)+(�u2, �w2) =(�u1 + �u2, �w1 + �w2) and r · (�u, �w) =(r�u, r �w) 

This is a vector space, the external direct sum of U and W ). 

(a) Checkthatitisavectorspace. 

(b) Find a basis for, and the dimension of, the external direct sum P2 × R 2 . 

(c) What is the relationship among dim(U), dim(W ), and dim(U × W )?


(d) Suppose that U and W are subspaces of a vector space V such that V = 

U ⊕ W . Show that the map f : U × W → V given by 

f 

(�u, �w) ↦−→ �u + �w 

is an isomorphism. Thus if the internal direct sum is defined then the internal 

and external direct sums are isomorphic. 

3.I.2 Dimension Characterizes Isomorphism 

In the prior subsection, after stating the definition of an isomorphism, we 

gave some results supporting the intuition that such a map describes spaces 

as “the same”. Here we will formalize this intuition. While two spaces that 

are isomorphic are not equal, we think of them as almost equal—as equivalent. 

In this subsection we shall show that the relationship ‘is isomorphic to’ is an 

equivalence relation. ∗ 

2.1 Theorem Isomorphism is an equivalence relation between vector spaces. 

Proof. We must prove that this relation has the three properties of being symmetric, 

reflexive, and transitive. For each of the three we will use item (2) of 

Lemma 1.9 and show that the map preserves structure by showing that the it 

preserves linear combinations of two members of the domain. 

To check reflexivity, that any space is isomorphic to itself, consider the identity 

map. It is clearly one-to-one and onto. The calculation showing that it 

preserves linear combinations is easy. 

id(c1 · �v1 + c2 · �v2) =c1�v1 + c2�v2 = c1 · id(�v1)+c2 · id(�v2) 

To check symmetry, that if V is isomorphic to W via some map f : V → W 

then there is an isomorphism going the other way, consider the inverse map 

f −1 : W → V . As stated in the appendix, the inverse of the correspondence 

f is also a correspondence, so we need only check that the inverse preserves 

linear combinations. Assume that �w1 = f(�v1), i.e., that f −1 ( �w1) =�v1, and also 

assume that �w2 = f(�v2). 

f −1 (c1 · �w1 + c2 · �w2) =f −1� c1 · f(�v1)+c2 · f(�v2) � 

= f −1 ( f � c1�v1 + c2�v2) � 

= c1�v1 + c2�v2 

= c1 · f −1 ( �w1)+c2 · f −1 ( �w2) 

Finally, to check transitivity, that if V is isomorphic to W via some map f 

and if W is isomorphic to U via some map g then also V is isomorphic to U, 

consider the composition map g ◦ f : V → U. As stated in the appendix, the 

∗ More information on equivalence relations is in the appendix.


composition of two correspondences is a correspondence, so we need only check 

that the composition preserves linear combinations. 

g ◦ f � � � 

c1 · �v1 + c2 · �v2 = g f(c1 · �v1 + c2 · �v2) � 

= g � c1 · f(�v1)+c2 · f(�v2) � 

= c1 · g � f(�v1)) + c2 · g(f(�v2) � 

= c1 · g ◦ f (�v1)+c2 · g ◦ f (�v2) 

Thus g ◦ f : V → U is an isomorphism. QED 

As a consequence of that result, we know that the universe of vector spaces 

is partitioned into classes: every space is in one and only one isomorphism class. 

Finite-dimensional 

vector spaces: 

.V 

✪ 

✥ 

✩ 

✦ ✜ 

✢ 

... 

W . 

V ∼ = W 

The next result gives a simple criteria describing which spaces are in each class. 

2.2 Theorem Vector spaces are isomorphic if and only if they have the same 

dimension. 

This theorem follows from the next two lemmas. 

2.3 Lemma If spaces are isomorphic then they have the same dimension. 

Proof. We shall show that an isomorphism of two spaces gives a correspondence 

between their bases. That is, where f : V → W is an isomorphism and a basis 

for the domain V is B = 〈 � β1,..., � βn〉, then the image set D = 〈f( � β1),...,f( � βn)〉 

is a basis for the codomain W . (The other half of the correspondence—that for 

any basis of W the inverse image is a basis for V —follows on recalling that if 

f is an isomorphism then f −1 is also an isomorphism, and applying the prior 

sentence to f −1 .) 

To see that D spans W ,fixa�w ∈ W , use the fact that f is onto and so there 

is a �v ∈ V with �w = f(�v), and expand �v as a combination of basis vectors. 

�w = f(�v) =f(v1 � β1 + ···+ vn � βn) =v1 · f( � β1)+···+ vn · f( � βn) 

For linear independence of D, if 

�0W = c1f( � β1)+···+ cnf( � βn) =f(c1 � β1 + ···+ cn � βn) 

then, since f is one-to-one and so the only vector sent to �0W is �0V ,wehave 

that �0V = c1 � β1 + ···+ cn � βn, implying that all the c’s are zero. QED


2.4 Lemma If spaces have the same dimension then they are isomorphic. 

Proof. To show that any two spaces of dimension n are isomorphic, we can 

simply show that any one is isomorphic to Rn . Then we will have shown that 

they are isomorphic to each other, by the transitivity of isomorphism (which 

was established in Theorem 2.1). 

Let V be an n-dimensional space. Fix a basis B = 〈 � β1,..., � βn〉 for the 

domain V and consider as a function the representation of the members of that 

domain with respect to the basis. 

⎛ ⎞ 

�v = v1 � β1 + ···+ vn � βn 

RepB 

↦−→ 

(This is well-defined since every �v has one and only one such representation—see 

Remark 2.5 below. ∗ ) 

This function is one-to-one because if 

then 

⎜ 

⎝ 

v1 

. 

vn 

⎟ 

⎠ 

RepB(u1 � β1 + ···+ un � βn) =RepB(v1 � β1 + ···+ vn � βn) 

⎛ 

⎜ 

⎝ 

u1 

. 

un 

⎞ 

⎟ 

⎠ = 

and so u1 = v1, ... , un = vn, and therefore the original arguments u1 � β1 + ···+ 

un � βn and v1 � β1 + ···+ vn � βn are equal. 

This function is onto; any n-tall vector 

⎛ ⎞ 

�w = 

is the image of some �v ∈ V , namely �w =Rep B(v1 � β1 + ···+ vn � βn). 

Finally, this function preserves structure. 

⎜ 

⎝ 

RepB(r · �u + s · �v) =RepB((ru1 + sv1) � β1 + ···+(run + svn) � βn ) 

⎛ ⎞ 

ru1 + sv1 

⎜ 

= 

. 

⎝ . 

⎟ 

. ⎠ 

run + svn 

⎛ ⎞ ⎛ ⎞ 

u1 v1 

⎜ 

= r · 

. ⎟ ⎜ 

⎝ . ⎠ + s · 

. ⎟ 

⎝ . ⎠ 

un 

⎛ 

⎜ 

⎝ 

w1 

. 

wn 

v1 

. 

vn 

⎟ 

⎠ 

⎞ 

⎟ 

⎠ 

vn 

= r · Rep B(�u)+s · Rep B(�v) 

∗ More information on well-definedness is in the appendix.


Thus the function is an isomorphism, and we can say that any n-dimensional 

space is isomorphic to the n-dimensional space R n . Consequently, as noted at 

the start, any two spaces with the same dimension are isomorphic. QED 

2.5 Remark The parenthetical comment in that proof about the role played 

by the ‘one and only one representation’ result requires some explanation. We 

need to show that each vector in the domain is associated by Rep B with one 

and only one vector in the codomain. 

A contrasting example, where an association doesn’t have this property, is 

illuminating. Consider this subset of P2, which is not a basis. 

A = {1+0x +0x 2 , 0+1x +0x 2 , 0+0x +1x 2 , 1+1x +2x 2 } 

Call those four polynomials �α1, ... , �α4. If, mimicing above proof, we try to 

write the members of P2 as �p = c1�α1 + c2�α2 + c3�α3 + c4�α4, and associate �p 

with the four-tall vector with components c1, ... , c4 then there is a problem. 

For, consider �p(x) =1+x + x 2 . The set A spans the space P2, so there is at 

least one four-tall vector associated with �p. But A is not linearly independent 

so vectors do not have unique decompositions. In this case, both 

�p(x) =1�α1 +1�α2 +1�α3 +0�α4 and �p(x) =0�α1 +0�α2 − 1�α3 +1�α4 

and so there is more than one four-tall vector associated with �p. 

⎛ ⎞ 

1 

⎜ 

⎜1 

⎟ 

⎝1⎠ 

0 

and 

⎛ ⎞ 

0 

⎜ 0 ⎟ 

⎝−1⎠ 

1 

If we are trying to think of this association as a function then the problem is 

that, for instance, with input �p the association does not have a well-defined 

output value. 

Any map whose definition appears possibly ambiguous must be checked to 

see that it is well-defined. For the above proof that check is Exercise 19. 

That ends the proof of Theorem 2.2. We say that the isomorphism classes 

are characterized by dimension because we can describe each class simply by 

giving the number that is the dimension of all of the spaces in that class. 

This subsection’s results give us a collection of representatives of the isomorphism 

classes. ∗ 

2.6 Corollary A finite-dimensional vector space is isomorphic to one and only 

one of the R n . 

2.7 Remark The proofs above pack many ideas into a small space. Through 

the rest of this chapter we’ll consider these ideas again, and fill them out. For a 

taste of this, we will close this section by indicating how we can expand on the 

proof of Lemma 2.4. 

∗ More information on equivalence class representatives is in the appendix.


2.8 Example The space M2×2 of 2×2 matrices is isomorphic to R4 . With this 

basis for the domain 

� � � � � � � � 

1 0 0 1 0 0 0 0 

B = 〈 , , , 〉 

0 0 0 0 1 0 0 1 

the isomorphism given in the lemma, the representation map, simply carries the 

entries over. 

⎛ ⎞ 

� � 

a 

a b f1 ⎜ 

↦−→ ⎜b 

⎟ 

c d ⎝c⎠ 

d 

One way to understand the map f1 is this: we fix the basis B for the domain 

and the basis E4 for the codomain, and associate � β1 with �e1, and � β2 with �e2, 

etc. We then extend this association to all of the vectors in two spaces. 

� � 

a b 

= a 

c d 

� β1 + b� β2 + c� β3 + d� β4 

f1 

↦−→ a�e1 + b�e2 + c�e3 + d�e4 = 

⎛ ⎞ 

a 

⎜ 

⎜b 

⎟ 

⎝c⎠ 

d 

We say that the map has been extended linearly from the bases to the spaces. 

We can do the same thing with different bases, for instance, taking this basis 

for the domain. 

� � � � � � � � 

2 0 0 2 0 0 0 0 

A = 〈 , , , 〉 

0 0 0 0 2 0 0 2 

Associating corresponding members of A and E4, and extending linearly, 

� � 

a b 

=(a/2)�α1 +(b/2)�α2 +(c/2)�α3 +(d/2)�α4 

c d 

f2 

↦−→ (a/2)�e1 +(b/2)�e2 +(c/2)�e3 +(d/2)�e4 = 

⎛ ⎞ 

a/2 

⎜ 

⎜b/2 

⎟ 

⎝c/2⎠ 

d/2 

gives rise to an isomorphism that is different than f1. 

We can also change the basis for the codomain. Starting with these bases, 

B and 

⎛ ⎞ 

1 

⎜ 

D = 〈 ⎜0 

⎟ 

⎝0⎠ 

0 

, 

⎛ ⎞ 

0 

⎜ 

⎜1 

⎟ 

⎝0⎠ 

0 

, 

⎛ ⎞ 

0 

⎜ 

⎜0 

⎟ 

⎝0⎠ 

1 

, 

⎛ ⎞ 

0 

⎜ 

⎜0 

⎟ 

⎝1⎠ 

0 

〉


associating � β1 with �δ1, etc., and then linearly extending that correspondence to 

all of the two spaces 

a� β1 + b� β2 + c� β3 + d� � 

a 

β4 = 

c 

� 

b 

d 

f3 

↦−→ a�δ1 + b�δ2 + c�δ3 + d� ⎛ ⎞ 

a 

⎜ 

δ4 = ⎜b 

⎟ 

⎝d⎠ 

c 

gives still another isomorphism. 

So there is a connection between the maps between spaces and bases for 

those spaces. We will explore that connection in later sections. 

We now finish this section with a summary. 

Recall that in the first chapter, we defined two matrices as row equivalent 

if they can be derived from each other by elementary row operations (this was 

the meaning of same-ness that was of interest there). We showed that is an 

equivalence relation and so the collection of matrices is partitioned into classes, 

where all the matrices that are row equivalent fall together into a single class. 

Then, for insight into which matrices are in each class, we gave representatives 

for the classes, the reduced echelon form matrices. 

In this section, except that the appropriate notion of same-ness here is vector 

space isomorphism, we have followed much the same outline. First we defined 

isomorphism, saw some examples, and established some basic properties. Then 

we showed that it is an equivalence relation, and now we have a set of class 

representatives, the real vector spaces R 1 , R 2 ,etc. 

Finite-dimensional 

vector spaces: 

⋆R 0 

.V 

⋆R 1 

✪ 

✥ 

✩ 

✦ ✜ 

✢ 

⋆R ... 

W . 

2 

⋆R 4 

⋆R 3 

One representative 

per class 

As before, the list of representatives helps us to understand the partition. It is 

simply a classification of spaces by dimension. 

In the second chapter, with the definition of vector spaces, we seemed to 

have opened up our studies to many examples of new structures besides the 

familiar R n ’s. We now know that isn’t the case. Any finite-dimensional vector 

space is actually “the same” as a real space. We are thus considering exactly 

the structures that we need to consider. 

In the next section, and in the rest of the chapter, we will fill out the work 

that we have done here. In particular, in the next section we will consider maps 

that preserve structure, but are not necessarily correspondences. 

Exercises 

� 2.9 Decide if the spaces are isomorphic.


(a) R 2 , R 4 

(b) P5, R 5 

(c) M2×3, R 6 

(d) P5, M2×3 (e) M2×k, C k 

� 2.10 Consider the isomorphism RepB(·): P1 → R 2 where B = 〈1, 1+x〉. Findthe 

image of each of these elements of the domain. 

(a) 3 − 2x; (b) 2+2x; (c) x 

� 2.11 Show that if m �= n then R m �∼ = R n . 

� 2.12 Is Mm×n ∼ = Mn×m? 

� 2.13 Are any two planes through the origin in R 3 isomorphic? 

2.14 Find a set of equivalence class representatives other than the set of R n ’s. 

2.15 True or false: between any n-dimensional space and R n there is exactly one 

isomorphism. 

2.16 Can a vector space be isomorphic to one of its (proper) subspaces? 

� 2.17 This subsection shows that for any isomorphism, the inverse map is also an isomorphism. 

This subsection also shows that for a fixed basis B of an n-dimensional 

vector space V ,themapRepB : V → R n is an isomorphism. Find the inverse of 

this map. 

� 2.18 Prove these facts about matrices. 

(a) The row space of a matrix is isomorphic to the column space of its transpose. 

(b) The row space of a matrix is isomorphic to its column space. 

2.19 Show that the function from Theorem 2.2 is well-defined. 

2.20 Is the proof of Theorem 2.2 valid when n =0? 

2.21 For each, decide if it is a set of isomorphism class representatives. 

(a) {C k � � 

� 

� k ∈ N} (b) {Pk � k ∈{−1, 0, 1,...}} (c) {Mm×n � m, n ∈ N} 

2.22 Let f be a correspondence between vector spaces V and W (that is, a map 

that is one-to-one and onto). Show that the spaces V and W are isomorphic via f 

if and only if there are bases B ⊂ V and D ⊂ W such that corresponding vectors 

have the same coordinates: Rep B (�v) =Rep D (f(�v)). 

2.23 Consider the isomorphism RepB : P3 → R 4 . 

(a) Vectors in a real space are orthogonal if and only if their dot product is zero. 

Give a definition of orthogonality for polynomials. 

(b) The derivative of a member of P3 is in P3. Give a definition of the derivative 

of a vector in R 4 . 

� 2.24 Does every correspondence between bases, when extended to the spaces, give 

an isomorphism? 

2.25 (Requires the subsection on Combining Subspaces, which is optional.) Suppose 

that V = V1 ⊕ V2 and that V is isomorphic to the space U under the map f. Show 

that U = f(V1) ⊕ f(U2). 

2.26 Show that this is not a well-defined function from the rational numbers to the 

integers: with each fraction, associate the value of its numerator.


3.II Homomorphisms 

The definition of isomorphism has two conditions. In this section we will consider 

the second one, that the map must preserve the algebraic structure of the 

space. We will focus on this condition by studying maps that are required only 

to preserve structure; that is, maps that are not required to be correspondences. 

Experience shows that this kind of map is tremendously useful in the study 

of vector spaces. For one thing, as we shall see in the second subsection below, 

while isomorphisms describe how spaces are the same, these maps describe how 

spaces can be thought of as alike. 

3.II.1 Definition 

1.1 Definition A function between vector spaces h: V → W that preserves 

the operations of addition 

and scalar multiplication 

is a homomorphism or linear map. 

if �v1,�v2 ∈ V then h(�v1 + �v2) =h(�v1)+h(�v2) 

if �v ∈ V and r ∈ R then h(r · �v) =r · h(�v) 

1.2 Example The projection map π : R 3 → R 2 

⎛ ⎞ 

x 

⎝y⎠ 

z 

π 

↦−→ 

is a homomorphism. It preserves addition 

⎛ 

π( ⎝ x1 

⎞ ⎛ 

⎠+ ⎝ x2 

⎞ ⎛ 

⎠) =π( ⎝ x1 

⎞ 

+ x2 

⎠) = 

y1 

z1 

y2 

z2 

y1 + y2 

z1 + z2 

� � 

x 

y 

� x1 + x2 

y1 + y2 

⎛ 

� 

= π( ⎝ x1 

⎞ ⎛ 

⎠)+π( ⎝ x2 

⎞ 

⎠) 

and it preserves scalar multiplication. 

⎛ 

π(r · ⎝ x1 

⎞ ⎛ 

y1⎠) 

=π( ⎝ 

z1 

rx1 

⎞ 

⎛ 

� � 

rx1 

ry1⎠) 

= = r · π( ⎝ 

ry1 

rz1 

x1 

⎞ 

y1⎠) 

z1 

Note that this map is not an isomorphism, since it is not one-to-one. For 

instance, both �0 and�e3 in R 3 are mapped to the zero vector in R 2 . 

1.3 Example The domain and codomain can be other than spaces of column 

vectors. Both of these maps are homomorphisms. 

y1 

z1 

y2 

z2

Section II. Homomorphisms 177 

(1) f1 : P2 →P3 given by 

(2) f2 : M2×2 → R given by 

The verifications are straightforward. 

a0 + a1x + a2x 2 ↦→ a0x +(a1/2)x 2 +(a2/3)x 3 

� � 

a b 

↦→ a + d 

c d 

1.4 Example Between any two spaces there is a zero homomorphism, sending 

every vector in the domain to the zero vector in the codomain. 

1.5 Example These two suggest why the term ‘linear map’ is used. 

(1) The map g : R3 → R given by 

⎛ ⎞ 

x 

⎝y⎠ 

z 

g 

↦−→ 3x +2y − 4.5z 

is linear (i.e., is a homomorphism). In contrast, the map ˆg : R3 → R given 

by 

⎛ ⎞ 

x 

⎝y⎠ 

z 

ˆg 

↦−→ 3x +2y − 4.5z +1 

is not linear; for instance, 

⎛ ⎞ ⎛ ⎞ 

0 1 

ˆg( ⎝0⎠ 

+ ⎝0⎠) 

= 4 while 

⎛ ⎞ ⎛ ⎞ 

0 1 

ˆg( ⎝0⎠)+ˆg( 

⎝0⎠) 

=5 

0 0 

0 0 

(to show that a map is not linear we need only produce one example of a 

linear combination that is not preserved). 

(2) The first of these two maps t1,t2 : R 3 → R 2 is linear while the second is 

not. 

⎛ 

⎝ x 

⎞ 

y⎠ 

z 

t1 

↦−→ 

� � 

5x − 2y 

x + y 

and 

⎛ 

⎝ x 

⎞ 

y⎠ 

z 

t2 

↦−→ 

� � 

5x − 2y 

xy 

(Finding an example that the second fails to preserve structure is easy.) 

What distinguishes the homomorphisms is that the coordinate functions are 

linear combinations of the arguments. See also Exercise 22.


Obviously, any isomorphism is a homomorphism—an isomorphism is a homomorphism 

that is also a correspondence. So, one way to think of the ‘homomorphism’ 

idea is that it is a generalization of ‘isomorphism’, motivated by the 

observation that many of the properties of isomorphisms have only to do with 

the map respecting structure and not to do with it being a correspondence. As 

examples, these two results from the prior section do not use one-to-one-ness or 

onto-ness in their proof, and therefore apply to any homomorphism. 

1.6 Lemma A homomorphism sends a zero vector to a zero vector. 

1.7 Lemma Each of these is a necessary and sufficient condition for f : V → W 

to be a homomorphism. 

(1) for any c1,c2 ∈ R and �v1,�v2 ∈ V , 

f(c1 · �v1 + c2 · �v2) =c1 · f(�v1)+c2 · f(�v2) 

(2) for any c1,...,cn ∈ R and �v1,... ,�vn ∈ V , 

f(c1 · �v1 + ···+ cn · �vn) =c1 · f(�v1)+···+ cn · f(�vn) 

This lemma simplifies the check that a function is linear since we can combine 

the check that addition is preserved with the one that scalar multiplication is 

preserved and since we need only check that combinations of two vectors are 

preserved. 

1.8 Example The map f : R 2 → R 4 given by 

� � 

x f 

↦−→ 

y 

⎛ ⎞ 

x/2 

⎜ 0 ⎟ 

⎝x 

+ y⎠ 

3y 

satisfies that check 

⎛ 

⎞ ⎛ 

r1(x1/2) + r2(x2/2) 

⎜ 

0 

⎟ ⎜ 

⎟ 

⎝r1(x1 

+ y1)+r2(x2 + y2) ⎠ = r1 

⎜ 

⎝ 

r1(3y1)+r2(3y2) 

and so it is a homomorphism. 

x1/2 

0 

x1 + y1 

3y1 

⎞ ⎛ 

⎟ ⎜ 

⎟ 

⎠ + r2 

⎜ 

⎝ 

x2/2 

0 

x2 + y2 

(Sometimes, such as with Lemma 1.15 below, it is less awkward to check preservation 

of addition and preservation of scalar multiplication separately, but this 

is purely a matter of taste.) 

However, some of the results that we have seen for isomorphisms fail to hold 

for homomorphisms in general. An isomorphism between spaces gives a correspondence 

between their bases, but a homomorphisms need not; Example 1.2 

shows this and another example is the zero map between any two nontrivial 

spaces. Instead, a weaker but still very useful result holds. 

3y2 

⎞ 

⎟ 

⎠


1.9 Theorem A homomorphism is determined by its action on a basis. That 

is, if 〈 � β1,..., � βn〉 is a basis of a vector space V and �w1,..., �wn are (perhaps 

not distinct) elements of a vector space W then there exists a homomorphism 

from V to W sending � β1 to �w1, ... ,and � βn to �wn, and that homomorphism is 

unique. 

Proof. We define the map h: V → W by associating � β1 with �w1, etc., and then 

extending linearly to all of the domain. That is, where �v = c1 � β1 + ···+ cn � βn, 

let h(�v) bec1 �w1 + ···+ cn �wn. This is well-defined because, with respect to the 

basis, the representation of each domain vector �v is unique. 

This map is a homomorphism since it preserves linear combinations; where 

�v1 = c1 � β1 + ···+ cn � βn and �v2 = d1 � β1 + ···+ dn � βn, wehavethis. 

h(r1�v1 + r2�v2) =h((r1c1 + r2d1) � β1 + ···+(r1cn + r2dn) � βn) 

=(r1c1 + r2d1) �w1 + ···+(r1cn + r2dn) �wn 

= r1h(�v1)+r2h(�v2) 

And, this map is unique since if ˆ h: V → W is another homomorphism such 

that ˆ h( � βi) = �wi for each i then h and ˆ h agree on all of the vectors in the domain. 

ˆh(�v) = ˆ h(c1 � β1 + ···+ cn � βn) 

= c1 ˆ h( � β1)+···+ cn ˆ h( � βn) 

= c1 �w1 + ···+ cn �wn 

= h(�v) 

Thus, h and ˆ h are the same map. QED 

1.10 Example This result says that we can construct homomorphisms by fixing 

a basis for the domain and specifying where the map sends those basis vectors. 

For instance, if we specify a map h: R 2 → R 2 that acts on the standard 

basis E2 in this way 

� � 

1 

h( )= 

0 

� � 

−1 

1 

� � 

0 

and h( )= 

1 

� � 

−4 

4 

then the action of h on any other member of the domain is also specified. For 

instance, the value of h on this argument 

� � � � � � � � � � � � 

3 

1 0 

1 

0 5 

h( )=h(3 · − 2 · )=3· h( ) − 2 · h( )= 

−2 

0 1 

0 

1 −5 

is a direct consiquence of the value of h on the basis vectors. (Later in this 

chapter we shall develop a scheme, using matrices, that is a convienent way to 

do computations like this one.) 

Just as isomorphisms of a space with itself are useful and interesting, so too 

are homomorphisms of a space with itself.


1.11 Definition A linear map from a space into itself t: V → V is a linear 

transformation. 

In this book we use ‘linear transformation’ only in the case where the codomain 

equals the domain, but it is also widely used as a general synonym for ‘homomorphism’. 

1.12 Example The map on R2 that projects all vectors down to the x-axis 

� � � � 

x x 

↦→ 

y 0 

is a linear transformation. 

1.13 Example The derivative map d/dx: Pn →Pn 

n d/dx 

a0 + a1x + ···+ anx ↦−→ a1 +2a2x +3a3x 2 + ···+ nanx n−1 

is a linear transformation by this result from calculus: d(c1f + c2g)/dx = 

c1 (df /dx)+c2 (dg/dx). 

1.14 Example The matrix transpose map 

� 

a 

c 

� 

b 

d 

↦→ 

� 

a 

b 

� 

c 

d 

is a linear transformation of M2×2. Note that this transformation is one-to-one 

and onto, and so in fact is an automorphism. 

We finish this subsection about maps by recalling that we can linearly combine 

maps. For instance, for these maps from R2 to itself 

� � � � � � � � 

x f 2x 

x g 0 

↦−→ 

and ↦−→ 

y 3x − 2y 

y 5x 

we can take the linear combination 5f − 2g to get this. 

� � � � 

x 5f−2g 10x 

↦−→ 

y 5x − 10y 

1.15 Lemma For vector spaces V and W , the set of linear functions from V 

to W is itself a vector space, a subspace of the space of all functions from V to 

W . It is denoted L(V,W). 

Proof. This set is non-empty because it contains the zero homomorphism. So 

to show that it is a subspace we need only check that it is closed under linear 

combinations. Let f,g: V → W be linear. Then their sum is linear 

(f + g)(c1�v1 + c2�v2) =c1f(�v1)+c2f(�v2)+c1g(�v1)+c2g(�v2) 

� � � � 

= c1 f + g (�v1)+c2 f + g (�v2)


and any scalar multiple is also linear. 

(r · f)(c1�v1 + c2�v2) =r(c1f(�v1)+c2f(�v2)) 

= c1(r · f)(�v1)+c2(r · f)(�v2) 

Hence L(V,W) is a subspace. QED 

We started this section by isolating the structure preservation property of 

isomorphisms. That is, we defined homomorphisms as a generalization of isomorphisms. 

Some of the properties that we studied for isomorphisms carried 

over unchanged, while others were adapted to this more general setting. 

It would be a mistake, though, to view this new notion of homomorphism as 

derived from or somehow secondary to that of isomorphism. In the rest of this 

chapter we shall work mostly with homomorphisms, partly because any statement 

made about homomorphisms is automatically true about isomorphisms, 

but more because, while the isomorphism concept is perhaps more natural, experience 

shows that the homomorphism concept is actually more fruitful and 

more central to further progress. 

Exercises 

� 1.16 Decide if each h: R 3 → R 2 � � 

is linear. 

x � � � � 

x 

x 

(a) h( y )= 

(b) h( y )= 

x + y + z 

z 

z 

� � 

x � � 

2x + y 

(d) h( y )= 

3y − 4z 

z 

� 1.17 Decide if each map h: M2×2 � � 

→ R is linear. 

a b 

(a) h( )=a + d 

c d 

� � 

a b 

(b) h( )=ad − bc 

c d 

� � 

a b 

(c) h( )=2a +3b + c − d 

c d 

� � 

a b 

(d) h( )=a 

c d 

2 + b 2 

� � 

0 

0 

� � 

x 

(c) h( y 

z 

)= 

� � 

1 

1 

� 1.18 Show that these two maps are homomorphisms. 

(a) d/dx: P3 →P2 given by a0 + a1x + a2x 2 + a3x 3 maps to a1 +2a2x +3a3x 2 

(b) � : P2 →P3 given by b0 + b1x + b2x 2 maps to b0x +(b1/2)x 2 +(b2/3)x 3 

Are these maps inverse to each other? 

1.19 Is (perpendicular) projection from R 3 to the xz-plane a homomorphism? Projection 

to the yz-plane? To the x-axis? The y-axis? The z-axis? Projection to the 

origin? 

1.20 Show that, while the maps from Example 1.3 preserve linear operations, they 

are not isomorphisms. 

1.21 Is an identity map a linear transformation?


� 1.22 Stating that a function is ‘linear’ is different than stating that its graph is a 

line. 

(a) The function f1 : R → R given by f1(x) =2x− 1 has a graph that is a line. 

Show that it is not a linear function. 

(b) The function f2 : R 2 → R given by 

� � 

x 

↦→ x +2y 

y 

does not have a graph that is a line. Show that it is a linear function. 

� 1.23 Part of the definition of a linear function is that it respects addition. Does a 

linear function respect subtraction? 

1.24 Assume that h is a linear transformation of V and that 〈 � β1,... , � βn〉 is a basis 

of V . Prove each statement. 

(a) If h( � βi) =�0 for each basis vector then h is the zero map. 

(b) If h( � βi) = � βi for each basis vector then h is the identity map. 

(c) If there is a scalar r such that h( � βi) = r · � βi for each basis vector then 

h(�v) =r · �v for all vectors in V . 

� 1.25 Consider the vector space R + where vector addition and scalar multiplication 

are not the ones inherited from R but rather are these: a + b is the product of 

a and b, andr · a is the r-th power of a. (This was shown to be a vector space 

in an earlier exercise.) Verify that the natural logarithm map ln: R + → R is a 

homomorphism between these two spaces. Is it an isomorphism? 

� 1.26 Consider this transformation of R 2 . 

� � 

x 

↦→ 

y 

� � 

x/2 

y/3 

Find the image under this map of this ellipse. 

� 

x 

{ 

y 

� �� (x 2 /4) + (y 2 /9) = 1} 

� 1.27 Imagine a rope wound around the earth’s equator so that it fits snugly (suppose 

that the earth is a sphere). How much extra rope must be added to raise the 

circle to a constant six feet off the ground? 

� 1.28 Verify that this map h: R 3 → R 

� � � � � � 

x x 3 

y ↦→ y −1 

z z −1 

is linear. Generalize. 

=3x − y − z 

1.29 Show that every homomorphism from R 1 to R 1 acts via multiplication by a 

scalar. Conclude that every nontrivial linear transformation of R 1 is an isomorphism. 

Is that true for transformations of R 2 ? R n ? 

1.30 (a) Show that for any scalars a1,1,...,am,n this map h: R n → R m is a ho- 

momorphism. 

⎛ 

⎜ 

⎝ 

x1 

. 

xn 

⎞ 

⎟ 

⎠ ↦→ 

⎛ 

⎞ 

a1,1x1 + ···+ a1,nxn 

⎜ . ⎟ 

⎝ . ⎠ 

am,1x1 + ···+ am,nxn


(b) Show that for each i, thei-th derivative operator d i /dx i is a linear transformation 

of Pn. Conclude that for any scalars ck,... ,c0 this map is a linear 

transformation of that space. 

f ↦→ dk d 

f + ck−1 

dxk k−1 

d 

f + ···+ c1 f + c0f 

dxk−1 dx 

1.31 Lemma 1.15 shows that a sum of linear functions is linear and that a scalar 

multiple of a linear function is linear. Show also that a composition of linear 

functions is linear. 

� 1.32 Where f : V → W is linear, suppose that f(�v1) = �w1, ... , f(�vn) = �wn for 

some vectors �w1, ... , �wn from W . 

(a) If the set of �w ’s is independent, must the set of �v’s also be independent? 

(b) If the set of �v ’s is independent, must the set of �w ’s also be independent? 

(c) If the set of �w ’s spans W , must the set of �v ’s span V ? 

(d) If the set of �v ’s spans V , must the set of �w ’s span W ? 

1.33 Generalize Example 1.14 by proving that the matrix transpose map is linear. 

What is the domain and codomain? 

1.34 (a) Where �u, �v ∈ R n , the line segment connecting them is defined to be 

the set ℓ = {t · �u +(1−t) · �v � � t ∈ [0..1]}. Show that the image, under a homomorphism 

h, of the segment between �u and �v is the segment between h(�u) and 

h(�v). 

(b) A subset of R n is convex if, for any two points in that set, the line segment 

joining them lies entirely in that set. (The inside of a sphere is convex while the 

skin of a sphere is not.) Prove that linear maps from R n to R m preserve the 

property of set convexity. 

� 1.35 Let h: R n → R m be a homomorphism. 

(a) Show that the image under h of a line in R n is a (possibly degenerate) line 

in R n . 

(b) What happens to a k-dimensional linear surface? 

1.36 Prove that the restriction of a homomorphism to a subspace of its domain is 

another homomorphism. 

1.37 Assume that h: V → W is linear. 

(a) Show that the rangespace of this map {h(�v) � � �v ∈ V } is a subspace of the 

codomain W . 

(b) Show that the nullspace of this map {�v ∈ V � � h(�v) =�0W } is a subspace of 

the domain V . 

(c) Show that if U is a subspace of the domain V then its image {h(�u) � � �u ∈ U} 

is a subspace of the codomain W . This generalizes the first item. 

(d) Generalize the second item. 

1.38 Consider the set of isomorphisms from a vector space to itself. Is this a 

subspace of the space L(V,V ) of homomorphisms from the space to itself? 

1.39 Does Theorem 1.9 need that 〈 � β1,... , � βn〉 is a basis? That is, can we still get 

a well-defined and unique homomorphism if we drop either the condition that the 

set of � β’s be linearly independent, or the condition that it span the domain? 

1.40 Let V be a vector space and assume that the maps f1,f2 : V → R 1 are linear. 

(a) Define a map F : V → R 2 whose component functions are the given linear 

ones. 

�v ↦→ 

� � 

f1(�v) 

f2(�v)


Show that F is linear. 

(b) Does the converse hold—is any linear map from V to R 2 made up of two 

linear component maps to R 1 ? 

(c) Generalize. 

3.II.2 Rangespace and Nullspace 

The difference between homomorphisms and isomorphisms is that while both 

kinds of map preserve structure, homomorphisms needn’t be onto and needn’t 

be one-to-one. Put another way, homomorphisms are a more general kind of 

map; they are subject to fewer conditions than isomorphisms. In this subsection, 

we will look at what can happen with homomorphisms that the extra conditions 

rule out happening with isomorphisms. 

We first consider the effect of dropping the onto requirement. Of course, 

any function is onto some set, its range. The next result says that when the 

function is a homomorphism, then this set is a vector space. 

2.1 Lemma Under a homomorphism, the image of any subspace of the domain 

is a subspace of the codomain. In particular, the image of the entire space, the 

range of the homomorphism, is a subspace of the codomain. 

Proof. Let h: V → W be linear and let S be a subspace of the domain V . 

The image h(S) is nonempty because S is nonempty. Thus, to show that h(S) 

is a subspace of the codomain W , we need only show that it is closed under 

linear combinations of two vectors. If h(�s1) andh(�s2) are members of h(S) then 

c1 · h(�s1)+c2 · h(�s2) =h(c1 ·�s1)+h(c2 ·�s2) =h(c1 ·�s1 + c2 ·�s2) isalsoamember 

of h(S) because it is the image of c1 · �s1 + c2 · �s2 from S. QED 

2.2 Definition The rangespace of h: V → W is 

R(h) ={h(�v) � � �v ∈ V } 

sometimes denoted h(V ). The dimension of the rangespace is the map’s rank . 

(We shall soon see the connection between the rank of a map and the rank of a 

matrix.) 

2.3 Example Recall that the derivative map d/dx: P3 →P3 given by a0 + 

a1x + a2x 2 + a3x 3 ↦→ a1 +2a2x +3a3x 2 is linear. The rangespace R(d/dx) is 

the set of quadratic polynomials {r + sx + tx 2 � � r, s, t ∈ R}. Thus, the rank of 

this map is three. 

2.4 Example With this homomorphism h: M2×2 →P3 

� � 

a b 

c d 

↦→ (a + b +2d)+0x + cx 2 + cx 3


an image vector in the range can have any constant term, must have an x 

coefficient of zero, and must have the same coefficient of x 2 as of x 3 . That is, 

the rangespace is R(h) ={r +0x + sx 2 + sx 3 � � r, s ∈ R} and so the rank is two. 

The prior result shows that, in passing from the definition of isomorphism to 

the more general definition of homomorphism, omitting the ‘onto’ requirement 

doesn’t make an essential difference. Any homomorphism is onto its rangespace. 

However, omitting the ‘one-to-one’ condition does make a difference. A 

homomorphism may have many elements of the domain map to a single element 

in the range. The general picture is below. There is a homomorphism and its 

domain, codomain, and range. The homomorphism is many-to-one, and two 

elements of the range are shown that are each the image of more than one 

member of the domain. 

domain V 

. 

. 

codomain W 

� 

R(h) 

(Recall that for a map h: V → W , the set of elements of the domain that are 

mapped to �w in the codomain {�v ∈ V � � h(�v) = �w} is the inverse image of �w. It 

is denoted h −1 ( �w); this notation is used even if h has no inverse function, that 

is, even if h is not one-to-one.) 

2.5 Example Consider the projection π : R 3 → R 2 

⎛ ⎞ 

x 

⎝y⎠ 

z 

π 

↦−→ 

� � 

x 

y 

which is a homomorphism but is not one-to-one. Picturing R 2 as the xy-plane 

inside of R 3 allows us to see π(�v) as the “shadow” of �v in the plane. In these 

terms, the preservation of addition property says that 

�v1 above (x1,y1) plus �v2 above (x2,y2) equals �v1 + �v2 above (x1 + x2,y1 + y2). 

Briefly, the shadow of a sum equals the sum of the shadows. (Preservation of 

scalar multiplication has a similar interpretation.)


This description of the projection in terms of shadows is is memorable, but 

strictly speaking, R 2 isn’t equal to the xy-plane inside of R 3 (it is composed of 

two-tall vectors, not three-tall vectors). Separating the two spaces by sliding 

R 2 over to the right gives an instance of the general diagram above. 

�w2 

�w1 

�w1 + �w2 

The vectors that map to �w1 on the right have endpoints that lie in a vertical 

line on the left. One such vector is shown, in gray. Call any such member 

of the inverse image of �w1 a“�w1 vector”. Similarly, there is a vertical line of 

“ �w2 vectors”, and a vertical line of “ �w1 + �w2 vectors”. 

We are interested in π because it is a homomorphism. In terms of the 

picture, this means that the classes add; any �w1 vector plus any �w2 vector 

equals a �w1 + �w2 vector, simply because if π(�v1) = �w1 and π(�v2) = �w2 then 

π(�v1 + �v2) =π(�v1) +π(�v2) = �w1 + �w2. (A similar statement holds about the 

classes under scalar multiplication.) Thus, although the two spaces R 3 and R 2 

are not isomorphic, π describes a way in which they are alike: vectors in R 3 add 

like the associated vectors in R 2 —vectors add as their shadows add. 

2.6 Example A homomorphism can be used to express an analogy between 

spaces that is more subtle than the prior one. For instance, this map from R 2 

to R 1 is a homomorphism. 

� � 

x h 

↦−→ x + y 

y 

Fix two numbers a and b in the range R. Then the preservation of addition 

condition says this for two vectors �u and �v from the domain. 

� � 

� � 

u1 

v1 

if h( )=a and h( )=b then h( 

u2 

v2 

� u1 + v1 

u2 + v2 

� 

)=a + b 

As in the prior example, we illustrate by showing the class of vectors in the 

domain that map to a, the class of vectors that map to b, and the class of 

vectors that map to a + b. Vectors that map to a have components that add 

to a, so a vector is in the inverse image h −1 (a) if its endpoint lies on the line 

x+y = a. We can call these the “a vectors”. Similarly, we have the “b vectors”, 

etc. Now the addition preservation statement becomes this.


(u1,u2) 

(v1,v2) 

an a vector plus a b vector equals an a + b vector 

(u1 + v1,u2 + v2) 

Restated, if an a vector is added to a b vector then the result is mapped by h to 

the real number a+b. Briefly, the image of a sum is the sum of the images. Even 

more briefly, h(�u + �v) =h(�u)+h(�v). (The preservation of scalar multiplication 

condition has a similar restatement.) 

2.7 Example Inverse images can be structures other than lines. For the linear 

map h: R 3 → R 2 

⎛ ⎞ 

x 

⎝y⎠ 

↦→ 

z 

� � 

x 

x 

the inverse image sets are planes perpendicular to the x-axis. 

2.8 Remark We won’t describe how every homomorphism that we will use in 

this book is an analogy, both because the formal sense we make of “alike in this 

way ... ” is ‘a homomorphism exists such that ... ’, and because many vector 

spaces are hard to draw (e.g., a space of polynomials). Nonetheless, the idea 

that a homomorphism between two spaces expresses how the domain’s vectors 

fall into classes that act like the the range’s vectors, is a good way to view 

homomorphisms. 

We derive two insights from examples 2.5, 2.6, and2.7. 

First, in all three, each inverse image shown is a linear surface. In particular, 

the inverse image of the range’s zero vector is a line or plane through the origin— 

a subspace of the domain. The next result shows that this insight extends to 

any vector space, not just spaces of column vectors (which are the only spaces 

where the term ‘linear surface’ is defined). 

2.9 Lemma For any homomorphism, the inverse image of a subspace of the 

range is a subspace of the domain. In particular, the inverse image of the trivial 

subspace of the range is a subspace of the domain. 

Proof. Let h: V → W be a homomorphism and let S be a subspace of the 

range of h. Consider {�v ∈ V � � h(�v) ∈ S}, the inverse image of S. It is nonempty


because it contains �0V , as S contains �0W . To show that it is closed under 

combinations, let �v1 and �v2 be elements of the inverse image, so that h(�v1) and 

h(�v2) are members of S. Then c1�v1 + c2�v2 is also in the inverse image because 

under h it is sent h(c1�v1 +c2�v2) =c1h(�v1)+c2h(�v2) to a member of the subspace 

S. QED 

2.10 Definition The nullspace or kernel of a linear map h: V → W is 

N (h) ={�v ∈ V � � h(�v) =�0W } = h −1 (�0W ). 

The dimension of the nullspace is the map’s nullity. 

2.11 Example The map from Example 2.3 has this nullspace N (d/dx) = 

{a0 +0x +0x 2 +0x 3 � � a0 ∈ R}. 

2.12 Example The map from Example 2.4 has this nullspace. 

� � 

a b �� 

N (h) ={ 

a, b ∈ R} 

0 −(a + b)/2 

Now for the second insight from the above pictures. In Example 2.5, each 

of the vertical lines is squashed down to a single point—π, in passing from the 

domain to the range, takes all of these one-dimensional vertical lines and “zeroes 

them out”, leaving the range one dimension smaller than the domain. Similarly, 

in Example 2.6, the two-dimensional domain is mapped to a one-dimensional 

range by breaking the domain into lines (here, they are diagonal lines), and 

compressing each of those lines to a single member of the range. Finally, in 

Example 2.7, the domain breaks into planes which get “zeroed out”, and so the 

map starts with a three-dimensional domain but ends with a one-dimensional 

range—this map “subtracts” two from the dimension. (Notice that, in this 

third example, the codomain is two-dimensional but the range of the map is 

only one-dimensional, and it is the dimension of the range that is of interest.) 

2.13 Theorem A linear map’s rank plus its nullity equals the dimension of its 

domain. 

Proof. Let h: V → W be linear and let BN = 〈 � β1,... , � βk〉 be a basis for 

the nullspace. Extend that to a basis BV = 〈 � β1,..., � βk, � βk+1,..., � βn〉 for the 

entire domain. We shall show that BR = 〈h( � βk+1),...,h( � βn)〉 is a basis for the 

rangespace. Then counting the size of these bases gives the result. 

To see that BR is linearly independent, consider the equation ck+1h( � βk+1)+ 

···+ cnh( � βn) =�0W . This gives that h(ck+1 � βk+1 + ···+ cn � βn) =�0W and so 

ck+1 � βk+1 +···+cn � βn is in the nullspace of h. AsBN is a basis for this nullspace, 

there are scalars c1,...,ck ∈ R satisfying this relationship. 

c1 � β1 + ···+ ck � βk = ck+1 � βk+1 + ···+ cn � βn 

But BV is a basis for V so each scalar equals zero. Therefore BR is linearly 

independent.


To show that BR spans the rangespace, consider h(�v) ∈ R(h) and write �v 

as a linear combination �v = c1 � β1 + ···+ cn � βn of members of BV . This gives 

h(�v) =h(c1 � β1+···+cn � βn) =c1h( � β1)+···+ckh( � βk)+ck+1h( � βk+1)+···+cnh( � βn) 

and since � β1, ... , � βk are in the nullspace, we have that h(�v) =�0 +···+ �0 + 

ck+1h( � βk+1)+···+ cnh( � βn). Thus, h(�v) is a linear combination of members of 

BR, andsoBR spans the space. QED 

2.14 Example Where h: R 3 → R 4 is 

⎛ 

⎝ x 

⎞ 

y⎠ 

z 

h 

⎛ ⎞ 

x 

⎜ 

↦−→ ⎜0 

⎟ 

⎝y⎠ 

0 

we have that the rangespace and nullspace are 

⎛ ⎞ 

a 

⎜ 

R(h) ={ ⎜0⎟ 

� 

⎟ � 

⎝b⎠ 

a, b ∈ R} 

0 

and 

⎛ ⎞ 

0 

N (h) ={ ⎝0⎠ 

z 

� � z ∈ R} 

and so the rank of h is two while the nullity is one. 

2.15 Example If t: R → R is the linear transformation x ↦→ −4x, then the 

range is R(t) =R 1 , and so the rank of t is one and the nullity is zero. 

2.16 Corollary The rank of a linear map is less than or equal to the dimension 

of the domain. Equality holds if and only if the nullity of the map is zero. 

We know that there an isomorphism exists between two spaces if and only 

if their dimensions are equal. Here we see that for a homomorphism to exist, 

the dimension of the range must be less than or equal to the dimension of the 

domain. For instance, there is no homomorphism from R 2 onto R 3 —there are 

many homomorphisms from R 2 into R 3 , but none has a range that is all of 

three-space. 

The rangespace of a linear map can be of dimension strictly less than the 

dimension of the domain (an example is that the derivative transformation on 

P3 has a domain of dimension four but a range of dimension three). Thus, under 

a homomorphism, linearly independent sets in the domain may map to linearly 

dependent sets in the range (for instance, the derivative sends {1,x,x 2 ,x 3 } to 

{0, 1, 2x, 3x 2 }). That is, under a homomorphism, independence may be lost. In 

contrast, dependence is preserved. 

2.17 Lemma Under a linear map, the image of a linearly dependent set is 

linearly dependent. 

Proof. Suppose that c1�v1 + ··· + cn�vn = �0V , with some ci nonzero. Then, 

because h(c1�v1 +···+cn�vn) =c1h(�v1)+···+cnh(�vn) and because h(�0V )=�0W , 

we have that c1h(�v1)+···+ cnh(�vn) =�0W with some nonzero ci. QED


When is independence not lost? One obvious sufficient condition is when 

the homomorphism is an isomorphism (this condition is also necessary; see 

Exercise 34.) We finish our comparison of homomorphisms and isomorphisms by 

observing that a one-to-one homomorphism is an isomorphism from its domain 

onto its range. 

2.18 Definition A linear map that is one-to-one is nonsingular. 

(In the next section we will see the connection between this use of ‘nonsingular’ 

for maps and its familiar use for matrices.) 

2.19 Example This nonsingular homomorphism ι: R 2 → R 3 

⎛ 

� � 

x ι 

↦−→ ⎝ 

y 

x 

⎞ 

y⎠ 

0 

gives the obvious correspondence between R 2 and the xy-plane inside of R 3 . 

We will close this section by adapting some results about isomorphisms to 

this setting. 

2.20 Theorem In an n-dimensional vector space V , then these 

(1) h is nonsingular, that is, one-to-one 

(2) h has a linear inverse 

(3) N (h) ={�0 }, that is, nullity(h) =0 

(4) rank(h) =n 

(5) if 〈 � β1,..., � βn〉 is a basis for V then 〈h( � β1),...,h( � βn)〉 is a basis for R(h) 

are equivalent statements about a linear map h: V → W . 

Proof. We will first show that (1) ⇐⇒ (2). We will then show that (1) =⇒ 

(3) =⇒ (4) =⇒ (5) =⇒ (2). 

For (1) =⇒ (2), suppose that the linear map h is one-to-one, and so has an 

inverse. The domain of that inverse is the range of h and so a linear combination 

of two members of that domain has the form c1h(�v1) +c2h(�v2). On that 

combination, the inverse h −1 gives this. 

h −1 (c1h(�v1)+c2h(�v2)) = h −1 (h(c1�v1 + c2�v2)) 

= h −1 ◦ h (c1�v1 + c2�v2) 

= c1�v1 + c2�v2 

= c1h −1 ◦ h (�v1)c2h −1 ◦ h (�v2) 

= c1 · h −1 (h(�v1)) + c2 · h −1 (h(�v2)) 

Thus the inverse of a one-to-one linear map is automatically linear. But this also 

gives the (1) =⇒ (2) implication, because the inverse itself must be one-to-one. 

Of the remaining implications, (1) =⇒ (3) holds because any homomorphism 

maps �0V to �0W , but a one-to-one map sends at most one member of V 

to �0W .


Next, (3) =⇒ (4) is true since rank plus nullity equals the dimension of the 

domain. 

For (4) =⇒ (5), to show that 〈h( � β1),...,h( � βn)〉 is a basis for the rangespace 

we need only show that it is a spanning set, because by assumption the range 

has dimension n. Consider h(�v) ∈ R(h). Expressing �v as a linear combination 

of basis elements produces h(�v) =h(c1 � β1 + c2 � β2 + ···+ cn � βn), which gives that 

h(�v) =c1h( � β1)+···+ cnh( � βn), as desired. 

Finally, for the (5) =⇒ (2) implication, assume that 〈 � β1,..., � βn〉 is a basis 

for V so that 〈h( � β1),...,h( � βn)〉 is a basis for R(h). Then every �w ∈ R(h) athe 

unique representation �w = c1h( � β1)+···+ cnh( � βn). Define a map from R(h) to 

V by 

�w ↦→ c1 � β1 + c2 � β2 + ···+ cn � βn 

(uniqueness of the representation makes this well-defined). Checking that it is 

linear and that it is the inverse of h are easy. QED 

We’ve now seen that a linear map shows how the structure of the domain is 

like that of the range. Such a map can be thought to organize the domain space 

into inverse images of points in the range. In the special case that the map is 

one-to-one, each inverse image is a single point and the map is an isomorphism 

between the domain and the range. 

Exercises 

� 2.21 Let h: P3 →P4 be given by p(x) ↦→ x · p(x). Which of these are in the 

nullspace? Which are in the rangespace? 

(a) x 3 

(b) 0 (c) 7 (d) 12x − 0.5x 3 

(e) 1+3x 2 − x 3 

� 2.22 Find the nullspace, nullity, rangespace, and rank of each map. 

(a) h: R 2 →P3 given by 

� � 

a 

↦→ a + ax + ax 

b 

2 

(b) h: M2×2 → R given by 

� � 

a b 

↦→ a + d 

c d 

(c) h: M2×2 →P2 given by 

� � 

a b 

↦→ a + b + c + dx 

c d 

2 

(d) the zero map Z : R 3 → R 4 

� 2.23 Find the nullity of each map.


(a) h: R 5 → R 8 of rank five (b) h: P3 →P3 of rank one 

(c) h: R 6 → R 3 ,anontomap (d) h: M3×3 →M3×3, onto 

� 2.24 What is the nullspace of the differentiation transformation d/dx: Pn →Pn? 

What is the nullspace of the second derivative, as a transformation of Pn? The 

k-th derivative? 

2.25 Example 2.5 restates the first condition in the definition of homomorphism as 

‘the shadow of a sum is the sum of the shadows’. Restate the second condition in 

the same style. 

2.26 For the homomorphism h: P3 →P3 given by h(a0 + a1x + a2x 2 + a3x 3 )= 

a0 +(a0 + a1)x +(a2 + a3)x 3 find these. 

(a) N (h) (b) h −1 (2 − x 3 ) (c) h −1 (1 + x 2 ) 

� 2.27 For the map f : R 2 → R given by 

� � 

x 

f( )=2x + y 

y 

sketch these inverse image sets: f −1 (−3), f −1 (0), and f −1 (1). 

� 2.28 Each of these transformations of P3 is nonsingular. Find the inverse function 

of each. 

(a) a0 + a1x + a2x 2 + a3x 3 ↦→ a0 + a1x +2a2x 2 +3a3x 3 

(b) a0 + a1x + a2x 2 + a3x 3 ↦→ a0 + a2x + a1x 2 + a3x 3 

(c) a0 + a1x + a2x 2 + a3x 3 ↦→ a1 + a2x + a3x 2 + a0x 3 

(d) a0+a1x+a2x 2 +a3x 3 ↦→ a0+(a0+a1)x+(a0+a1+a2)x 2 +(a0+a1+a2+a3)x 3 

2.29 Describe the nullspace and rangespace of a transformation given by �v ↦→ 2�v. 

2.30 List all pairs (rank(h), nullity(h)) that are possible for linear maps from R 5 

to R 3 . 

2.31 Does the differentiation map d/dx: Pn →Pn have an inverse? 

� 2.32 Find the nullity of the map h: Pn → R given by 

a0 + a1x + ···+ anx n � x=1 

↦→ a0 + a1x + ···+ anx n dx. 

x=0 

2.33 (a) Prove that a homomorphism is onto if and only if its rank equals the 

dimension of its codomain. 

(b) Conclude that a homomorphism between vector spaces with the same dimension 

is one-to-one if and only if it is onto. 

2.34 Show that a linear map is nonsingular if and only if it preserves linear independence. 

2.35 Corollary 2.16 says that for there to be an onto homomorphism from a vector 

space V to a vector space W , it is necessary that the dimension of W be less 

than or equal to the dimension of V . Prove that this condition is also sufficient; 

use Theorem 1.9 to show that if the dimension of W is less than or equal to the 

dimension of V , then there is a homomorphism from V to W that is onto. 

2.36 Let h: V → R be a homomorphism, but not the zero homomorphism. Prove 

that if 〈 � β1,..., � βn〉 is a basis for the nullspace and if �v ∈ V is not in the nullspace 

then 〈�v, � β1,..., � βn〉 is a basis for the entire domain V . 

� 2.37 Recall that the nullspace is a subset of the domain and the rangespace is a 

subset of the codomain. Are they necessarily distinct? Is there a homomorphism 

that has a nontrivial intersection of its nullspace and its rangespace?


2.38 Prove that the image of a span equals the span of the images. That is, where 

h: V → W is linear, prove that if S is a subset of V then h([S]) equals [h(S)]. This 

generalizes Lemma 2.1 since it shows that if U is any subspace of V then its image 

{h(�u) � � �u ∈ U} is a subspace of W , because the span of the set U is U. 

� 2.39 (a) Prove that for any linear map h: V → W and any �w ∈ W , the set 

h −1 ( �w) has the form 

{�v + �n � � �n ∈ N (h)} 

for �v ∈ V with h(�v) = �w (if h is not onto then this set may be empty). Such a 

set is a coset of N (h) and is denoted �v + N (h). 

(b) Consider the map t: R 2 → R 2 � � 

given 

� 

by 

� 

x t ax + by 

↦−→ 

y cx + dy 

for some scalars a, b, c, andd. Provethattis linear. 

(c) Conclude from the prior two items that for any linear system of the form 

ax + by = e 

cx + dy = f 

the solution set can be written (the vectors are members of R 2 ) 

{�p + �h � � �h satisfies the associated homogeneous system} 

where �p is a particular solution of that linear system (if there is no particular 

solution then the above set is empty). 

(d) Show that this map h: R n → R m ⎛ ⎞ ⎛ 

is linear 

⎞ 

⎜ 

⎝ 

x1 

. 

⎟ 

⎠ ↦→ 

⎜ 

⎝ 

a1,1x1 + ···+ a1,nxn 

xn am,1x1 + ···+ am,nxn 

for any scalars a1,1, ... , am,n. Extend the conclusion made in the prior item. 

(e) Show that the k-th derivative map is a linear transformation of Pn for each 

k. Prove that this map is a linear transformation of that space 

f ↦→ dk d 

f + ck−1 

dxk k−1 

d 

f + ···+ c1 f + c0f 

dxk−1 dx 

for any scalars ck, ... , c0. Draw a conclusion as above. 

2.40 Prove that for any transformation t: V → V that is rank one, the map given 

by composing the operator with itself t ◦ t: V → V satisfies t ◦ t = r · t for some 

real number r. 

2.41 Show that for any space V of dimension n, thedual space 

L(V,R) ={h: V → R � � h is linear} 

is isomorphic to R n . It is often denoted V ∗ . Conclude that V ∗ ∼ = V . 

2.42 Show that any linear map is the sum of maps of rank one. 

2.43 Is ‘is homomorphic to’ an equivalence relation? (Hint: the difficulty is to 

decide on an appropriate meaning for the quoted phrase.) 

2.44 Show that the rangespaces and nullspaces of powers of linear maps t: V → V 

form descending 

V ⊇ R(t) ⊇ R(t 2 ) ⊇ ... 

and ascending 

{�0} ⊆N (t) ⊆ N (t 2 ) ⊆ ... 

chains. Also show that if k is such that R(t k ) = R(t k+1 ) then all following 

rangespaces are equal: R(t k )=R(t k+1 )=R(t k+2 ) .... Similarly, if N (t k )= 

N (t k+1 )thenN (t k )=N (t k+1 )=N (t k+2 )=.... 

. 

⎟ 

⎠


3.III Computing Linear Maps 

The prior section shows that a linear map is determined by its action on a basis. 

In fact, the equation 

h(�v) =h(c1 · � β1 + ···+ cn · � βn) =c1 · h( � β1)+···+ cn · h( � βn) 

shows that, if we know the value of the map on the vectors in a basis, then we 

can compute the value of the map on any vector �v at all just by finding the c’s 

to express �v with respect to the basis. 

This section gives the scheme that computes, from the representation of a 

vector in the domain Rep B(�v), the representation of that vector’s image in the 

codomain Rep D(h(�v)), using the representations of h( � β1), ... , h( � βn). 

3.III.1 Representing Linear Maps with Matrices 

1.1 Example Consider a map h with domain R 2 and codomain R 3 (fixing 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

� � � � 

1 0 1 

2 1 

B = 〈 , 〉 and D = 〈 ⎝0⎠ 

, ⎝−2⎠ 

, ⎝0⎠〉 

0 4 

0 0 1 

as the bases for these spaces) that is determined by this action on the vectors 

in the domain’s basis. 

⎛ ⎞ 

� � 1 

2 h 

↦−→ ⎝1⎠ 

0 

1 

⎛ ⎞ 

� � 1 

1 h 

↦−→ ⎝2⎠ 

4 

0 

To compute the action of this map on any vector at all from the domain, we 

first express h( � β1) andh( � β2) with respect to the codomain’s basis: 

and 

⎛ 

⎝ 1 

⎞ ⎛ 

1⎠ 

=0⎝ 

1 

1 

⎞ 

0⎠ 

− 

0 

1 

⎛ 

⎝ 

2 

0 

⎞ ⎛ 

−2⎠ 

+1⎝ 

0 

1 

⎞ 

0⎠ 

so RepD(h( 1 

� ⎛ 

β1)) = ⎝ 0 

⎞ 

−1/2⎠ 

1 

⎛ 

⎝ 1 

⎞ ⎛ 

2⎠ 

=1⎝ 

0 

1 

⎞ ⎛ 

0⎠ 

− 1 ⎝ 

0 

0 

⎞ ⎛ 

−2⎠ 

+0⎝ 

0 

1 

⎞ 

0⎠ 

so RepD(h( 1 

� ⎛ 

β2)) = ⎝ 1 

⎞ 

−1⎠ 

0 

D 

D

Section III. Computing Linear Maps 195 

(these are easy to check). Then, as described in the preamble, for any member 

�v of the domain, we can express the image h(�v) in terms of the h( � β)’s. 

� � � � 

2 1 

h(�v) =h(c1 · + c2 · ) 

0 4 

� � � � 

2 

1 

= c1 · h( )+c2 · h( ) 

0 

4 

⎛ 

= c1 · (0 ⎝ 1 

⎞ 

0⎠− 

0 

1 

⎛ 

⎝ 

2 

0 

⎞ ⎛ 

−2⎠+1⎝ 

0 

1 

⎞ ⎛ 

0⎠)+c2 

· (1 ⎝ 

1 

1 

⎞ ⎛ 

0⎠− 

1 ⎝ 

0 

0 

⎞ ⎛ 

−2⎠+0⎝ 

0 

1 

⎞ 

0⎠) 

1 

⎛ 

=(0c1 +1c2) · ⎝ 1 

⎞ 

0⎠ 

+(− 

0 

1 

2 c1 

⎛ 

− 1c2) · ⎝ 0 

⎞ 

⎛ 

−2⎠ 

+(1c1 +0c2) · ⎝ 

0 

1 

⎞ 

0⎠ 

1 

Thus, 

For instance, 

⎛ 

⎞ 

� � 

0c1 +1c2 

c1 

with RepB(�v) = then Rep 

c2 

D( h(�v))= ⎝−(1/2)c1 

− 1c2⎠. 

1c1 +0c2 

⎛ ⎞ 

� � � � 

� � 2 

4 1 

4 

with RepB( )= then Rep 

8 2 

D( h( ))= ⎝−5/2⎠. 

8 

B 

1 

This is a formula that computes how h acts on any argument. 

We will express computations like the one above with a matrix notation. 

⎛ 

0 

⎝−1/2 

1 

⎞ 

1 

−1⎠ 

0 

⎛ 

⎞ 

� � 0c1 +1c2 

c1 

= ⎝(−1/2)c1 

− 1c2⎠ 

c2 B 1c1 +0c2 

B,D 

In the middle is the argument �v to the map, represented with respect to the 

domain’s basis B by a column vector with components c1 and c2. On the right 

is the value h(�v) of the map on that argument, represented with respect to the 

codomain’s basis D by a column vector with components 0c1 +1c2, etc. The 

matrix on the left is the new thing. It consists of the coefficients from the vector 

on the right, 0 and 1 from the first row, −1/2 and−1 from the second row, and 

1 and 0 from the third row. 

This notation simply breaks the parts from the right, the coefficients and the 

c’s, out separately on the left, into a vector that represents the map’s argument 

and a matrix that we will take to represent the map itself. 

D


1.2 Definition Suppose that V and W are vector spaces of dimensions n and 

m with bases B and D, and that h: V → W is a linear map. If 

RepD(h( � ⎛ ⎞ 

h1,1 

⎜ h2,1 ⎟ 

⎜ ⎟ 

β1)) = ⎜ . ⎟ , ... ,Rep 

⎝ . ⎠ 

D(h( � ⎛ ⎞ 

h1,n 

⎜ h2,n ⎟ 

⎜ ⎟ 

βn)) = ⎜ . ⎟ 

⎝ . ⎠ 

then 

hm,1 

⎛ 

⎜ 

RepB,D(h) = ⎜ 

⎝ 

D 

h1,1 h1,2 ... h1,n 

h2,1 h2,2 ... h2,n 

. 

hm,1 hm,2 ... hm,n 

is the matrix representation of h with respect to B,D. 

⎞ 

⎟ 

⎠ 

hm,n 

Briefly, the vectors representing the h( � β)’s are adjoined to make the matrix 

representing the map. 

⎛ 

. 

⎜ 

. 

. 

RepB,D(h) = ⎜ 

⎝ RepD( h( � β1)) ··· RepD( h( � ⎞ 

⎟ 

βn)) ⎟ 

⎠ 

. 

. 

. 

. 

Observe that the number of columns of the matrix is the dimension of the 

domain of the map, and the number of rows is the dimension of the codomain. 

1.3 Example If h: R3 →P1 is given by 

⎛ ⎞ 

a1 

⎝ ⎠ h 

↦−→ (2a1 + a2)+(−a3)x 

then where 

a2 

a3 

B,D 

⎛ ⎞ 

0 

⎛ ⎞ 

0 

⎛ ⎞ 

2 

B = 〈 ⎝0⎠ 

, ⎝2⎠ 

, ⎝0⎠〉 

and D = 〈1+x, −1+x〉 

1 0 0 

the action of h on B is given by 

⎛ 

⎝ 0 

⎞ 

0⎠ 

1 

h 

↦−→ −x 

and a simple calculation gives 

� � 

−1/2 

RepD(−x) = 

−1/2 

D 

⎛ 

⎝ 0 

⎞ 

2⎠ 

0 

h 

↦−→ 2 

Rep D(2) = 

� � 

1 

−1 

D 

⎛ 

⎝ 2 

⎞ 

0⎠ 

0 

h 

↦−→ 4 

Rep D(4) = 

D 

� 2 

−2 

� 

D


showing that this is the matrix representing h with respect to the bases. 

Rep B,D(h) = 

� 

−1/2 1 

� 

2 

−1/2 −1 −2 

B,D 

We will use lower case letters for a map, upper case for the matrix, and 

lower case again for the entries of the matrix. Thus for the map h, the matrix 

representing it is H, with entries hi,j. 

1.4 Theorem Assume that V and W are vector spaces of dimensions m and 

n with bases B and D, and that h: V → W is a linear map. If h is represented 

by 

⎛ 

⎞ 

h1,1 h1,2 ... h1,n 

⎜ h2,1 h2,2 ⎜ 

... h2,n ⎟ 

RepB,D(h) = ⎜ . 

⎝ . 

⎟ 

. 

⎠ 

hm,1 hm,2 ... hm,n 

and �v ∈ V is represented by 

⎛ ⎞ 

c1 

⎜c2⎟ 

⎜ ⎟ 

RepB(�v) = ⎜ 

⎝ . 

⎟ 

. ⎠ 

cn 

B 

B,D 

then the representation of the image of �v is this. 

⎛ 

⎞ 

h1,1c1 + h1,2c2 + ···+ h1,ncn 

⎜ h2,1c1 ⎜ + h2,2c2 + ···+ h2,ncn ⎟ 

RepD( h(�v))= ⎜ 

⎝ 

. 

⎟ 

. 

⎠ 

hm,1c1 + hm,2c2 + ···+ hm,ncn 

Proof. Exercise 28. QED 

We will think of the matrix Rep B,D(h) and the vector Rep B(�v) as combining 

to make the vector Rep D(h(�v)). 

1.5 Definition The matrix-vector product of a m×n matrix and a n×1 vector 

is this. 

⎛ 

⎞ ⎛ 

⎞ 

a1,1 a1,2 ... a1,n ⎛ ⎞ a1,1c1 + a1,2c2 + ···+ a1,ncn 

c1 

⎜ a2,1 a2,2 ⎜ 

... a2,n ⎟ ⎜ a2,1c1 ⎟ ⎜ . 

⎜ . 

⎝ . 

⎟ ⎝ . 

⎟ ⎜ + a2,2c2 + ···+ a2,ncn ⎟ 

. 

⎠ . ⎠ = ⎜ 

. 

⎝ 

. 

⎟ 

. 

⎠ 

cn 

am,1 am,2 ... am,n 

am,1c1 + am,2c2 + ···+ am,ncn 

D


The point of Definition 1.2 is to generalize Example 1.1, that is, the point 

of the definition is Theorem 1.4, that the matrix describes how to get from 

the representation of a domain vector with respect to the domain’s basis to 

the representation of its image in the codomain with respect to the codomain’s 

basis. With Definition 1.5, we can restate this as: application of a linear map is 

represented by the matrix-vector product of the map’s representative and the 

vector’s representative. 

1.6 Example With the matrix from Example 1.3 we can calculate where that 

map sends this vector. 

⎛ 

�v = ⎝ 4 

⎞ 

1⎠ 

0 

This vector is represented, with respect to the domain basis B, by 

⎛ ⎞ 

0 

RepB(�v) = ⎝1/2⎠ 

2 

and so this is the representation of the value h(�v) with respect to the codomain 

basis D. 

⎛ ⎞ 

� � 0 

−1/2 1 2 

RepD(h(�v)) = 

⎝1/2⎠ 

−1/2 −1 −2 

B,D 2 

B 

� � � � 

(−1/2) · 0+1· (1/2) + 2 · 2 9/2 

= 

= 

(−1/2) · 0 − 1 · (1/2) − 2 · 2 −9/2 

D 

D 

To find h(�v) itself, not its representation, take (9/2)(1 + x) − (9/2)(−1+x) =9. 

1.7 Example Let π : R 3 → R 2 be projection onto the xy-plane. To give a 

matrix representing this map, we first fix bases. 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

1 1 −1 

� � � � 

B = 〈 ⎝0⎠ 

, ⎝1⎠ 

, ⎝ 0 ⎠〉 

2 1 

D = 〈 , 〉 

1 1 

0 0 1 

For each vector in the domain’s basis, we find its image under the map. 

⎛ 

⎝ 1 

⎞ 

0⎠ 

0 

π 

� � 

1 

↦−→ 

0 

⎛ 

⎝ 1 

⎞ 

1⎠ 

0 

π 

� � 

1 

↦−→ 

1 

⎛ 

⎝ −1 

⎞ 

0 ⎠ 

1 

π 

� � 

−1 

↦−→ 

0 

Then we find the representation of each image with respect to the codomain’s 

basis 

� � � � � � � � � � � � 

1 1 

1 0 

−1 −1 

RepD( )= Rep 

0 −1 

D( )= Rep 

1 1 

D( )= 

0 1 

B


(these are easily checked). Finally, adjoining these representations gives the 

matrix representing π with respect to B,D. 

� 

1 

RepB,D(π) = 

−1 

0 

1 

� 

−1 

1 

B,D 

We can illustrate Theorem 1.4 by computing the matrix-vector product representing 

the following statement about the projection map. 

⎛ 

π( ⎝ 2 

⎞ 

2⎠) 

= 

1 

� � 

2 

2 

Representing this vector from the domain with respect to the domain’s basis 

⎛ ⎞ ⎛ ⎞ 

2 1 

RepB( ⎝2⎠) 

= ⎝2⎠ 

1 1 

gives this matrix-vector product. 

⎛ ⎞ 

2 � 

RepD( π( ⎝1⎠))= 

1 

−1 

1 

0 

1 

⎛ ⎞ 

� 1 

−1 

⎝2⎠ 

1 

B,D 1 

B 

B 

= 

� � 

0 

2 

D 

Expanding this representation into a linear combination of vectors from D 

0 · 

� � 

2 

+2· 

1 

� � 

1 

= 

1 

� � 

2 

2 

checks that the map’s action is indeed reflected in the operation of the matrix. 

(We will sometimes compress these three displayed equations into one 

⎛ ⎞ ⎛ ⎞ 

2 1 

⎝2⎠ 

= ⎝2⎠ 

1 1 

in the course of a calculation.) 

B 

h 

↦−→ H 

� � 

0 

= 

2 

D 

� � 

2 

2 

We now have two ways to compute the effect of projection, the straightforward 

formula that drops each three-tall vector’s third component to make 

a two-tall vector, and the above formula that uses representations and matrixvector 

multiplication. Compared to the first way, the second way might seem 

complicated. However, it has advantages. The next example shows that giving 

a formula for some maps is simplified by this new scheme. 

1.8 Example To represent a rotation map tθ : R 2 → R 2 that turns all vectors 

in the plane counterclockwise through an angle θ


�v 

t θ 

↦−→ 

tθ(�v) 

we start by fixing bases. Using E2 both as a domain basis and as a codomain 

basis is natural, Now, we find the image under the map of each vector in the 

domain’s basis. 

� � 

1 tθ 

↦−→ 

0 

� cos θ 

sin θ 

� � 0 

1 

� 

tθ 

↦−→ 

� � 

− sin θ 

cos θ 

Then we represent these images with respect to the codomain’s basis. Because 

this basis is E2, vectors are represented by themselves. Finally, adjoining the 

representations gives the matrix representing the map. 

� � 

cos θ − sin θ 

RepE2,E2 (tθ) = 

sin θ cos θ 

The advantage of this scheme is that just by knowing how to represent the image 

of the two basis vectors, we get a formula that tells us the image of any vector 

at all; here a vector rotated by θ = π/6. 

� � 

3 tπ/6 

↦−→ 

−2 

�√ 

3/2 −1/2 

1/2 √ �� 

3 

3/2 −2 

≈ 

� � 

3.598 

−0.232 

(Again, we are using the fact that, with respect to E2, vectors represent themselves.) 

We have already seen the addition and scalar multiplication operations of 

matrices and the dot product operation of vectors. Matrix-vector multiplication 

is a new operation in the arithmetic of vectors and matrices. Nothing in Definition 

1.5 requires us to view it in terms of representations. We can get some 

insight into this operation by turning away from what is being represented, and 

instead focusing on how the entries combine. 

1.9 Example In the definition the width of the matrix equals the height of 

the vector. Hence, the first product below is defined while the second is not. 

� � 

1 0 0 

4 3 1 

⎛ 

⎝ 1 

⎞ 

� � � �� 

0⎠ 

1 1 0 0 1 

= 

6 4 3 1 0 

2 

One reason that this product is not defined is purely formal: the definition requires 

that the sizes match, and these sizes don’t match. (Behind the formality, 

though, we have a reason why it is left undefined—the matrix represents a map 

with a three-dimensional domain while the vector represents a member of a 

two-dimensional space.)


A good way to view a matrix-vector product is as the dot products of the 

rows of the matrix with the column vector. 

⎛ 

⎞ ⎛ ⎞ 

. 

c1 

. 

⎜ . 

⎟ ⎜c2⎟ 

⎜ai,1 

ai,2 ⎝ 

... ai,n⎟ 

⎜ ⎟ 

⎠ ⎜ . 

. 

⎝ . 

⎟ 

. ⎠ 

. 

= 

⎛ 

⎞ 

. 

⎜ 

. 

⎟ 

⎜ai,1c1 

⎝ 

+ ai,2c2 + ...+ ai,ncn⎟ 

⎠ 

. 

cn 

Looked at in this row-by-row way, this new operation generalizes dot product. 

Matrix-vector product can also be viewed column-by-column. 

⎛ 

⎜ 

⎝ 

h1,1 

h2,1 

h1,2 

h2,2 

. 

... 

... 

h1,n 

h2,n 

⎞ ⎛ ⎞ 

c1 

⎟ ⎜c2⎟ 

⎟ ⎜ ⎟ 

⎟ ⎜ . ⎟ 

⎠ ⎝ . ⎠ 

hm,1 hm,2 ... hm,n cn 

= 

⎛ 

h1,1c1 + h1,2c2 + ···+ h1,ncn 

⎜ h2,1c1 ⎜ + h2,2c2 + ···+ h2,ncn 

⎜ 

. 

⎝ 

. 

⎞ 

⎟ 

⎠ 

hm,1c1 + hm,2c2 + ···+ hm,ncn 

⎛ ⎞ ⎛ ⎞ 

h1,1 

h1,n 

⎜ h2,1 ⎟ ⎜ h2,n ⎟ 

⎜ ⎟ ⎜ ⎟ 

= c1 ⎜ . ⎟ + ···+ cn ⎜ . ⎟ 

⎝ . ⎠ ⎝ . ⎠ 

1.10 Example 

� 

1 

2 

0 

0 

� 

−1 

3 

⎛ ⎞ 

2 

⎝−1 

1 

⎠ =2 

� � 

1 

− 1 

2 

hm,1 

� � 

0 

+1 

0 

� � 

−1 

= 

3 

hm,n 

� � 

1 

7 

The result has the columns of the matrix weighted by the entries of the 

vector. This way of looking at it brings us back to the objective stated at the 

start of this section, to compute h(c1 � β1 + ···+ cn � βn) asc1h( � β1)+···+ cnh( � βn). 

We began this section by noting that the equality of these two enables us 

to compute the action of h on any argument knowing only h( � β1), ... , h( � βn). 

We have developed this into a scheme to compute the action of the map by 

taking the matrix-vector product of the matrix representing the map and the 

vector representing the argument. In this way, any linear map is represented 

with respect to some bases by a matrix. In the next subsection, we will show 

the converse, that any matrix represents a linear map. 

Exercises 

� 1.11 Multiply, where it is defined, the matrix 

� � 

1 3 1 

0 −1 2 

1 1 0 

by each vector. 

� � 

2 

(a) 1 (b) 

0 

� � 

−2 

−2 

� � 

0 

(c) 0 

0 

1.12 Perform, if possible, each matrix-vector multiplication.


(a) 

� 

2 

3 

�� 

1 4 

−1/2 2 

(b) 

� 

1 

−2 

1 

1 

� 

0 

0 

� � 

1 

3 

1 

(c) 

� 1.13 Solve this matrix equation. 

� 

2 1 

�� 

1 x 

� � 

8 

0 1 3 y = 4 

1 −1 2 z 4 

� 1.14 For a homomorphism from P2 to P3 that sends 

1 ↦→ 1+x, x ↦→ 1+2x, and x 2 ↦→ x − x 3 

where does 1 − 3x +2x 2 go? 

� 1.15 Assume that h: R 2 → R 3 is determined by this action. 

� � 

1 

0 

↦→ 

� � 

2 � � 

0 

2 

1 

0 

↦→ 

� � 

0 

1 

−1 

� � 

1 1 

−2 1 

� � 

1 

3 

1 

Using the standard bases, find 

(a) the matrix representing this map; 

(b) a general formula for h(�v). 

� 1.16 Let d/dx: P3 →P3 be the derivative transformation. 

(a) Represent d/dx with respect to B,B where B = 〈1,x,x 2 ,x 3 〉. 

(b) Represent d/dx with respect to B,D where D = 〈1, 2x, 3x 2 , 4x 3 〉. 

� 1.17 Represent each linear map with respect to each pair of bases. 

(a) d/dx: Pn →Pn with respect to B,B where B = 〈1,x,...,x n 〉,givenby 

a0 + a1x + a2x 2 + ···+ anx n ↦→ a1 +2a2x + ···+ nanx n−1 

(b) � : Pn →Pn+1 with respect to Bn,Bn+1 where Bi = 〈1,x,...,x i 〉,givenby 

a0 + a1x + a2x 2 + ···+ anx n ↦→ a0x + a1 

2 x2 + ···+ an 

n +1 xn+1 

(c) � 1 

0 : Pn → R with respect to B,E1 where B = 〈1,x,...,xn 〉 and E1 = 〈1〉, 

given by 

a0 + a1x + a2x 2 + ···+ anx n ↦→ a0 + a1 

2 

+ ···+ an 

n +1 

(d) eval3 : Pn → R with respect to B,E1 where B = 〈1,x,...,x n 〉 and E1 = 〈1〉, 

given by 

a0 + a1x + a2x 2 + ···+ anx n ↦→ a0 + a1 · 3+a2 · 3 2 + ···+ an · 3 n 

(e) slide−1 : Pn →Pn with respect to B,B where B = 〈1,x,... ,x n 〉,givenby 

a0 + a1x + a2x 2 + ···+ anx n ↦→ a0 + a1 · (x +1)+···+ an · (x +1) n 

1.18 Represent the identity map on any nontrivial space with respect to B,B, 

where B is any basis. 

1.19 Represent, with respect to the natural basis, the transpose transformation on 

the space M2×2 of 2×2 matrices. 

1.20 Assume that B = 〈 � β1, � β2, � β3, � β4〉 is a basis for a vector space. Represent with 

respect to B,B the transformation that is determined by each. 

(a) � β1 ↦→ � β2, � β2 ↦→ � β3, � β3 ↦→ � β4, � β4 ↦→ �0 

(b) � β1 ↦→ � β2, � β2 ↦→ �0, � β3 ↦→ � β4, � β4 ↦→ �0


(c) � β1 ↦→ � β2, � β2 ↦→ � β3, � β3 ↦→ �0, � β4 ↦→ �0 

1.21 Example 1.8 shows how to represent the rotation transformation of the plane 

with respect to the standard basis. Express these other transformations also with 

respect to the standard basis. 

(a) the dilation map ds, which multiplies all vectors by the same scalar s 

(b) the reflection map fℓ, which reflects all all vectors across a line ℓ through the 

origin 

� 1.22 Consider a linear transformation of R 2 determined by these two. 

� � � � � � � � 

1 2 1 −1 

↦→ 

↦→ 

1 0 0 0 

(a) Represent this transformation with respect to the standard bases. 

(b) Where does the transformation send this vector? 

� � 

0 

5 

(c) Represent this transformation with respect to these bases. 

� � � � � � � � 

1 1 

2 −1 

B = 〈 , 〉 D = 〈 , 〉 

−1 1 

2 1 

(d) Using B from the prior item, represent the transformation with respect to 

B,B. 

1.23 Suppose that h: V → W is nonsingular so that by Theorem 2.20, for any 

basis B = 〈 � β1,..., � βn〉 ⊂V the image h(B) =〈h( � β1),...,h( � βn)〉 is a basis for 

W . 

(a) Represent the map h with respect to B,h(B). 

(b) For a member �v of the domain, where the representation of �v has components 

c1, ... , cn, represent the image vector h(�v) with respect to the image basis h(B). 

1.24 Give a formula for the product of a matrix and �ei, the column vector that is 

all zeroes except for a single one in the i-th position. 

� 1.25 For each vector space of functions of one real variable, represent the derivative 

transformation with respect to B,B. 

(a) {a cos x + b sin x � � a, b ∈ R}, B = 〈cos x, sin x〉 

(b) {ae x + be 2x � � a, b ∈ R}, B = 〈e x ,e 2x 〉 

(c) {a + bx + ce x + dxe 2x � � a, b, c, d ∈ R}, B = 〈1,x,e x ,xe x 〉 

1.26 Find the range of the linear transformation of R 2 represented with respect to 

the standard � �bases 

by each � matrix. � 

� � 

1 0 

0 0 

a b 

(a) 

(b) 

(c) a matrix of the form 

0 0 

3 2 

2a 2b 

� 1.27 Can one matrix represent two different linear maps? That is, can RepB,D(h) = 

RepB, ˆ D ˆ ( ˆ h)? 

1.28 Prove Theorem 1.4. 

� 1.29 Example 1.8 shows how to represent rotation of all vectors in the plane through 

an angle θ about the origin, with respect to the standard bases. 

(a) Rotation of all vectors in three-space through an angle θ about the x-axis is a 

transformation of R 3 . Represent it with respect to the standard bases. Arrange 

the rotation so that to someone whose feet are at the origin and whose head is 

at (1, 0, 0), the movement appears clockwise.


(b) Repeat the prior item, only rotate about the y-axis instead. (Put the person’s 

head at �e2.) 

(c) Repeat, about the z-axis. 

(d) Extend the prior item to R 4 .(Hint: ‘rotate about the z-axis’ can be restated 

as ‘rotate parallel to the xy-plane’.) 

1.30 (Schur’s Triangularization Lemma) 

(a) Let U be a subspace of V and fix bases BU ⊆ BV . What is the relationship 

between the representation of a vector from U with respect to BU and the 

representation of that vector (viewed as a member of V )withrespecttoBV ? 

(b) What about maps? 

(c) Fix a basis B = 〈 � β1,..., � βn〉 for V and observe that the spans 

[{�0}] ={�0} ⊂[{ � β1}] ⊂ [{ � β1, � β2}] ⊂ ··· ⊂[B] =V 

form a strictly increasing chain of subspaces. Show that for any linear map 

h: V → W there is a chain W0 = {�0} ⊆W1 ⊆ ··· ⊆ Wm = W of subspaces of 

W such that 

h([{ � β1,..., � βi}]) ⊂ Wi 

for each i. 

(d) Conclude that for every linear map h: V → W there are bases B,D so the 

matrix representing h with respect to B,D is upper-triangular (that is, each 

entry hi,j with i>jis zero). 

(e) Is an upper-triangular representation unique? 

3.III.2 Any Matrix Represents a Linear Map 

The prior subsection shows that the action of a linear map h is described by 

a matrix H, with respect to appropriate bases, in this way. 

⎛ ⎞ ⎛ 

⎞ 

�v = 

⎜ 

⎝ 

v1 

. 

vn 

⎟ 

⎠ 

B 

h 

↦−→ H 

⎜ 

⎝ 

h1,1v1 + ···+ h1,nvn 

. 

hm,1v1 + ···+ hm,nvn 

⎟ 

⎠ 

D 

= h(�v) 

In this subsection, we will show the converse, that each matrix represents a 

linear map. 

Recall that, in the definition of the matrix representation of a linear map, 

the number of columns of the matrix is the dimension of the map’s domain and 

the number of rows of the matrix is the dimension of the map’s codomain. Thus, 

for instance, a 2×3 matrix cannot represent a map from R 5 to R 4 . The next 

result says that, beyond this restriction on the dimensions, there are no other 

limitations: the 2×3 matrix represents a map from any three-dimensional space 

to any two-dimensional space. 

2.1 Theorem Any matrix represents a homomorphism between vector spaces 

of appropriate dimensions, with respect to any pair of bases.


Proof. For the matrix 

⎛ 

⎜ 

H = ⎜ 

⎝ 

h1,1 h1,2 ... h1,n 

h2,1 h2,2 ... h2,n 

. 

hm,1 hm,2 ... hm,n 

fix any n-dimensional domain space V and any m-dimensional codomain space 

W . Also fix bases B = 〈 � β1,..., � βn〉 and D = 〈 �δ1,..., �δm〉 for those spaces. 

Define a function h: V → W by: where �v in the domain is represented as 

⎛ ⎞ 

Rep B(�v) = 

then its image h(�v) is the member the codomain represented by 

⎛ 

⎞ 

h1,1v1 + ···+ h1,nvn 

⎜ 

RepD( h(�v))= 

. 

⎝ . 

⎟ 

. ⎠ 

hm,1v1 + ···+ hm,nvn 

that is, h(�v) =h(v1 � β1 + ···+ vn � βn) is defined to be (h1,1v1 + ···+ h1,nvn) · �δ1 + 

···+(hm,1v1 + ···+ hm,nvn) · �δm. (This is well-defined by the uniqueness of the 

representation RepB(�v).) Observe that h has simply been defined to make it the map that is represented 

with respect to B,D by the matrix H. So to finish, we need only check 

that h is linear. If �v,�u ∈ V are such that 

⎛ ⎞ 

⎛ ⎞ 

Rep B(�v) = 

⎜ 

⎝ 

v1 

. 

vn 

and c, d ∈ R then the calculation 

⎜ 

⎝ 

v1 

. 

vn 

⎟ 

⎠ 

B 

⎟ 

⎠ and Rep B(�u) = 

h(c�v + d�u) = � h1,1(cv1 + du1)+···+ h1,n(cvn + dun) � · �δ1 + 

···+ � hm,1(cv1 + du1)+···+ hm,n(cvn + dun) � · �δm = c · h(�v)+d · h(�u) 

provides this verification. QED 

2.2 Example Which map the matrix represents depends on which bases are 

used. If 

� � 

� � � � 

� � � � 

1 0 

1 0 

0 1 

H = , B1 = D1 = 〈 , 〉, and B2 = D2 = 〈 , 〉, 

0 0 

0 1 

1 0 

⎞ 

⎟ 

⎠ 

⎜ 

⎝ 

u1 

. 

un 

D 

⎟ 

⎠


then h1 : R 2 → R 2 represented by H with respect to B1,D1 maps 

� � � � 

c1 c1 

= 

c2 

c2 

B1 

↦→ 

� � 

c1 

0 

D1 

= 

� � 

c1 

0 

while h2 : R 2 → R 2 represented by H with respect to B2,D2 is this map. 

� � � � 

c1 c2 

= 

c2 

c1 

B2 

↦→ 

� � 

c2 

0 

D2 

� � 

0 

= 

These two are different. The first is projection onto the x axis, while the second 

is projection onto the y axis. 

So not only is any linear map described by a matrix but any matrix describes 

a linear map. This means that we can, when convenient, handle linear maps 

entirely as matrices, simply doing the computations, without have to worry that 

a matrix of interest does not represent a linear map on some pair of spaces of 

interest. (In practice, when we are working with a matrix but no spaces or 

bases have been specified, we will often take the domain and codomain to be R n 

and R m and use the standard bases. In this case, because the representation 

is transparent—the representation with respect to the standard basis of �v is 

�v—the column space of the matrix equals the range of the map. Consequently, 

the column space of H is often denoted by R(H).) 

With the theorem, we have characterized linear maps as those maps that act 

in this matrix way. Each linear map is described by a matrix and each matrix 

describes a linear map. We finish this section by illustrating how a matrix can 

be used to tell things about its maps. 

2.3 Theorem The rank of a matrix equals the rank of any map that it represents. 

Proof. Suppose that the matrix H is m×n. Fix domain and codomain spaces 

V and W of dimension n and m, with bases B = 〈 � β1,..., � βn〉 and D. Then H 

represents some linear map h between those spaces with respect to these bases 

whose rangespace 

{h(�v) � � �v ∈ V } = {h(c1 � β1 + ···+ cn � βn) � � c1,...,cn ∈ R} 

= {c1h( � β1)+···+ cnh( � βn) � � c1,...,cn ∈ R} 

is the span [{h( � β1),...,h( � βn)}]. The rank of h is the dimension of this rangespace. 

The rank of the matrix is its column rank (or its row rank; the two are 

equal). This is the dimension of the column space of the matrix, which is the 

span of the set of column vectors [{Rep D(h( � β1)),...,Rep D(h( � βn))}]. 

To see that the two spans have the same dimension, recall that a representation 

with respect to a basis gives an isomorphism Rep D : W → R m . Under 

this isomorphism, there is a linear relationship among members of the rangespace 

if and only if the same relationship holds in the column space, e.g, �0 = 

c2


c1h( � β1)+···+ cnh( � βn) if and only if �0 =c1Rep D(h( � β1)) + ···+ cnRep D(h( � βn)). 

Hence, a subset of the rangespace is linearly independent if and only if the corresponding 

subset of the column space is linearly independent. This means that 

the size of the largest linearly independent subset of the rangespace equals the 

size of the largest linearly independent subset of the column space, and so the 

two spaces have the same dimension. QED 

2.4 Example Any map represented by 

⎛ 

1 

⎜ 

⎜1 

⎝0 

2 

2 

0 

⎞ 

2 

1 ⎟ 

3⎠ 

0 0 2 

must, by definition, be from a three-dimensional domain to a four-dimensional 

codomain. In addition, because the rank of this matrix is two (we can spot this 

by eye or get it with Gauss’ method), any map represented by this matrix has 

a two-dimensional rangespace. 

2.5 Corollary Let h be a linear map represented by a matrix H. Then h 

is onto if and only if the rank of H equals the number of its rows, and h is 

one-to-one if and only if the rank of H equals the number of its columns. 

Proof. For the first half, the dimension of the rangespace of h is the rank of 

h, which equals the rank of H by the theorem. Since the dimension of the 

codomain of h is the number of columns in H, if the rank of H equals the 

number of columns, then the dimension of the rangespace equals the dimension 

of the codomain. But a subspace with the same dimension as its superspace 

must equal that superspace (a basis for the rangespace is a linearly independent 

subset of the codomain, whose size is equal to the dimension of the codomain, 

and so this set is a basis for the codomain). 

For the second half, a linear map is one-to-one if and only if it is an isomorphism 

between its domain and its range, that is, if and only if its domain has the 

same dimension as its range. But the number of columns in h is the dimension 

of h’s domain, and by the theorem the rank of H equals the dimension of h’s 

range. QED 

The above results end any confusion caused by our use of the word ‘rank’ to 

mean apparently different things when applied to matrices and when applied to 

maps. We can also justify the dual use of ‘nonsingular’. We’ve defined a matrix 

to be nonsingular if it is square and is the matrix of coefficients of a linear system 

with a unique solution, and we’ve defined a linear map to be nonsingular if it is 

one-to-one. 

2.6 Corollary A square matrix represents nonsingular maps if and only if it 

is a nonsingular matrix. Thus, a matrix represents an isomorphism if and only 

if it is square and nonsingular. 

Proof. Immediate from the prior result. QED


2.7 Example Any map from R2 to P1 represented with respect to any pair of 

bases by 

� � 

1 2 

0 3 

is nonsingular because this matrix has rank two. 

2.8 Example Any map g : V → W represented by 

� � 

1 2 

3 6 

is not nonsingular because this matrix is not nonsingular. 

We’ve now seen that the relationship between maps and matrices goes both 

ways: fixing bases, any linear map is represented by a matrix and any matrix 

describes a linear map. That is, by fixing spaces and bases we get a correspondence 

between maps and matrices. In the rest of this chapter we will explore 

this correspondence. For instance, we’ve defined for linear maps the operations 

of addition and scalar multiplication and we shall see what the corresponding 

matrix operations are. We shall also see the matrix operation that represent 

the map operation of composition. And, we shall see how to find the matrix 

that represents a map’s inverse. 

Exercises 

� 2.9 Decide if the vector is in the column space of the matrix. 

� � � � � � � � � 

1 

2 1 1 

4 −8 0 

(a) , (b) 

, (c) 1 

2 5 −3 

2 −4 1 

−1 

−1 

1 

−1 

� 

1 

−1 , 

1 

� � 

2 

0 

0 

� 2.10 Decide if each vector lies in the range of the map from R 3 to R 2 represented 

with respect � to the � standard � � bases �by the matrix. � � � 

1 1 3 1 

2 0 3 1 

(a) 

, (b) 

, 

0 1 4 3 

4 0 6 1 

� 2.11 Consider this matrix, representing a transformation of R 2 , and these bases for 

that space. 

1 

2 · 

� 

1 

� 

1 

−1 1 

� � � � � � � � 

0 1 

1 1 

B = 〈 , 〉 D = 〈 , 〉 

1 0 

1 −1 

(a) To what vector in the codomain is the first member of B mapped? 

(b) The second member? 

(c) Where is a general vector from the domain (a vector with components x and 

y) mapped? That is, what transformation of R 2 is represented with respect to 

B,D by this matrix? 

2.12 What transformation of F = {a cos θ + b sin θ � � a, b ∈ R} is represented with 

respect to B = 〈cos θ − sin θ, sin θ〉 and D = 〈cos θ +sinθ, cos θ〉 by this matrix? 

� � 

0 0 

1 0


� 2.13 Decide if 1 + 2x is in the range of the map from R 3 to P2 represented with 

respect to E3 and 〈1, 1+x 2 ,x〉 by this matrix. 

� � 

1 3 0 

0 1 0 

1 0 1 

2.14 Example 2.8 gives a matrix that is nonsingular, and is therefore associated 

with maps that are nonsingular. 

(a) Find the set of column vectors representing the members of the nullspace of 

any map represented by this matrix. 

(b) Find the nullity of any such map. 

(c) Find the set of column vectors representing the members of the rangespace 

of any map represented by this matrix. 

(d) Find the rank of any such map. 

(e) Check that rank plus nullity equals the dimension of the domain. 

� 2.15 Because the rank of a matrix equals the rank of any map it represents, if 

one matrix represents two different maps H =Rep B,D(h) =Rep ˆ B, ˆ D ( ˆ h)(where 

h, ˆ h: V → W ) then the dimension of the rangespace of h equals the dimension of 

the rangespace of ˆ h. Must these equal-dimensioned rangespaces actually be the 

same? 

� 2.16 Let V be an n-dimensional space with bases B and D. Consider a map that 

sends, for �v ∈ V , the column vector representing �v with respect to B to the column 

vector representing �v with respect to D. Show that is a linear transformation of 

R n . 

2.17 Example 2.2 shows that changing the pair of bases can change the map that 

a matrix represents, even though the domain and codomain remain the same. 

Could the map ever not change? Is there a matrix H, vectorspacesV and W , 

and associated pairs of bases B1,D1 and B2,D2 (with B1 �= B2 or D1 �= D2 or 

both) such that the map represented by H with respect to B1,D1 equals the map 

represented by H with respect to B2,D2? 

� 2.18 A square matrix is a diagonal matrix if it is all zeroes except possibly for the 

entries on its upper-left to lower-right diagonal—its 1, 1entry,its2, 2entry,etc. 

Show that a linear map is an isomorphism if there are bases such that, with respect 

to those bases, the map is represented by a diagonal matrix with no zeroes on the 

diagonal. 

2.19 Describe geometrically the action on R 2 of the map represented with respect 

to the standard bases E2, E2 by this matrix. 

� � 

3 0 

0 2 

Do the same for these. �1 � 

0 

� 

0 

� 

1 

� 

1 

� 

3 

0 0 1 0 0 1 

2.20 The fact that for any linear map the rank plus the nullity equals the dimension 

of the domain shows that a necessary condition for the existence of a homomorphism 

between two spaces, onto the second space, is that there be no gain in 

dimension. That is, where h: V → W is onto, the dimension of W must be less 

than or equal to the dimension of V . 

(a) Show that this (strong) converse holds: no gain in dimension implies that


there is a homomorphism and, further, any matrix with the correct size and 

correct rank represents such a map. 

(b) Are there bases for R 3 such that this matrix 

� � 

1 0 0 

H = 2 0 0 

0 1 0 

represents a map from R 3 to R 3 whose range is the xy plane subspace of R 3 ? 

2.21 Let V be an n-dimensional space and suppose that �x ∈ R n . Fix a basis 

B for V and consider the map h�x : V → R given �v ↦→ �x RepB(�v) by the dot 

product. 

(a) Show that this map is linear. 

(b) Show that for any linear map g : V → R there is an �x ∈ R n such that g = h�x. 

(c) In the prior item we fixed the basis and varied the �x to get all possible linear 

maps. Can we get all possible linear maps by fixing an �x and varying the basis? 

2.22 Let V,W,X be vector spaces with bases B,C,D. 

(a) Suppose that h: V → W is represented with respect to B,C by the matrix 

H. Give the matrix representing the scalar multiple rh (where r ∈ R) with 

respect to B,C by expressing it in terms of H. 

(b) Suppose that h, g : V → W are represented with respect to B,C by H and 

G. Give the matrix representing h + g with respect to B,C by expressing it in 

terms of H and G. 

(c) Suppose that h: V → W is represented with respect to B,C by H and 

g : W → X is represented with respect to C, D by G. Give the matrix representing 

g ◦ h with respect to B,D by expressing it in terms of H and G.

Section IV. Matrix Operations 211 

3.IV Matrix Operations 

The prior section shows how matrices represent linear maps. A good strategy, on 

seeing a new idea, is to explore how it interacts with some already-established 

ideas. In the first subsection we will ask how the representation of the sum 

of two maps f + g related to the representations F and G of the two maps, 

and how the representation of a scalar product r · h of a map is related to the 

representation H of that map. In later subsections we will see how to represent 

map composition and map inverse. 

3.IV.1 Sums and Scalar Products 

Recall that for two maps f and g with the same domain and codomain, the 

map sum f + g has the natural definition. 

�v f+g 

↦−→ f(�v)+g(�v) 

The easiest way to see how the representations of the maps combine to represent 

the map sum is with an example. 

1.1 Example Suppose that f,g: R 2 → R 3 are represented with respect to the 

bases B and D by these matrices. 

⎛ ⎞ 

1 3 

F =RepB,D(f) = ⎝2 

0⎠ 

1 0 

B,D 

⎛ ⎞ 

0 0 

G =RepB,D(g) = ⎝−1 

−2⎠ 

2 4 

Then, for any �v ∈ V represented with respect to B, computation of the representation 

of f(�v)+g(�v) 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

1 3 � � 0 0 � � 1v1 +3v2 0v1 +0v2 

v1 

v1 

⎝2 

0⎠ 

+ ⎝−1 

−2⎠ 

= ⎝2v1 

+0v2⎠ 

+ ⎝−1v1 

− 2v2⎠ 

v2 

v2 

1 0 

2 4 

1v1 +0v2 2v1 +4v2 

gives this representation of f + g (�v). 

⎛ 

⎝ (1 + 0)v1 

⎞ ⎛ 

+(3+0)v2 

(2 − 1)v1 +(0− 2)v2⎠ 

= ⎝ 

(1 + 2)v1 +(0+4)v2 

1v1 

⎞ 

+3v2 

1v1 − 2v2⎠ 

3v1 +4v2 

Thus, the action of f + g is described by this matrix-vector product. 

⎛ 

1 

⎝1 3 

⎞ 

3 

−2⎠ 

4 

⎛ 

� � 

v1 

= ⎝ 

v2 B 

1v1 

⎞ 

+3v2 

1v1 − 2v2⎠ 

3v1 +4v2 

B,D 

This matrix is the entry-by-entry sum of original matrices, e.g., the 1, 1entry 

of Rep B,D(f + g) is the sum of the 1, 1entryofF and the 1, 1entryofG. 

D 

B,D


Representing a scalar multiple of a map works the same way. 

1.2 Example If t is a transformation represented by 

� 

1 

RepB,D(t) = 

1 

� 

0 

1 

so that 

� � 

v1 

�v = 

� 

↦→ 

v1 

B,D 

then the scalar multiple map 5t acts in this way. 

� � 

v1 

�v = 

v2 

↦−→ 

� � 

5v1 

5v1 +5v2 

Therefore, this is the matrix representing 5t. 

� � 

5 0 

RepB,D(5t) = 

5 5 

B 

v2 

D 

B 

B,D 

=5· t(�v) 

v1 + v2 

� 

D 

= t(�v) 

1.3 Definition The sum of two same-sized matrices is their entry-by-entry 

sum. The scalar multiple of a matrix is the result of entry-by-entry scalar 

multiplication. 

1.4 Remark These extend the vector addition and scalar multiplication operations 

that we defined in the first chapter. 

1.5 Theorem Let h, g : V → W be linear maps represented with respect to 

bases B,D by the matrices H and G, and let r be a scalar. Then the map 

h + g : V → W is represented with respect to B,D by H + G, and the map 

r · h: V → W is represented with respect to B,D by rH. 

Proof. Exercise 8; generalize the examples above. QED 

A notable special case of scalar multiplication is multiplication by zero. For 

any map 0 · h is the zero homomorphism and for any matrix 0 · H is the zero 

matrix. 

1.6 Example The zero map from any three-dimensional space to any twodimensional 

space is represented by the 2×3 zero matrix 

� � 

0 0 0 

Z = 

0 0 0 

no matter which domain and codomain bases are used. 

Exercises 

� 1.7 Perform � the indicated � � operations, � if defined. 

5 −1 2 2 1 4 

(a) 

+ 

6 1 1 3 0 5 

� � 

2 −1 −1 

(b) 6 · 

1 2 3


� � � � 

2 1 2 1 

(c) + 

0 3 0 3 

� � � � 

1 2 −1 4 

(d) 4 +5 

3 −1 −2 1 

� � � � 

2 1 1 1 4 

(e) 3 +2 

3 0 3 0 5 

1.8 Prove Theorem 1.5. 

(a) Prove that matrix addition represents addition of linear maps. 

(b) Prove that matrix scalar multiplication represents scalar multiplication of 

linear maps. 

� 1.9 Prove each, where the operations are defined, where G, H, andJare matrices, 

where Z is the zero matrix, and where r and s are scalars. 

(a) Matrix addition is commutative G + H = H + G. 

(b) Matrix addition is associative G +(H + J) =(G + H)+J. 

(c) The zero matrix is an additive identity G + Z = G. 

(d) 0 · G = Z 

(e) (r + s)G = rG + sG 

(f) Matrices have an additive inverse G +(−1) · G = Z. 

(g) r(G + H) =rG + rH 

(h) (rs)G = r(sG) 

1.10 Fix domain and codomain spaces. In general, one matrix can represent many 

different maps with respect to different bases. However, prove that a zero matrix 

represents only a zero map. Are there other such matrices? 

� 1.11 Let V and W be vector spaces of dimensions n and m. Show that the space 

L(V,W) of linear maps from V to W is isomorphic to Mm×n. 

� 1.12 Show that it follows from the prior questions that for any six transformations 

t1,...,t6 : R 2 → R 2 there are scalars c1,...,c6 ∈ R such that c1t1 + ···+ c6t6 is 

the zero map. (Hint: this is a bit of a misleading question.) 

1.13 The trace of a square matrix is the sum of the entries on the main diagonal 

(the 1, 1 entry plus the 2, 2 entry, etc.; we will see the significance of the trace in 

Chapter Five). Show that trace(H + G) =trace(H)+trace(G). Is there a similar 

result for scalar multiplication? 

1.14 Recall that the transpose of a matrix M is another matrix, whose i, j entry is 

the j, i entry of M. Verifiy these identities. 

(a) (G + H) trans = G trans + H trans 

(b) (r · H) trans = r · H trans 

� 1.15 A square matrix is symmetric if each i, j entry equals the j, i entry, that is, if 

the matrix equals its transpose. 

(a) Prove that for any H, the matrix H + H trans is symmetric. Does every 

symmetric matrix have this form? 

(b) Prove that the set of n×n symmetric matrices is a subspace of Mn×n. 

� 1.16 (a) How does matrix rank interact with scalar multiplication—can a scalar 

product of a rank n matrix have rank less than n? Greater? 

(b) How does matrix rank interact with matrix addition—can a sum of rank n 

matrices have rank less than n? Greater?


3.IV.2 Matrix Multiplication 

After representing addition and scalar multiplication of linear maps in the 

prior subsection, the natural next map operation to consider is composition. 

2.1 Lemma A composition of linear maps is linear. 

Proof. (This argument has appeared earlier, as part of the proof that isomorphism 

is an equivalence relation between spaces.) Let h: V → W and g : W → U 

be linear. The natural calculation 

g ◦ h � � � 

c1 · �v1 + c2 · �v2 = g h(c1 · �v1 + c2 · �v2) � = g � c1 · h(�v1)+c2 · h(�v2) � 

= c1 · g � h(�v1)) + c2 · g(h(�v2) � = c1 · (g ◦ h)(�v1)+c2 · (g ◦ h)(�v2) 

shows that g ◦ h: V → U preserves linear combinations. QED 

To see how the representation of the composite arises out of the representations 

of the two compositors, consider an example. 

2.2 Example Let h: R 4 → R 2 and g : R 2 → R 3 ,fixbasesB ⊂ R 4 , C ⊂ R 2 , 

D ⊂ R3 , and let these be the representations. 

� 

4 

H =RepB,C(h) = 

5 

6 

7 

8 

9 

� 

2 

3 

B,C 

⎛ 

1 

G =RepC,D(g) = ⎝0 

1 

⎞ 

1 

1⎠ 

0 

To represent the composition g ◦ h: R 4 → R 3 we fix a �v, represent h of �v, and 

then represent g of that. The representation of h(�v) is the product of h’s matrix 

and �v’s vector. 

Rep C( h(�v))= 

� 

4 6 8 

� 

2 

5 7 9 3 

B,C 

⎛ 

⎜ 

⎝ 

v1 

v2 

v3 

v4 

⎞ 

⎟ 

⎠ 

B 

C,D 

� � 

4v1 +6v2 +8v3 +2v4 

= 

5v1 +7v2 +9v3 +3v4 

The representation of g( h(�v) ) is the product of g’s matrix and h(�v)’s vector. 

⎛ ⎞ 

1 1 � � 

4v1 +6v2 +8v3 +2v4 

RepD( g(h(�v)) )= ⎝0 1⎠ 

5v1 +7v2 +9v3 +3v4 

1 0 

C 

C,D 

⎛ 

= ⎝ 1 · (4v1 

⎞ 

+6v2 +8v3 +2v4)+1· (5v1 +7v2 +9v3 +3v4) 

0 · (4v1 +6v2 +8v3 +2v4)+1· (5v1 +7v2 +9v3 +3v4) ⎠ 

1 · (4v1 +6v2 +8v3 +2v4)+0· (5v1 +7v2 +9v3 +3v4) 

Distributing and regrouping on the v’s gives 

⎛ 

= ⎝ (1 · 4+1· 5)v1 

⎞ 

+(1· 6+1· 7)v2 +(1· 8+1· 9)v3 +(1· 2+1· 3)v4 

(0 · 4+1· 5)v1 +(0· 6+1· 7)v2 +(0· 8+1· 9)v3 +(0· 2+1· 3)v4⎠ 

(1 · 4+0· 5)v1 +(1· 6+0· 7)v2 +(1· 8+0· 9)v3 +(1· 2+0· 3)v4 

C 

D 

D


which we recognizing as the result of this matrix-vector product. 

⎛ 

1 · 4+1· 5 

= ⎝0 · 4+1· 5 

1 · 4+0· 5 

1· 6+1· 7 

0· 6+1· 7 

1· 6+0· 7 

1· 8+1· 9 

0· 8+1· 9 

1· 8+0· 9 

⎞ 

1· 2+1· 3 

0· 2+1· 3⎠ 

1· 2+0· 3 

⎛ 

⎜ 

⎝ 

Thus, the matrix representing g◦h has the rows of G combined with the columns 

of H. 

2.3 Definition The matrix-multiplicative product of the m×r matrix G and 

the r×n matrix H is the m×n matrix P , where 

pi,j = gi,1h1,j + gi,2h2,j + ···+ gi,rhr,j 

that is, the i, j-th entry of the product is the dot product of the i-th row and 

the j-th column. 

⎛ 

⎞ ⎛ 

⎞ 

. 

h1,j 

. 

⎜ . 

⎟ ⎜ 

GH = ⎜gi,1 

gi,2 ⎝ 

... gi,r⎟ 

⎜... 

h2,j ... ⎟ 

⎠ ⎜ . 

⎝ . 

⎟ 

. ⎠ 

. 

= 

⎛ 

⎞ 

. 

⎜ . ⎟ 

⎜ 

⎝ 

... pi,j ... ⎟ 

⎠ 

. 

2.4 Example The matrices from Example 2.2 combine in this way. 

⎛ 

1 · 4+1· 5 

⎝0 

· 4+1· 5 

1· 6+1· 7 

0· 6+1· 7 

1· 8+1· 9 

0· 8+1· 9 

⎞ ⎛ 

1· 2+1· 3 9 

0· 2+1· 3⎠ 

= ⎝5 

13 

7 

17 

9 

⎞ 

5 

3⎠ 

1 · 4+0· 5 1· 6+0· 7 1· 8+0· 9 1· 2+0· 3 4 6 8 2 

2.5 Example 

⎛ 

2 

⎝4 

8 

⎞ 

0 � 

6⎠ 

1 

5 

2 

⎛ 

� 2 · 1+0· 5 

3 

= ⎝4 

· 1+6· 5 

7 

8 · 1+2· 5 

⎞ ⎛ 

2· 3+0· 7 2 

4· 3+6· 7⎠ 

= ⎝34 

8· 3+2· 7 18 

⎞ 

6 

54⎠ 

38 

2.6 Theorem A composition of linear maps is represented by the matrix product 

of the representatives. 

Proof. (This argument parallels Example 2.2.) Let h: V → W and g : W → X 

be represented by H and G with respect to bases B ⊂ V , C ⊂ W ,andD ⊂ X, 

of sizes n, r, andm. For any �v ∈ V ,thek-th component of Rep C( h(�v)) is 

hr,j 

hk,1v1 + ···+ hk,nvn 

and so the i-th component of Rep D( g ◦ h (�v) ) is this. 

gi,1 · (h1,1v1 + ···+ h1,nvn)+gi,2 · (h2,1v1 + ···+ h2,nvn) 

+ ···+ gi,r · (hr,1v1 + ···+ hr,nvn) 

B,D 

v1 

v2 

v3 

v4 

⎞ 

⎟ 

⎠ 

D


Distribute and regroup on the v’s. 

=(gi,1h1,1 + gi,2h2,1 + ···+ gi,rhr,1) · v1 

Finish by recognizing that the coefficient of each vj 

+ ···+(gi,1h1,n + gi,2h2,n + ···+ gi,rhr,n) · vn 

gi,1h1,j + gi,2h2,j + ···+ gi,rhr,j 

matches the definition of the i, j entry of the product GH. QED 

The theorem is an example of a result that supports a definition. We can 

picture what the definition and theorem together say with this arrow diagram 

(‘w.r.t.’ abbreviates ‘with respect to’). 

↗h H 

Vw.r.t. B 

Ww.r.t. C 

g◦h 

−→ 

GH 

↘ g 

G 

Xw.r.t. D 

Above the arrows, the maps show that the two ways of going from V to X, 

straight over via the composition or else by way of W , have the same effect 

�v g◦h 

↦−→ g(h(�v)) �v h 

↦−→ h(�v) 

g 

↦−→ g(h(�v)) 

(this is just the definition of composition). Below the arrows, the matrices 

indicate that the product does the same thing—multiplying GH into the column 

vector Rep B(�v) has the same effect as multiplying the column first by H and 

then multiplying the result by G. 

Rep B,D(g ◦ h) =GH =Rep C,D(g)Rep B,C(h) 

The definition of the matrix-matrix product operation does not restrict us 

to view it as a representation of a linear map composition. We can get insight 

into this operation by studying it as a mechanical procedure. The striking thing 

is the way that rows and columns combine. 

One aspect of that combination is that the sizes of the matrices involved is 

significant. Briefly, m×r times r×n equals m×n. 

2.7 Example This product is not defined 

� 

−1 2 

�� 

0 0 

� 

0 

0 10 1.1 0 2 

because the number of columns on the left does not equal the number of rows 

on the right.


In terms of the underlying maps, the fact that the sizes must match up reflects 

the fact that matrix multiplication is defined only when a corresponding function 

composition 

dimension n space 

h 

−→ dimension r space 

g 

−→ dimension m space 

is possible. 

2.8 Remark The order in which these things are written can be confusing. In 

the ‘m×r times r×n equals m×n’ equation, the number written first m is the 

dimension of g’s codomain and is thus the number that appears last in the map 

dimension description above. The explanation is that while f is done first and 

then g is applied, that composition is written g ◦ f, from the notation ‘g(f(�v))’. 

(Some people try to lessen confusion by reading ‘g ◦ f’ aloud as “g following 

f”.) That order then carries over to matrices: g ◦ f is represented by GF . 

Another aspect of the way that rows and columns combine in the matrix 

product operation is that in the definition of the i, j entry 

pi,j = g i, 1 h 1 ,j + g i, 2 h 2 ,j + ···+ g i, r h r ,j 

the boxed subscripts on the g’s are column indicators while those on the h’s 

indicate rows. That is, summation takes place over the columns of G but over 

the rows of H; left is treated differently than right, so GH may be unequal to 

HG. Matrix multiplication is not commutative. 

2.9 Example Matrix multiplication hardly ever commutes. Test that by mul- 

tiplying randomly chosen matrices both ways. 

� 

1 

3 

�� 

2 5 

4 7 

� � 

6 19 

= 

8 43 

� 

22 

50 

� 

5 

7 

�� 

6 1 

8 3 

� 

2 

= 

4 

2.10 Example Commutativity can fail more dramatically: 

� 

5 

7 

�� 

6 1 

8 3 

2 

4 

� � 

0 23 

= 

0 31 

34 

46 

� 

0 

0 

while 

isn’t even defined. 

� 

1 2 

�� 

0 5 

� 

6 

3 4 0 7 8 

� 

23 

� 

34 

31 46 

2.11 Remark The fact that matrix multiplication is not commutative may 

be puzzling at first sight, perhaps just because most algebraic operations in 

elementary mathematics are commutative. But on further reflection, it isn’t 

so surprising. After all, matrix multiplication represents function composition, 

which is not commutative—if f(x) =2x and g(x) =x + 1 then g ◦ f(x) =2x +1 

while f ◦ g(x) =2(x +1) = 2x + 2. True, this g is not linear and we might 

have hoped that linear functions commute, but this perspective shows that the 

failure of commutativity for matrix multiplication fits into a larger context.


Except for the lack of commutativity, matrix multiplication is algebraically 

well-behaved. Below are some nice properties and more are in Exercise 23 and 

Exercise 24. 

2.12 Theorem If F , G, andH are matrices, and the matrix products are 

defined, then the product is associative (FG)H = F (GH) and distributes over 

matrix addition F (G + H) =FG+ FH and (G + H)F = GF + HF. 

Proof. Associativity holds because matrix multiplication represents function 

composition, which is associative: the maps (f ◦ g) ◦ h and f ◦ (g ◦ h) are equal 

as both send �v to f(g(h(�v))). 

Distributivity is similar. For instance, the first one goes f ◦ (g + h)(�v) = 

f � (g + h)(�v) � = f � g(�v)+h(�v) � = f(g(�v)) + f(h(�v)) = f ◦ g(�v)+f ◦ h(�v) (the 

third equality uses the linearity of f). QED 

2.13 Remark We could alternatively prove that result by slogging through 

the indices. For example, associativity goes: the i, j-th entry of (FG)H is 

(fi,1g1,1 + fi,2g2,1 + ···+ fi,rgr,1)h1,j 

+(fi,1g1,2 + fi,2g2,2 + ···+ fi,rgr,2)h2,j 

. 

+(fi,1g1,s + fi,2g2,s + ···+ fi,rgr,s)hs,j 

(where F , G, andH are m×r, r×s, ands×n matrices), distribute 

and regroup around the f’s 

fi,1g1,1h1,j + fi,2g2,1h1,j + ···+ fi,rgr,1h1,j 

+ fi,1g1,2h2,j + fi,2g2,2h2,j + ···+ fi,rgr,2h2,j 

. 

+ fi,1g1,shs,j + fi,2g2,shs,j + ···+ fi,rgr,shs,j 

fi,1(g1,1h1,j + g1,2h2,j + ···+ g1,shs,j) 

+ fi,2(g2,1h1,j + g2,2h2,j + ···+ g2,shs,j) 

. 

+ fi,r(gr,1h1,j + gr,2h2,j + ···+ gr,shs,j) 

to get the i, j entry of F (GH). 

Contrast these two ways of verifying associativity, the one in the proof and 

the one just above. The argument just above is hard to understand in the sense 

that, while the calculations are easy to check, the arithmetic seems unconnected 

to any idea (it also essentially repeats the proof of Theorem 2.6 and so is inefficient). 

The argument in the proof is shorter, clearer, and says why this property 

“really” holds. This illustrates the comments made in the preamble to the chapter 

on vector spaces—at least some of the time an argument from higher-level 

constructs is clearer.


We have now seen how the representation of the composition of two linear 

maps is derived from the representations of the two maps. We have called 

the combination the product of the two matrices. This operation is extremely 

important. Before we go on to study how to represent the inverse of a linear 

map, we will explore it some more in the next subsection. 

Exercises 

� 2.14 Compute, or state ‘not defined’. 

� �� 

3 1 0 5 

1 1 

(a) 

(b) 

−4 2 0 0.5 

4 0 

� 

−1 

3 

� � � 

2 −7 

(c) 

7 4 

2 

3 

3 

� 

−1 −1 

1 1 

1 1 

� � 

1 0 5 � 

5 

−1 1 1 (d) 

3 

3 8 4 

�� 

2 −1 

1 3 

� 

2 

−5 

� 2.15 Where 

A = 

� � 

1 −1 

2 0 

B = 

� � 

5 2 

4 4 

C = 

� 

−2 

� 

3 

−4 1 

compute or state ‘not defined’. 

(a) AB (b) (AB)C (c) BC (d) A(BC) 

2.16 Which products are defined? 

(a) 3×2 times2×3 (b) 2×3 times 3 ×2 (c) 2×2 times3×3 

(d) 3×3 times 2×2 

� 2.17 Give the size of the product or state ‘not defined’. 

(a) a2×3 matrix times a 3×1 matrix 

(b) a1×12 matrix times a 12×1 matrix 

(c) a2×3 matrix times a 2×1 matrix 

(d) a2×2 matrix times a 2×2 matrix 

� 2.18 Find the system of equations resulting from starting with 

h1,1x1 + h1,2x2 + h1,3x3 = d1 

h2,1x1 + h2,2x2 + h2,3x3 = d2 

and making this change of variable (i.e., substitution). 

x1 = g1,1y1 + g1,2y2 

x2 = g2,1y1 + g2,2y2 

x3 = g3,1y1 + g3,2y2 

2.19 As Definition 2.3 points out, the matrix product operation generalizes the dot 

product. Is the dot product of a 1×n row vector and a n×1 column vector the 

same as their matrix-multiplicative product? 

� 2.20 Represent the derivative map on Pn with respect to B,B where B is the 

natural basis 〈1,x,... ,x n 〉. Show that the product of this matrix with itself is 

defined; what the map does it represent? 

2.21 Show that composition of linear transformations on R 1 is commutative. Is 

this true for any one-dimensional space? 

2.22 Why is matrix multiplication not defined as entry-wise multiplication? That 

would be easier, and commutative too. 

� 2.23 (a) Prove that H p H q = H p+q and (H p ) q = H pq for positive integers p, q. 

(b) Prove that (rH) p = r p · H p for any positive integer p and scalar r ∈ R.


� 2.24 (a) How does matrix multiplication interact with scalar multiplication: is 

r(GH) =(rG)H? IsG(rH) =r(GH)? 

(b) How does matrix multiplication interact with linear combinations: is F (rG+ 

sH) =r(FG)+s(FH)? Is (rF + sG)H = rFH + sGH? 

2.25 We can ask how the matrix product operation interacts with the transpose 

operation. 

(a) Show that (GH) trans = H trans G trans . 

(b) A square matrix is symmetric if each i, j entry equals the j, i entry, that is, 

if the matrix equals its own transpose. Show that the matrices HH trans and 

H trans H are symmetric. 

� 2.26 Rotation of vectors in R 3 about an axis is a linear map. Show that linear 

maps do not commute by showing geometrically that rotations do not commute. 

2.27 In the proof of Theorem 2.12 some maps are used. What are the domains and 

codomains? 

2.28 How does matrix rank interact with matrix multiplication? 

(a) Can the product of rank n matrices have rank less than n? Greater? 

(b) Show that the rank of the product of two matrices is less than or equal to 

the minimum of the rank of each factor. 

2.29 Is ‘commutes with’ an equivalence relation among n×n matrices? 

� 2.30 (This will be used in the Matrix Inverses exercises.) Here is another property 

of matrix multiplication that might be puzzling at first sight. 

(a) Prove that the composition of the projections πx,πy : R 3 → R 3 onto the x 

and y axes is the zero map despite that neither one is itself the zero map. 

(b) Prove that the composition of the derivatives d 2 /dx 2 ,d 3 /dx 3 : P4 →P4 is 

the zero map despite that neither is the zero map. 

(c) Give a matrix equation representing the first fact. 

(d) Give a matrix equation representing the second. 

When two things multiply to give zero despite that neither is zero, each is said to 

be a zero divisor. 

2.31 Show that, for square matrices, (S + T )(S − T ) need not equal S 2 − T 2 . 

� 2.32 Represent the identity transformation id: V → V with respect to B,B for any 

basis B. This is the identity matrix I. Show that this matrix plays the role in 

matrix multiplication that the number 1 plays in real number multiplication: HI = 

IH = H (for all matrices H for which the product is defined). 

2.33 In real number algebra, quadratic equations have at most two solutions. That 

is not so with matrix algebra. Show that the 2×2 matrix equation T 2 = I has 

more than two solutions, where I is the identity matrix (this matrix has ones in 

its 1, 1and2, 2 entries and zeroes elsewhere; see Exercise 32). 

2.34 (a) Prove that for any 2×2 matrixT there are scalars c0,...,c4 such that the 

combination c4T 4 + c3T 3 + c2T 2 + c1T + I isthezeromatrix(whereI is the 2×2 

identity matrix, with ones in its 1, 1and2, 2 entries and zeroes elsewhere; see 

Exercise 32). 

(b) Let p(x) be a polynomial p(x) =cnx n + ··· + c1x + c0. If T is a square 

matrix we define p(T ) to be the matrix cnT n + ···+ c1T + I (where I is the 

appropriately-sized identity matrix). Prove that for any square matrix there is 

a polynomial such that p(T ) is the zero matrix. 

(c) The minimal polynomial m(x) of a square matrix is the polynomial of least 

degree, and with leading coefficient 1, such that m(T ) is the zero matrix. Find


the minimal polynomial of this matrix. 

�√ 

3/2 −1/2 

1/2 

√ � 

3/2 

(This is the representation with respect to E2, E2, the standard basis, of a rotation 

through π/6 radians counterclockwise.) 

2.35 The infinite-dimensional space P of all finite-degree polynomials gives a memorable 

example of the non-commutativity of linear maps. Let d/dx: P→Pbe the 

usual derivative and let s: P→Pbe the shift map. 

a0 + a1x + ···+ anx n 

s 

↦−→ 0+a0x + a1x 2 + ···+ anx n+1 

Show that the two maps don’t commute d/dx ◦ s �= s ◦ d/dx; in fact, not only is 

(d/dx ◦ s) − (s ◦ d/dx) not the zero map, it is the identity map. 

2.36 Recall the notation for the sum of the sequence of numbers a1,a2,...,an. 

n� 

i=1 

ai = a1 + a2 + ···+ an 

In this notation, the i, j entry of the product of G and H is this. 

r� 

pi,j = 

gi,khk,j 

k=1 

Using this notation, 

(a) reprove that matrix multiplication is associative; 

(b) reprove Theorem 2.6. 

3.IV.3 Mechanics of Matrix Multiplication 

In this subsection we consider matrix multiplication as a mechanical process, 

putting aside for the moment any implications about the underlying maps. As 

described earlier, the striking thing about matrix multiplication is the way rows 

and columns combine. The i, j entry of the matrix product is the dot product 

of the row i of the left matrix with column j of the right one. For instance, here 

is a second row and a third column combining to make a 2, 3entry. 

⎛ ⎞ 

1 1 � 

⎜ ⎟ 4 

⎝ 0 1⎠ 

5 

1 0 

6 

7 

8 

9 

� ⎛ 

⎞ 

9 13 17 5 

2 

= ⎝5 7 9 3⎠ 

3 

4 6 8 2 

We can view this as the left matrix acting by multiplying its rows, one at a time, 

into the columns of the right matrix. Of course, another perspective is that the 

right matrix uses its columns to act on the left matrix’s rows. Below, we will 

examine actions from the left and from the right for some simple matrices. 

The first case, the action of a zero matrix, is very easy.


3.1 Example Multiplying by an appropriately-sized zero matrix from the left 

or from the right 

� �� 

0 0 1 3 2 

= 

0 0 −1 1 −1 

results in a zero matrix. 

� 0 0 0 

0 0 0 

� � �� 

2 3 0 0 

= 

1 4 0 0 

� � 

0 0 

0 0 

After zero matrices, the matrices whose actions are easiest to understand 

are the ones with a single nonzero entry. 

3.2 Definition A matrix with all zeroes except for a one in the i, j entry is an 

i, j unit matrix. 

3.3 Example This is the 1, 2 unit matrix with three rows and two columns, 

multiplying from the left. 

⎛ ⎞ ⎛ ⎞ 

0 1 � � 7 8 

⎝0 

0⎠ 

5 6 

= ⎝0 

0⎠ 

7 8 

0 0 

0 0 

Acting from the left, an i, j unit matrix copies row j of the multiplicand into 

row i of the result. From the right an i, j unit matrix copies column i of the 

multiplicand into column j of the result. 

⎛ 

1 

⎝4 

2 

5 

⎞ ⎛ 

3 0 

6⎠⎝0 

⎞ ⎛ 

1 0 

0⎠ 

= ⎝0 

⎞ 

1 

4⎠ 

7 8 9 0 0 0 7 

3.4 Example Rescaling these matrices simply rescales the result. This is the 

action from the left of the matrix that is twice the one in the prior example. 

⎛ ⎞ ⎛ ⎞ 

0 2 � � 14 16 

⎝0 

0⎠ 

5 6 

= ⎝ 0 0⎠ 

7 8 

0 0 

0 0 

And this is the action of the matrix that is minus three times the one from the 

prior example. 

⎛ 

1 

⎝4 2 

5 

⎞ ⎛ 

3 0 

6⎠⎝0 

⎞ ⎛ 

−3 0 

0 ⎠ = ⎝0 ⎞ 

−3 

−12⎠ 

7 8 9 0 0 0 −21 

Next in complication are matrices with two nonzero entries. There are two 

cases. If a left-multiplier has entries in different rows then their actions don’t 

interact.


3.5 Example 

⎛ 

1 

⎝0 0 

0 

⎞ ⎛ 

0 1 

2⎠⎝4 

2 

5 

⎞ ⎛ 

3 1 

6⎠ 

=( ⎝0 0 

0 

⎞ ⎛ 

0 0 

0⎠ 

+ ⎝0 0 

0 

⎞ ⎛ 

0 1 

2⎠) 

⎝4 2 

5 

⎞ 

3 

6⎠ 

0 0 0 7 8 9 0 

⎛ 

1 

= ⎝0 0 

2 

0 

0 0 

⎞ ⎛ 

3 0 

0⎠ 

+ ⎝14 0 0 7 

⎞ 

0 0 

16 18⎠ 

8 9 

0 

⎛ 

1 

= ⎝14 0 0 

⎞ 

2 3 

16 18⎠ 

0 0 0 

0 0 0 

But if the left-multiplier’s nonzero entries are in the same row then that row of 

the result is a combination. 

3.6 Example 

⎛ 

1 

⎝0 

0 

0 

⎞ ⎛ 

2 1 

0⎠⎝4 

2 

5 

⎞ ⎛ 

3 1 

6⎠ 

=( ⎝0 

0 

0 

⎞ ⎛ 

0 0 

0⎠ 

+ ⎝0 

0 

0 

⎞ ⎛ 

2 1 

0⎠) 

⎝4 

2 

5 

⎞ 

3 

6⎠ 

0 0 0 7 8 9 0 

⎛ 

1 

= ⎝0 

0 

2 

0 

0 0 

⎞ ⎛ 

3 14 

0⎠ 

+ ⎝ 0 

0 0 7 

⎞ 

16 18 

0 0⎠ 

8 9 

0 

⎛ 

15 

0 0 

⎞ 

18 21 

0 0 0 

= ⎝ 0 0 0⎠ 

0 0 0 

Right-multiplication acts in the same way, with columns. 

These observations about matrices that are mostly zeroes extend to arbitrary 

matrices. 

3.7 Lemma In a product of two matrices G and H, the columns of GH are 

formed by taking G times the columns of H 

⎛ 

⎞ 

. 

⎜ 

. 

. 

⎟ 

G · ⎜� 

⎝h1 

··· �hn ⎟ 

⎠ 

. 

. 

. 

. 

= 

⎛ 

. 

⎜ 

. 

. 

⎜ 

⎝G 

· �h1 ··· G · � ⎞ 

⎟ 

hn 

⎟ 

⎠ 

. 

. 

. 

. 

and the rows of GH are formed by taking the rows of G times H 

⎛ ⎞ ⎛ 

⎞ 

··· �g1 ··· ··· �g1 · H ··· 

⎜ ⎟ ⎜ 

⎟ 

⎜ . 

⎝ . 

⎟ 

⎠ · H = ⎜ . 

⎝ . 

⎟ 

⎠ 

··· �gr ··· 

(ignoring the extra parentheses). 

··· �gr · H ···


Proof. We will exhibit the 2×2 case, and leave the general case as an exercise. 

� 

g1,1 

GH = 

�� 

g1,2 h1,1 

� � 

h1,2 g1,1h1,1 + g1,2h2,1 

= 

� 

g1,1h1,2 + g1,2h2,2 

g2,1 g2,2 

h2,1 h2,2 

The right side of the first equation in the result 

� � � � �� 

h1,1 h1,2 

G G = 

h2,1 

h2,2 

g2,1h1,1 + g2,2h2,1 g2,1h1,2 + g2,2h2,2 

�� 

g1,1h1,1 + g1,2h2,1 g1,1h1,2 + g1,2h2,2 

g2,1h1,1 + g2,2h2,1 

g2,1h1,2 + g2,2h2,2 

is indeed the same as the right side of GH, except for the extra parentheses (the 

ones marking the columns as column vectors). The other equation is similarly 

easy to recognize. QED 

An application of those observations is that there is a matrix that just copies 

out the rows and columns. 

3.8 Definition The main diagonal (or principle diagonal or diagonal) ofa 

square matrix goes from the upper left to the lower right. 

3.9 Definition An identity matrix is square and has with all entries zero except 

for ones in the main diagonal. 

⎛ 

1 

⎜ 

⎜0 

In×n = ⎜ 

⎝ 

0 

1 

. 

... 

... 

⎞ 

0 

0 ⎟ 

⎠ 

0 0 ... 1 

3.10 Example The 3×3 identity leaves its multiplicand unchanged both from 

the left 

⎛ 

1 

⎝0 

0 

1 

⎞ ⎛ 

0 2 

0⎠⎝1 

3 

3 

⎞ ⎛ 

6 2 

8⎠ 

= ⎝ 1 

3 

3 

⎞ 

6 

8⎠ 

0 0 1 −7 1 0 −7 1 0 

and from the right. 

⎛ 

2 

⎝ 1 

3 

3 

⎞ ⎛ 

6 1 

8⎠⎝0 

0 

1 

⎞ ⎛ 

0 2 

0⎠ 

= ⎝ 1 

3 

3 

⎞ 

6 

8⎠ 

−7 1 0 0 0 1 −7 1 0 

3.11 Example So does the 2×2 identity matrix. 

⎛ 

1 

⎜ 

⎜0 

⎝1 

⎞ 

−2 

� 

−2 ⎟ 1 

−1⎠ 

0 

⎛ 

� 

1 

0 ⎜ 

= ⎜0 

1 ⎝1 

⎞ 

−2 

−2 ⎟ 

−1⎠ 

4 3 

4 3 

��


In short, an identity matrix is the identity element of the set of n×n matrices, 

with respect to the operation of matrix multiplication. 

We next see two ways to generalize the identity matrix. 

The first is that if the ones are relaxed to arbitrary reals, the resulting matrix 

will rescale whole rows or columns. 

3.12 Definition A diagonal matrix is square and has zeros off the main diagonal. 

⎛ 

a1,1 

⎜ 0 

⎜ 

⎝ 

0 

a2,2 

. 

... 

... 

0 

0 

⎞ 

⎟ 

⎠ 

0 0 ... an,n 

3.13 Example From the left, the action of multiplication by a diagonal matrix 

is to rescales the rows. 

� �� 

2 0 2 1 4 −1 4 2 8 −2 

= 

0 −1 −1 3 4 4 1 −3 −4 −4 

From the right such a matrix rescales the columns. 

� � 

1 2 1 

2 2 2 

⎛ ⎞ 

3 0 0 � � 

⎝0 

2 0 ⎠ 

3 4 −2 

= 

6 4 −4 

0 0 −2 

The second generalization of identity matrices is that we can put a single one 

in each row and column in ways other than putting them down the diagonal. 

3.14 Definition A permutation matrix is square and is all zeros except for a 

single one in each row and column. 

3.15 Example From the left these matrices permute rows. 

⎛ 

0 

⎝1 0 

0 

⎞ ⎛ 

1 1 

0⎠⎝4 

2 

5 

⎞ ⎛ 

3 7 

6⎠ 

= ⎝1 8 

2 

⎞ 

9 

3⎠ 

0 1 0 7 8 9 4 5 6 

From the right they permute columns. 

⎛ 

1 

⎝4 2 

5 

⎞ ⎛ 

3 0 

6⎠⎝1 

0 

0 

⎞ ⎛ 

1 2 

0⎠ 

= ⎝5 3 

6 

⎞ 

1 

4⎠ 

7 8 9 0 1 0 8 9 7 

We finish this subsection by applying these observations to get matrices that 

perform Gauss’ method and Gauss-Jordan reduction.


3.16 Example We have seen how to produce a matrix that will rescale rows. 

Multiplying by this diagonal matrix rescales the second row of the other by a 

factor of three. 

⎛ 

1 

⎝0 0 

3 

⎞ ⎛ 

0 0 

0⎠⎝0 

2 

1/3 

1 

1 

⎞ ⎛ 

1 0 

−1⎠ 

= ⎝0 2 

1 

1 

3 

⎞ 

1 

−3⎠ 

0 0 1 1 0 2 0 1 0 2 0 

We have seen how to produce a matrix that will swap rows. Multiplying by this 

permutation matrix swaps the first and third rows. 

⎛ 

0 

⎝0 0 

1 

⎞ ⎛ 

1 0 

0⎠⎝0 

2 

1 

1 

3 

⎞ ⎛ 

1 1 

−3⎠ 

= ⎝0 0 

1 

2 

3 

⎞ 

0 

−3⎠ 

1 0 0 1 0 2 0 0 2 1 1 

To see how to perform a pivot, we observe something about those two examples. 

The matrix that rescales the second row by a factor of three arises in 

this way from the identity. 

⎛ 

1 

⎝0 

0 

1 

⎞ 

0 

0⎠ 

0 0 1 

3ρ2 

⎛ 

1 

−→ ⎝0 

0 

3 

⎞ 

0 

0⎠ 

0 0 1 

Similarly, the matrix that swaps first and third rows arises in this way. 

⎛ 

1 

⎝0 

0 

1 

⎞ 

0 

0⎠ 

0 0 1 

ρ1↔ρ3 

⎛ 

0 

−→ ⎝0 

0 

1 

⎞ 

1 

0⎠ 

1 0 0 

3.17 Example The 3×3 matrix that arises as 

⎛ 

1 

⎝0 

0 

1 

⎞ 

0 

0⎠ 

0 0 1 

−2ρ2+ρ3 

⎛ 

1 

−→ ⎝0 

0 

1 

⎞ 

0 

0⎠ 

0 −2 1 

will, when it acts from the left, perform the pivot operation −2ρ2 + ρ3. 

⎛ 

1 0 

⎞ ⎛ 

0 1 0 2 

⎞ 

0 

⎛ 

1 0 2 

⎞ 

0 

⎝0 

1 0⎠⎝0 

1 3 −3⎠ 

= ⎝0 

1 3 −3⎠ 

0 −2 1 0 2 1 1 0 0 −5 7 

3.18 Definition The elementary reduction matrices are obtained from identity 

matrices with one Gaussian operation. We denote them: 

(1) I kρi 

−→ Mi(k) fork �= 0; 

(2) I ρi↔ρj 

−→ Pi,j for i �= j; 

(3) I kρi+ρj 

−→ Ci,j(k) fori �= j.


3.19 Lemma Gaussian reduction can be done through matrix multiplication. 

(1) If H kρi 

−→ G then Mi(k)H = G. 

(2) If H ρi↔ρj 

−→ G then Pi,jH = G. 

(3) If H kρi+ρj 

−→ G then Ci,j(k)H = G. 

Proof. Clear. QED 

3.20 Example This is the first system, from the first chapter, on which we 

performed Gauss’ method. 

3x3 =9 

x1 +5x2 − 2x3 =2 

(1/3)x1 +2x2 =3 

It can be reduced with matrix multiplication. Swap the first and third rows, 

⎛ 

0 

⎝0 

0 

1 

⎞ ⎛ 

1 0 

0⎠⎝1 

0 

5 

3 

−2 

⎞ ⎛ 

9 1/3 

2⎠ 

= ⎝ 1 

2 

5 

0 

−2 

⎞ 

3 

2⎠ 

1 0 0 1/3 2 0 3 0 0 3 9 

triple the first row, 

⎛ 

3 

⎝0 

0 

1 

⎞ ⎛ 

0 1/3 

0⎠⎝1 

2 

5 

0 

−2 

⎞ ⎛ 

3 1 

2⎠ 

= ⎝1 

6 

5 

0 

−2 

⎞ 

9 

2⎠ 

0 0 1 0 0 3 9 0 0 3 9 

and then add −1 times the first row to the second. 

⎛ 

1 

⎝−1 

0 

1 

⎞ ⎛ 

0 1 

0⎠⎝1 

6 

5 

0 

−2 

⎞ ⎛ 

9 1 

2⎠ 

= ⎝0 

6 

−1 

0 

−2 

⎞ 

9 

−7⎠ 

0 0 1 0 0 3 9 0 0 3 9 

Now back substitution will give the solution. 

3.21 Example Gauss-Jordan reduction works the same way. For the matrix 

ending the prior example, first adjust the leading entries 

⎛ 

1 

⎝0 0 

−1 

⎞ ⎛ 

0 1 

0 ⎠ ⎝0 6 

−1 

0 

−2 

⎞ ⎛ 

9 1 

−7⎠ 

= ⎝0 6 

1 

0 

2 

⎞ 

9 

7⎠ 

0 0 1/3 0 0 3 9 0 0 1 3 

and to finish, clear the third column and then the second column. 

⎛ 

1 

⎝0 −6 

1 

⎞ ⎛ 

0 1 

0⎠⎝0 

0 

1 

⎞ ⎛ 

0 1 

−2⎠ 

⎝0 6 

1 

0 

2 

⎞ ⎛ 

9 1 

7⎠ 

= ⎝0 0 

1 

0 

0 

⎞ 

3 

1⎠ 

0 0 1 0 0 1 0 0 1 3 0 0 1 3


We have observed the following result, which we shall use in the next subsection. 

3.22 Corollary For any matrix H there are elementary reduction matrices 

R1, ... , Rr such that Rr · Rr−1 ···R1 · H is in reduced echelon form. 

Until now we have taken the point of view that our primary objects of study 

are vector spaces and the maps between them, and have adopted matrices only 

for computational convenience. This subsection show that this point of view 

isn’t the whole story. Matrix theory is a fascinating and fruitful area. 

In the rest of this book we shall continue to focus on maps as the primary 

objects, but we will be pragmatic—if the matrix point of view gives some clearer 

idea then we shall use it. 

Exercises 

� 3.23 Predict the result of each multiplication by an elementary reduction matrix, 

and then � check ��by multiplying � it�out. �� 

3 0 1 2 

4 0 1 2 

1 0 1 2 

(a) 

(b) 

(c) 

0 0 3 4 

0 2 3 4 

−2 1 3 4 

� �� 

1 2 1 −1 

1 2 0 1 

(d) 

(e) 

3 4 0 1 

3 4 1 0 

� 3.24 The need to take linear combinations of rows and columns in tables of numbers 

arises often in practice. For instance, this is a map of part of Vermont and New 

York. 

In part because of Lake Champlain, 

there are no roads connecting some 

pairs of towns. For instance, there 

isnowaytogofromWinooskito 

Grand Isle without going through 

Colchester. (Of course, many other 

roads and towns have been left off 

to simplify the graph. From top to 

bottom of this map is about forty 

miles.) 

Grand Isle 

Swanton 

Colchester 

Winooski 

Burlington 

(a) The incidence matrix of a map is the square matrix whose i, j entry is the 

number of roads from city i to city j. Produce the incidence matrix of this map 

(take the cities in alphabetical order). 

(b) Amatrixissymmetric if it equals its transpose. Show that an incidence 

matrix is symmetric. (These are all two-way streets. Vermont doesn’t have 

many one-way streets.) 

(c) What is the significance of the square of the incidence matrix? The cube?


� 3.25 The need to take linear combinations of rows and columns in tables of numbers 

arises often in practice. For instance, this table gives the number of hours of each 

type done by each worker, and the associated pay rates. Use matrices to compute 

the wages due. 

regular overtime 

Alan 40 12 

Betty 35 6 

Catherine 40 18 

Donald 28 0 

3.26 Find the product of this matrix with its transpose. 

� � 

cos θ − sin θ 

sin θ cos θ 

wage 

regular $25.00 

overtime $45.00 

� 3.27 Prove that the diagonal matrices form a subspace of Mn×n. What is its 

dimension? 

3.28 Does the identity matrix represent the identity map if the bases are unequal? 

3.29 Show that every multiple of the identity commutes with every square matrix. 

Are there other matrices that commute with all square matrices? 

3.30 Prove or disprove: nonsingular matrices commute. 

� 3.31 Show that the product of a permutation matrix and its transpose is an identity 

matrix. 

3.32 Show that if the first and second rows of G are equal then so are the first and 

second rows of GH. Generalize. 

3.33 Describe the product of two diagonal matrices. 

3.34 Write � � 

1 0 

−3 3 

as the product of two elementary reduction matrices. 

� 3.35 Show that if G hasarowofzerosthenGH (if defined) has a row of zeros. 

Does that work for columns? 

3.36 Show that the set of unit matrices forms a basis for Mn×m. 

3.37 Find the formula for the n-th power of this matrix. 

� � 

1 1 

1 0 

� 3.38 The trace of a square matrix is the sum of the entries on its diagonal (its 

significance appears in Chapter Five). Show that trace(GH) =trace(HG). 

� 3.39 A square matrix is upper triangular if its only nonzero entries lie above, or 

on, the diagonal. Show that the product of two upper triangular matrices is upper 

triangular. Does this hold for lower triangular also? 

3.40 A square matrix is a Markov matrix if each entry is between zero and one 

and the sum along each row is one. Prove that a product of Markov matrices is 

Markov. 

� 3.41 Give an example of two matrices of the same rank with squares of differing 

rank. 

3.42 Combine the two generalizations of the identity matrix, the one allowing entires 

to be other than ones, and the one allowing the single one in each row and 

column to be off the diagonal. What is the action of this type of matrix?


3.43 On a computer multiplications are more costly than additions, so people are 

interested in reducing the number of multiplications used to compute a matrix 

product. 

(a) How many real number multiplications are needed in formula we gave for the 

product of a m×r matrix and a r×n matrix? 

(b) Matrix multiplication is associative, so all associations yield the same result. 

The cost in number of multiplications, however, varies. Find the association 

requiring the fewest real number multiplications to compute the matrix product 

of a 5×10 matrix, a 10×20 matrix, a 20×5 matrix, and a 5×1 matrix. 

(c) (Very hard.) Find a way to multiply two 2 × 2 matrices using only seven 

multiplications instead of the eight suggested by the naive approach. 

3.44 [Putnam, 1990, A-5] IfA and B are square matrices of the same size such 

that ABAB = 0, does it follow that BABA =0? 

3.45 [Am. Math. Mon., Dec. 1966] Demonstrate these four assertions to get an alternate 

proof that column rank equals row rank. 

(a) �y · �y = �0 iff�y = �0. 

(b) A�x = �0 iffA trans A�x = �0. 

(c) dim(R(A)) = dim(R(A trans A)). 

(d) col rank(A) =colrank(A trans )=rowrank(A). 

3.46 [Ackerson] Prove (where A is an n×n matrix and so defines a transformation 

of any n-dimensional space V with respect to B,B where B is a basis) dim(R(A)∩ 

N (A)) = dim(R(A)) − dim(R(A 2 )). Conclude 

(a) N (A) ⊂ R(A) iffdim(N (A)) = dim(R(A)) − dim(R(A 2 )); 

(b) R(A) ⊆ N (A) iffA 2 =0; 

(c) R(A) =N (A) iffA 2 =0anddim(N (A)) = dim(R(A)) ; 

(d) dim(R(A) ∩ N (A)) = 0 iff dim(R(A)) = dim(R(A 2 )) ; 

(e) (Requires the Direct Sum subsection, which is optional.) V = R(A) ⊕ N (A) 

iff dim(R(A)) = dim(R(A 2 )). 

3.IV.4 Inverses 

We now consider how to represent the inverse of a linear map. 

We start by recalling some facts about function inverses. ∗ Some functions 

have no inverse, or have an inverse on one side only. 

4.1 Example Where π : R 3 → R 2 is the projection map 

and η : R 2 → R 3 is the embedding 

⎛ 

⎝ x 

⎞ 

y⎠ 

↦→ 

z 

� � 

x 

y 

⎛ 

� � 

x 

↦→ ⎝ 

y 

x 

⎞ 

y⎠ 

0 

∗ More information on function inverses is in the appendix.


the composition π ◦ η is the identity map on R2 . 

⎛ 

� � 

x η 

π ◦ η : ↦−→ ⎝ 

y 

x 

⎞ 

y⎠ 

0 

π 

↦−→ 

� � 

x 

y 

We say π is a left inverse map of η or, what is the same thing, that η is a right 

inverse map of π. However, composition in the other order η ◦π doesn’t give the 

identity map—here is a vector that is not sent to itself under this composition. 

⎛ 

η ◦ π : ⎝ 0 

⎞ 

0⎠ 

1 

π 

⎛ 

� � 

0 η 

↦−→ ↦−→ ⎝ 

0 

0 

⎞ 

0⎠ 

0 

In fact, the projection π has no left inverse at all. For, if f were to be a left 

inverse of π then we would have 

⎛ ⎞ 

x 

f ◦ π : ⎝y⎠ 

z 

π 

⎛ ⎞ 

� � x 

x f 

↦−→ ↦−→ ⎝y⎠ 

y 

z 

for all of the infinitely many z’s. But no function f can send a single argument 

to more than one value. 

(An example of a function with no inverse on either side, is the zero transformation 

on R 2 .) Some functions have a two-sided inverse map, another function 

that is the inverse of the first, both from the left and from the right. For instance, 

the map given by �v ↦→ 2 · �v has the two-sided inverse �v ↦→ (1/2) · �v. In 

this subsection we will focus on two-sided inverses. The appendix shows that a 

function (linear or not) has a two-sided inverse if and only if it is both one-to-one 

and onto. The appendix also shows that if a function f has a two-sided inverse 

then it is unique, and so it is called ‘the’ inverse, and is denoted f −1 . So our 

purpose in this subsection is, where a linear map h has an inverse, to find the 

relationship between Rep B,D(h) and Rep D,B(h −1 ). 

4.2 Definition A matrix G is a left inverse matrix of the matrix H if GH is 

the identity matrix. It is a right inverse matrix if HG is the identity. A matrix 

H with a two-sided inverse is an invertible matrix. That two-sided inverse is 

called the inverse matrix and is denoted H −1 . 

Because of the correspondence between linear maps and matrices, statements 

about map inverses translate into statements about matrix inverses. 

4.3 Lemma If a matrix has both a left inverse and a right inverse then the 

two are equal. 

4.4 Theorem A matrix is invertible if and only if it is nonsingular.


Proof. (For both results.) Given a matrix H, fix spaces of appropriate dimension 

for the domain and codomain. Fix bases for these spaces. With respect to 

these bases, H represents a map h. The statements are true about the map and 

therefore they are true about the matrix. QED 

4.5 Lemma A product of invertible matrices is invertible—if G and H are 

invertible and if GH is defined then GH is invertible and (GH) −1 = H −1 G −1 . 

Proof. (This is just like the prior proof except that it requires two maps.) Fix 

appropriate spaces and bases and consider the represented maps h and g. Note 

that h −1 g −1 is a two-sided map inverse of gh since (h −1 g −1 )(gh) =h −1 (id)h = 

h −1 h = id and (gh)(h −1 g −1 )=g(id)g −1 = gg −1 = id. This equality is reflected 

in the matrices representing the maps, as required. QED 

Here is the arrow diagram giving the relationship between map inverses and 

matrix inverses. It is a special case of the diagram for function composition and 

matrix multiplication. 

↗h H 

Vw.r.t. B 

Ww.r.t. D 

id 

−→ I 

↘ h−1 

H −1 

Vw.r.t. B 

Beyond its place in our general program of seeing how to represent map 

operations, another reason for our interest in inverses comes from solving linear 

systems. A linear system is equivalent to a matrix equation, as here. 

x1 + x2 =3 

2x1 − x2 =2 ⇐⇒ 

� �� 

1 1 x1 3 

= 

2 −1 

2 

By fixing spaces and bases (e.g., R 2 , R 2 and E2, E2), we take the matrix H to 

represent some map h. Then solving the system is the same as asking: what 

domain vector �x is mapped by h to the result � d ? If we could invert h then we 

could solve the system by multiplying Rep D,B(h −1 ) · Rep D( � d)togetRep B(�x). 

4.6 Example We can find a left inverse for the matrix just given 

� �� 

m n 1 1 1 0 

= 

p q 2 −1 0 1 

by using Gauss’ method to solve the resulting linear system. 

m +2n =1 

m − n =0 

p +2q =0 

p − q =1 

Answer: m =1/3, n =1/3, p =2/3, and q = −1/3. This matrix is actually 

the two-sided inverse of H, as can easily be checked. With it we can solve the 

system (∗) above by applying the inverse. 

� � � �� 

x 1/3 1/3 3 5/3 

= 

= 

y 2/3 −1/3 2 4/3 

x2 

(∗)


4.7 Remark Why solve systems this way, when Gauss’ method takes less 

arithmetic (this assertion can be made precise by counting the number of arithmetic 

operations, as computer algorithm designers do)? Beyond its conceptual 

appeal of fitting into our program of discovering how to represent the various 

map operations, solving linear systems by using the matrix inverse has at least 

two advantages. 

First, once the work of finding an inverse has been done, solving a system 

with the same coefficients but different constants is easy and fast: if we change 

the entries on the right of the system (∗) then we get a related problem 

� 

1 

�� 

1 x 

2 −1 y 

= 

� � 

5 

1 

wtih a related solution method. 

� � � �� 

x 1/3 1/3 5 

= 

y 2/3 −1/3 1 

x2 

= 

� � 

2 

3 

In applications, solving many systems having the same matrix of coefficients is 

common. 

Another advantage of inverses is that we can explore a system’s sensitivity 

to changes in the constants. For example, tweaking the 3 on the right of the 

system (∗) to 

� �� 

1 1 x1 3.01 

= 

2 −1 

2 

can be solved with the inverse. 

� �� 

1/3 1/3 3.01 

= 

2/3 −1/3 2 

� � 

(1/3)(3.01) + (1/3)(2) 

(2/3)(3.01) − (1/3)(2) 

to show that x1 changes by 1/3 of the tweak while x2 moves by 2/3 of that 

tweak. This sort of analysis is used, for example, to decide how accurately data 

must be specified in a linear model to ensure that the solution has a desired 

accuracy. 

We finish by describing the computational procedure usually used to find 

the inverse matrix. 

4.8 Lemma A matrix is invertible if and only if it can be written as the product 

of elementary reduction matrices. The inverse can be computed by applying 

to the identity matrix the same row steps, in the same order, as are used to 

Gauss-Jordan reduce the invertible matrix. 

Proof. A matrix H is invertible if and only if it is nonsingular and thus Gauss- 

Jordan reduces to the identity. By Corollary 3.22 this reduction can be done 

with elementary matrices Rr · Rr−1 ...R1 · H = I. This equation gives the two 

halves of the result.


First, elementary matrices are invertible and their inverses are also elementary. 

Applying R−1 r to the left of both sides of that equation, then R −1 

r−1 , etc., 

gives H as the product of elementary matrices H = R −1 

1 ···R−1 r · I (the I is 

here to cover the trivial r =0case). 

Second, matrix inverses are unique and so comparison of the above equation 

with H−1H = I shows that H−1 = Rr · Rr−1 ...R1 · I. Therefore, applying R1 

to the identity, followed by R2, etc., yields the inverse of H. QED 

4.9 Example To find the inverse of 

� � 

1 1 

2 −1 

we do Gauss-Jordan reduction, meanwhile performing the same operations on 

the identity. For clerical convenience we write the matrix and the identity sideby-side, 

and do the reduction steps together. 

� � � � 

1 1 1 0 −2ρ1+ρ2 1 1 1 0 

−→ 

2 −1 0 1 

0 −3 −2 1 

� � 

−1/3ρ2 1 1 1 0 

−→ 

0 1 2/3 −1/3 

� � 

−ρ2+ρ1 1 0 1/3 1/3 

−→ 

0 1 2/3 −1/3 

This calculation has found the inverse. 

� �−1 1 1 

= 

2 −1 

� 

1/3 

� 

1/3 

2/3 −1/3 

4.10 Example This one happens to start with a row swap. 

⎛ 

0 

⎝1 

3 

0 

−1 

1 

1 

0 

0 

1 

⎞ 

0 

0⎠ 

1 −1 0 0 0 1 

ρ1↔ρ2 

−→ 

⎛ 

1 

⎝0 

0 

3 

1 

−1 

0 

1 

1 

0 

⎞ 

0 

0⎠ 

−ρ1+ρ3 

−→ 

1 

⎛ 

1 

⎝0 

−1 

0 

3 

0 

1 

−1 

0 

0 

1 

0 

1 

0 

1 

⎞ 

0 

0⎠ 

0 −1 −1 0 −1 1 

. 

−→ 

⎛ 

1 

⎝0 0 

1 

0 

0 

1/4 

1/4 

1/4 

1/4 

⎞ 

3/4 

−1/4⎠ 

0 0 1 −1/4 3/4 −3/4 

4.11 Example A non-invertible matrix is detected by the fact that the left 

half won’t reduce to the identity. 

� � � � 

1 1 1 0 −2ρ1+ρ2 1 1 1 0 

−→ 

2 2 0 1 

0 0 −2 1


This procedure will find the inverse of a general n×n matrix. The 2×2 case 

is handy. 

4.12 Corollary The inverse for a 2×2 matrix exists and equals 

if and only if ad − bc �= 0. 

� �−1 a b 

= 

c d 

1 

ad − bc 

� 

d 

� 

−b 

−c a 

Proof. This computation is Exercise 22. QED 

We have seen here, as in the Mechanics of Matrix Multiplication subsection, 

that we can exploit the correspondence between linear maps and matrices. So 

we can fruitfully study both maps and matrices, translating back and forth to 

whichever helps us the most. 

Over the entire four subsections of this section we have developed an algebra 

system for matrices. We can compare it with the familiar algebra system for 

the real numbers. Here we are working not with numbers but with matrices. 

We have matrix addition and subtraction operations, and they work in much 

the same way as the real number operations, except that they only combine 

same-sized matrices. We also have a matrix multiplication operation and an 

operation inverse to multiplication. These are somewhat like the familiar real 

number operations (associativity, and distributivity over addition, for example), 

but there are differences (failure of commutativity, for example). And, we have 

scalar multiplication, which is in some ways another extension of real number 

multiplication. This matrix system provides an example that algebra systems 

other than the elementary one can be interesting and useful. 

Exercises 

4.13 Supply the intermediate steps in Example 4.10. 

� 4.14 Use � Corollary � 4.12 to� decide �if 

each matrix � has an � inverse. 

2 1 

0 4 

2 −3 

(a) 

(b) 

(c) 

−1 1 

1 −3 

−4 6 

� 4.15 For each invertible matrix in the prior problem, use Corollary 4.12 to find its 

inverse. 

� 4.16 Find the inverse, if it exists, by using the Gauss-Jordan method. Check the 

answers for the 2×2 matrices with Corollary 4.12. 

� � � � � � 

3 1 

2 1/2 

2 −4 

(a) 

(b) 

(c) 

0 2 

3 1 

−1 2 

� 

0 1 

� 

5 

� 

2 2 

� 

3 

(e) 0 −2 4 (f) 1 −2 −3 

2 3 −2 

4 −2 −3 

� 4.17 What matrix has this one for its inverse? 

� � 

1 3 

2 5 

(d) 

� 

1 1 

� 

3 

0 2 4 

−1 1 0


4.18 How does the inverse operation interact with scalar multiplication and addition 

of matrices? 

(a) What is the inverse of rH? 

(b) Is (H + G) −1 = H −1 + G −1 ? 

� 4.19 Is (T k ) −1 =(T −1 ) k ? 

4.20 Is H −1 invertible? 

4.21 For each real number θ let tθ : R 2 → R 2 be represented with respect to the 

standard bases by this matrix. 

�cos 

θ 

� 

− sin θ 

sin θ cos θ 

Show that tθ1+θ2 = tθ1 · tθ2. Show also that tθ −1 = t−θ. 

4.22 Do the calculations for the proof of Corollary 4.12. 

4.23 Show that this matrix 

H = 

� 

1 0 

� 

1 

0 1 0 

has infinitely many right inverses. Show also that it has no left inverse. 

4.24 In Example 4.1, how many left inverses has η? 

4.25 If a matrix has infinitely many right-inverses, can it have infinitely many 

left-inverses? Must it have? 

� 4.26 Assume that H is invertible and that HG is the zero matrix. Show that G is 

a zero matrix. 

4.27 Prove that if H is invertible then the inverse commutes with a matrix GH −1 = 

H −1 G if and only if H itself commutes with that matrix GH = HG. 

� 4.28 Show that if T is square and if T 4 is the zero matrix then (I − T ) −1 = 

I + T + T 2 + T 3 . Generalize. 

� 4.29 Let D be diagonal. Describe D 2 , D 3 , ... , etc. Describe D −1 , D −2 , ... ,etc. 

Define D 0 appropriately. 

4.30 Prove that any matrix row-equivalent to an invertible matrix is also invertible. 

4.31 The first question below appeared as Exercise 28. 

(a) Show that the rank of the product of two matrices is less than or equal to 

the minimum of the rank of each. 

(b) Show that if T and S are square then TS = I if and only if ST = I. 

4.32 Show that the inverse of a permutation matrix is its transpose. 

4.33 The first two parts of this question appeared as Exercise 25. 

(a) Show that (GH) trans = H trans G trans . 

(b) A square matrix is symmetric if each i, j entry equals the j, i entry (that is, if 

the matrix equals its transpose). Show that the matrices HH trans and H trans H 

are symmetric. 

(c) Show that the inverse of the transpose is the transpose of the inverse. 

(d) Show that the inverse of a symmetric matrix is symmetric. 

� 4.34 The items starting this question appeared as Exercise 30. 

(a) Prove that the composition of the projections πx,πy : R 3 → R 3 is the zero 

map despite that neither is the zero map. 

(b) Prove that the composition of the derivatives d 2 /dx 2 ,d 3 /dx 3 : P4 →P4 is 

the zero map despite that neither map is the zero map. 

(c) Give matrix equations representing each of the prior two items.


When two things multiply to give zero despite that neither is zero, each is said to 

be a zero divisor. Prove that no zero divisor is invertible. 

4.35 In real number algebra, there are exactly two numbers, 1 and −1, that are 

their own multiplicative inverse. Does H 2 = I have exactly two solutions for 2×2 

matrices? 

4.36 Is the relation ‘is a two-sided inverse of’ transitive? Reflexive? Symmetric? 

4.37 [Am. Math. Mon., Nov. 1951] Prove: if the sum of the elements of a square 

matrix is k, then the sum of the elements in each row of the inverse matrix is 1/k.


3.V Change of Basis 

Representations, whether of vectors or of maps, vary with the bases. For instance, 

with respect to the two bases E2 and 

� � � � 

1 1 

B = 〈 , 〉 

1 −1 

for R2 , the vector �e1 has two different representations. 

� � 

� � 

1 

1/2 

RepE2 (�e1) = Rep 

0 

B(�e1) = 

1/2 

Similarly, with respect to E2, E2 and E2,B, the identity map has two different 

representations. 

� � 

� � 

1 0 

1/2 1/2 

RepE2,E2 (id) = 

Rep 

0 1 

E2,B(id) = 

1/2 −1/2 

With our point of view that the objects of our studies are vectors and maps, in 

fixing bases we are adopting a scheme of tags or names for these objects, that 

are convienent for computation. We will now see how to translate among these 

names—we will see exactly how representations vary as the bases vary. 

3.V.1 Changing Representations of Vectors 

In converting Rep B(�v) toRep D(�v) the underlying vector �v doesn’t change. 

Thus, this translation is accomplished by the identity map on the space, described 

so that the domain space vectors are represented with respect to B and 

the codomain space vectors are represented with respect to D. 

Vw.r.t. B 

⏐ 

id� 

Vw.r.t. D 

(The diagram is vertical to fit with the ones in the next subsection.) 

1.1 Definition The change of basis matrix for bases B,D ⊂ V is the representation 

of the identity map id: V → V with respect to those bases. 

⎛ 

. 

. 

⎜ 

. 

. 

RepB,D(id) = ⎜ 

⎝RepD( 

� β1) ··· RepD( � ⎞ 

⎟ 

βn) ⎟ 

⎠ 

. 

. 

. 

.

Section V. Change of Basis 239 

1.2 Lemma Left-multiplication by the change of basis matrix for B,D converts 

a representation with respect to B to one with respect to D. Conversly, if 

left-multiplication by a matrix changes bases M · Rep B(�v) =Rep D(�v) then M 

is a change of basis matrix. 

Proof. For the first sentence, for each �v, as matrix-vector multiplication represents 

a map application, Rep B,D(id) · Rep B(�v) =Rep D(id(�v))=Rep D(�v). For 

the second sentence, with respect to B,D the matrix M represents some linear 

map, whose action is �v ↦→ �v, and is therefore the identity map. QED 

1.3 Example With these bases for R2 , 

� � � � � � � � 

2 1 

−1 1 

B = 〈 , 〉 D = 〈 , 〉 

1 0 

1 1 

because 

� � 

2 

RepD(id( )) = 

1 

� � 

−1/2 

3/2 

D 

the change of basis matrix is this. 

Rep B,D(id) = 

� � 

1 

RepD(id( )) = 

0 

� 

−1/2 

� 

−1/2 

3/2 1/2 

� � 

−1/2 

1/2 

D 

We can see this matrix at work by finding the two representations of �e2 

� � � � 

� � � � 

0 1 

0 1/2 

RepB( )= 

Rep 

1 −2 

D( )= 

1 1/2 

and checking that the conversion goes as expected. 

� �� 

−1/2 −1/2 1 1/2 

= 

3/2 1/2 −2 1/2 

We finish this subsection by recognizing that the change of basis matrices 

are familiar. 

1.4 Lemma A matrix changes bases if and only if it is nonsingular. 

Proof. For one direction, if left-multiplication by a matrix changes bases then 

the matrix represents an invertible function, simply because the function is 

inverted by changing the bases back. Such a matrix is itself invertible, and so 

nonsingular. 

To finish, we will show that any nonsingular matrix M performs a change of 

basis operation from any given starting basis B to some ending basis. Because 

the matrix is nonsingular, it will Gauss-Jordan reduce to the identity, so there 

are elementatry reduction matrices such that Rr ···R1 · M = I. Elementary 

matrices are invertible and their inverses are also elementary, so multiplying 

from the left first by Rr −1 , then by Rr−1 −1 , etc., gives M as a product of


elementary matrices M = R1 −1 ···Rr −1 . Thus, we will be done if we show 

that elementary matrices change a given basis to another basis, for then Rr −1 

changes B to some other basis Br, and Rr−1 −1 changes Br to some Br−1, 

... , and the net effect is that M changes B to B1. We will prove this about 

elementary matrices by covering the three types as separate cases. 

Applying a row-multiplication matrix 

⎛ ⎞ ⎛ ⎞ 

c1 c1 

⎜ . ⎟ ⎜ . ⎟ 

⎜ . ⎟ ⎜ 

⎟ ⎜ . ⎟ 

Mi(k) ⎜ci 

⎟ 

⎜ ⎟ = ⎜kci⎟ 

⎜ ⎟ 

⎜ . ⎟ ⎜ 

⎝ . 

. ⎟ 

. ⎠ ⎝ . ⎠ 

cn 

changes a representation with respect to 〈 � β1,..., � βi,..., � βn〉 to one with respect 

to 〈 � β1,...,(1/k) � βi,..., � βn〉 in this way. 

�v = c1 · � β1 + ···+ ci · � βi + ···+ cn · � βn 

cn 

↦→ c1 · � β1 + ···+ kci · (1/k) � βi + ···+ cn · � βn = �v 

Similarly, left-multiplication by a row-swap matrix Pi,j changes a representation 

with respect to the basis 〈 � β1,..., � βi,..., � βj,..., � βn〉 into one with respect to the 

basis 〈 � β1,..., � βj,..., � βi,..., � βn〉 in this way. 

�v = c1 · � β1 + ···+ ci · � βi + ···+ cj � βj + ···+ cn · � βn 

↦→ c1 · � β1 + ···+ cj · � βj + ···+ ci · � βi + ···+ cn · � βn = �v 

And, a representation with respect to 〈 � β1,..., � βi,..., � βj,..., � βn〉 changes via 

left-multiplication by a row-combination matrix Ci,j(k) into a representation 

with respect to 〈 � β1,..., � βi − k � βj,..., � βj,..., � βn〉 

�v = c1 · � β1 + ···+ ci · � βi + cj � βj + ···+ cn · � βn 

↦→ c1 · � β1 + ···+ ci · ( � βi − k � βj)+···+(kci + cj) · � βj + ···+ cn · � βn = �v 

(the definition of reduction matrices specifies that i �= k and k �= 0 and so this 

last one is a basis). QED 

1.5 Corollary A matrix is nonsingular if and only if it represents the identity 

map with respect to some pair of bases. 

In the next subsection we will see how to translate among representations 

of maps, that is, how to change Rep B,D(h) toRepˆ B, ˆ D (h). The above corollary 

is a special case of this, where the domain and range are the same space, and 

where the map is the identity map.


Exercises 

� 1.6 In R 2 ,where 

� � � � 

2 −2 

D = 〈 , 〉 

1 4 

find the change of basis matrices from D to E2 and from E2 to D. Multiply the 

two. 

� 1.7 Find the change of basis matrix for B,D ⊆ R 2 . � � � � 

1 1 

(a) B = E2, D = 〈�e2,�e1〉 (b) B = E2, D = 〈 , 〉 

2 4 

� � � � 

� � � � � � � � 

1 1 

−1 2 

0 1 

(c) B = 〈 , 〉, D = E2 (d) B = 〈 , 〉, D = 〈 , 〉 

2 4 

1 2 

4 3 

1.8 ForthebasesinExercise7, find the change of basis matrix in the other direction, 

from D to B. 

� 1.9 Find the change of basis matrix for each B,D ⊆P2. 

(a) B = 〈1,x,x 2 〉,D = 〈x 2 , 1,x〉 (b) B = 〈1,x,x 2 〉,D = 〈1, 1+x, 1+x+x 2 〉 

(c) B = 〈2, 2x, x 2 〉,D = 〈1+x 2 , 1 − x 2 ,x+ x 2 〉 

� 1.10 Decide if each changes bases on R 2 . To what basis is E2 changed? 

(a) 

� � 

5 0 

0 4 

(b) 

� � 

2 1 

3 1 

(c) 

� � 

−1 4 

2 −8 

(d) 

� � 

1 −1 

1 1 

1.11 Find bases such that this matrix represents the identity map with respect to 

those bases. � 

3 1 

� 

4 

2 −1 1 

0 0 4 

1.12 Conside the vector space of real-valued functions with basis 〈sin(x), cos(x)〉. 

Show that 〈2sin(x)+cos(x), 3cos(x)〉 is also a basis for this space. Find the change 

of basis matrix in each direction. 

1.13 Wheredoesthismatrix� 

cos(2θ) sin(2θ) 

� 

sin(2θ) − cos(2θ) 

send the standard basis for R 2 ? Any other bases? Hint. Consider the inverse. 

� 1.14 What is the change of basis matrix with respect to B,B? 

1.15 Prove that a matrix changes bases if and only if it is invertible. 

1.16 Finish the proof of Lemma 1.4. 

� 1.17 Let H be a n×n nonsingular matrix. What basis of R n does H change to the 

standard basis? 

� 1.18 (a) In P3 with basis B = 〈1+x, 1 − x, x 2 + x 3 ,x 2 − x 3 〉 we have this repre- 

senatation. 

RepB(1 − x +3x 2 − x 3 ⎛ ⎞ 

0 

⎜1⎟ 

)= ⎝ 

1 

⎠ 

2 

B 

Find a basis D giving this different representation for the same polynomial. 

RepD(1 − x +3x 2 − x 3 ⎛ ⎞ 

1 

⎜0⎟ 

)= ⎝ 

2 

⎠ 

0 

D


(b) State and prove that any nonzero vector representation can be changed to 

any other. 

Hint. The proof of Lemma 1.4 is constructive—it not only says the bases change, 

it shows how they change. 

1.19 Let V,W be vector spaces, and let B, ˆ B be bases for V and D, ˆ D be bases for 

W .Whereh: V → W is linear, find a formula relating RepB,D(h) toRepˆ B, D ˆ (h). 

� 1.20 Show that the columns of an n×n change of basis matrix form a basis for 

R n . Do all bases appear in that way: can the vectors from any R n basis make the 

columns of a change of basis matrix? 

� 1.21 Find a matrix having this effect. 

� � 

1 

↦→ 

3 

� � 

4 

−1 

That is, find a M that left-multiplies the starting vector to yield the ending vector. 

Is there� a�matrix � having � � these � two � effects? � � � � � � � � � 

1 1 2 −1 

1 1 2 −1 

(a) ↦→ 

↦→ (b) ↦→ 

↦→ 

3 1 −1 −1 

3 1 6 −1 

Give a necessary and sufficient condition for there to be a matrix such that �v1 ↦→ �w1 

and �v2 ↦→ �w2. 

3.V.2 Changing Map Representations 

The first subsection shows how to convert the representation of a vector with 

respect to one basis to the representation of that same vector with respect to 

another basis. Here we will see how to convert the representation of a map with 

respect to one pair of bases to the representation of that map with respect to 

a different pair. That is, we want the relationship between the matrices in this 

arrow diagram. 

Vw.r.t. B 

h 

−−−−→ 

H 

Ww.r.t. D 

⏐ 

⏐ 

⏐ 

⏐ 

id� 

id� 

V w.r.t. ˆ B 

h 

−−−−→ 

ˆH 

W w.r.t. ˆ D 

To move from the lower-left of this diagram to the lower-right we can either go 

straight over, or else up to VB then over to WD and then down. Restated in 

terms of the matrices, we can calculate ˆ H =Repˆ B, ˆ D (h) either by simply using 

ˆB and ˆ D, or else by first changing bases with Rep ˆ B,B (id) then multiplying 

by H =Rep B,D(h) and then changing bases with Rep D, ˆ D (id). This equation 

summarizes. 

ˆH =Rep D, ˆ D (id) · H · Rep ˆ B,B (id) (∗)


(To compare this equation with the sentence before it, remember that the equation 

is read from right to left because function composition is read right to left 

and matrix multiplication represent the composition.) 

2.1 Example The matrix 

� � 

cos(π/6) − sin(π/6) 

T = 

= 

sin(π/6) cos(π/6) 

�√ 

3/2 −1/2 

1/2 √ � 

3/2 

represents, with respect to E2, E2, the transformation t: R 2 → R 2 that rotates 

vectors π/6 radians counterclockwise. 

� � 

1 

3 

t 

↦−→ 

� √ 

(−3+ 3)/2 

(1 + 3 √ � 

3)/2 

We can translate that representation with respect to E2, E2 to one with respect 

to 

� �� 

ˆB 

1 0 

= 〈 〉 D ˆ −1 2 

= 〈 〉 

1 2 

0 3 

by using the arrow diagram and formula (∗). 

t 

−−−−→ 

T 

R2 w.r.t. R E2 

2 w.r.t. E2 

⏐ 

⏐ 

⏐ 

⏐ 

id� 

id� 

R 2 

w.r.t. ˆ B 

t 

−−−−→ 

ˆT 

R 2 

w.r.t. ˆ D 

ˆT =Rep E2, ˆ D (id) · T · Rep ˆ B,E2 (id) 

Note that Rep E2, ˆ D (id) can be calculated as the matrix inverse of Rep ˆ D,E2 (id). 

Rep ˆ B, ˆ D (t) = 

� �−1 �√ 

−1 2 3/2 −1/2 

0 3 1/2 √ �� 

1 0 

3/2 1 2 

� √ √ 

(5 − 3)/6 (3+2 3)/3 

= 

(1 + √ � 

√ 

3)/6 3/3 

Although the new matrix is messier-appearing, the map that it represents is the 

same. For instance, to replicate the effect of t in the picture, start with ˆ B, 

� � � � 

1 1 

RepB ˆ( )= 

3 1 

apply ˆ T , 

� √ √ 

(5 − 3)/6 (3+2 3)/3 

(1 + √ � 

√ 

3)/6 3/3 

ˆB, ˆ D 

ˆB 

� � 

1 

= 

1 ˆB 

� (11 + 3 √ 3)/6 

(1 + 3 √ 3)/6 

� 

ˆD


and check it against ˆ D 

11 + 3 √ 3 

6 

· 

� � 

−1 

+ 

0 

1+3√3 · 

6 

to see that it is the same result as above. 

� � 

2 

= 

3 

2.2 Example On R3 the map 

⎛ 

⎝ x 

⎞ 

y⎠ 

z 

t 

⎛ ⎞ 

y + z 

↦−→ ⎝x + z⎠ 

x + y 

� √ 

(−3+ 3)/2 

(1 + 3 √ � 

3)/2 

that is represented with respect to the standard basis in this way 

⎛ 

0 1 

⎞ 

1 

RepE3,E3 (t) = ⎝1 

1 

0 

1 

1⎠ 

0 

can also be represented with respect to another basis 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

1 1 1 

if B = 〈 ⎝−1⎠ 

, ⎝ 1 ⎠ , ⎝1⎠〉 

0 −2 1 

⎛ 

−1 

then RepB,B(t) = ⎝ 0 

0 

0 

−1 

0 

⎞ 

0 

0⎠ 

2 

in a way that is simpler, in that the action of a diagonal matrix is easy to 

understand. 

Naturally, we usually prefer basis changes that make the representation easier 

to understand. When the representation with respect to equal starting and 

ending bases is a diagonal matrix we say the map or matrix has been diagonalized. 

In Chaper Five we shall see which maps and matrices are diagonalizable, 

and where one is not, we shall see how to get a representation that is nearly 

diagonal. 

We finish this subsection by considering the easier case where representations 

are with respect to possibly different starting and ending bases. Recall 

that the prior subsection shows that a matrix changes bases if and only if it 

is nonsingular. That gives us another version of the above arrow diagram and 

equation (∗). 

2.3 Definition Same-sized matrices H and ˆ H are matrix equivalent if there 

are nonsingular matrices P and Q such that ˆ H = PHQ. 

2.4 Corollary Matrix equivalent matrices represent the same map, with respect 

to appropriate pairs of bases. 

Exercise 19 checks that matrix equivalence is an equivalence relation. Thus 

it partitions the set of matrices into matrix equivalence classes.


All matrices: 

.H 

✥ 

✪ 

✩ 

✦ ✜ 

✢ 

... 

ˆH . 

H matrix equivalent 

to ˆ H 

We can get some insight into the classes by comparing matrix equivalence with 

row equivalence (recall that matrices are row equivalent when they can be reduced 

to each other by row operations). In ˆ H = PHQ, the matrices P and 

Q are nonsingular and thus each can be written as a product of elementary 

reduction matrices (Lemma 4.8). Left-multiplication by the reduction matrices 

making up P has the effect of performing row operations. Right-multiplication 

by the reduction matrices making up Q performs column operations. Therefore, 

matrix equivalence is a generalization of row equivalence—two matrices are row 

equivalent if one can be converted to the other by a sequence of row reduction 

steps, while two matrices are matrix equivalent if one can be converted to the 

other by a sequence of row reduction steps followed by a sequence of column 

reduction steps. 

Thus, if matrices are row equivalent then they are also matrix equivalent 

(since we can take Q to be the identity matrix and so perform no column 

operations). The converse, however, does not hold. 

2.5 Example These two 

� 

1 

� 

0 

� 

1 

� 

1 

0 0 0 0 

are matrix equivalent because the second can be reduced to the first by the 

column operation of taking −1 times the first column and adding to the second. 

They are not row equivalent because they have different reduced echelon forms 

(in fact, both are already in reduced form). 

We will close this section by finding a set of representatives for the matrix 

equivalence classes. ∗ 

2.6 Theorem Any m×n matrix of rank k is matrix equivalent to the m×n 

matrix that is all zeros except that the first k diagonal entries are ones. 

⎛ 

1 0 ... 0 0 ... 

⎞ 

0 

⎜ 

0 

⎜ 

⎜0 

⎜ 

⎜0 

⎜ 

⎝ 

1 

. 

0 

0 

. 

... 

... 

... 

0 

1 

0 

0 

0 

0 

... 

... 

... 

0 ⎟ 

0 ⎟ 

0 ⎟ 

⎠ 

0 0 ... 0 0 ... 0 

∗ More information on class representatives is in the appendix.


Sometimes this is described as a block partial-identity form. 

� � 

I Z 

Z Z 

Proof. As discussed above, Gauss-Jordan reduce the given matrix and combine 

all the reduction matrices used there to make P . Then use the leading entries to 

do column reduction and finish by swapping columns to put the leading ones on 

the diagonal. Combine the reduction matrices used for those column operations 

into Q. QED 

2.7 Example We illustrate the proof by finding the P and Q for this matrix. 

⎛ 

1 

⎝0 2 

0 

1 

1 

⎞ 

−1 

−1⎠ 

2 4 2 −2 

First Gauss-Jordan row-reduce. 

⎛ 

1 −1 

⎞ ⎛ 

0 1 0 

⎞ ⎛ 

0 1 2 1 

⎞ 

−1 

⎛ 

1 2 0 

⎞ 

0 

⎝0 

1 0⎠⎝0 

1 0⎠⎝0 

0 1 −1⎠ 

= ⎝0 

0 1 −1⎠ 

0 0 1 −2 0 1 2 4 2 −2 0 0 0 0 

Then column-reduce, which involves right-multiplication. 

⎛ 

1 

⎝0 

0 

2 

0 

0 

0 

1 

0 

⎛ 

⎞ 1 

0 ⎜ 

−1⎠ 

⎜0 

⎝0 

0 

0 

−2 

1 

0 

0 

0 

0 

1 

0 

⎞ ⎛ 

0 1 

0⎟⎜ 

⎟ ⎜0 

0⎠⎝0 

1 0 

0 

1 

0 

0 

0 

0 

1 

0 

⎞ 

0 

0 ⎟ 

1⎠ 

1 

= 

⎛ 

1 

⎝0 

0 

0 

0 

0 

0 

1 

0 

⎞ 

0 

0⎠ 

0 

Finish by swapping columns. 

⎛ 

1 

⎝0 

0 

0 

0 

0 

0 

1 

0 

⎛ 

⎞ 1 

0 ⎜ 

0⎠⎜0 

⎝0 

0 

0 

0 

0 

1 

0 

0 

1 

0 

0 

⎞ 

0 

0 ⎟ 

0⎠ 

1 

= 

⎛ 

1 

⎝0 

0 

0 

1 

0 

0 

0 

0 

⎞ 

0 

0⎠ 

0 

Finally, combine the left-multipliers together as P and the right-multipliers 

together as Q to get the PHQ equation. 

⎛ 

⎞ 

⎛ 

⎞ ⎛ 

⎞ 1 0 −2 0 

1 −1 0 1 2 1 −1 ⎜ 

⎝ 0 1 0⎠⎝0 

0 1 −1⎠ 

⎜0 

0 1 0 ⎟ 

⎝0 

1 0 1⎠ 

−2 0 1 2 4 2 −2 

0 0 0 1 

= 

⎛ ⎞ 

1 0 0 0 

⎝0 1 0 0⎠ 

0 0 0 0 

2.8 Corollary Two same-sized matrices are matrix equivalent if and only if 

they have the same rank. That is, the matrix equivalence classes are characterized 

by rank. 

Proof. Two same-sized matrices with the same rank are equivalent to the same 

block partial-identity matrix. QED


2.9 Example Now that we know that the block partial-identity matrices form 

canonical representatives of the matrix-equivalence classes, we can see what the 

classes look like and how many classes there are. Consider the 2×2 matrices. 

There are only three possible ranks: zero, one, or two. Thus the 2×2 matrices 

fall into three matrix-equivalence classes. 

All 2×2 

matrices: 

⋆ 

� � 

00 

00 

⋆ 

. 

✪ 

✄ ✄✄✄✄✄ 

� . 

. 

. . 

� 10 

00 

⋆ 

� � 

10 

01 

the three 

equivalence 

classes 

Each class just consists of all the 2×2 matrices with the same rank. 

In this subsection we have seen how to change the representation of a map 

with respect to a first pair of bases to one with respect to a second pair. That 

led to a definition describing when matrices are equivalent in this way. Finally 

we noted that, with the proper choice of (possibly different) starting and ending 

bases, any map can be represented in block partial-identity form. 

One of the nice things about this representation is that, in some sense, we 

can completely understand the map when it is expressed in this way: if the 

bases are B = 〈 � β1,..., � βn〉 and D = 〈 � δ1,..., � δm〉 then the map sends 

c1 � β1 + ···+ ck � βk + ck+1 � βk+1 + ···+ cn � βn ↦−→ c1 � δ1 + ···+ ck � δk + �0+···+ �0 

where k is the map’s rank. Thus, we can understand any linear map as a kind 

of projection. 

⎛ 

⎜ 

⎝ 

c1 

. 

ck 

ck+1 

. 

cn 

⎞ 

⎟ 

⎠ 

B 

↦→ 

⎛ ⎞ 

c1 

⎜ . ⎟ 

⎜ . ⎟ 

⎜ck⎟ 

⎜ ⎟ 

⎜ 0 ⎟ 

⎜ ⎟ 

⎝ . ⎠ 

0 

Of course, “understanding” a map expressed in this way requires that we understand 

the relationship between B and D. However, despite that difficulty, 

this is a good classification of linear maps. 

Exercises 

� 2.10 Decide � if these � �matrices are �matrix 

equivalent. 

1 3 0 2 2 1 

(a) 

, 

2 3 0 0 5 −1 

� � � � 

0 3 4 0 

(b) , 

1 1 0 5 

D


� 

1 

(c) 

2 

� � 

3 1 

, 

6 2 

� 

3 

−6 

� 2.11 Find the canonical representative of the matrix-equivalence class of each matrix. 

(a) 

� 

2 

4 

1 

2 

� 

0 

0 

(b) 

� 

0 

1 

3 

1 

1 

3 

0 

0 

3 

� 

2 

4 

−1 

2.12 Suppose that, with respect to 

B = E2 

� � � � 

1 1 

D = 〈 , 〉 

1 −1 

the transformation t: R 2 → R 2 is represented by this matrix. 

� � 

1 2 

3 4 

Use change of basis matrices to represent t with respect to each pair. 

(a) ˆ � � � � 

0 1 

B = 〈 , 〉, 

1 1 

ˆ � � � � 

−1 2 

D = 〈 , 〉 

0 1 

(b) ˆ � � � � 

1 1 

B = 〈 , 〉, 

2 0 

ˆ � � � � 

1 2 

D = 〈 , 〉 

2 1 

� 2.13 What size are P and Q? 

� 2.14 Use Theorem 2.6 to show that a square matrix is nonsingular if and only if it 

is equivalent to an identity matrix. 

� 2.15 Show that, where A is a nonsingular square matrix, if P and Q are nonsingular 

square matrices such that PAQ = I then QP = A −1 . 

� 2.16 Why does Theorem 2.6 not show that every matrix is diagonalizable (see 

Example 2.2)? 

2.17 Must matrix equivalent matrices have matrix equivalent transposes? 

2.18 What happens in Theorem 2.6 if k =0? 

� 2.19 Show that matrix-equivalence is an equivalence relation. 

� 2.20 Show that a zero matrix is alone in its matrix equivalence class. Are there 

other matrices like that? 

2.21 What are the matrix equivalence classes of matrices of transformations on R 1 ? 

R 3 ? 

2.22 How many matrix equivalence classes are there? 

2.23 Are matrix equivalence classes closed under scalar multiplication? Addition? 

2.24 Let t: R n → R n represented by T with respect to En, En. 

(a) Find RepB,B(t) in this specific case. 

� � � � � � 

1 1 

1 −1 

T = 

B = 〈 , 〉 

3 −1 

2 −1 

(b) Describe Rep B,B(t) in the general case where B = 〈 � β1,... , � βn〉. 

2.25 (a) Let V have bases B1 and B2 and suppose that W has the basis D. Where 

h: V → W , find the formula that computes Rep B2,D(h) fromRep B1,D(h). 

(b) Repeat the prior question with one basis for V and two bases for W . 

2.26 (a) If two matrices are matrix-equivalent and invertible, must their inverses 

be matrix-equivalent? 

(b) If two matrices have matrix-equivalent inverses, must the two be matrixequivalent?


(c) If two matrices are square and matrix-equivalent, must their squares be 

matrix-equivalent? 

(d) If two matrices are square and have matrix-equivalent squares, must they be 

matrix-equivalent? 

� 2.27 Square matrices are similar if they represent the same transformation, but 

each with respect to the same ending as starting basis. That is, RepB1,B1 (t) is 

similar to Rep B2,B2 (t). 

(a) Give a definition of matrix similarity like that of Definition 2.3. 

(b) Prove that similar matrices are matrix equivalent. 

(c) Show that similarity is an equivalence relation. 

(d) Show that if T is similar to ˆ T then T 2 is similar to ˆ T 2 , the cubes are similar, 

etc. Contrast with the prior exercise. 

(e) Prove that there are matrix equivalent matrices that are not similar.


3.VI Projection 

The prior section describes the matrix equivalence canonical form as expressing a 

projection and so this section takes the natural next step of studying projections. 

However, this section is optional; only the last two sections of Chapter Five 

require this material. In addition, this section requires some optional material 

from the subsection on length and angle measure in n-space. 

We have described the projection π from R 3 into its xy plane subspace as 

a ‘shadow map’. This shows why, but it also shows that some shadows fall 

upward. 

� � 

1 

2 

2 

� � 

1 

2 

0 

� � 

1 

2 

0 

� � 

1 

2 

−1 

So perhaps a better description is: the projection of �v is the �p in the plane with 

the property that someone standing on �p and looking straight up or down sees 

�v. In this section we will generalize this to other projections, both orthogonal 

(i.e., ‘straight up and down’) and nonorthogonal. 

3.VI.1 Orthogonal Projection Into a Line 

We first consider orthogonal projection into a line. To orthogonally project 

a vector �v into a line ℓ, darken a point on the line if someone on that line and 

looking straight up or down (from that person’s point of view) sees �v. 

�v 

The picture shows someone who has walked out on the line until the tip of 

�v is straight overhead. That is, where the line is described as the span of 

some nonzero vector ℓ = {c · �s � � c ∈ R}, the person has walked out to find the 

coefficient c�p with the property that �v − c�p · �s is orthogonal to c�p · �s. 

�p 

ℓ

Section VI. Projection 251 

�v 

c�p�s 

�v − c�p�s 

We can solve for this coefficient by noting that because �v − c�p�s is orthogonal to 

a scalar multiple of �s it must be orthogonal to �s itself, and then the consequent 

fact that the dot product (�v − c�p�s) �s is zero gives that c�p = �v �s/�s �s. 

1.1 Definition The orthogonal projection of �v into the line spanned by a 

nonzero �s is this vector. 

�v �s 

proj [�s ](�v) = · �s 

�s �s 

Exercise 19 checks that the outcome of the calculation depends only on the line 

and not on which vector �s happens to be used to describe that line. 

1.2 Remark The wording of that definition says ‘spanned by �s ’ instead the 

more formal ‘the span of the set {�s }’. This casual first phrase is common. 

1.3 Example In R2 , to orthogonally project into the line y =2x, wefirstpick 

a direction vector for this line. For instance, 

� � 

1 

�s = 

2 

will do. With that, calculation of a projection is routine. 

� � � � 

� � 

2 1 

2 

� � 

�v = 

3 

3 2 

� � � � 

1 

· = 

1 1 2 

2 2 

8 

5 · 

� � � � 

1 8/5 

= 

2 16/5 

1.4 Example In R3 , the orthogonal projection of a general vector 

⎛ ⎞ 

x 

⎝y⎠ 

z 

into the y-axis is 

⎛ 

⎝ x 

⎞ 

y⎠ 

z 

⎛ 

⎝ 0 

⎞ 

1⎠ 

0 

which matches our intuitive expectation. 

⎛ 

⎝ 0 

⎞ 

1⎠ 

0 

⎛ 

⎝ 0 

⎛ 

⎞ · ⎝ 

1⎠ 

0 

0 

⎞ ⎛ 

1⎠ 

= ⎝ 

0 

0 

⎞ 

y⎠ 

0


The picture above with the stick figure walking out on the line until �v’s tip 

is overhead is one way to think of the orthogonal projection of a vector into a 

line. We finish this subsection with two other ways. 

1.5 Example A railroad car left on an east-west track without its brake is 

pushed by a wind blowing toward the northeast at fifteen miles per hour; what 

speed will the car reach? 

For the wind we use a vector of length 15 that points toward the northeast. 

�v = 

� � 

15 1/2 

15 � � 

1/2 

The car can only be affected by the part of the wind blowing in the east-west 

direction—the part of �v in the direction of the x-axis is this (the picture has the 

same perspective as the railroad car picture above). 

north 

� � � 

15 1/2 

�p = 

0 

�p 

east 

So the car will reach a velocity of 15 � 1/2 miles per hour toward the east. 

Thus, another way to think of the picture that precedes the definition is that 

it shows �v as decomposed into two parts, the part with the line (here, the part 

with the tracks, �p), and the part that is orthogonal to the line (shown here lying 

on the north-south axis). These two are “not interacting” or “independent”, in 

the sense that the east-west car is not at all affected by the north-south part 

of the wind (see Exercise 11). So the orthogonal projection of �v into the line 

spanned by �s can be thought of as the part of �v that lies in the direction of �s. 

Finally, another useful way to think of the orthogonal projection is to have 

the person stand not on the line, but on the vector that is to be projected to the 

line. This person has a rope over the line and pulls it tight, naturally making 

the rope orthogonal to the line.


That is, we can think of the projection �p as being the vector in the line that is 

closest to �v (see Exercise 17). 

1.6 Example A submarine is tracking a ship moving along the line y =3x+2. 

Torpedo range is one-half mile. Can the sub stay where it is, at the origin on 

the chart below, or must it move to reach a place where the ship will pass within 

range? 

north 

The formula for projection into a line does not immediately apply because the 

line doesn’t pass through the origin, and so isn’t the span of any �s. To adjust 

for this, we start by shifting the entire map down two units. Now the line is 

y =3x, which is a subspace, and we can project to get the point �p of closest 

approach, the point on the line through the origin closest to 

the sub’s shifted position. 

�v = 

� � � � 

0 1 

−2 3 

�p = � � � � · 

1 1 

3 3 

east 

� � 

0 

−2 

� � 

1 

= 

3 

� � 

−3/5 

−9/5 

The distance between �v and �p is approximately 0.63 miles and so the sub must 

move to get in range. 

This subsection has developed a natural projection map, orthogonal projection 

into a line. As suggested by the examples, it is often called for in applications. 

The next subsection shows how the definition of orthogonal projection 

into a line gives us a way to calculate especially convienent bases for vector 

spaces, again something that is common in applications. The final subsection 

completely generalizes projection, orthogonal or not, into any subspace at all. 

Exercises 

� 1.7 Project the first vector orthogonally into the line spanned by the second vector. 

� � � � � � � � � � � � � � � � 

1 1 

1 3 

2 3 

2 3 

(a) , (b) , (c) 1 , 2 (d) 1 , 3 

1 −2 

1 0 

4 −1 

4 12 

� 1.8 Project the vector orthogonally into the line.


(a) 

� � 

2 

−1 

4 

, {c 

� � 

−3 �� 

1 c ∈ R} (b) 

−3 

� � 

−1 

, the line y =3x 

−1 

1.9 Although the development of Definition 1.1 is guided by the pictures, we are 

not restricted to spaces that we can draw. In R 4 project this vector into this line. 

⎛ ⎞ 

⎛ ⎞ 

1 

−1 

⎜2⎟ 

⎜ 1 ⎟ 

�v = ⎝ 

1 

⎠ ℓ = {c · ⎝ 

−1 

⎠ 

3 

1 

� � c ∈ R} 

� 1.10 Definition 1.1 uses two vectors �s and �v. Consider the transformation of R 2 

resulting from fixing 

� � 

3 

�s = 

1 

and projecting �v into the line that is the span of �s. Apply it to these vectors. 

� � � � 

1 

0 

(a) (b) 

2 

4 

Show that in general the projection tranformation is this. 

� � � � 

x1 (x1 +3x2)/10 

↦→ 

x2 (3x1 +9x2)/10 

Express the action of this transformation with a matrix. 

1.11 Example 1.5 suggests that projection breaks �v into two parts, proj [�s ](�v )and 

�v − proj [�s ](�v ), that are “not interacting”. Recall that the two are orthogonal. 

Show that any two nonzero orthogonal vectors make up a linearly independent 

set. 

1.12 (a) What is the orthogonal projection of �v into a line if �v isamemberof 

that line? 

(b) Show that if �v is not a member of the line then the set {�v,�v − proj [�s ] (�v )} is 

linearly independent. 

1.13 Definition 1.1 requires that �s be nonzero. Why? What is the right definition 

of the orthogonal projection of a vector into the (degenerate) line spanned by the 

zero vector? 

1.14 Are all vectors the projection of some other vector into some line? 

� 1.15 Show that the projection of �v into the line spanned by �s haslengthequalto 

the absolute value of the number �v �s divided by the length of the vector �s . 

1.16 Find the formula for the distance from a point to a line. 

1.17 Find the scalar c such that (cs1,cs2) is a minimum distance from the point 

(v1,v2) by using calculus (i.e., consider the distance function, set the first derivative 

equal to zero, and solve). Generalize to R n . 

� 1.18 Prove that the orthogonal projection of a vector into a line is shorter than the 

vector. 

� 1.19 Show that the definition of orthogonal projection into a line does not depend 

on the spanning vector: if �s is a nonzero multiple of �q then (�v �s/�s �s ) · �s equals 

(�v �q/�q �q ) · �q. 

� 1.20 Consider the function mapping to plane to itself that takes a vector to its 

projection into the line y = x. These two each show that the map is linear, the


first one in a way that is bound to the coordinates (that is, it fixes a basis and 

then computes) and the second in a way that is more conceptual. 

(a) Produce a matrix that describes the function’s action. 

(b) Show also that this map can be obtained by first rotating everything in the 

plane π/4 radians clockwise, then projecting into the x-axis, and then rotating 

π/4 radians counterclockwise. 

1.21 For �a, �b ∈ R n let �v1 be the projection of �a into the line spanned by �b,let�v2 be 

the projection of �v1 into the line spanned by �a, let�v3 be the projection of �v2 into 

the line spanned by �b, etc., back and forth between the spans of �a and �b.Thatis, �vi+1 is the projection of �vi into the span of �a if i + 1 is even, and into the span of �b if i + 1 is odd. Must that sequence of vectors eventually settle down—must there 

be a sufficiently large i such that �vi+2 equals �vi and �vi+3 equals �vi+1? If so, what 

is the earliest such i? 

3.VI.2 Gram-Schmidt Orthogonalization 

This subsection is optional. It requires material from the prior, also optional, 

subsection. The work done here will only be needed in the final two sections of 

Chapter Five. 

The prior subsection suggests that projecting into the line spanned by �s 

decomposes a vector �v into two parts 

�v 

�v − proj [�s ] (�v ) 

proj [�s ] (�v ) 

�v =proj [�s ](�v) + � �v − proj [�s ](�v) � 

that are orthogonal and so are “not interacting”. We now make that phrase 

precise. 

2.1 Definition Vectors �v1,...,�vk ∈ R n are mutually orthogonal when any two 

are orthogonal: if i �= j then the dot product �vi �vj is zero. 

2.2 Theorem If the vectors in a set {�v1,...,�vk} ⊂R n are mutually orthogonal 

and nonzero then that set is linearly independent. 

Proof. Consider a linear relationship c1�v1 + c2�v2 + ···+ ck�vk = �0. If i ∈ [1..k] 

then taking the dot product of �vi with both sides of the equation 

�vi (c1�v1 + c2�v2 + ···+ ck�vk) =�vi �0 

ci · (�vi �vi) =0 

shows, since �vi is nonzero, that ci is zero. QED


2.3 Corollary If the vectors in a size k subset of an k dimensional space are 

mutually orthogonal and nonzero then that set is a basis for the space. 

Proof. Any linearly independent size k subset of an k dimensional space is a 

basis. QED 

Of course, the converse of Corollary 2.3 does not hold—not every basis of 

every subspace of R n is made of mutually orthogonal vectors. However, we can 

get the partial converse that for every subspace of R n there is at least one basis 

consisting of mutually orthogonal vectors. 

2.4 Example The members � β1 and � β2 of this basis for R 2 are not orthogonal. 

� � � � 

�β2 

4 1 

B = 〈 , 〉 � 

2 3 

β1 

However, we can derive from B a new basis for the same space that does have 

mutually orthogonal members. For the first member of the new basis we simply 

use � β1. 

�κ1 = 

� � 

4 

2 

For the second member of the new basis, we take away from � β2 its part in the 

direction of �κ1, 

� � � � 

1 

1 

�κ2 = − proj 

3 

[�κ1]( )= 

3 

� � 

1 

− 

3 

� � 

2 

= 

1 

� � 

−1 

2 

which leaves the part, �κ2 pictured above, of � β2 that is orthogonal to �κ1 (it is 

orthogonal by the definition of the projection into the span of �κ1). Note that, 

by the corollary, {�κ1,�κ2} is a basis for R 2 . 

2.5 Definition An orthogonal basis for a vector space is a basis of mutually 

orthogonal vectors. 

2.6 Example To turn this basis for R 3 

⎛ 

〈 ⎝ 1 

⎞ ⎛ 

1⎠ 

, ⎝ 

1 

0 

⎞ ⎛ 

2⎠ 

, ⎝ 

1 

1 

⎞ 

0⎠〉 

3 

into an orthogonal basis, we take the first vector as it is given. 

⎛ 

�κ1 = ⎝ 1 

⎞ 

1⎠ 

1 

�κ2


We get �κ2 by starting with the given second vector � β2 and subtracting away the 

part of it in the direction of �κ1. 

⎛ 

�κ2 = ⎝ 0 

⎞ ⎛ 

2⎠ 

− proj [�κ1]( ⎝ 

0 

0 

⎞ ⎛ 

2⎠) 

= ⎝ 

0 

0 

⎞ ⎛ 

2⎠ 

− ⎝ 

0 

2/3 

⎞ ⎛ 

2/3⎠ 

= ⎝ 

2/3 

−2/3 

⎞ 

4/3 ⎠ 

−2/3 

Finally, we get �κ3 by taking the third given vector and subtracting the part of 

it in the direction of �κ1, and also the part of it in the direction of �κ2. 

⎛ 

�κ3 = ⎝ 1 

⎞ ⎛ 

0⎠ 

− proj [�κ1]( ⎝ 

3 

1 

⎞ 

⎛ 

0⎠) 

− proj [�κ2]( ⎝ 

3 

1 

⎞ ⎛ 

0⎠) 

= ⎝ 

3 

−1 

⎞ 

0 ⎠ 

1 

Again the corollary gives that 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

1 −2/3 −1 

〈 ⎝1⎠ 

, ⎝ 4/3 ⎠ , ⎝ 0 ⎠〉 

1 −2/3 1 

is a basis for the space. 

The next result verifies that the process used in those examples works with 

any basis for any subspace of an R n (we are restricted to R n only because we 

have not given a definition of orthogonality for other vector spaces). 

2.7 Theorem (Gram-Schmidt orthogonalization) If 〈 � β1,... � βk〉 is a basis 

for a subspace of R n then, where 

�κ1 = � β1 

�κ2 = � β2 − proj [�κ1]( � β2) 

�κ3 = � β3 − proj [�κ1]( � β3) − proj [�κ2]( � β3) 

. 

�κk = � βk − proj [�κ1]( � βk) −···−proj [�κk−1]( � βk) 

the �κ ’s form an orthogonal basis for the same subspace. 

Proof. We will use induction to check that each �κi is nonzero, is in the span of 

〈 � β1,... � βi〉 and is orthogonal to all preceding vectors: �κ1 �κi = ···= �κi−1 �κi =0. 

With those, and with Corollary 2.3, we will have that 〈�κ1,...�κk〉 is a basis for 

the same space as 〈 � β1,... � βk〉. 

We shall cover the cases up to i = 3, which give the sense of the argument. 

Completing the details is Exercise 23. 

The i = 1 case is trivial—setting �κ1 equal to � β1 makes it a nonzero vector 

since � β1 is a member of a basis, it is obviously in the desired span, and the 

‘orthogonal to all preceding vectors’ condition is vacuously met.


For the i = 2 case, expand the definition of �κ2. 

�κ2 = � β2 − proj [�κ1]( � β2) = � β2 − � β2 �κ1 

· �κ1 = 

�κ1 �κ1 

� β2 − � β2 �κ1 

· 

�κ1 �κ1 

� β1 

This expansion shows that �κ2 is nonzero or else this would be a non-trivial linear 

dependence among the � β’s (it is nontrivial because the coefficient of � β2 is 1) and 

also shows that �κ2 is in the desired span. Finally, �κ2 is orthogonal to the only 

preceding vector 

�κ1 �κ2 = �κ1 ( � β2 − proj [�κ1]( � β2)) = 0 

because this projection is orthogonal. 

The i = 3 case is the same as the i = 2 case except for one detail. As in the 

i = 2 case, expanding the definition 

�κ3 = � β3 − � β3 �κ1 

· �κ1 − 

�κ1 �κ1 

� β3 �κ2 

· �κ2 

�κ2 �κ2 

= � β3 − � β3 �κ1 

· 

�κ1 �κ1 

� β1 − � β3 �κ2 

· 

�κ2 �κ2 

� β2 

� − � β2 �κ1 

· 

�κ1 �κ1 

� β1 

shows that �κ3 is nonzero and is in the span. A calculation shows that �κ3 is 

orthogonal to the preceding vector �κ1. 

� �β3 − proj [�κ1]( � β3) − proj [�κ2]( � β3) � 

�κ1 �κ3 = �κ1 

= �κ1 

� 

�β3 − proj [�κ1]( � β3) � − �κ1 proj [�κ2]( � =0 

β3) 

(Here’s the difference from the i = 2 case—the second line has two kinds of 

terms. The first term is zero because this projection is orthogonal, as in the 

i = 2 case. The second term is zero because �κ1 is orthogonal to �κ2 and so is 

orthogonal to any vector in the line spanned by �κ2.) The check that �κ3 is also 

orthogonal to the other preceding vector �κ2 is similar. QED 

Beyond having the vectors in the basis be orthogonal, we can do more; we 

can arrange for each vector to have length one by dividing each by its own length 

(we can normalize the lengths). 

2.8 Example Normalizing the length of each vector in the orthogonal basis of 

Example 2.6 produces this orthonormal basis. 

⎛ 

1/ 

〈 ⎝ 

√ 3 

1/ √ 3 

1/ √ ⎞ ⎛ 

−1/ 

⎠ , ⎝ 

3 

√ 6 

2/ √ 6 

−1/ √ ⎞ ⎛ 

⎠ , ⎝ 

6 

−1/√2 0 

1/ √ ⎞ 

⎠〉 

2 

Besides its intuitive appeal, and its analogy with the standard basis En for R n , 

an orthonormal basis also simplifies some computations. See Exercise 17, for 

example. 

Exercises 

2.9 Perform the Gram-Schmidt process on each of these bases for R 2 . 

�


� � � � � � � � � � � � 

1 2 

0 −1 

0 −1 

(a) 〈 , 〉 (b) 〈 , 〉 (c) 〈 , 〉 

1 1 

1 3 

1 0 

Then turn those orthogonal bases into orthonormal bases. 

� 2.10 Perform the Gram-Schmidt process on each of these bases for R 3 . 

� � � � � � � � � � � � 

2 1 0 

1 0 2 

(a) 〈 2 , 0 , 3 〉 (b) 〈 −1 , 1 , 3 〉 

2 −1 1 

0 0 1 

Then turn those orthogonal bases into orthonormal bases. 

� 2.11 Find an orthonormal basis for this subspace of R 3 : the plane x − y + z =0. 

2.12 Find an orthonormal basis for this subspace of R 4 . 

⎛ ⎞ 

x 

⎜y 

⎟ 

{ ⎝ 

z 

⎠ 

w 

� � x − y − z + w =0andx + z =0} 

2.13 Show that any linearly independent subset of R n can be orthogonalized without 

changing its span. 

� 2.14 What happens if we apply the Gram-Schmidt process to a basis that is already 

orthogonal? 

2.15 Let 〈�κ1,...,�κk〉 be a set of mutually orthogonal vectors in R n . 

(a) Prove that for any �v in the space, the vector �v−(proj [�κ1](�v )+···+proj [�vk](�v )) 

is orthogonal to each of �κ1, ... , �κk. 

(b) Illustrate the prior item in R 3 by using �e1 as �κ1, using�e2 as �κ2, and taking 

�v to have components 1, 2, and 3. 

(c) Show that proj [�κ1](�v )+···+proj [�vk](�v ) is the vector in the span of the set 

of �κ’s that is closest to �v. Hint. To the illustration done for the prior part, 

add a vector d1�κ1 + d2�κ2 and apply the Pythagorean Theorem to the resulting 

triangle. 

2.16 Find a vector in R 3 that is orthogonal to both of these. 

� � � � 

1 2 

5 2 

−1 0 

� 2.17 One advantage of orthogonal bases is that they simplify finding the representation 

of a vector with respect to that basis. 

(a) For this vector and this non-orthogonal basis for R 2 

� � � � � � 

2 

1 1 

�v = B = 〈 , 〉 

3 

1 0 

first represent the vector with respect to the basis. Then project the vector into 

the span of each basis vector [ � β1] and[ � β2]. 

(b) With this orthogonal basis for R 2 

� � � � 

1 1 

K = 〈 , 〉 

1 −1 

represent the same vector �v with respect to the basis. Then project the vector 

into the span of each basis vector. Note that the coefficients in the representation 

and the projection are the same. 

(c) Let K = 〈�κ1,... ,�κk〉 be an orthogonal basis for some subspace of R n .Prove 

that for any �v in the subspace, the i-th component of the representation RepK(�v ) 

is the scalar coefficient (�v �κi)/(�κi �κi) from proj [�κi] (�v ).


(d) Prove that �v =proj [�κ1](�v )+···+proj [�κk](�v ). 

2.18 Bessel’s Inequality. Consider these orthonormal sets 

B1 = {�e1} B2 = {�e1,�e2} B3 = {�e1,�e2,�e3} B4 = {�e1,�e2,�e3,�e4} 

along with the vector �v ∈ R 4 whose components are 4, 3, 2, and 1. 

(a) Find the coefficient c1 for the projection of �v into the span of the vector in 

B1. Checkthat��v � 2 ≥|c1| 2 . 

(b) Find the coefficients c1 and c2 for the projection of �v into the spans of the 

two vectors in B2. Checkthat��v � 2 ≥|c1| 2 + |c2| 2 . 

(c) Find c1, c2, andc3 associated with the vectors in B3, andc1, c2, c3, andc4 

for the vectors in B4. Check that ��v � 2 ≥|c1| 2 + ···+ |c3| 2 and that ��v � 2 ≥ 

|c1| 2 + ···+ |c4| 2 . 

Show that this holds in general: where {�κ1,... ,�κk} is an orthonormal set and ci is 

coefficient of the projection of a vector �v from the space then ��v � 2 ≥|c1| 2 + ···+ 

|ck| 2 . Hint. One way is to look at the inequality 0 ≤��v − (c1�κ1 + ···+ ck�κk)� 2 

and expand the c’s. 

2.19 Prove or disprove: every vector in R n is in some orthogonal basis. 

2.20 Show that the columns of an n×n matrix form an orthonormal set if and only 

if the inverse of the matrix is its transpose. Produce such a matrix. 

2.21 Does the proof of Theorem 2.2 fail to consider the possibility that the set of 

vectors is empty (i.e., that k =0)? 

2.22 Theorem 2.7 describes a change of basis from any basis B = 〈 � β1,... , � βk〉 to 

one that is orthogonal K = 〈�κ1,... ,�κk〉. Consider the change of basis matrix 

Rep B,K(id). 

(a) Prove that the matrix Rep K,B(id) changing bases in the direction opposite 

to that of the theorem has an upper triangular shape—all of its entries below 

the main diagonal are zeros. 

(b) Prove that the inverse of an upper triangular matrix is also upper triangular 

(if the matrix is invertible, that is). This shows that the matrix Rep B,K(id) 

changing bases in the direction described in the theorem is upper triangular. 

2.23 Complete the induction argument in the proof of Theorem 2.7. 

3.VI.3 Projection Into a Subspace 

This subsection, like the others in this section, is optional. It also requires 

material from the optional earlier subsection on Direct Sums. 

The prior subsections project a vector into a line by decomposing it into two 

parts: the part in the line proj [�s ](�v ) and the rest �v − proj [�s ](�v ). To generalize 

projection to arbitrary subspaces, we follow this idea. 

3.1 Definition For any direct sum V = M ⊕ N and any �v ∈ V ,theprojection 

of �v into M along N is 

where �v = �m + �n with �m ∈ M, �n ∈ N. 

proj M,N(�v )=�m


This definition doesn’t involve a sense of ‘orthogonal’ so we can apply it to 

spaces other than subspaces of an R n . (Definitions of orthogonality for other 

spaces are perfectly possible, but we haven’t seen any in this book.) 

3.2 Example The space M2×2 of 2×2 matrices is the direct sum of these two. 

� 

a 

M = { 

0 

� 

b �� 

a, b ∈ R} 

0 

� 

0 

N = { 

c 

� 

0 �� 

c, d ∈ R} 

d 

To project 

A = 

� � 

3 1 

0 4 

into M along N, we first fix bases for the two subspaces. 

� 

1 

BM = 〈 

0 

� � 

0 0 

, 

0 0 

� 

1 

〉 

0 

� 

0 

BN = 〈 

1 

� � 

0 0 

, 

0 0 

� 

0 

〉 

1 

The concatenation of these 

� 

⌢ 1 

B = BM BN = 〈 

0 

� � 

0 0 

, 

0 0 

� � 

1 0 

, 

0 1 

� � 

0 0 

, 

0 0 

� 

0 

〉 

1 

is a basis for the entire space, because the space is the direct sum, so we can 

use it to represent A. 

� � � � � � � � � � 

3 1 1 0 0 1 0 0 0 0 

=3· +1· +0· +4· 

0 4 0 0 0 0 1 0 0 1 

Now the projection of A into M along N is found by keeping the M part of this 

sum and dropping the N part. 

� � � � � � � � 

3 1 1 0 0 1 3 1 

projM,N( )=3· +1· = 

0 4 0 0 0 0 0 0 

3.3 Example Both subscripts on projM,N(�v ) are significant. The first subscript 

M matters because the result of the projection is an �m ∈ M, and changing 

this subspace would change the possible results. For an example showing that 

the second subscript matters, fix this plane subspace of R3 and its basis 

⎛ 

M = { ⎝ x 

⎞ 

y⎠ 

z 

� ⎛ 

� y − 2z =0} BM = 〈 ⎝ 1 

⎞ ⎛ 

0⎠ 

, ⎝ 

0 

0 

⎞ 

2⎠〉 

1 

and compare the projections along two different subspaces. 

⎛ 

N = {k ⎝ 0 

⎞ 

0⎠ 

1 

� ⎛ 

� k ∈ R} N ˆ = {k ⎝ 0 

⎞ 

1 ⎠ 

−2 

� � k ∈ R}


(Verification that R 3 = M ⊕ N and R 3 = M ⊕ ˆ N is routine.) We will check 

that these projections are different by checking that they have different effects 

on this vector. 

⎛ 

�v = ⎝ 2 

⎞ 

2⎠ 

5 

For the first one we find a basis for N 

⎛ 

BN = 〈 ⎝ 0 

⎞ 

0⎠〉 

1 

⌢ 

and represent �v with respect to the concatenation BM BN . 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

2 1 0 0 

⎝2⎠ 

=2· ⎝0⎠ 

+1· ⎝2⎠ 

+4· ⎝0⎠ 

5 0 1 1 

The projection of �v into M along N is found by keeping the M part and dropping 

the N part. 

⎛ ⎞ 

1 

⎛ ⎞ 

0 

⎛ ⎞ 

2 

projM,N(�v )=2· ⎝0⎠ 

+1· ⎝2⎠ 

= ⎝2⎠ 

0 1 1 

For the other subspace ˆ N, this basis is natural. 

⎛ ⎞ 

0 

BN ˆ = 〈 ⎝ 1 ⎠〉 

−2 

Representing �v with respect to the concatenation 

⎛ ⎞ 

2 

⎛ ⎞ 

1 

⎛ ⎞ 

0 

⎛ ⎞ 

0 

⎝2⎠ 

=2· ⎝0⎠ 

+(9/5) · ⎝2⎠ 

− (8/5) · ⎝ 1 ⎠ 

5 0 

1 

−2 

and then keeping only the M part gives this. 

⎛ 

projM, N ˆ (�v )=2· ⎝ 1 

⎞ ⎛ 

0⎠ 

+(9/5) · ⎝ 

0 

0 

⎞ ⎛ 

2⎠ 

= ⎝ 

1 

2 

⎞ 

18/5⎠ 

9/5 

Therefore projection along different subspaces may yield different results. 

These pictures compare the two maps. Both show that the projection is 

indeed ‘into’ the plane and ‘along’ the line.


N 

M 

Notice that the projection along N is not orthogonal—there are members of the 

plane M that are not orthogonal to the dotted line. But the projection along 

ˆN is orthogonal. 

A natural question is: what is the relationship between the projection operation 

defined above, and the operation of orthogonal projection into a line? 

The second picture above suggests the answer—orthogonal projection into a 

line is a special case of the projection defined above; it is just projection along 

a subspace perpendicular to the line. 

N 

In addition to pointing out that projection along a subspace is a generalization, 

this scheme shows how to define orthogonal projection into any subspace of R n , 

of any dimension. 

3.4 Definition The orthogonal complement of a subspace M of R n is 

M ⊥ = {�v ∈ R n � � �v is perpendicular to all vectors in M} 

(read “M perp”). The orthogonal projection proj M (�v ) of a vector is its projection 

into M along M ⊥ . 

3.5 Example In R3 , to find the orthogonal complement of the plane 

⎛ 

P = { ⎝ x 

⎞ 

y⎠ 

z 

� � 3x +2y − z =0} 

we start with a basis for P . 

ˆN 

M 

⎛ 

B = 〈 ⎝ 1 

⎞ ⎛ 

0⎠ 

, ⎝ 

3 

0 

⎞ 

1⎠〉 

2 

Any �v perpendicular to every vector in B is perpendicular to every vector in the 

span of B (the proof of this assertion is Exercise 19). Therefore, the subspace 

M


P ⊥ consists of the vectors that satisfy these two conditions. 

⎛ 

⎝ 1 

⎞ 

0⎠ 

⎛ 

⎝ 

3 

v1 

⎞ 

v2⎠ 

=0 

⎛ 

⎝ 0 

⎞ 

1⎠ 

⎛ 

⎝ 

2 

v1 

⎞ 

v2⎠ 

=0 

v3 

We can express those conditions more compactly as a linear system. 

P ⊥ ⎛ 

= { ⎝ v1 

⎞ 

v2⎠ 

� � 

� 1 

0 

0 

1 

� 

3 

2 

⎛ 

⎝ v1 

⎞ 

� � 

v2⎠ 

0 

= } 

0 

v3 

v3 

We are thus left with finding the nullspace of the map represented by the matrix, 

that is, with calculating the solution set of a homogeneous linear system. 

P ⊥ ⎛ ⎞ 

v1 

= { ⎝v2⎠ 

� � v1 

⎛ ⎞ 

−3 

+3v3 =0 

} = {k ⎝−2⎠ 

v2 +2v3 =0 

1 

� � k ∈ R} 

3.6 Example Where M is the xy-plane subspace of R 3 , what is M ⊥ ? A 

common first reaction is that M ⊥ is the yz-plane, but that’s not right. Some 

vectors from the yz-plane are not perpendicular to every vector in the xy-plane. 

� � 

1 

1 �⊥ 

0 

� � 

0 

3 

2 

cos θ = 

v3 

v3 

1 · 0+1· 3+0· 2 

√ 1+1+0· √ 0+9+4 so θ ≈ 0.94 rad 

Instead M ⊥ is the z-axis, since proceeding as in the prior example and taking 

the natural basis for the xy-plane gives this. 

M ⊥ ⎛ ⎞ 

x 

= { ⎝y⎠ 

z 

� � � 

� 1 0 0 

0 1 0 

⎛ ⎞ 

⎛ ⎞ 

x � � x 

⎝y⎠ 

0 

= } = { ⎝y⎠ 

0 

z 

z 

� � x =0andy =0} 

The two examples that we’ve seen since Definition 3.4 illustrate the first 

sentence in that definition. The next result justifies the second sentence. 

3.7 Lemma Let M be a subspace of R n . The orthogonal complement of M is 

also a subspace. The space is the direct sum of the two R n = M ⊕ M ⊥ . And, 

for any �v ∈ R n , the vector �v − proj M (�v ) is perpendicular to every vector in M. 

Proof. First, the orthogonal complement M ⊥ is a subspace of R n because, as 

noted in the prior two examples, it is a nullspace. 

Next, we can start with any basis BM = 〈�µ1,...,�µk〉 for M and expand it to 

a basis for the entire space. Apply the Gram-Schmidt process to get an orthogonal 

basis K = 〈�κ1,...,�κn〉 for R n . This K is the concatenation of two bases 

〈�κ1,...,�κk〉 (with the same number of members as BM) and〈�κk+1,...,�κn〉. 

ThefirstisabasisforM, so if we show that the second is a basis for M ⊥ then 

we will have that the entire space is the direct sum of the two subspaces.


Exercise 17 from the prior subsection proves this about any orthogonal basis: 

each vector �v in the space is the sum of its orthogonal projections onto the 

lines spanned by the basis vectors. 

�v =proj [�κ1](�v )+···+proj [�κn](�v ) (∗) 

To check this, represent the vector �v = r1�κ1 + ···+ rn�κn, apply �κi to both sides 

�v �κi =(r1�κ1 + ···+ rn�κn) �κi = r1 · 0+···+ ri · (�κi �κi) +···+ rn · 0, and 

solve to get ri =(�v �κi)/(�κi �κi), as desired. 

Since obviously any member of the span of 〈�κk+1,...,�κn〉 is orthogonal to 

any vector in M, to show that this is a basis for M ⊥ we need only show the 

other containment—that any �w ∈ M ⊥ is in the span of this basis. The prior 

paragraph does this. On projections into basis vectors from M, any�w ∈ M ⊥ 

gives proj [�κ1]( �w )=�0,..., proj [�κk]( �w )=�0 and therefore (∗) gives that �w is a 

linear combination of �κk+1,...,�κn. ThusthisisabasisforM ⊥ and R n is the 

direct sum of the two. 

The final sentence is proved in much the same way. Write �v =proj [�κ1](�v )+ 

···+proj [�κn](�v ). Then proj M (�v ) is gotten by keeping only the M part and 

dropping the M ⊥ part proj M (�v )=proj [�κk+1](�v )+···+proj [�κk](�v ). Therefore 

�v − proj M (�v ) consists of a linear combination of elements of M ⊥ and so is 

perpendicular to every vector in M. QED 

We can find the orthogonal projection into a subspace by following the steps 

of the proof, but the next result gives a convienent formula. 

3.8 Theorem Let �v be a vector in R n and let M be a subspace of R n 

with basis 〈 � β1,..., � βk〉. If A is the matrix whose columns are the � β’s then 

proj M (�v )=c1 � β1 + ···+ ck � βk where the coefficients ci are the entries of the 

vector (A trans A)A trans · �v. That is, proj M (�v )=A(A trans A) −1 A trans · �v. 

Proof. The vector proj M(�v) is a member of M and so it is a linear combination 

of basis vectors c1 · � β1 + ···+ ck · � βk. Since A’s columns are the � β’s, that can 

be expressed as: there is a �c ∈ R k such that proj M(�v )=A�c (this is expressed 

compactly with matrix multiplication as in Example 3.5 and 3.6). Because 

�v − proj M(�v ) is perpendicular to each member of the basis, we have this (again, 

expressed compactly). 

�0 =A trans� �v − A�c � = A trans �v − A trans A�c 

Solving for �c (showing that A trans A is invertible is an exercise) 

�c = � A trans A � −1 A trans · �v 

gives the formula for the projection matrix as proj M (�v )=A · �c. QED 

3.9 Example To orthogonally project this vector into this subspace 

⎛ 

�v = ⎝ 1 

⎞ ⎛ 

−1⎠ 

P = { ⎝ 

1 

x 

⎞ 

y⎠ 

z 

� � x + z =0}


first make a matrix whose columns are a basis for the subspace 

⎛ ⎞ 

0 1 

A = ⎝1 0 ⎠ 

0 −1 

and then compute. 

A � A trans A � −1 A trans = 

⎛ ⎞ 

0 1 � �� 

⎝1 0 ⎠ 

0 1 1 

1/2 0 0 

0 −1 

⎛ 

⎞ 

1/2 0 −1/2 

= ⎝ 0 1 0 ⎠ 

0 

1 

� 

−1 

0 

−1/2 0 1/2 

With the matrix, calculating the orthogonal projection of any vector into P is 

easy. 

⎛ 

⎞ ⎛ ⎞ ⎛ ⎞ 

1/2 0 −1/2 1 0 

projP (�v) = ⎝ 0 1 0 ⎠ ⎝−1⎠ 

= ⎝−1⎠ 

−1/2 0 1/2 1 0 

Exercises 

� 3.10 Project � � the vectors �into � M along N. � � 

3 

x �� x �� 

(a) , M = { x + y =0}, N = { −x − 2y =0} 

−2 

y 

y 

� � � � � � 

1 

x �� x �� 

(b) , M = { x − y =0}, N = { 2x + y =0} 

2 

y 

y 

� � � � � � 

3 

x �� 1 �� 

(c) 0 , M = { y x + y =0}, N = {c · 0 c ∈ R} 

1 

z 

1 

� 3.11 Find M ⊥ . � � � � 

x �� x �� 

(a) M = { x + y =0} (b) M = { −2x +3y =0} 

y 

y 

� � � � 

x �� x �� 

(c) M = { x − y =0} (d) M = {�0 } (e) M = { x =0} 

y 

y 

� � � � 

x �� x �� 

(f) M = { y −x +3y + z =0} (g) M = { y x =0andy + z =0} 

z 

z 

3.12 This subsection shows how to project orthogonally in two ways, the method of 

Example 3.2 and 3.3, and the method of Theorem 3.8. To compare them, consider 

the plane P specified by 3x +2y − z =0inR 3 . 

(a) Find a basis for P . 

(b) Find P ⊥ and a basis for P ⊥ . 

(c) Represent this vector with respect to the concatenation of the two bases from 

the prior item. 

�v = 

� � 

1 

1 

2


(d) Find the orthogonal projection of �v into P by keeping only the P part from 

the prior item. 

(e) Check that against the result from applying Theorem 3.8. 

� 3.13 We have three ways to find the orthogonal projection of a vector into a line, 

the Definition 1.1 way from the first subsection of this section, the Example 3.2 

and 3.3 way of representing the vector with respect to a basis for the space and 

then keeping the M part, and the way of Theorem 3.8. For these cases, do all 

three ways. � � � � 

1 

x �� 

(a) �v = , M = { x + y =0} 

−3 

y 

� � � � 

0 

x �� 

(b) �v = 1 , M = { y x + z = 0 and y =0} 

2 

z 

3.14 Check that the operation of Definition 3.1 is well-defined. That is, in Example 

3.2 and 3.3, doesn’t the answer depend on the choice of bases? 

3.15 What is the orthogonal projection into the trivial subspace? 

3.16 What is the projection of �v into M along N if �v ∈ M? 

3.17 Show that if M ⊆ R n is a subspace with orthonormal basis 〈�κ1,... ,�κn〉 then 

the orthogonal projection of �v into M is this. 

(�v �κ1) · �κ1 + ···+(�v �κn) · �κn 

� 3.18 Prove that the map p: V → V is the projection into M along N if and only 

if the map id − p is the projection into N along M. (Recall the definition of the 

difference of two maps: (id − p)(�v) =id(�v) − p(�v) =�v − p(�v).) 

� 3.19 Show that if a vector is perpendicular to every vector in a set then it is 

perpendicular to every vector in the span of that set. 

3.20 True or false: the intersection of a subspace and its orthogonal complement is 

trivial. 

3.21 Show that the dimensions of orthogonal complements add to the dimension 

oftheentirespace. 

� 3.22 Suppose that �v1,�v2 ∈ R n are such that for all complements M,N ⊆ R n ,the 

projections of �v1 and �v2 into M along N are equal. Must �v1 equal �v2? (If so, what 

if we relax the condition to: all orthogonal projections of the two are equal?) 

� 3.23 Let M,N be subspaces of R n . The perp operator acts on subspaces; we can 

ask how it interacts with other such operations. 

(a) Show that two perps cancel: (M ⊥ ) ⊥ = M. 

(b) Prove that M ⊆ N implies that N ⊥ ⊆ M ⊥ . 

(c) Show that (M + N) ⊥ = M ⊥ ∩ N ⊥ . 

� 3.24 The material in this subsection allows us to express a geometric relationship 

that we have not yet seen between the rangespace and the nullspace of a linear 

map. 

(a) Represent f : R 3 → R given by 

� � 

v1 

v2 

v3 

↦→ 1v1 +2v2 +3v3


with respect to the standard bases and show that 

� � 

1 

2 

3 

is a member of the perp of the nullspace. Prove that N (f) ⊥ is equal to the 

span of this vector. 

(b) Generalize that to apply to any f : R n → R. 

(c) Represent f : R 3 → R 2 

� � 

v1 � � 

1v1 +2v2 +3v3 

v2 ↦→ 

4v1 +5v2 +6v3 

with respect to the standard bases and show that 

� � � � 

1 4 

2 , 5 

3 6 

v3 

are both members of the perp of the nullspace. Prove that N (f) ⊥ is the span 

of these two. (Hint. See the third item of Exercise 23.) 

(d) Generalize that to apply to any f : R n → R m . 

This, and related results, is called the Fundamental Theorem of Linear Algebra in 

[Strang 93]. 

3.25 Define a projection to be a linear transformation t: V → V with the property 

that repeating the projection does nothing more than does the projection alone: (t◦ 

t)(�v) =t(�v) for all �v ∈ V . 

(a) Show that orthogonal projection into a line has that property. 

(b) Show that projection along a subspace has that property. 

(c) Show that for any such t there is a basis B = 〈 � β1,... , � βn〉 for V such that 

t( � � 

�βi i =1, 2,..., r 

βi) = 

�0 i = r +1,r+2,..., n 

where r is the rank of t. 

(d) Conclude that every projection is a projection along a subspace. 

(e) Also conclude that every projection has a representation 

� � 

I Z 

RepB,B(t) = 

Z Z 

in block partial-identity form. 

3.26 A square matrix is symmetric if each i, j entry equals the j, i entry (i.e., if the 

matrix equals its transpose). Show that the projection matrix A(A trans A) −1 A trans 

is symmetric. Hint. Find properties of transposes by looking in the index under 

‘transpose’.

Topic: Line of Best Fit 269 

Topic: Line of Best Fit 

This Topic requires the formulas from the subsections on Orthogonal Projection 

IntoaLine,andProjectionIntoaSubspace. 

Scientists are often presented with a system that has no solution and they 

must find an answer anyway, that is, they must find a value that is as close as 

possible to being an answer. An often-encountered example is in finding a line 

that, as closely as possible, passes through experimental data. 

For instance, suppose that we have a coin to flip, and want to know: is it 

fair? This question means that a coin has some proportion m of heads to flips, 

determined by how it is balanced beween the two sides, and we want to know 

if m =1/2. We can get experimental information about it by flipping the coin 

many times. This is the result a penny experiment, including some intermediate 

numbers. 

number of flips 30 60 90 

number of heads 16 34 51 

Naturally, because of randomness, the exact proportion is not found with this 

sample — indeed, there is no solution to this system. 

30m =16 

60m =34 

90m =51 

That is, the vector of experimental data is not in the subspace of solutions. 

⎛ ⎞ ⎛ ⎞ 

16 30 

⎝34⎠ 

�∈ {m ⎝60⎠ 

51 90 

� � m ∈ R} 

However, as described above, we expect that there is an m that nearly works. 

An orthogonal projection of the data vector into the line subspace gives our best 

guess at m. 

⎛ ⎞ ⎛ ⎞ 

16 30 

⎝34⎠ 

⎝60⎠ 

⎛ ⎞ 

51 90 

30 

⎛ ⎞ ⎛ ⎞ · ⎝60⎠ 

= 

30 30 

90 

⎝60⎠ 

⎝60⎠ 

90 90 

7110 

12600 · 

⎛ ⎞ 

30 

⎝60⎠ 

90 

The estimate (m = 7110/12600 ≈ 0.56) is higher than 1/2, but not by much, so 

probably the penny is fair enough for flipping purposes. 

The line with the slope m ≈ 0.56 is called the line of best fit for this data. 

heads 

60 

30 

�� 

�� 

�� 

30 60 90 

flips


Minimizing the distance between the given vector and the vector used as the 

right-hand side minimizes the total of these vertical lengths (these have been 

distorted, exaggerated by a factor of ten, to make them more visible). 

�� 

Because it involves minimizing this total distance, we say that the line has been 

obtained through fitting by least-squares. 

In the previous example, the line that we use, whose slope is our best guess 

of the true ratio of heads to flips, must pass through (0, 0). We can also handle 

cases where the line is not required to pass through the origin. 

For example, the different denominations of U.S. money have different average 

times in circulation (the $2 bill is left off as a special case). How long should 

we expect a $25 bill to last? 

denomination 1 5 10 20 50 100 

average life (years) 1.5 2 3 5 9 20 

The plot (see below) looks roughly linear. It isn’t a perfect line, i.e., the linear 

system with equations b +1m =1.5, ... , b + 100m = 20 has no solution, but 

we can again use orthogonal projection to find a best approximation. Consider 

the matrix of coefficients of that linear system and also its vector of constants, 

the experimentally-determined values. 

⎛ ⎞ ⎛ ⎞ 

1 1 

⎜ 

⎜1 

5 ⎟ 

⎜ 

A = ⎜1 

10 ⎟ 

⎜ 

⎜1 

20 ⎟ 

⎝1 

50 ⎠ 

1 100 

�� 

�� 

1.5 

⎜ 2 ⎟ 

⎜ 

�v = ⎜ 3 ⎟ 

⎜ 5 ⎟ 

⎝ 9 ⎠ 

20 

The ending result in the subsection on Projection into a Subspace says that 

coefficients b and m so that the linear combination of the columns of A is as 

close as possible to the vector �v are the entries of (A trans A) −1 A trans · �v. Some 

calculation gives an intercept of b =1.05 and a slope of m =0.18. 

avg life 

15 

5 

�� 

�� 

�� 

10 30 50 70 90 

�� 

�� 

denom 

Plugging x = 25 into the equation of the line shows that such a bill should last 

between five and six years.


We close with an example [Oakley & Baker] that cautions about overusing 

least-squares fitting. These are the world record times for the men’s mile race 

that were in force on January first of the given years. We want to project when 

a 3:40 mile will be run. 

year 1870 1880 1890 1900 1910 1920 1930 

seconds 268.8 264.5 258.4 255.6 255.6 252.6 250.4 

1940 1950 1960 1970 1980 1990 

246.4 241.4 234.5 231.1 229.0 226.3 

The plot below shows that the data is surprisingly linear. With this input 

⎛ 

1 

⎜ 

⎜1 

⎜ 

A = 

. 

⎜ 

⎜. 

⎝1 

⎞ 

1860 

1870 ⎟ 

. ⎟ 

. ⎟ 

1980⎠ 

⎛ ⎞ 

280.0 

⎜ 268.8 ⎟ 

⎜ 

�v = 

. ⎟ 

⎜ . ⎟ 

⎝ 229.0 ⎠ 

1 1990 

226.32 

MAPLE gives b = 970.68 and m = −0.37 (rounded to two places). 

secs 

290 

270 

250 

230 

�� 

�� 

�� 

�� 

�� 

�� 

1870 1890 1910 1930 1950 1970 1990 

�� 

When will a 220 second mile be run? Solving 220 = 970.68 − 0.37x gives an 

estimate of the year 2027. 

This example is amusing, but serves as a caution because the linearity of the 

data will break down someday — the tool of fitting by orthogonal projection 

should be applied judicioulsy. 

Exercises 

The calculations here are most practically done on a computer. In addition, some 

of the problems require more data, available in your library, on the net, or in the 

Answers to the Exercises. 

1 Use least-squares to judge if the coin in this experiment is fair. 

flips 8 16 24 32 40 

heads 4 9 13 17 20 

2 For the men’s mile record, rather than give each of the many records and its 

exact date, we’ve “smoothed” the data somewhat by taking a periodic sample. Do 

the longer calculation and compare the conclusions. 

3 Find the line of best fit for the men’s 1500 meter run. How does the slope compare 

with that for the men’s mile (the distances are close; a mile is about 1609 meters)? 

4 Find the line of best fit for the records for women’s mile. 

5 Do the lines of best fit for the men’s and women’s miles cross? 

�� 

�� 

�� 

�� 

�� 

�� 

year


6 When the space shuttle Challenger exploded in 1986, one of the criticisms made of 

NASA’s decision to launch was in the way the analysis of number of O-ring failures 

versus temperature was made (of course, O-ring failure caused the explosion). Four 

O-ring failures will cause the rocket to explode. NASA had data from 24 previous 

flights. 

temp ◦ F 53 75 57 58 63 70 70 66 67 67 67 

failures 3 2 1 1 1 1 1 0 0 0 0 

68 69 70 70 72 73 75 76 76 78 79 80 81 

0 0 0 0 0 0 0 0 0 0 0 0 0 

The temperature that day was forecast to be 31 ◦ F. 

(a) NASA based the decision to launch partially on a chart showing only the 

flights that had at least one O-ring failure. Find the line that best fits these 

seven flights. On the basis of this data, predict the number of O-ring failures 

when the temperature is 31, and when the number of failures will exceed four. 

(b) Find the line that best fits all 24 flights. On the basis of this extra data, 

predict the number of O-ring failures when the temperature is 31, and when the 

number of failures will exceed four. 

Which do you think is the more accurate method of predicting? (An excellent 

discussion appears in [Dalal, et. al.].) 

7 This table lists the average distance from the sun to each of the first seven planets, 

using earth’s average as a unit. 

Mercury Venus Earth Mars Jupiter Saturn Uranus 

0.39 0.72 1.00 1.52 5.20 9.54 19.2 

(a) Plot the number of the planet (Mercury is 1, etc.) versus the distance. Note 

that it does not look like a line, and so finding the line of best fit is not fruitful. 

(b) It does, however look like an exponential curve. Therefore, plot the number 

of the planet versus the logarithm of the distance. Does this look like a line? 

(c) The asteroid belt between Mars and Jupiter is thought to be what is left of a 

planet that broke apart. Renumber so that Jupiter is 6, Saturn is 7, and Uranus 

is 8, and plot against the log again. Does this look better? 

(d) Use least squares on that data to predict the location of Neptune. 

(e) Repeat to predict where Pluto is. 

(f) Is the formula accurate for Neptune and Pluto? 

This method was used to help discover Neptune (although the second item is misleading 

about the history; actually, the discovery of Neptune in position 9 prompted 

people to look for the “missing planet” in position 5). See [Gardner, 1970] 

8 William Bennett has proposed an Index of Leading Cultural Indicators for the 

US ([Bennett], in 1993). Among the statistics cited are the average daily hours 

spent watching TV, and the average combined SAT scores. 

1960 1965 1970 1975 1980 1985 1990 1992 

TV 5:06 5:29 5:56 6:07 6:36 7:07 6:55 7:04 

SAT 975 969 948 910 890 906 900 899 

Suppose that a cause and effect relationship is proposed between the time spent 

watching TV and the decline in SAT scores (in this article, Mr. Bennett does not 

argue that there is a direct connection). 

(a) Find the line of best fit relating the independent variable of average daily 

TV hours to the dependent variable of SAT scores.


(b) Find the most recent estimate of the average daily TV hours (Bennett’s cites 

Neilsen Media Research as the source of these estimates). Estimate the associated 

SAT score. How close is your estimate to the actual average? (Warning: a 

change has been made recently in the SAT, so you should investigate whether 

some adjustment needs to be made to the reported average to make a valid 

comparison.)


Topic: Geometry of Linear Maps 

The geometric effect of linear maps h: R n → R m is appealing both for its simplicity 

and for its usefulness. 

Even just in the case of linear transformations of R 1 , the geometry is quite 

nice. The pictures below contrast two nonlinear maps with two linear maps. 

Each picture shows the domain R 1 on the left mapped to the codomain R 1 on 

the right (the usual cartesian view, with the codomain drawn perpendicular to 

the domain, doesn’t make the point as well as this one). The first two show the 

nonlinear functions f1(x) =e x and f2(x) =x 2 . Arrows trace out where each 

map sends x =0,x =1,x =2,x = −1, and x = −2. Note how these nonlinear 

maps distort the domain in transforming it into the range. In the left picture, 

for instance, the top three arrows show that f1(1) is much further from f1(2) 

than it is from f1(0) — the map is spreading the domain out unevenly so that 

in being carried over to the range, an interval from the domain near x =2is 

spread apart more than is an interval near x =0. 

5 

0 

−5 

Contrast those with the linear maps h1(x) =2x and h2(x) =−x. 

5 

0 

−5 

5 

0 

−5 

5 

0 

−5 

These maps are nicer, more regular, in that for each map all of the domain is 

spread out by the same factor. 

Because the only transformations of R 1 are multiplications by a scalar, these 

pictures are possibly misleading by being too simple. In higher-dimensional 

spaces more can happen. For instance, this linear transformation of R 2 ,which 

rotates all vectors counterclockwise, is not a simple scalar multiplication. 

� � 

x 

y 

↦→ 

−→ 

5 

0 

−5 

5 

0 

−5 

� � 

x cos θ − y sin θ 

x sin θ + y cos θ 

5 

0 

−5 

5 

0 

−5 

θ

Topic: Geometry of Linear Maps 275 

And neither is this transformation of R3 , which projects vectors into the xzplane. 

� � 

x 

y 

z 

↦→ 

−→ 

� � 

x 

0 

z 

But even in higher-dimensional spaces, the situation isn’t complicated. Of 

course, any linear map h: R n → R m can be represented with respect to, say, 

the standard bases by a matrix H. Recall that any matrix H can be factored as 

H = PBQ where P and Q are nonsingular and B is a partial-identity matrix. 

And, recall that nonsingular matrices factor into elementary matrices, matrices 

that are obtained from the identity matrix with one Gaussian step 

I kρi 

−→ Mi(k) I ρi↔ρj 

−→ Pi,j 

I kρi+ρj 

−→ Ci,j(k) 

(i �= j, k �= 0). Thus we have the factorization H = TnTn−1 ...TjBTj−1 ...T1 

where the T ’s are elementary. Geometrically, a partial-identity matrix acts as a 

projection, as here. (That is, the map that this matrix represents with respect 

to the standard bases is a projection. We say that this is the map induced by 

the matrix.) 

⎛ ⎞ 

x 

⎝y⎠ 

z 

� 

1 0 

� 

0 

0 1 0 

0 0 0 

−→ 

E3 , E3 ⎛ ⎞ 

x 

⎝y⎠ 

0 

Therefore, we will have a description of the geometric action of h if we just 

describe the geometric actions of the three kinds of elementary matrices. The 

pictures below sticks to the elementary transformations of R 2 only, for ease of 

drawing. 

The action of a matrix of the form Mi(k) (that is, the action of the transformation 

of R 2 that is induced by this matrix) is to stretch vectors by a factor 

of k along the i-th axis. This is a dilation. This map stretches by a factor of 3 

along the x-axis. 

� � 

x 

y 

↦→ 

−→ 

� � 

3x 

y 

Note that if 0 ≤ k


� � 

x 

y 

↦→ 

−→ 

� � 

y 

x 

(In higher dimensions, permutations involving many axes can be decomposed 

into a combination of swaps of pairs of axes—see Exercise 5.) 

The remaining case is the action of matrices of the form Ci,j(k). Recall that, 

for instance, C1,2(2) does this. 

The picture 

�u 

�v 

� � 

x 

y 

� 

1 

� 

0 

2 1 

−→ 

E2 , E2 � � 

x 

y 

↦→ 

−→ 

� 

x 

2x + y 

� 

� 

x 

2x + y 

� 

shows that any Ci,j(k) affects vectors depending on their i-th component; in 

this example, the vector �v with the larger first component is affected more—it 

is pushed further vertically, since h(�v) is 4 higher than �v while h(�u) isonly2 

higher than �u. Another way to see the action of this map is to see where it 

sends the unit square. 

�u �v 

�w 

� � 

x 

y 

↦→ 

−→ 

� 

x 

2x + y 

� 

In this picture, vectors with a first component of 0, like �u, are not pushed 

vertically at all but vectors with a positive first component are slid up. In 

general, for any Ci,j(k), the sliding happens in such a way that vectors with the 

same i-th component are slid by the same amount. Here, �v and �w are each slid 

up by 2. The resulting shape, a rhombus, has the same base and height as the 

square (and thus the same area) but the right angles are gone. Because of this 

action, this kind of map is called a skew. 

Recall that under a linear map, the image of a subspace is a subspace. Thus 

a linear transformation maps lines through the origin to lines through the origin 

(the dimension of the image space cannot be greater than the dimension of the 

domain space, so a line can’t map onto, say, a plane). Note, however, that all 

four sides of the above rhombus are straight, not just the two sides lying in lines 

through the origin. A skew — in fact a linear map of any kind — maps any 

line to a line. Exercise 6 asks for a proof of this. That is, linear transformations 

respect the linear structures of a space. This is the reason for the assertion 

made above that, even on higher-dimensional spaces, linear maps are “nice” or 

“regular”. 

h(�u) 

h(�u) 

h(�v) 

h(�v) 

h( �w)


To finish, we will consider a familiar application, in calculus. On the left 

below is a picture, like the ones that started this Topic, of the action of the nonlinear 

function y(x) =x 2 + x. As described at that start, overall the geometric 

effect of this map is irregular in that at different domain points it has different 

effects (e.g., as the domain point x goes from 2 to −2, the associated range point 

f(x) at first decreases, then pauses instantaneously, and then increases). 

5 

0 

5 

0 

But in calculus we don’t focus on the map overall, we focus on the local effect 

of the map. The picture on the right looks more closely at what this map does 

near x =1. Atx = 1 the derivative is y ′ (1) = 3, so that near x =1wehave 

that ∆y ≈ 3 · ∆x; in other words, (1.001 2 +1.001) − (1 2 +1)≈ 3 · (0.001). That 

is, in a neighborhood of x = 1, this map carries the domain to the codomain by 

stretching by a factor of 3 — it is, locally, approximately, a dilation. This shows 

a small interval in the domain (x − ∆x..x+∆x) carried over to an interval in 

the codomain (y − ∆y..y+∆y) that is three times as wide: ∆y ≈ 3 · ∆x. 

x =1 

(When the above picture is drawn in the traditional cartesian way then the prior 

sentence is usually rephrased as: the derivative y ′ (1) = 3 gives the slope of the 

line tangent to the graph at the point (1, 2).) 

Calculus considers the map that locally approximates the change ∆x ↦→ 

3 · ∆x, instead of the actual change map ∆x ↦→ y(1 + ∆x) − y(1), because the 

local map is easy to work with. Specifically, if the input change is doubled, or 

tripled, etc., then the resulting output change will double, or triple, etc. 

5 

0 

y =2 

3(r ∆x) =r (3∆x) 

(for r ∈ R) and adding changes in input adds the resulting output changes. 

3(∆x1 +∆x2) =3∆x1 +3∆x2 

In short, what’s easy to work with about ∆x ↦→ 3 · ∆x is that it is linear. 

5 

0


This point of view makes clear an often-misunderstood, but very important, 

result about derivatives: the derivative of the composition of two functions is 

computed by using the Chain Rule for combining their derivatives. Recall that 

(with suitable conditions on the two functions) 

d (g ◦ f) 

dx 

(x) = dg df 

(f(x)) · 

dx dx (x) 

so that, for instance, the derivative of sin(x 2 +3x) iscos(x 2 +3x)·(2x+3). How 

does this combination arise? From this picture of the action of the composition. 

x 

f(x) 

g(f(x)) 

The first map f dilates the neighborhood of x by a factor of 

df 

dx (x) 

and the second map g dilates some more, this time dilating a neighborhood of 

f(x) by a factor of 

dg 

( f(x)) 

dx 

and as a result, the composition dilates by the product of these two. 

Extending from the calculus of one-variable functions to more variables starts 

with taking the natural next step: for a function y : R n → R m and a point 

�x ∈ R n , the derivative is defined to be the linear map h: R n → R m best approximating 

how y changes near y(�x). Then, for instance, the geometric description 

given earlier of transformations of R 2 characterizes how these derivatives of 

functions y : R 2 → R 2 can act. (Another example of how the extension steps 

are natural is that when there is a composition, the Chain Rule just involves 

multiplying the matrices expressing those derivatives.) 

Exercises 

1 Let h: R 2 → R 2 be the transformation that rotates vectors clockwise by π/4 radians. 

(a) Find the matrix H representing h with respect to the standard bases. Use 

Gauss’ method to reduce H to the identity. 

(b) Translate the row reduction to to a matrix equation TjTj−1 ···T1H = I (the 

prior item shows both that H is similar to I, and that no column operations are 

needed to derive I from H). 

(c) Solve this matrix equation for H.


(d) Sketch the geometric effect matrix, that is, sketch how H is expressed as a 

combination of dilations, flips, skews, and projections (the identity is a trivial 

projection). 

2 What combination of dilations, flips, skews, and projections produces a rotation 

counterclockwise by 2π/3 radians? 

3 What combination of dilations, flips, skews, and projections produces the map 

h: R 3 → R 3 represented with respect to the standard bases by this matrix? 

� 

1 2 

� 

1 

3 6 0 

1 2 2 

4 Show that any linear transformation of R 1 is the map that multiplies by a scalar 

x ↦→ kx. 

5 Show that for any permutation (that is, reordering) p of the numbers 1, ... , n, 

the map 

⎛ ⎞ ⎛ ⎞ 

x1 

⎜x2⎟ 

⎜ 

⎝ . ⎟ 

. ⎠ 

. 

↦→ 

⎜ 

⎝ 

xp(1) 

xp(2) 

xn xp(n) 

can be accomplished with a composition of maps, each of which only swaps a single 

pair of coordinates. Hint: it can be done by induction on n. (Remark: in the fourth 

chapter we will show this and we will also show that the parity of the number of 

swaps used is determined by p. That is, although a particular permutation could 

be accomplished in two different ways with two different numbers of swaps, either 

both ways use an even number of swaps, or both use an odd number.) 

6 Show that linear maps preserve the linear structures of a space. 

(a) Show that for any linear map from R n to R m , the image of any line is a line. 

The image may be a degenerate line, that is, a single point. 

(b) Show that the image of any linear surface is a linear surface. This generalizes 

the result that under a linear map the image of a subspace is a subspace. 

(c) Linear maps preserve other linear ideas. Show that linear maps preserve 

“betweeness”: if the point B is between A and C then the image of B is between 

the image of A and the image of C. 

7 Use a picture like the one that appears in the discussion of the Chain Rule to 

answer: if a function f : R → R has an inverse, what’s the relationship between how 

the function —locally, approximately — dilates space, and how its inverse dilates 

space (assuming, of course, that it has an inverse)? 

. 

⎟ 

⎠


Topic: Markov Chains 

Here is a simple game. A player bets on coin tosses, a dollar each time, and the 

game ends either when the player has no money left or is up to five dollars. If 

the player starts with three dollars, what is the chance the game takes at least 

five flips? Twenty five flips? 

At any point in the game, this player has either $0, or $1, ... , or $5. We 

say that the player is the state s0, s1, ... ,ors5. A game consists of moves, 

with, for instance, a player in state s3 having on the next flip a .5 chance of 

moving to state s2 and a .5 chance of moving to s4. Once in either state s0 or 

state s5, the player never leaves that state. Writing pi,n for the probability that 

the player is in state si after n flips, this equation sumarizes. 

⎛ 

⎞ ⎛ ⎞ ⎛ ⎞ 

1 .5 0 0 0 0 p0,n p0,n+1 

⎜ 

⎜0 

0 .5 0 0 0⎟⎜p1,n⎟ 

⎜p1,n+1⎟ 

⎟ ⎜ ⎟ ⎜ ⎟ 

⎜ 

⎜0 

.5 0 .5 0 0⎟⎜p2,n⎟ 

⎜p2,n+1⎟ 

⎟ ⎜ ⎟ 

⎜ 

⎜0 

0 .5 0 .5 0⎟⎜p3,n⎟ 

= ⎜ ⎟ 

⎜p3,n+1⎟ 

⎟ ⎜ ⎟ ⎜ ⎟ 

⎝0 

0 0 .5 0 0⎠⎝p4,n⎠ 

⎝p4,n+1⎠ 

0 0 0 0 .5 1 

p5,n 

p5,n+1 

For instance, the probability of being in state s0 after flip n +1 is p0,n+1 = 

p0,n +0.5 · p1,n. With the initial condition that the player starts with three 

dollars, calculation gives this. 

⎛ 

n =0 

⎞ 

0 

⎜ 

⎜0 

⎟ 

⎜ 

⎜0 

⎟ 

⎜ 

⎜1 

⎟ 

⎝0 

⎠ 

⎛ 

n =1 

⎞ 

0 

⎜ 

⎜0 

⎟ 

⎜ .5 ⎟ 

⎜ 

⎜0 

⎟ 

⎝ .5⎠ 

⎛ 

n =2 

⎞ 

0 

⎜ .25 ⎟ 

⎜ 

⎜0 

⎟ 

⎜ .5 ⎟ 

⎝0 

⎠ 

⎛ 

n =3 

⎞ 

.125 

⎜ 

⎜0 

⎟ 

⎜ .375 ⎟ 

⎜ 

⎜0 

⎟ 

⎝ .25 ⎠ 

⎛ 

n =4 

⎞ 

.125 

⎜ .1875 ⎟ 

⎜ 

⎜0 

⎟ 

⎜ .3125 ⎟ 

⎝0 

⎠ 

··· 

··· 

⎛ 

n =24 

⎞ 

.39600 

⎜ .00276 ⎟ 

⎜ 

⎜0 

⎟ 

⎜ .00447 ⎟ 

⎝0 

⎠ 

0 0 .25 .25 .375 

.59676 

For instance, after the fourth flip there is a probability of 0.50 that the game 

is already over — the player either has no money left or has won five dollars. 

As this computational exploration suggests, the game is not likely to go on for 

long, with the player quickly ending in either state s0 or state s5. (Because a 

player who enters either of these two states never leaves, they are said to be 

absorbtive. An argument that involves taking the limit as n goes to infinity will 

show that when the player starts with $3, there is a probability of 0.60 that the 

player eventually ends with $5 and consequently a probability of 0.40 that the 

player ends the game with $0. That argument is beyond the scope of this Topic, 

however; here we will just look at a few computations for applications.) 

This game is an example of a Markov chain, named for work by A.A. Markov 

at the start of this century. The vectors of p’s are probability vectors. The 

matrix is a transition matrix. A Markov chain is historyless in that, with a 

fixed transition matrix, the next state depends only on the current state and 

not on any states that came before. Thus a player, say, who starts in state s3,

Topic: Markov Chains 281 

then goes to state s2, then to s1, and then to s2 has exactly the same chance 

at this point of moving next to state s3 as does a player whose history was to 

start in s3, then go to s4, then to s3, and then to s2. 

Here is a Markov chain from sociology. A study ([Macdonald & Ridge], 

p. 202) divided occupations in the United Kingdom into upper level (executives 

and professionals), middle level (supervisors and skilled manual workers), and 

lower level (unskilled). To determine the mobility across these levels in a generation, 

about two thousand men were asked, “At which level are you, and at 

which level was your father when you were fourteen years old?” This equation 

summarizes the results. 

⎛ 

⎞ ⎛ 

.60 .29 .16 

⎝ .26 .37 .27⎠ 

⎝ 

.14 .34 .57 

pU,n 

⎞ ⎛ 

pM,n⎠ 

= ⎝ pU,n+1 

⎞ 

pM,n+1⎠ 

pL,n 

pL,n+1 

For instance, a child of a lower class worker has a .27 probability of growing up to 

be middle class. Notice that the Markov model assumption about history seems 

reasonable—we expect that while a parent’s occupation has a direct influence 

on the occupation of the child, the grandparent’s occupation has no such direct 

influence. With the initial distribution of the respondents’s fathers given below, 

this table lists the distributions for the next five generations. 

⎛ 

n =0 

⎞ 

.12 

⎝.32⎠ 

⎛ 

n =1 

⎞ 

.23 

⎝.34⎠ 

⎛ 

n =2 

⎞ 

.29 

⎝.34⎠ 

⎛ 

n =3 

⎞ 

.31 

⎝.34⎠ 

⎛ 

n =4 

⎞ 

.32 

⎝.33⎠ 

⎛ 

n =5 

⎞ 

.33 

⎝.33⎠ 

.56 .42 .37 .35 .34 .34 

One more example, from a very important subject, indeed. The World Series 

of American baseball is played between the team winning the American League 

and the team winning the National League (we follow [Brunner] but see also 

[Woodside]). The series is won by the first team to win four games. That means 

that a series is in one of twenty-four states: 0-0 (no games won yet by either 

team), 1-0 (one game won for the American League team and no games for the 

National League team), etc. If we assume that there is a probability p that the 

American League team wins each game then we have the following transition 

matrix. 

⎛ 

⎞ ⎛ ⎞ ⎛ ⎞ 

0 0 0 0 ... p0-0,n p0-0,n+1 

⎜ p 0 0 0 ... ⎟ ⎜p1-0,n⎟ 

⎜p1-0,n+1⎟ 

⎟ ⎜ ⎟ ⎜ ⎟ 

⎜ 

⎜1 

− p 0 0 0 ... ⎟ ⎜p0-1,n⎟ 

⎜p0-1,n+1⎟ 

⎟ ⎜ ⎟ ⎜ ⎟ 

⎜ 0 p 0 0 ... ⎟ ⎜p2-0,n⎟ 

⎜ 

⎟ ⎜ ⎟ = p2-0,n+1⎟ 

⎜ ⎟ 

⎜ 0 1−p p 0 ... ⎟ ⎜p1-1,n⎟ 

⎜p1-1,n+1⎟ 

⎟ ⎜ ⎟ ⎜ ⎟ 

⎜ 0 0 1−p0 ... ⎟ ⎜ ⎟ ⎜ ⎟ 

⎝ 

. 

. 

. 

. 

⎠ ⎝ 

p0-2,n 

. 

⎠ 

⎝ 

p0-2,n+1 

An especially interesting special case is p =0.50; this table lists the resulting 

components of the n = 0 through n = 7 vectors. (The code to generate this 

table in the computer algebra system Octave follows the exercises.) 

. 

⎠


0 − 0 

1 − 0 

0 − 1 

2 − 0 

1 − 1 

0 − 2 

3 − 0 

2 − 1 

1 − 2 

0 − 3 

4 − 0 

3 − 1 

2 − 2 

1 − 3 

0 − 4 

4 − 1 

3 − 2 

2 − 3 

1 − 4 

4 − 2 

3 − 3 

2 − 4 

4 − 3 

3 − 4 

n =0 n =1 n =2 n =3 n =4 n =5 n =6 n =7 

1 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0.5 

0.5 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0.25 

0.5 

0.25 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0.125 

0.375 

0.375 

0.125 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0.0625 

0.25 

0.375 

0.25 

0.0625 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0.0625 

0 

0 

0 

0.0625 

0.125 

0.3125 

0.3125 

0.125 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0.0625 

0 

0 

0 

0.0625 

0.125 

0 

0 

0.125 

0.15625 

0.3125 

0.15625 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0.0625 

0 

0 

0 

0.0625 

0.125 

0 

0 

0.125 

0.15625 

0 

0.15625 

0.15625 

0.15625 

Note that evenly-matched teams are likely to have a long series—there is a 

probability of 0.625 that the series goes at least six games. 

One reason for the inclusion of this Topic is that Markov chains are one 

of the most widely-used applications of matrix operations. Another reason is 

that it provides an example of the use of matrices where we do not consider 

the significance of any of the maps represented by the matrices. For more on 

Markov chains, there are many sources such as [Kemeny & Snell] and [Iosifescu]. 

Exercises 

Most of these problems need enough computation that a computer should be used. 

1 These questions refer to the coin-flipping game. 

(a) Check the computations in the table at the end of the first paragraph. 

(b) Consider the second row of the vector table. Note that this row has alternating 

0’s. Must p1,j be 0 when j is odd? Prove that it must be, or produce a 

counterexample. 

(c) Perform a computational experiment to estimate the chance that the player 

ends at five dollars, starting with one dollar, two dollars, and four dollars. 

2 ([Feller], p. 424) We consider throws of a die, and say the system is in state si if 

the largest number yet appearing on the die was i. 

(a) Give the transition matrix. 

(b) Start the system in state s1, and run it for five throws. What is the vector 

at the end? 

3 There has been much interest in whether industries in the United States are 

moving from the Northeast and North Central regions to the South and West,


motivated by the warmer climate, by lower wages, and by less unionization. Here is 

the transition matrix for large firms in Electric and Electronic Equipment ([Kelton], 

p. 43) 

NE NC S W Z 

NE 

NC 

S 

W 

Z 

0.787 

0 

0 

0 

0.021 

0 

0.966 

0.063 

0 

0.009 

0 

0.034 

0.937 

0.074 

0.005 

0.111 

0 

0 

0.612 

0.010 

0.102 

0 

0 

0.314 

0.954 

For example, a firm in the Northeast region will be in the West region next year 

with probability 0.111. (The Z entry is a “birth-death” state. For instance, with 

probability 0.102 a large Electric and Electronic Equipment firm from the Northeast 

will move out of this system next year: go out of business, move abroad, or 

move to another category of firm. There is a 0.021 probability that a firm in the 

National Census of Manufacturers will move into Electronics, or be created, or 

move in from abroad, into the Northeast. Finally, with probability 0.954 a firm 

out of the categories will stay out, according to this research.) 

(a) Does the Markov model assumption of lack of history seem justified? 

(b) Assume that the initial distribution is even, except that the value at Z is 

0.9. Compute the vectors for n =1throughn =4. 

(c) Suppose that the initial distribution is this. 

NE NC S W Z 

0.0000 0.6522 0.3478 0.0000 0.0000 

Calculate the distributions for n =1throughn =4. 

(d) Find the distribution for n =50andn = 51. Has the system settled down 

to an equilibrium? 

4 This model has been suggested for some kinds of learning ([Wickens], p. 41). The 

learner starts in an undecided state sU . Eventually the learner has to decide to do 

either response A (that is, end in state sA) orresponseB (ending in sB). However, 

the learner doesn’t jump right from being undecided to being sure A is the correct 

thingtodo(orB). Instead, the learner spends some time in a “tentative-A” 

state, or a “tentative-B” state, trying the response out (denoted here tA and tB). 

Imagine that once the learner has decided, it is final, so once sA or sB is entered 

it is never left. For the other state changes, imagine a transition is made with 

probability p, in either direction. 

(a) Construct the transition matrix. 

(b) Take p =0.25 and take the initial vector to be 1 at sU . Run this for five 

steps. What is the chance of ending up at sA? 

(c) Do the same for p =0.20. 

(d) Graph p versus the chance of ending at sA. Is there a threshold value for p, 

above which the learner is almost sure not to take longer than five steps? 

5 A certain town is in a certain country (this is a hypothetical problem). Each year 

ten percent of the town dwellers move to other parts of the country. Each year 

one percent of the people from elsewhere move to the town. Assume that there 

are two states sT , living in town, and sC, living elsewhere. 

(a) Construct the transistion matrix. 

(b) Starting with an initial distribution sT =0.3 andsC =0.7, get the results 

for the first ten years. 

(c) Do the same for sT =0.2.


(d) Are the two outcomes alike or different? 

6 For the World Series application, use a computer to generate the seven vectors 

for p =0.55 and p =0.6. 

(a) What is the chance of the National League team winning it all, even though 

they have only a probability of 0.45 or 0.40 of winning any one game? 

(b) Graph the probability p against the chance that the American League team 

wins it all. Is there a threshold value—a p above which the better team is 

essentially ensured of winning? 

(Some sample code is included below.) 

7 A Markov matrix has each entry positive, and each columns sums to 1. 

(a) Check that the three transistion matrices shown in this Topic meet these two 

conditions. Must any transition matrix do so? 

(b) Observe that if A�v0 = �v1 and A�v1 = �v2 then A 2 is a transition matrix from 

�v0 to �v2. Show that a power of a Markov matrix is also a Markov matrix. 

(c) Generalize the prior item by proving that the product of two appropriatelysized 

Markov matrices is a Markov matrix. 

Computer Code 

This is the code for the computer algebra system Octave that was used 

to generate the table of World Series outcomes. First, this script is kept 

in the file markov.m. (The sharp character # marks the rest of a line as a 

comment.) 

# Octave script file to compute chance of World Series outcomes. 

function w = markov(p,v) 

q = 1-p; 

A=[0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 0-0 

p,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 1-0 

q,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 0-1_ 

0,p,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 2-0 

0,q,p,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 1-1 

0,0,q,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 0-2__ 

0,0,0,p,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 3-0 

0,0,0,q,p,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 2-1 

0,0,0,0,q,p, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 1-2_ 

0,0,0,0,0,q, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 0-3 

0,0,0,0,0,0, p,0,0,0,1,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 4-0 

0,0,0,0,0,0, q,p,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 3-1__ 

0,0,0,0,0,0, 0,q,p,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 2-2 

0,0,0,0,0,0, 0,0,q,p,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 1-3 

0,0,0,0,0,0, 0,0,0,q,0,0, 0,0,1,0,0,0, 0,0,0,0,0,0; # 0-4_ 

0,0,0,0,0,0, 0,0,0,0,0,p, 0,0,0,1,0,0, 0,0,0,0,0,0; # 4-1 

0,0,0,0,0,0, 0,0,0,0,0,q, p,0,0,0,0,0, 0,0,0,0,0,0; # 3-2 

0,0,0,0,0,0, 0,0,0,0,0,0, q,p,0,0,0,0, 0,0,0,0,0,0; # 2-3__ 

0,0,0,0,0,0, 0,0,0,0,0,0, 0,q,0,0,0,0, 1,0,0,0,0,0; # 1-4 

0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,p,0, 0,1,0,0,0,0; # 4-2 

0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,q,p, 0,0,0,0,0,0; # 3-3_ 

0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,q, 0,0,0,1,0,0; # 2-4 

0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,p,0,1,0; # 4-3 

0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,q,0,0,1]; # 3-4


w = A * v; 

endfunction 

Then the Octave session was this. 

> v0=[1;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0] 

> p=.5 

> v1=markov(p,v0) 

> v2=markov(p,v1) 

... 

Translating to another computer algebra system should be easy—all have 

commands similar to these.


Topic: Orthonormal Matrices 

In The Elements, Euclid considers two figures to be the same if they have the 

same size and shape. That is, the triangles below are not equal because they are 

not the same set of points. But they are congruent—essentially indistinguishable 

for Euclid’s purposes—because we can imagine picking up the plane up, sliding 

it over and turning it a bit (although not bending it or stretching it), and then 

putting it back down, to superimpose the first figure on the second. 

P1 

P2 

P3 

(Euclid never explicitly states this principle but he uses it often [Casey].) In 

modern terms, “picking the plane up ... ” means taking a map from the plane 

to itself. We, and Euclid, are considering only certain transformations of the 

plane, ones that may possibly slide or turn the plane but not bend or stretch 

it. Accordingly, we define a function f : R 2 → R 2 to be distance-preserving (or 

a rigid motion, orisometry) if for all points P1,P2 ∈ R 2 , the map satisfies the 

condition that the distance from f(P1) tof(P2) equals the distance from P1 to 

P2. We define a plane figure to be a set of points in the plane and we say that 

two figures are congruent if there is a distance-preserving map from the plane 

to itself that carries one figure onto the other. 

Many statements from Euclidean geometry follow easily from these definitions. 

Some are: (i) collinearity is invariant under any distance-preserving map 

(that is, if P1, P2, andP3 are collinear then so are f(P1), f(P2), and f(P3)), 

(ii) betweeness is invariant under any distance-preserving map (if P2 is between 

P1 and P3 then so is f(P2) between f(P1) andf(P3)), (iii) the property of 

being a triangle is invariant under any distance-preserving map (if a figure is a 

triangle then the image of that figure is also a triangle), (iv) and the property of 

being a circle is invariant under any distance-preserving map. In 1872, F. Klein 

suggested that Euclidean geometry can be characterized as the study of properties 

that are invariant under distance-preserving maps. (This forms part of 

Klein’s Erlanger Program, which proposes the organizing principle that each 

kind of geometry—Euclidean, projective, etc.—can be described as the study 

of the properties that are invariant under some group of transformations. The 

word ‘group’ here means more than just ‘collection’, but that lies outside of our 

scope.) 

We can use linear algebra to characterize the distance-preserving maps of 

the plane. 

First, there are distance-preserving transformations of the plane that are not 

linear. The obvious example is this translation. 

� � � � � � 

x 

x 1 

↦→ + = 

y 

y 0 

Q1 

Q2 

Q3 

� � 

x +1 

y 

However, this example turns out to be the only example, in the sense that if f 

is distance-preserving and sends �0 to�v0 then the map �v ↦→ f(�v) − �v0 is linear.

Topic: Orthonormal Matrices 287 

That will follow immediately from this statement: a map t that is distancepreserving 

and sends �0 to itself is linear. To prove this statement, let 

� � 

� � 

a 

c 

t(�e1) = t(�e2) = 

b 

d 

for some a, b, c, d ∈ R. Then to show that t is linear, it suffices to show that it 

can be represented by a matrix, that is, that t acts in this way for all x, y ∈ R. 

� � � � 

x t ax + cy 

�v = ↦−→ 

(∗) 

y bx + dy 

Recall that if we fix three non-collinear points then any point in the plane can 

be described by giving its distance from those three. So any point �v in the 

domain is determined by its distance from the three fixed points �0, �e1, and�e2. 

Similarly, any point t(�v) in the codomain is determined by its distance from the 

three fixed points t(�0), t(�e1), and t(�e2) (these three are not collinear because, as 

mentioned above, collinearity is invariant and �0, �e1, and�e2 are not collinear). 

In fact, because t is distance-preserving, we can say more: for the point �v in the 

plane that is determined by being the distance d0 from �0, the distance d1 from 

�e1, and the distance d2 from �e2, its image t(�v) must be the unique point in the 

codomain that is determined by being d0 from t(�0), d1 from t(�e1), and d2 from 

t(�e2). Because of the uniqueness, checking that the action in (∗) works in the 

d0, d1, andd2cases � � 

� � 

� � 

x 

x 

ax + cy 

dist( ,�0) = dist(t( ),t(�0)) = dist( ,�0) 

y 

y 

bx + dy 

(t is assumed to send �0 toitself) 

� � 

� � 

� � � � 

x 

x 

ax + cy a 

dist( ,�e1) = dist(t( ),t(�e1)) = dist( , ) 

y 

y 

bx + dy b 

and 

� � 

� � 

� � � � 

x 

x 

ax + cy c 

dist( ,�e2) = dist(t( ),t(�e2)) = dist( , ) 

y 

y 

bx + dy d 

suffices to show that (∗) describes t. Those checks are routine. 

Thus, any distance-preserving f : R 2 → R 2 can be written f(�v) =t(�v)+�v0 

for some constant vector �v0 and linear map t that is distance-preserving. 

Not every linear map is distance-preserving, for example, �v ↦→ 2�v does not 

preserve distances. But there is a neat characterization: a linear transformation 

t of the plane is distance-preserving if and only if both �t(�e1)� = �t(�e2)� =1and 

t(�e1) is orthogonal to t(�e2). The ‘only if’ half of that statement is easy—because 

t is distance-preserving it must preserve the lengths of vectors, and because t 

is distance-preserving the Pythagorean theorem shows that it must preserve 

orthogonality. For the ‘if’ half, it suffices to check that the map preserves


lengths of vectors, because then for all �p and �q the distance between the two is 

preserved �t(�p − �q )� = �t(�p) − t(�q )� = ��p − �q �. For that check, let 

�v = 

� � 

x 

y 

t(�e1) = 

� � 

a 

b 

t(�e2) = 

� � 

c 

d 

and, with the ‘if’ assumptions that a 2 + b 2 = c 2 + d 2 =1andac + bd =0we 

have this. 

�t(�v )� 2 =(ax + cy) 2 +(bx + dy) 2 

= a 2 x 2 +2acxy + c 2 y 2 + b 2 x 2 +2bdxy + d 2 y 2 

= x 2 (a 2 + b 2 )+y 2 (c 2 + d 2 )+2xy(ac + bd) 

= x 2 + y 2 

= ��v � 2 

One thing that is neat about this characterization is that we can easily 

recognize matrices that represent such a map with respect to the standard bases. 

Those matrices have that when the columns are written as vectors then they 

are of length one and are mutually orthogonal. Such a matrix is called an 

orthonormal matrix or orthogonal matrix (the second term is commonly used to 

mean not just that the columns are orthogonal, but also that they have length 

one). 

We can use this insight to delimit the geometric actions possible in distancepreserving 

maps. Because �t(�v )� = ��v �, any�v is mapped by t to lie somewhere 

on the circle about the origin that has radius equal to the length of �v, and 

so in particular �e1 and �e2 are mapped to vectors on the unit circle. What’s 

more, because of the orthogonality restriction, once we fix the unit vector �e1 as 

mapped to the vector with components a and b then there are only two places 

where �e2 can go. 

� � −b 

a 

� � a 

b 

Thus, only two types of maps are possible. 

Rep E2,E2 (t) = 

� � 

a −b 

b a 

� � a 

b 

� � b 

−a 

Rep E2,E2 (t) = 

� � 

a b 

b −a 

We can geometrically describe these two cases. Let θ be the angle between the 

x-axis and the image of �e1, measured counterclockwise. 

The first matrix above represents, with respect to the standard bases, a 

rotation of the plane by θ radians.


� � −b 

a 

� � a 

b 

� � 

x t 

↦−→ 

y 

� � 

x cos θ − y sin θ 

x sin θ + y cos θ 

The second matrix above represents a reflection of the plane through the line 

bisecting the angle between �e1 and t(�e1). (This picture shows �e1 reflected up 

into the first quadrant and �e2 reflected down into the fourth quadrant.) 

� � a 

b 

� � b 

−a 

� � 

x t 

↦−→ 

y 

� � 

x cos θ + y sin θ 

x sin θ − y cos θ 

Note that in this second case, the right angle from �e1 to �e2 has a counterclockwise 

sense but the right angle between the images of these two has a clockwise sense, 

so the sense gets reversed. Geometers speak of a distance-preserving map as 

direct if it preserves sense and as opposite if it reverses sense. 

So, we have characterized the Euclidean study of congruence into the consideration 

of the properties that are invariant under combinations of (i) a rotation 

followed by a translation (possibly the trivial translation), or (ii) a reflection 

followed by a translation (a reflection followed by a non-trivial translation is a 

glide reflection). 

Another idea, besides congruence of figures, encountered in elementary geometry 

is that figures are similar if they are congruent after a change of scale. 

These two triangles are similar since the second is the same shape as the first, 

but 3/2-ths the size. 

P1 

P2 

P3 

From the above work, we have that figures are similar if there is an orthonormal 

matrix T such that the points �q on one are derived from the points �p by �q = 

(kT)�v + �p0 for some nonzero real number k and constant vector �p0. 

Although many of these ideas were first explored by Euclid, mathematics is 

timeless and they are very much in use today. One application of rigid motions 

is in computer graphics. We can, for example, take this top view of a cube 

and animate it by putting together film frames of it rotating. 

Q1 

Q2 

Q3


Frame 1: Frame 2: Frame 3: 

−.2 radians −.4 radians −.6 radians 

We could also make the cube appear to be coming closer to us by producing 

film frames of it gradually enlarging. 

Frame 1: Frame 2: Frame 3: 

110 percent 120 percent 130 percent 

In practice, computer graphics incorporate many interesting techniques from 

linear algebra (see Exercise 4). 

So the analysis above of distance-preserving maps is useful as well as interesting. 

For instance, it shows that to include in graphics software all possible 

rigid motions of the plane, we need only include a few cases. It is not possible 

that we’ve somehow ovelooked some rigid motions. 

A beautiful book that explores more in this area is [Weyl]. More on groups, 

of transformations and otherwise, can be found in any book on Modern Algebra, 

for instance [Birkhoff & MacLane]. More on Klein and the Erlanger Program is 

in [Yaglom]. 

Exercises 

1 Decide � if √each of these √ is an orthonormal matrix. 

1/ 2 −1/ 2 

(a) 

−1/ √ 2 −1/ √ � 

2 

� √ √ 

1/ 3 −1/ 3 

(b) 

−1/ √ 3 −1/ √ � 

3 

� √ √ √ 

1/ 3 − 2/ 3 

(c) 

− √ 2/ √ 3 −1/ √ � 

3 

2 Write down the formula for each of these distance-preserving maps. 

(a) the map that rotates π/6 radians, and then translates by �e2 

(b) the map that reflects about the line y =2x 

(c) the map that reflects about y = −2x andtranslatesover1andup1 

3 (a) The proof that a map that is distance-preserving and sends the zero vector 

to itself incidentally shows that such a map is one-to-one and onto (the 

point in the domain determined by d0, d1, and d2 corresponds to the point 

in the codomain determined by those three numbers). Therefore any distancepreserving 

map has an inverse. Show that the inverse is also distance-preserving. 

(b) Using the definitions given in this Topic, prove that congruence is an equivalence 

relation between plane figures.


4 In practice the matrix for the distance-preserving linear transformation and the 

translation are often combined into one. Check that these two computations yield 

the same first two components. 

� �� 

a c x e 

+ 

b d y f 

� 

a 

b 

0 

c 

d 

0 

�� 

e x 

f y 

1 1 

(These are homogeneous coordinates; see the Topic on Projective Geometry). 

5 (a) Verify that the properties described in the second paragraph of this Topic 

as invariant under distance-preserving maps are indeed so. 

(b) Give two more properties that are of interest in Euclidean geometry from 

your experience in studying that subject that are also invariant under distancepreserving 

maps. 

(c) Give a property that is not of interest in Euclidean geometry and is not 

invariant under distance-preserving maps.

Chapter 4 

Determinants 

In the first chapter of this book we considered linear systems and we picked out 

the special case of systems with the same number of equations as unknowns, 

those of the form T�x = � b where T is a square matrix. We noted a distinction 

between two classes of T ’s. While such systems may have a unique solution or 

no solutions or infinitely many solutions, if a particular T is associated with a 

unique solution in any system, such as the homogeneous system � b = �0, then 

T is associated with a unique solution for every � b. We call such a matrix of 

coefficients ‘nonsingular’. The other kind of T , where every linear system for 

which it is the matrix of coefficients has either no solution or infinitely many 

solutions, we call ‘singular’. 

Through the second and third chapters the value of this distinction has been 

a theme. For instance, we now know that nonsingularity of an n×n matrix T 

is equivalent to each of these: 

• asystemT�x = � b has a solution, and that solution is unique; 

• Gauss-Jordan reduction of T yields an identity matrix; 

• the rows of T form a linearly independent set; 

• the columns of T form a basis for R n ; 

• any map that T represents is an isomorphism; 

• an inverse matrix T −1 exists. 

So when we look at a particular square matrix, the question of whether it 

is nonsingular is one of the first things that we ask. This chapter develops 

a formula to determine this. (Since we will restrict the discussion to square 

matrices, in this chapter we will usually simply say ‘matrix’ in place of ‘square 

matrix’.) 

More precisely, we will develop infinitely many formulas, one for 1×1 matrices, 

one for 2×2 matrices, etc. Of course, these formulas are related — that 

is, we will develop a family of formulas, a scheme that describes the formula for 

each size. 

293

294 Chapter 4. Determinants 

4.I Definition 

For 1×1 matrices, determining nonsingularity is trivial. 

� a � is nonsingular iff a �= 0 

The 2×2 formula came out in the course of developing the inverse. 

� � 

a b 

is nonsingular iff ad − bc �= 0 

c d 

The 3×3 formula can be produced similarly (see Exercise 9). 

⎛ 

a 

⎝d b 

e 

⎞ 

c 

f⎠is 

nonsingular iff aei + bfg + cdh − hfa − idb − gec �= 0 

g h i 

With these cases in mind, we posit a family of formulas, a, ad−bc, etc. For each 

n the formula gives rise to a determinant function detn×n : Mn×n → R such that 

an n×n matrix T is nonsingular if and only if detn×n(T ) �= 0. (We usually omit 

the subscript because if T is n×n then ‘det(T )’ could only mean ‘detn×n(T )’.) 

4.I.1 Exploration 

This subsection is optional. It briefly describes how an investigator might 

come to a good general definition, which is given in the next subsection. 

The three cases above don’t show an evident pattern to use for the general 

n×n formula. We may spot that the 1×1 term a has one letter, that the 2×2 

terms ad and bc have two letters, and that the 3×3 termsaei, etc., have three 

letters. We may also observe that in those terms there is a letter from each row 

and column of the matrix, e.g., the letters in the cdh term 

⎛ ⎞ 

c 

⎝d 

⎠ 

h 

come one from each row and one from each column. But these observations 

perhaps seem more puzzling than enlightening. For instance, we might wonder 

why some of the terms are added while others are subtracted. 

A good problem solving strategy is to see what properties a solution must 

have and then search for something with those properties. So we shall start by 

asking what properties we require of the formulas. 

At this point, our primary way to decide whether a matrix is singular is 

to do Gaussian reduction and then check whether the diagonal of resulting 

echelon form matrix has any zeroes (that is, to check whether the product 

down the diagonal is zero). So, we may expect that the proof that a formula

Section I. Definition 295 

determines singularity will involve applying Gauss’ method to the matrix, to 

show that in the end the product down the diagonal is zero if and only if the 

determinant formula gives zero. This suggests our initial plan: we will look for 

a family of functions with the property of being unaffected by row operations 

and with the property that a determinant of an echelon form matrix is the 

product of its diagonal entries. Under this plan, a proof that the functions 

determine singularity would go, “Where T →···→ ˆ T is the Gaussian reduction, 

the determinant of T equals the determinant of ˆ T (because the determinant is 

unchanged by row operations), which is the product down the diagonal, which 

is zero if and only if the matrix is singular”. In the rest of this subsection we 

will test this plan on the 2×2 and 3×3 determinants that we know. We will end 

up modifying the “unaffected by row operations” part, but not by much. 

The first step in checking the plan is to test whether the 2 × 2 and 3 × 3 

formulas are unaffected by the row operation of pivoting: if 

T kρi+ρj 

−→ ˆ T 

then is det( ˆ T ) = det(T )? This check of the 2×2 determinant after the kρ1 + ρ2 

operation 

� 

a 

det( 

ka + c 

� 

b 

)=a(kb + d) − (ka + c)b = ad − bc 

kb+ d 

shows that it is indeed unchanged, and the other 2×2 pivot kρ2 + ρ1 gives the 

same result. The 3×3 pivot kρ3 + ρ2 leaves the determinant unchanged 

⎛ 

⎞ 

a b c 

det( ⎝kg 

+ d kh+ e ki+ f⎠) 

=a(kh + e)i + b(ki + f)g + c(kg + d)h 

g h i − h(ki + f)a − i(kg + d)b − g(kh + e)c 

= aei + bfg + cdh − hfa − idb − gec 

as do the other 3×3 pivot operations. 

So there seems to be promise in the plan. Of course, perhaps the 4 × 4 

determinant formula is affected by pivoting. We are exploring a possibility here 

and we do not yet have all the facts. Nonetheless, so far, so good. 

The next step is to compare det( ˆ T ) with det(T ) for the operation 

T ρi↔ρj 

−→ ˆ T 

of swapping two rows. The 2×2 rowswapρ1↔ρ2 � � 

c d 

det( )=cb − ad 

a b 

does not yield ad − bc. This ρ1 ↔ ρ3 swap inside of a 3×3 matrix 

⎛ ⎞ 

g h i 

det( ⎝d e f⎠) 

=gec + hfa + idb − bfg − cdh − aei 

a b c


also does not give the same determinant as before the swap — again there is a 

sign change. Trying a different 3×3 swapρ1↔ρ2 ⎛ ⎞ 

d e f 

det( ⎝a b c⎠) 

=dbi + ecg + fah− hcd − iae − gbf 

g h i 

also gives a change of sign. 

Thus, row swaps appear to change the sign of a determinant. This modifies 

our plan, but does not wreck it. We intend to decide nonsingularity by 

considering only whether the determinant is zero, not by considering its sign. 

Therefore, instead of expecting determinants to be entirely unaffected by row 

operations, will look for them to change sign on a swap. 

To finish, we compare det( ˆ T ) to det(T ) for the operation 

T kρi 

−→ ˆ T 

of multiplying a row by a scalar k �= 0. One of the 2×2 cases is 

� � 

a b 

det( )=a(kd) − (kc)b = k · (ad − bc) 

kc kd 

and the other case has the same result. Here is one 3×3 case 

⎛ 

⎞ 

a b c 

det( ⎝ d e f ⎠) =ae(ki)+bf(kg)+cd(kh) 

kg kh ki −(kh)fa− (ki)db − (kg)ec 

= k · (aei + bfg + cdh − hfa − idb − gec) 

and the other two are similar. These lead us to suspect that multiplying a row 

by k multiplies the determinant by k. This fits with our modified plan because 

we are asking only that the zeroness of the determinant be unchanged and we 

are not focusing on the determinant’s sign or magnitude. 

In summary, to develop the scheme for the formulas to compute determinants, 

we look for determinant functions that remain unchanged under the 

pivoting operation, that change sign on a row swap, and that rescale on the 

rescaling of a row. In the next two subsections we will find that for each n such 

a function exists and is unique. 

For the next subsection, note that, as above, scalars come out of each row 

without affecting other rows. For instance, in this equality 

⎛ ⎞ ⎛ ⎞ 

3 3 9 

1 1 3 

det( ⎝2 1 1 ⎠) =3· det( ⎝2 1 1 ⎠) 

5 10 −5 

5 10 −5 

the 3 isn’t factored out of all three rows, only out of the top row. The determinant 

acts on each row of independently of the other rows. When we want to use 

this property of determinants, we shall write the determinant as a function of 

the rows: ‘det(�ρ1,�ρ2,...�ρn)’, instead of as ‘det(T )’ or ‘det(t1,1,...,tn,n)’. The 

definition of the determinant that starts the next subsection is written in this 

way.


Exercises 

� 1.1 Evaluate the determinant of each. 

� � � � 

2 0 1 

3 1 

(a) 

(b) 3 1 1 

−1 1 

−1 0 1 

1.2 Evaluate the determinant of each. 

� � � � 

2 1 1 

2 0 

(a) 

(b) 0 5 −2 

−1 3 

1 −3 4 

(c) 

(c) 

� 

4 0 

� 

1 

0 0 1 

1 3 −1 

� 

2 3 

� 

4 

5 6 7 

8 9 1 

� 1.3 Verify that the determinant of an upper-triangular 3×3 matrix is the product 

down the diagonal. 

� 

a b 

� 

c 

det( 0 e f )=aei 

0 0 i 

Do lower-triangular matrices work the same way? 

� 1.4 Use�the determinant � �to decide � if each �is singular � or nonsingular. 

2 1 

0 1 

4 2 

(a) 

(b) 

(c) 

3 1 

1 −1 

2 1 

1.5 Singular or nonsingular? Use the determinant to decide. 

� � � � � � 

2 1 1 

1 0 1 

2 1 0 

(a) 3 2 2 (b) 2 1 1 (c) 3 −2 0 

0 1 4 

4 1 3 

1 0 0 

� 1.6 Each pair of matrices differ by one row operation. Use this operation to compare 

det(A) with � det(B). � � � 

1 2 1 2 

(a) A = B = 

2 3 0 −1 

� � � 

3 1 0 3 1 

� 

0 

(b) A = 0 0 1 B = 0 1 2 

0 

� 

1 

1 2 

−1 

� 

3 

0 0 

� 

1 

1 

−1 

� 

3 

(c) A = 2 2 −6 B = 1 1 −3 

1 0 4 1 0 4 

1.7 Show this. 

� 

1 1 1 

det( a b c 

a 2 

b 2 

c 2 

� 

)=(b− a)(c − a)(c − b) 

� 1.8 Which real numbers x make this matrix singular? 

� � 

12 − x 4 

−8 8−x 1.9 Do the Gaussian reduction to check the formula for 3×3 matrices stated in the 

preamble to this section. 

� � 

a b c 

d e f is nonsingular iff aei + bfg + cdh − hfa − idb − gec �= 0 

g h i


1.10 Show that the equation of a line in R 2 thru (x1,y1) and(x2,y2) is expressed 

by this determinant. 

� 

x y 

� 

1 

det( 

)=0 x1 �= x2 

x1 y1 1 

x2 y2 1 

� 1.11 Many people know this mnemonic for the determinant of a 3×3 matrix: first 

repeat the first two columns and then sum the products on the forward diagonals 

and subtract the products on the backward diagonals. That is, first write 

� 

and then calculate this. 

� h1,1 h1,2 h1,3 h1,1 h1,2 

h2,1 h2,2 h2,3 h2,1 h2,2 

h3,1 h3,2 h3,3 h3,1 h3,2 

h1,1h2,2h3,3 + h1,2h2,3h3,1 + h1,3h2,1h3,2 

−h3,1h2,2h1,3 − h3,2h2,3h1,1 − h3,3h2,1h1,2 

(a) Check that this agrees with the formula given in the preamble to this section. 

(b) Does it extend to other-sized determinants? 

1.12 The cross product of the vectors 

� � � � 

x1 

y1 

�x = 

�y = 

is the vector computed as this determinant. 

� 

�e1 �e2 

� 

�e3 

�x × �y =det( 

) 

x2 

x3 

y2 

y3 

x1 x2 x3 

y1 y2 y3 

Note that the first row is composed of vectors, the vectors from the standard basis 

for R 3 . Show that the cross product of two vectors is perpendicular to each vector. 

1.13 Prove that each statement holds for 2×2 matrices. 

(a) The determinant of a product is the product of the determinants det(ST)= 

det(S) · det(T ). 

(b) If T is invertible then the determinant of the inverse is the inverse of the 

determinant det(T −1 )=(det(T )) −1 . 

Matrices T and T ′ are similar if there is a nonsingular matrix P such that T ′ = 

PTP −1 . (This definition is in Chapter Five.) Show that similar 2×2 matrices have 

the same determinant. 

� 1.14 Prove that the area of this region in the plane 

� � 

x2 

is equal to the value of this determinant. 

Compare with this. 

y2 

det( 

det( 

� 

x1 x2 

y1 y2 

� 

x2 x1 

y2 y1 

� 

) 

� 

) 

� � 

x1 

y1


1.15 Prove that for 2×2 matrices, the determinant of a matrix equals the determinant 

of its transpose. Does that also hold for 3×3 matrices? 

� 1.16 Is the determinant function linear — is det(x·T +y·S) =x·det(T )+y·det(S)? 

1.17 Show that if A is 3×3 thendet(c · A) =c 3 · det(A) for any scalar c. 

1.18 Which real numbers θ make 

�cos � 

θ − sin θ 

sin θ cos θ 

singular? Explain geometrically. 

1.19 [Am. Math. Mon., Apr. 1955] If a third order determinant has elements 1, 2, 

... , 9, what is the maximum value it may have? 

4.I.2 Properties of Determinants 

As described above, we want a formula to determine whether an n×n matrix 

is nonsingular. We will not begin by stating such a formula. Instead, we will 

begin by considering the function that such a formula calculates. We will define 

the function by its properties, then prove that the function with these properties 

exist and is unique and also describe formulas that compute this function. 

(Because we will show that the function exists and is unique, from the start we 

will say ‘det(T )’ instead of ‘if there is a determinant function then det(T )’ and 

‘the determinant’ instead of ‘any determinant’.) 

2.1 Definition A n×n determinant is a function det: Mn×n → R such that 

(1) det(�ρ1,...,k· �ρi + �ρj,...,�ρn) = det(�ρ1,...,�ρj,...,�ρn) fori �= j 

(2) det(�ρ1,... ,�ρj,...,�ρi,...,�ρn) =− det(�ρ1,...,�ρi,...,�ρj,...,�ρn) fori �= j 

(3) det(�ρ1,...,k�ρi,...,�ρn) =k · det(�ρ1,...,�ρi,...,�ρn) fork �= 0 

(4) det(I) = 1 where I is an identity matrix 

(the �ρ ’s are the rows of the matrix). We often write |T | for det(T ). 

2.2 Remark Property (2) is redundant since 

T ρi+ρj 

−→ −ρj+ρi 

−→ ρi+ρj 

−→ −ρi 

−→ ˆ T 

swaps rows i and j. It is listed only for convenience. 

The first result shows that a function satisfying these conditions gives a 

criteria for nonsingularity. (Its last sentence is that, in the context of the first 

three conditions, (4) is equivalent to the condition that the determinant of an 

echelon form matrix is the product down the diagonal.)


2.3 Lemma A matrix with two identical rows has a determinant of zero. A 

matrix with a zero row has a determinant of zero. A matrix is nonsingular if 

and only if its determinant is nonzero. The determinant of an echelon form 

matrix is the product down its diagonal. 

Proof. To verify the first sentence, swap the two equal rows. The sign of the 

determinant changes, but the matrix is unchanged and so its determinant is 

unchanged. Thus the determinant is zero. 

The second sentence is clearly true if the matrix is 1×1. If it has at least 

two rows then apply property (1) of the definition with the zero row as row j 

and with k =1. 

det(...,�ρi,...,�0,...) = det(...,�ρi,...,�ρi + �0,...) 

The first sentence of this lemma gives that the determinant is zero. 

For the third sentence, where T → ··· → ˆ T is the Gauss-Jordan reduction, 

by the definition the determinant of T is zero if and only if the determinant of 

ˆT is zero (although they could differ in sign or magnitude). A nonsingular T 

Gauss-Jordan reduces to an identity matrix and so has a nonzero determinant. 

A singular T reduces to a ˆ T with a zero row; by the second sentence of this 

lemma its determinant is zero. 

Finally, for the fourth sentence, if an echelon form matrix is singular then it 

has a zero on its diagonal, that is, the product down its diagonal is zero. The 

third sentence says that if a matrix is singular then its determinant is zero. So 

if the echelon form matrix is singular then its determinant equals the product 

down its diagonal. 

If an echelon form matrix is nonsingular then none of its diagonal entries is 

zero so we can use property (3) of the definition to factor them out (again, the 

vertical bars |···| indicate the determinant operation). 

� 

� 

� 

� 

�t1,1 

t1,2 t1,n� 

� 

� 

� 

�1 

t1,2/t1,1 t1,n/t1,1� 

� 

� 

� 0 t2,2 t2,n� 

� 

� 

�0 

1 t2,n/t2,2� 

� 

� 

� 

. .. 

� = t1,1 · t2,2 ···tn,n · � 

� 

� 

. .. 

� 

� 

� 

� 

� 

� 

� 0 tn,n 

� 

�0 

1 � 

Next, the Jordan half of Gauss-Jordan elimination, using property (1) of the 

definition, leaves the identity matrix. 

� 

� 

� 

�1 

0 0� 

� 

� 

�0 

1 0� 

� 

= t1,1 · t2,2 ···tn,n · � 

� 

. .. 

� = t1,1 · t2,2 ···tn,n · 1 

� 

� 

� 

�0 

1� 

Therefore, if an echelon form matrix is nonsingular then its determinant is the 

product down its diagonal. QED


That result gives us a way to compute the value of a determinant function on 

a matrix. Do Gaussian reduction, keeping track of any changes of sign caused by 

row swaps and any scalars that are factored out, and then finish by multiplying 

down the diagonal of the echelon form result. This procedure takes the same 

time as Gauss’ method and so is sufficiently fast to be practical on the size 

matrices that we see in this book. 

2.4 Example Doing 2×2 determinants 

� � 

� 

� 2 4� 

� 

�−1 3� 

= 

� � 

� 

�2 

4� 

� 

�0 5� 

=10 

with Gauss’ method won’t give a big savings because the 2 × 2 determinant 

formula is so easy. However, a 3×3 determinant is usually easier to calculate 

with Gauss’ method than with the formula given earlier. 

� � 

� 

�2 

2 6� 

� 

� 

�4 

4 3� 

� 

�0 

−3 5� 

= 

� 

� � 

� 

� 

�2 

2 6 � � 

� �2 

2 6 � 

� 

� 

�0 

0 −9� 

� = − � 

�0 

−3 5 � 

� = −54 

�0 

−3 5 � �0 

0 −9� 

2.5 Example Determinants of matrices any bigger than 3×3 are almost always 

most quickly done with this Gauss’ method procedure. 

� 

� � 

� � 

� 

� 

�1 

0 1 3� 

� 

� �1 

0 1 3 � � 

� �1 

0 1 3 � 

� 

� 

�0 

1 1 4� 

� 

� 

� 

�0 

0 0 5�= 

− �0 

1 1 4 � � 

� 

� 

� �0 

0 0 5 � = − �0 

1 1 4 � 

� 

� 

� �0 

0 −1 −3� 

= −(−5) = 5 

� 

�0 

1 0 1� 

�0 

0 −1 −3� 

�0 

0 0 5 � 

The prior example illustrates an important point. Although we have not yet 

found a 4×4 determinant formula, if one exists then we know what value it gives 

to the matrix — if there is a function with properties (1)-(4) then on the above 

matrix the function must return 5. 

2.6 Lemma For each n, if there is an n×n determinant function then it is 

unique. 

Proof. For any n×n matrix we can perform Gauss’ method on the matrix, 

keeping track of how the sign alternates on row swaps, and then multiply down 

the diagonal of the echelon form result. By the definition and the lemma, all n×n 

determinant functions must return this value on this matrix. Thus all n×n determinant 

functions are equal, that is, there is only one input argument/output 

value relationship satisfying the four conditions. QED 

The ‘if there is an n×n determinant function’ emphasizes that, although we 

can use Gauss’ method to compute the only value that a determinant function 

could possibly return, we haven’t yet shown that such a determinant function 

exists for all n. In the rest of the section we will produce determinant functions. 

Exercises 

For these, assume that an n×n determinant function exists for all n. 

� 2.7 Use Gauss’ method to find each determinant.


� 

�3 

� 

(a) �3 

� 

0 

1 

1 

1 

� 

2� 

� 

0� 

4 

� 

� 

� 1 

� 

� 2 

(b) � 

�−1 

� 1 

0 

1 

0 

1 

0 

1 

1 

1 

� 

1� 

� 

0� 

� 

0� 

0� 

2.8 Use Gauss’ method to find each. 

� � � � 

� 

(a) � 2 −1� 

�1 

1 0� 

� � � 

�−1 −1� 

(b) �3 

0 2� 

� 

5 2 2 

� 

2.9 For which values of k does this system have a unique solution? 

x + z − w =2 

y − 2z =3 

x + kz =4 

z − w =2 

� 2.10 Expresseachoftheseintermsof|H|. 

� 

� 

�h3,1 

h3,2 h3,3� 

� 

� 

(a) �h2,1 

h2,2 h2,3� 

� 

� 

h1,1 h1,2 h1,3 

� 

� 

� −h1,1 −h1,2 −h1,3 � 

� 

� 

(b) �−2h2,1 

−2h2,2 −2h2,3� 

� 

� 

−3h3,1 −3h3,2 −3h3,3 

� 

� 

�h1,1 

+ h3,1 h1,2 + h3,2 h1,3 + h3,3� 

� 

� 

(c) � h2,1 h2,2 h2,3 � 

� 

� 

5h3,1 5h3,2 5h3,3 

� 2.11 Find the determinant of a diagonal matrix. 

2.12 Describe the solution set of a homogeneous linear system if the determinant 

of the matrix of coefficients is nonzero. 

� 2.13 Show that this determinant is zero. 

� 

� 

�y 

+ z x+ z x+ y� 

� 

� 

� x y z � 

� 

1 1 1 

� 

2.14 (a) Find the 1×1, 2×2, and 3×3 matrices with i, j entry given by (−1) i+j . 

(b) Find the determinant of the square matrix with i, j entry (−1) i+j . 

2.15 (a) Find the 1×1, 2×2, and 3×3 matrices with i, j entry given by i + j. 

(b) Find the determinant of the square matrix with i, j entry i + j. 

� 2.16 Show that determinant functions are not linear by giving a case where |A + 

B| �= |A| + |B|. 

2.17 The second condition in the definition, that row swaps change the sign of a 

determinant, is somewhat annoying. It means we have to keep track of the number 

of swaps, to compute how the sign alternates. Can we get rid of it? Can we replace 

it with the condition that row swaps leave the determinant unchanged? (If so then 

we would need new 1 ×1, 2×2, and 3×3 formulas, but that would be a minor 

matter.) 

2.18 Prove that the determinant of any triangular matrix, upper or lower, is the 

product down its diagonal. 

2.19 Refer to the definition of elementary matrices in the Mechanics of Matrix 

Multiplication subsection. 

(a) What is the determinant of each kind of elementary matrix?


(b) Prove that if E is any elementary matrix then |ES| = |E||S| for any appropriately 

sized S. 

(c) (This question doesn’t involve determinants.) Prove that if T is singular then 

a product TS is also singular. 

(d) Show that |TS| = |T ||S|. 

(e) Show that if T is nonsingular then |T −1 | = |T | −1 . 

2.20 Prove that the determinant of a product is the product of the determinants 

|TS| = |T ||S| in this way. Fix the n × n matrix S and consider the function 

d: Mn×n → R given by T ↦→ |TS|/|S|. 

(a) Check that d satisfies property (1) in the definition of a determinant function. 

(b) Check property (2). 

(c) Check property (3). 

(d) Check property (4). 

(e) Conclude the determinant of a product is the product of the determinants. 

2.21 A submatrix of a given matrix A is one that can be obtained by deleting some 

of the rows and columns of A. Thus, the first matrix here is a submatrix of the 

second. 

� � � � 

3 4 1 

3 1 

0 9 −2 

2 5 

2 −1 5 

Prove that for any square matrix, the rank of the matrix is r if and only if r is the 

largest integer such that there is an r×r submatrix with a nonzero determinant. 

� 2.22 Prove that a matrix with rational entries has a rational determinant. 

2.23 [Am. Math. Mon., Feb. 1953] Find the element of likeness in (a) simplifying a 

fraction, (b) powdering the nose, (c) building new steps on the church, (d) keeping 

emeritus professors on campus, (e) putting B, C, D in the determinant 

� 

� 1 a a 

� 

� 

� 

� 

� 

2 

a 3 

a 3 

1 a a 2 

B a 3 

1 a 

C D a 3 

� 

� 

� 

� 

� . 

� 

1 � 

4.I.3 The Permutation Expansion 

The prior subsection defines a function to be a determinant if it satisfies four 

conditions and shows that there is at most one n×n determinant function for 

each n. Whatisleftistoshowthatforeachn such a function exists. 

How could such a function not exist? After all, we have done computations 

that start with a square matrix, follow the conditions, and end with a number. 

The difficulty is that, as far as we know, the computation might not give a 

well-defined result. To illustrate this possibility, suppose that we were to change 

the second condition in the definition of determinant to be that the value of a 

determinant does not change on a row swap. By Remark 2.2 we know that 

this conflicts with the first and third conditions. Here is an instance of the


conflict: here are two Gauss’ method reductions of the same matrix, the first 

without any row swap 

� � � � 

1 2 −3ρ1+ρ2 1 2 

−→ 

3 4 

0 −2 

and the second with a swap. 

� � 

1 2 ρ1↔ρ2 

−→ 

3 4 

� � � � 

3 4 −(1/3)ρ1+ρ2 3 4 

−→ 

1 2 

0 2/3 

Following Definition 2.1 gives that both calculations yield the determinant −2 

since in the second one we keep track of the fact that the row swap changes 

the sign of the result of multiplying down the diagonal. But if we follow the 

supposition and change the second condition then the two calculations yield 

different values, −2 and 2. That is, under the supposition the outcome would not 

be well-defined — no function exists that satisfies the changed second condition 

along with the other three. 

Of course, observing that Definition 2.1 does the right thing in this one 

instance is not enough; what we will do in the rest of this section is to show 

that there is never a conflict. The natural way to try this would be to define 

the determinant function with: “The value of the function is the result of doing 

Gauss’ method, keeping track of row swaps, and finishing by multiplying down 

the diagonal”. (Since Gauss’ method allows for some variation, such as a choice 

of which row to use when swapping, we would have to fix an explicit algorithm.) 

Then we would be done if we verified that this way of computing the determinant 

satisfies the four properties. For instance, if T and ˆ T are related by a row swap 

then we would need to show that this algorithm returns determinants that are 

negatives of each other. However, how to verify this is not evident. So the 

development below will not proceed in this way. Instead, in this subsection we 

will define a different way to compute the value of a determinant, a formula, 

and we will use this way to prove that the conditions are satisfied. 

The formula that we shall use is based on an insight gotten from property (2) 

of the definition of determinants. This property shows that determinants are 

not linear. 

3.1 Example For this matrix det(2A) �= 2· det(A). 

� � 

2 1 

A = 

−1 3 

Instead, the scalar comes out of each of the two rows. 

� 

� 

� 4 

�−2 � 

2� 

� 

6� 

=2· 

� 

� 

� 2 

�−2 � 

1� 

� 

6� 

=4· 

� 

� 

� 2 

�−1 � 

1� 

� 

3� 

Since scalars come out a row at a time, we might guess that determinants 

are linear a row at a time.


3.2 Definition Let V be a vector space. A map f : V n → R is multilinear if 

(1) f(�ρ1,...,�v + �w,... ,�ρn) =f(�ρ1,...,�v,...,�ρn)+f(�ρ1,..., �w,...,�ρn) 

(2) f(�ρ1,...,k�v,...,�ρn) =k · f(�ρ1,...,�v,...,�ρn) 

for �v, �w ∈ V and k ∈ R. 

3.3 Lemma Determinants are multilinear. 

Proof. The definition of determinants gives property (2) (Lemma 2.3 following 

that definition covers the k = 0 case) so we need only check property (1). 

det(�ρ1,...,�v + �w,...,�ρn) = det(�ρ1,...,�v,...,�ρn) + det(�ρ1,..., �w,...,�ρn) 

If the set {�ρ1,...,�ρi−1,�ρi+1,...,�ρn} is linearly dependent then all three matrices 

are singular and so all three determinants are zero and the equality is trivial. 

Therefore assume that the set is linearly independent. This set of n-wide row 

vectors has n − 1 members, so we can make a basis by adding one more vector 

〈�ρ1,...,�ρi−1, � β, �ρi+1,...,�ρn〉. Express�v and �w with respect to this basis 

giving this. 

�v = v1�ρ1 + ···+ vi−1�ρi−1 + vi � β + vi+1�ρi+1 + ···+ vn�ρn 

�w = w1�ρ1 + ···+ wi−1�ρi−1 + wi � β + wi+1�ρi+1 + ···+ wn�ρn 

�v + �w =(v1 + w1)�ρ1 + ···+(vi + wi) � β + ···+(vn + wn)�ρn 

By the definition of determinant, the value of det(�ρ1,...,�v + �w,...,�ρn) is unchanged 

by the pivot operation of adding −(v1 + w1)�ρ1 to �v + �w. 

�v + �w − (v1 + w1)�ρ1 =(v2 + w2)�ρ2 + ···+(vi + wi) � β + ···+(vn + wn)�ρn 

Then, to the result, we can add −(v2 + w2)�ρ2, etc.Thus 

det(�ρ1,...,�v + �w,...,�ρn) 

= det(�ρ1,...,(vi + wi) · � β,...,�ρn) 

=(vi + wi) · det(�ρ1,..., � β,...,�ρn) 

= vi · det(�ρ1,..., � β,...,�ρn)+wi · det(�ρ1,..., � β,...,�ρn) 

(using (2) for the second equality). To finish, bring vi and wi back inside in 

front of � β and use pivoting again, this time to reconstruct the expressions of �v 

and �w in terms of the basis, e.g., start with the pivot operations of adding v1�ρ1 

to vi � β and w1�ρ1 to wi�ρ1, etc. QED 

Multilinearity allows us to expand a determinant into a sum of determinants, 

each of which involves a simple matrix.


3.4 Example We can use multilinearity to split this determinant into two, 

first breaking up the first row 

� � 

� 

�2 

1� 

� 

�4 3� 

= 

� � 

� 

�2 

0� 

� 

�4 3� 

+ 

� � 

� 

�0 

1� 

� 

�4 3� 

and then separating each of those two, breaking along the second rows. 

� 

� 

= �2 

�4 � 

0� 

� 

0� 

+ 

� 

� 

�2 

�0 � 

0� 

� 

3� 

+ 

� 

� 

�0 

�4 � 

1� 

� 

0� 

+ 

� 

� 

�0 

�0 � 

1� 

� 

3� 

We are left with four determinants, such that in each row of each matrix there 

is a single entry from the original matrix. 

3.5 Example In the same way, a 3×3 determinant separates into a sum of 

many simpler determinants. We start by splitting along the first row, producing 

three determinants (the zero in the 1, 3 position is underlined to set it off visually 

from the zeroes that appear in the splitting). 

� � 

� 

�2 

1 −1� 

� 

� 

�4 

3 0 � 

� 

�2 

1 5 � = 

� � 

� 

�2 

0 0� 

� 

� 

�4 

3 0� 

� 

�2 

1 5� 

+ 

� � 

� 

�0 

1 0� 

� 

� 

�4 

3 0� 

� 

�2 

1 5� 

+ 

� � 

� 

�0 

0 −1� 

� 

� 

�4 

3 0 � 

� 

�2 

1 5 � 

Each of these three will itself split in three along the second row. Each of 

the resulting nine splits in three along the third row, resulting in twenty seven 

determinants 

� 

� 

�2 

= � 

�4 

�2 

0 

0 

0 

� 

0� 

� 

0� 

� 

0� 

+ 

� 

� 

�2 

� 

�4 

�0 

0 

0 

1 

� 

0� 

� 

0� 

� 

0� 

+ 

� 

� 

�2 

� 

�4 

�0 

0 

0 

0 

� 

0� 

� 

0� 

� 

5� 

+ 

� 

� 

�2 

� 

�0 

�2 

0 

3 

0 

� � 

0� 

� 

� �0 

0� 

� + ···+ � 

�0 

0� 

�0 

0 

0 

0 

� 

−1� 

� 

0 � 

� 

5 � 

such that each row contains a single entry from the starting matrix. 

So an n×n determinant expands into a sum of n n determinants where each 

row of each summands contains a single entry from the starting matrix. However, 

many of these summand determinants are zero. 

3.6 Example In each of these three matrices from the above expansion, two 

of the rows have their entry from the starting matrix in the same column, e.g., 

in the first matrix, the 2 and the 4 both come from the first column. 

� � � � � � 

� 

�2 

0 0� 

� 

� �0 

0 −1� 

� 

� �0 

1 0� 

� 

� 

�4 

0 0� 

� 

� �0 

3 0 � � 

� �0 

0 0� 

� 

�0 

1 0� 

�0 

0 5 � �0 

0 5� 

Any such matrix is singular, because in each, one row is a multiple of the other 

(or is a zero row). Thus, any such determinant is zero, by Lemma 2.3.


Therefore, the above expansion of the 3×3 determinant into the sum of the 

twenty seven determinants simplifies to the sum of these six. 

� � 

� 

�2 

1 −1� 

� 

� 

�4 

3 0 � 

� 

�2 

1 5 � = 

� � 

� 

�2 

0 0� 

� 

� 

�0 

3 0� 

� 

�0 

0 5� 

+ 

� � 

� 

�2 

0 0� 

� 

� 

�0 

0 0� 

� 

�0 

1 0� 

� � 

� 

�0 

1 0� 

� 

+ � 

�4 

0 0� 

� 

�0 

0 5� 

+ 

� � 

� 

�0 

1 0� 

� 

� 

�0 

0 0� 

� 

�2 

0 0� 

� � 

� 

�0 

0 −1� 

� 

+ � 

�4 

0 0 � 

� 

�0 

1 0 � + 

� � 

� 

�0 

0 −1� 

� 

� 

�0 

3 0 � 

� 

�2 

0 0 � 

We can bring out the scalars. 

� � 

� � 

� 

�1 

0 0� 

� 

� 

�1 

0 0� 

� 

= (2)(3)(5) � 

�0 

1 0� 

� + (2)(0)(1) � 

�0 

0 1� 

� 

�0 

0 1� 

�0 

1 0� 

� � 

� � 

� 

�0 

1 0� 

� 

� 

�0 

1 0� 

� 

+ (1)(4)(5) � 

�1 

0 0� 

� + (1)(0)(2) � 

�0 

0 1� 

� 

�0 

0 1� 

�1 

0 0� 

� � 

� 

�0 

0 1� 

� 

+(−1)(4)(1) � 

�1 

0 0� 

� 

�0 

1 0� 

+(−1)(3)(2) 

� � 

� 

�0 

0 1� 

� 

� 

�0 

1 0� 

� 

�1 

0 0� 

To finish, we evaluate those six determinants by row-swapping them to the 

identity matrix, keeping track of the resulting sign changes. 

=30· (+1) + 0 · (−1) 

+20· (−1) + 0 · (+1) 

− 4 · (+1) − 6 · (−1) = 12 

That example illustrates the key idea. We’ve applied multilinearity to a 3×3 

determinant to get 3 3 separate determinants, each with one distinguished entry 

per row. We can drop most of these new determinants because the matrices 

are singular, with one row a multiple of another. We are left with the oneentry-per-row 

determinants also having only one entry per column (one entry 

from the original determinant, that is). And, since we can factor scalars out, we 

can further reduce to only considering determinants of one-entry-per-row-andcolumn 

matrices where the entries are ones. 

These are permutation matrices. Thus, the determinant can be computed 

in this three-step way (Step 1) for each permutation matrix, multiply together 

the entries from the original matrix where that permutation matrix has ones, 

(Step 2) multiply that by the determinant of the permutation matrix and 

(Step 3) do that for all permutation matrices and sum the results together.


To state this as a formula, we introduce a notation for permutation matrices. 

Let ιj be the row vector that is all zeroes except for a one in its j-th entry, so 

that the four-wide ι2 is � 0 1 0 0 � . We can construct permutation matrices 

by permuting — that is, scrambling — the numbers 1, 2, ... , n, and using them 

as indices on the ι’s. For instance, to get a 4×4 permutation matrix matrix, we 

can scramble the numbers from 1 to 4 into this sequence 〈3, 2, 1, 4〉 and take the 

corresponding row vector ι’s. 

⎛ ⎞ 

ι3 

⎜ι2⎟ 

⎜ ⎟ 

⎝ι1⎠ 

= 

⎛ ⎞ 

0 0 1 0 

⎜ 

⎜0 

1 0 0 ⎟ 

⎝1 

0 0 0⎠ 

0 0 0 1 

ι4 

3.7 Definition An n-permutation is a sequence consisting of an arrangement 

of the numbers 1, 2, ... , n. 

3.8 Example The 2-permutations are φ1 = 〈1, 2〉 and φ2 = 〈2, 1〉. These are 

the associated permutation matrices. 

Pφ1 = 

� � � � 

ι1 1 0 

= 

Pφ2 0 1 

= 

� � � � 

ι2 0 1 

= 

1 0 

ι2 

We sometimes write permutations as functions, e.g., φ2(1) = 2, and φ2(2) = 1. 

Then the rows of Pφ2 are ι φ2(1) = ι2 and ι φ2(2) = ι1. 

The 3-permutations are φ1 = 〈1, 2, 3〉, φ2 = 〈1, 3, 2〉, φ3 = 〈2, 1, 3〉, φ4 = 

〈2, 3, 1〉, φ5 = 〈3, 1, 2〉, andφ6 = 〈3, 2, 1〉. Here are two of the associated permu- 

tation matrices. 

Pφ2 = 

⎛ 

⎝ 

ι1 

ι3 

ι2 

⎞ ⎛ ⎞ 

1 0 0 

⎠ = ⎝0 

0 1⎠ 

Pφ5 

0 1 0 

= 

⎛ 

⎝ 

ι1 

ι3 

ι1 

ι2 

⎞ ⎛ 

0 

⎠ = ⎝1 

0 

0 

⎞ 

1 

0⎠ 

0 1 0 

For instance, the rows of Pφ5 are ι φ5(1) = ι3, ι φ5(2) = ι1, andι φ5(3) = ι2. 

3.9 Definition The permutation expansion for determinants is 

� 

�t1,1 

� 

�t2,1 

� 

� 

� 

� 

� 

t1,2 

t2,2 

. 

... 

... 

� 

t1,n� 

� 

t2,n� 

� 

� 

� 

� 

� 

tn,1 tn,2 ... tn,n 

where φ1,... ,φk are all of the n-permutations. 

= t 1,φ1(1)t 2,φ1(2) ···t n,φ1(n)|Pφ1 | 

+ t 1,φ2(1)t 2,φ2(2) ···t n,φ2(n)|Pφ2 | 

. 

+ t 1,φk(1)t 2,φk(2) ···t n,φk(n)|Pφk | 

This formula is often written in summation notation 

|T | = 

� 

permutations φ 

t 1,φ(1)t 2,φ(2) ···t n,φ(n) |Pφ|


read aloud as “the sum, over all permutations φ, of terms having the form 

t1,φ(1)t2,φ(2) ···tn,φ(n)|Pφ|”. This phrase is just a restating of the three-step 

process (Step 1) for each permutation matrix, compute t1,φ(1)t2,φ(2) ···tn,φ(n) (Step 2) multiply that by |Pφ| and (Step 3) sum all such terms together. 

3.10 Example The familiar formula for the determinant of a 2×2 matrix can 

be derived in this way. 

� 

� 

� 

� t1,1 

� 

t1,2� 

� 

t2,1 t2,2� 

= t1,1t2,2 ·|Pφ1 | + t1,2t2,1 ·|Pφ2 | 

� � 

� 

= t1,1t2,2 · �1 

0� 

� 

�0 1� 

+ t1,2t2,1 

� � 

� 

· �0 

1� 

� 

�1 0� 

= t1,1t2,2 − t1,2t2,1 

(the second permutation matrix takes one row swap to pass to the identity). 

Similarly, the formula for the determinant of a 3×3 matrix is this. 

� 

�t1,1 

� 

�t2,1 

� 

�t3,1 

t1,2 

t2,2 

t3,2 

� 

t1,3� 

� 

t2,3� 

� 

t3,3� 

= t1,1t2,2t3,3 |Pφ1 | + t1,1t2,3t3,2 |Pφ2 | + t1,2t2,1t3,3 |Pφ3 | 

+ t1,2t2,3t3,1 |Pφ4 | + t1,3t2,1t3,2 |Pφ5 | + t1,3t2,2t3,1 |Pφ6 | 

= t1,1t2,2t3,3 − t1,1t2,3t3,2 − t1,2t2,1t3,3 

+ t1,2t2,3t3,1 + t1,3t2,1t3,2 − t1,3t2,2t3,1 

Computing a determinant by permutation expansion usually takes longer 

than Gauss’ method. However, here we are not trying to do the computation 

efficiently, we are instead trying to give a determinant formula that we can 

prove to be well-defined. While the permutation expansion is impractical for 

computations, it is useful in proofs. In particular, we can use it for the result 

that we are after. 

3.11 Theorem For each n there is a n×n determinant function. 

The proof is deferred to the following subsection. Also there is the proof of 

the next result (they share some features). 

3.12 Theorem The determinant of a matrix equals the determinant of its 

transpose. 

The consequence of this theorem is that, while we have so far stated results 

in terms of rows (e.g., determinants are multilinear in their rows, row swaps 

change the signum, etc.), all of the results also hold in terms of columns. The 

final result gives examples. 

3.13 Corollary A matrix with two equal columns is singular. Column swaps 

change the sign of a determinant. Determinants are multilinear in their columns. 

Proof. For the first statement, transposing the matrix results in a matrix with 

the same determinant, and with two equal rows, and hence a determinant of 

zero. The other two are proved in the same way. QED


We finish with a summary (although the final subsection contains the unfinished 

business of proving the two theorems). Determinant functions exist, 

are unique, and we know how to compute them. As for what determinants are 

about, perhaps these lines [Kemp] help make it memorable. 

Determinant none, 

Solution: lots or none. 

Determinant some, 

Solution: just one. 

Exercises 

These summarize the notation used in this book for the 2- and3-permutations. i 1 2 i 1 2 3 

φ1(i) 1 2 φ1(i) 1 2 3 

φ2(i) 2 1 φ2(i) 1 3 2 

φ3(i) 2 1 3 

φ4(i) 2 3 1 

φ5(i) 3 1 2 

φ6(i) 3 2 1 

� 3.14 Compute 

� 

the 

� 

determinant 

� 

by using 

� 

the permutation expansion. 

�1 

2 3� 

� 2 2 1� 

� � � � 

(a) �4 

5 6� 

(b) � 3 −1 0� 

� 

7 8 9 

� � 

−2 0 5 

� 

� 3.15 Compute these both with Gauss’ method and with the permutation expansion 

formula. 

� � � � 

� 

(a) �2 

1� 

�0 

1 4� 

� � � 

�3 1� 

(b) �0 

2 3� 

� 

1 5 1 

� 

� 3.16 Use the permutation expansion formula to derive the formula for 3×3 determinants. 

3.17 List all of the 4-permutations. 

3.18 A permutation, regarded as a function from the set {1, .., n} to itself, is oneto-one 

and onto. Therefore, each permutation has an inverse. 

(a) Find the inverse of each 2-permutation. 

(b) Find the inverse of each 3-permutation. 

3.19 Prove that f is multilinear if and only if for all �v, �w ∈ V and k1,k2 ∈ R, this 

holds. 

f(�ρ1,...,k1�v1 + k2�v2,...,�ρn) =k1f(�ρ1,...,�v1,...,�ρn)+k2f(�ρ1,...,�v2,...,�ρn) 

3.20 Find the only nonzero term in the permutation expansion of this matrix. 

� � 

�0 

1 0 0� 

� � 

�1 

0 1 0� 

� � 

�0 

1 0 1� 

�0 

0 1 0� 

Compute that determinant by finding the signum of the associated permutation. 

3.21 How would determinants change if we changed property (4) of the definition 

to read that |I| =2? 

3.22 Verify the second and third statements in Corollary 3.13.


� 3.23 Show that if an n×n matrix has a nonzero determinant then any column vector 

�v ∈ R n can be expressed as a linear combination of the columns of the matrix. 

3.24 True or false: a matrix whose entries are only zeros or ones has a determinant 

equal to zero, one, or negative one. 

3.25 (a) Show that there are 120 terms in the permutation expansion formula of 

a5×5 matrix. 

(b) How many are sure to be zero if the 1, 2 entry is zero? 

3.26 How many n-permutations are there? 

3.27 AmatrixAis skew-symmetric if A trans � 

= 

� 

−A, asinthismatrix. 

0 3 

A = 

−3 0 

Show that n×n skew-symmetric matrices with nonzero determinants exist only for 

even n. 

� 3.28 What is the smallest number of zeros, and the placement of those zeros, needed 

to ensure that a 4×4 matrix has a determinant of zero? 

� 3.29 If we have n data points (x1,y1), (x2,y2),... ,(xn,yn) and want to find a 

polynomial p(x) =an−1x n−1 + an−2x n−2 + ···+ a1x + a0 passing through those 

points then we can plug in the points to get an n equation/n unknown linear 

system. The matrix of coefficients for that system is called the Vandermonde 

matrix. Prove that the determinant of the transpose of that matrix of coefficients 

� 

� 1 1 ... 1 

� 

� x1 x2 ... xn 

� 

� x1 

� 

� 

� 

� 

2 

x2 2 

... xn 2 

. 

x1 n−1 

x2 n−1 

... xn n−1 

� 

� 

� 

� 

� 

� 

� 

� 

� 

� 

equals the product, over all indices i, j ∈{1,...,n} with i


3.34 [Am. Math. Mon., Jan. 1949] LetSbe the sum of the integer elements of a 

magic square of order three and let D be the value of the square considered as a 

determinant. Show that D/S is an integer. 

3.35 [Am. Math. Mon., Jun. 1931] Show that the determinant of the n 2 elements 

in the upper left corner of the Pascal triangle 

1 1 1 1 . . 

1 2 3 . . 

1 3 . . 

1 

. 

. 

. . 

has the value unity. 

4.I.4 Determinants Exist 

This subsection is optional. It consists of proofs of two results from the prior 

subsection. These proofs involve the properties of permutations, which will not 

be used later, except in the optional Jordan Canonical Form subsection. 

The prior subsection attacks the problem of showing that for any size there 

is a determinant function on the set of square matrices of that size by using 

multilinearity to develop the permutation expansion. 

� 

� 

�t1,1 

t1,2 � 

... t1,n� 

� 

�t2,1 

t2,2 � 

... t2,n� 

� 

� . 

� = t 

� 

� . 

� 

1,φ1(1)t2,φ1(2) ···tn,φ1(n)|Pφ1 � 

�tn,1 

tn,2 ... tn,n 

� 

| 

+ t1,φ2(1)t2,φ2(2) ···tn,φ2(n)|Pφ2 | 

. 

= 

+ t 1,φk(1)t 2,φk(2) ···t n,φk(n)|Pφk | 

� 


t 1,φ(1)t 2,φ(2) ···t n,φ(n) |Pφ| 

This reduces the problem to showing that there is a determinant function on 

the set of permutation matrices of that size. 

Of course, a permutation matrix can be row-swapped to the identity matrix 

and to calculate its determinant we can keep track of the number of row swaps. 

However, the problem is still not solved. We still have not shown that the result 

is well-defined. For instance, the determinant of 

⎛ ⎞ 

0 1 0 0 

⎜ 

Pφ = ⎜1 

0 0 0 ⎟ 

⎝0 

0 1 0⎠ 

0 0 0 1


could be computed with one swap 

Pφ 

or with three. 

⎛ 

0 

ρ3↔ρ1 ⎜ 

Pφ −→ ⎜1 

⎝0 

0 

0 

1 

1 

0 

0 

⎞ 

0 

0 ⎟ 

0⎠ 

0 0 0 1 

ρ1↔ρ2 

−→ 

ρ2↔ρ3 

−→ 

⎛ 

1 

⎜ 

⎜0 

⎝0 

0 

1 

0 

0 

0 

1 

⎞ 

0 

0 ⎟ 

0⎠ 

0 0 0 1 

⎛ 

0 

⎜ 

⎜0 

⎝1 

0 

1 

0 

1 

0 

0 

⎞ 

0 

0 ⎟ 

0⎠ 

0 0 0 1 

ρ1↔ρ3 

−→ 

⎛ 

1 

⎜ 

⎜0 

⎝0 

0 

1 

0 

0 

0 

1 

⎞ 

0 

0 ⎟ 

0⎠ 

0 0 0 1 

Both reductions have an odd number of swaps so we figure that |Pφ| = −1 

but how do we know that there isn’t some way to do it with an even number of 

swaps? Corollary 4.6 below proves that there is no permutation matrix that can 

be row-swapped to an identity matrix in two ways, one with an even number of 

swaps and the other with an odd number of swaps. 

4.1 Definition Two rows of a permutation matrix 

⎛ ⎞ 

. 

⎜ . ⎟ 

⎜ιk⎟ 

⎜ ⎟ 

⎜ . 

⎜ . 

⎟ 

⎜ . ⎟ 

⎜ιj 

⎟ 

⎝ ⎠ 

. 

such that k>jareinaninversion of their natural order. 

4.2 Example This permutation matrix 

⎛ ⎞ ⎛ 

ι3 0 

⎝ι2⎠ 

= ⎝0 

0 

1 

⎞ 

1 

0⎠ 

1 0 0 

ι1 

has three inversions: ι3 precedes ι1, ι3 precedes ι2, andι2 precedes ι1. 

4.3 Lemma A row-swap in a permutation matrix changes the number of inversions 

from even to odd, or from odd to even. 

Proof. Consider a swap of rows j and k, where k>j. If the two rows are 

adjacent 

⎛ ⎞ ⎛ ⎞ 

. 

⎜ . 

. 

⎟ ⎜ . ⎟ 

⎜ 

Pφ = ⎜ι 

⎟ 

φ(j) ρk↔ρj ⎜ 

⎟ 

⎜ 

⎝ 

ι ⎟ −→ ⎜ι 

⎟ 

φ(k) ⎟ 

⎜ 

φ(k) ⎠ ⎝ 

ι ⎟ 

φ(j) ⎠ 

. 

. 

. 

.


then the swap changes the total number of inversions by one — either removing 

or producing one inversion, depending on whether φ(j) >φ(k) or not, since 

inversions involving rows not in this pair are not affected. Consequently, the 

total number of inversions changes from odd to even or from even to odd. 

If the rows are not adjacent then they can be swapped via a sequence of 

adjacent swaps, first bringing row k up 

⎛ ⎞ 

. 

⎜ . ⎟ 

⎜ ι ⎟ φ(j) ⎟ 

⎜ 

⎜ι 

⎟ φ(j+1) ⎟ 

⎜ 

⎜ι 

⎟ φ(j+2) ⎟ 

⎜ . ⎟ 

⎜ . ⎟ 

⎜ 

⎝ ι ⎟ 

φ(k) ⎠ 

. 

and then bringing row j down. 

ρj+1↔ρj+2 

−→ 

ρk↔ρk−1 

−→ ρk−1↔ρk−2 

−→ ... ρj+1↔ρj 

−→ 

ρj+2↔ρj+3 

−→ ... ρk−1↔ρk 

−→ 

⎛ ⎞ 

. 

⎜ . ⎟ 

⎜ ι ⎟ φ(k) ⎟ 

⎜ ι ⎟ φ(j) ⎟ 

⎜ 

⎜ι 

⎟ φ(j+1) ⎟ 

⎜ . ⎟ 

⎜ . ⎟ 

⎜ 

⎝ι 

⎟ 

φ(k−1) ⎠ 

. 

⎛ ⎞ 

. 

⎜ . ⎟ 

⎜ ι ⎟ φ(k) ⎟ 

⎜ 

⎜ι 

⎟ φ(j+1) ⎟ 

⎜ 

⎜ι 

⎟ φ(j+2) ⎟ 

⎜ . 

⎜ . 

⎟ 

⎜ . ⎟ 

⎜ 

⎝ ι ⎟ 

φ(j) ⎠ 

. 

Each of these adjacent swaps changes the number of inversions from odd to even 

or from even to odd. There are an odd number (k − j)+(k − j − 1) of them. 

The total change in the number of inversions is from even to odd or from odd 

to even. QED 

4.4 Definition The signum of a permutation sgn(φ) is+1ifthenumberof 

inversions in Pφ is even, and is −1 if the number of inversions is odd. 

4.5 Example With the subscripts from Example 3.8 for the 3-permutations, 

sgn(φ1) = 1 while sgn(φ2) =−1. 

4.6 Corollary If a permutation matrix has an odd number of inversions then 

swapping it to the identity takes an odd number of swaps. If it has an even 

number of inversions then swapping to the identity takes an even number of 

swaps. 

Proof. The identity matrix has zero inversions. To change an odd number to 

zero requires an odd number of swaps, and to change an even number to zero 

requires an even number of swaps. QED


We still have not shown that the permutation expansion is well-defined because 

we have not considered row operations on permutation matrices other than 

row swaps. We will finesse this problem: we will define a function d: Mn×n → R 

by altering the permutation expansion formula, replacing |Pφ| with sgn(φ) 

d(T )= 

� 

t1,φ(1)t2,φ(2) ...tn,φ(n) sgn(φ) 


(this gives the same value as the permutation expansion because the prior result 

shows that det(Pφ) =sgn(φ)). This formula’s advantage is that the number of 

inversions is clearly well-defined — just count them. Therefore, we will show 

that a determinant function exists for all sizes by showing that d is it, that is, 

that d satisfies the four conditions. 

4.7 Lemma The function d is a determinant. Hence determinants exist for 

every n. 

Proof. We’ll must check that it has the four properties from the definition. 

Property (4) is easy; in 

d(I) = � 

ι1,φ(1)ι2,φ(2) ···ιn,φ(n) sgn(φ) 

perms φ 

all of the summands are zero except for the product down the diagonal, which 

is one. 

For property (3) consider d( ˆ T ) where T kρi 

−→ ˆ T . 

� 

ˆt 1,φ(1) ···ˆt i,φ(i) ···ˆt n,φ(n) sgn(φ) = 

perms φ 

� 

t1,φ(1) ···kti,φ(i) ···tn,φ(n) sgn(φ) 

φ 

Factor the k out of each term to get the desired equality. 

= k · � 

For (2), let T ρi↔ρj 

−→ ˆ T . 

d( ˆ T )= � 

φ 

t 1,φ(1) ···t i,φ(i) ···t n,φ(n) sgn(φ) =k · d(T ) 

ˆt 1,φ(1) ···ˆt i,φ(i) ···ˆt j,φ(j) ···ˆt n,φ(n) sgn(φ) 

perms φ 

To convert to unhatted t’s, for each φ consider the permutation σ that equals φ 

except that the i-th and j-th numbers are interchanged, σ(i) =φ(j) andσ(j) = 

φ(i). Replacing the φ in ˆt 1,φ(1) ···ˆt i,φ(i) ···ˆt j,φ(j) ···ˆt n,φ(n) with this σ gives 

t1,σ(1) ···tj,σ(j) ···ti,σ(i) ···tn,σ(n). Now sgn(φ) = − sgn(σ) (by Lemma 4.3) 

andsoweget 

t1,σ(1) ···tj,σ(j) ···ti,σ(i) ···tn,σ(n) · � − sgn(σ) � 

= � 

σ 

= − � 

σ 

t 1,σ(1) ···t j,σ(j) ···t i,σ(i) ···t n,σ(n) · sgn(σ)


where the sum is over all permutations σ derived from another permutation φ 

by a swap of the i-th and j-th numbers. But any permutation can be derived 

from some other permutation by such a swap, in one and only one way, so this 

summation is in fact a sum over all permutations, taken once and only once. 

Thus d( ˆ T )=−d(T ). 

To do property (1) let T kρi+ρj 

−→ ˆ T and consider 

d( ˆ T )= � 

ˆt 1,φ(1) ···ˆt i,φ(i) ···ˆt j,φ(j) ···ˆt n,φ(n) sgn(φ) 

perms φ 

= � 

t1,φ(1) ···ti,φ(i) ···(kti,φ(j) + tj,φ(j)) ···tn,φ(n) sgn(φ) 

φ 

(notice: that’s kt i,φ(j), notkt j,φ(j)). Distribute, commute, and factor. 

= � 

φ 

= � 

φ 

� t1,φ(1) ···t i,φ(i) ···kt i,φ(j) ···t n,φ(n) sgn(φ) 

+ t 1,φ(1) ···t i,φ(i) ···t j,φ(j) ···t n,φ(n) sgn(φ) � 

t 1,φ(1) ···t i,φ(i) ···kt i,φ(j) ···t n,φ(n) sgn(φ) 

+ � 

= k · � 

φ 

φ 

+ d(T ) 

t 1,φ(1) ···t i,φ(i) ···t j,φ(j) ···t n,φ(n) sgn(φ) 

t 1,φ(1) ···t i,φ(i) ···t i,φ(j) ···t n,φ(n) sgn(φ) 

We finish by showing that the terms t 1,φ(1) ···t i,φ(i) ···t i,φ(j) ...t n,φ(n) sgn(φ) 

add to zero. This sum represents d(S) where S is a matrix equal to T except 

that row j of S is a copy of row i of T (because the factor is t i,φ(j), nott j,φ(j)). 

Thus, S has two equal rows, rows i and j. Since we have already shown that d 

changes sign on row swaps, as in Lemma 2.3 we conclude that d(S) =0. QED 

We have now shown that determinant functions exist for each size. We 

already know that for each size there is at most one determinant. Therefore, 

the permutation expansion computes the one and only determinant value of a 

square matrix. 

We end this subsection by proving the other result remaining from the prior 

subsection, that the determinant of a matrix equals the determinant of its transpose. 

4.8 Example Writing out the permutation expansion of the general 3×3 matrix 

and of its transpose, and comparing corresponding terms 

� � 

� � 

� 

�a 

b c� 

� 

� 

�0 

0 1� 

� 

� 

�d 

e f� 

� = ··· + cdh · � 

�1 

0 0� 

� + ··· 

�g 

h i� 

�0 

1 0�


(terms with the same letters) 

� 

� 

�a 

� 

�b 

�c 

d 

e 

f 

� 

� 

g� 

� 

� 

�0 

h� 

� = ··· + dhc · � 

�0 

i� 

�1 

1 

0 

0 

� 

0� 

� 

1� 

� + ··· 

0� 

shows that the corresponding permutation matrices are transposes. That is, 

there is a relationship between these corresponding permutations. Exercise 15 

shows that they are inverses. 

4.9 Theorem The determinant of a matrix equals the determinant of its transpose. 

Proof. Call the matrix T and denote the entries of T trans with s’s so that 

ti,j = sj,i. Substitution gives this 

|T | = � 

t1,φ(1) ...tn,φ(n) sgn(φ) = � 

sφ(1),1 ...sφ(n),n sgn(φ) 

perms φ 

and we can finish the argument by manipulating the expression on the right 

to be recognizable as the determinant of the transpose. We have written all 

permutation expansions (as in the middle expression above) with the row indices 

ascending. To rewrite the expression on the right in this way, note that because 

φ is a permutation, the row indices in the term on the right φ(1), ... , φ(n) are 

just the numbers 1, ... , n, rearranged. We can thus commute to have these 

ascend, giving s 1,φ −1 (1) ···s n,φ −1 (n) (if the column index is j and the row index 

is φ(j) then, where the row index is i, the column index is φ −1 (i)). Substituting 

on the right gives 

= � 

φ −1 

s 1,φ −1 (1) ···s n,φ −1 (n) sgn(φ −1 ) 

(Exercise 14 shows that sgn(φ −1 )=sgn(φ)). Since every permutation is the 

inverse of another, a sum over all φ −1 is a sum over all permutations φ 

= � 

perms σ 

s 1,σ ( 1) ...s n,σ(n) sgn(σ) = � � T trans � � 

as required. QED 

Exercises 

These summarize the notation used in this book for the 2- and3-permutations. i 1 2 i 1 2 3 

φ1(i) 1 2 φ1(i) 1 2 3 

φ2(i) 2 1 φ2(i) 1 3 2 

φ3(i) 2 1 3 

φ4(i) 2 3 1 

φ5(i) 3 1 2 

φ6(i) 3 2 1 

φ


4.10 Give the permutation expansion of a general 2×2 matrix and its transpose. 

� 4.11 This problem appears also in the prior subsection. 

(a) Find the inverse of each 2-permutation. 

(b) Find the inverse of each 3-permutation. 

� 4.12 (a) Find the signum of each 2-permutation. 

(b) Find the signum of each 3-permutation. 

4.13 What is the signum of the n-permutation φ = 〈n, n − 1,...,2, 1〉? 

4.14 Prove these. 

(a) Every permutation has an inverse. 

(b) sgn(φ −1 ) = sgn(φ) 

(c) Every permutation is the inverse of another. 

4.15 Prove that the matrix of the permutation inverse is the transpose of the matrix 

of the permutation Pφ−1 = Pφ trans , for any permutation φ. 

� 4.16 Show that a permutation matrix with m inversions can be row swapped to 

the identity in m steps. Contrast this with Corollary 4.6. 

� 4.17 For any permutation φ let g(φ) be the integer defined in this way. 

� 

g(φ) = [φ(j) − φ(i)] 

i

Section II. Geometry of Determinants 319 

4.II Geometry of Determinants 

The prior section develops the determinant algebraically, by considering what 

formulas satisfy certain properties. This section complements that with a geometric 

approach. One advantage of this approach is that, while we have so far 

only considered whether or not a determinant is zero, here we shall give a meaning 

to the value of that determinant. (The prior section handles determinants 

as functions of the rows, but in this section columns are more convenient. The 

final result of the prior section says that we can make the switch.) 

4.II.1 Determinants as Size Functions 

This parallelogram picture 

� � 

x2 

y2 

� � 

x1 

is familiar from the construction of the sum of the two vectors. One way to 

compute the area that it encloses is to draw this rectangle and subtract the 

area of each subregion. 

y 2 

y 1 

A 

C 

x 2 

B 

E 

x 1 

D 

F 

y1 

area of parallelogram 

= area of rectangle − area of A − area of B 

−···−area of F 

=(x1 + x2)(y1 + y2) − x2y1 − x1y1/2 

− x2y2/2 − x2y2/2 − x1y1/2 − x2y1 

= x1y2 − x2y1 

The fact that the area equals the value of the determinant 

� � 

� � 

� � 

� = x1y2 − x2y1 

� x1 x2 

y1 y2 

is no coincidence. The properties in the definition of determinants make reasonable 

postulates for a function that measures the size of the region enclosed 

by the vectors in the matrix. 

For instance, this shows the effect of multiplying one of the box-defining 

vectors by a scalar (the scalar used is k =1.4). 

�w 

�v 

�w 

k�v


Compared to the shaded region enclosed by �v and �w, the region formed by 

k�v and �w is bigger by a factor of k. This illustrates that size(k�v, �w) =k · 

size(�v, �w). Generalized, we expect of the size measure that size(...,k�v,...)= 

k · size(...,�v,...). Of course, this postulate is already familiar as one of the 

properties in the defintion of determinants. 

Another property of determinants is that they are unaffected by pivoting. 

Here are before-pivoting and after-pivoting boxes (the scalar used is k =0.35). 

�w 

�v 

k�v + �w 

Although the region on the right, the box formed by v and k�v + �w, is more 

slanted than the shaded region, the two have the same base and the same height 

and hence the same area. This illustrates that size(�v, k�v + �w) = size(�v, �w). 

Generalized, size(...,�v,..., �w,...) = size(...,�v,...,k�v + �w,...), which is a 

restatement of the determinant postulate. 

Of course, this picture 

�e2 

shows that size(�e1,�e2) = 1, and we naturally extend that to any number of 

dimensions size(�e1,...,�en) = 1, which is a restatement of the property that the 

determinant of the identity matrix is one. 

With that, because property (2) of determinants is redundant (as remarked 

right after the definition), we have that all of the properties of determinants are 

reasonable to expect of a function that gives the size of boxes. We can now cite 

the work done in the prior section to show that the determinant exists and is 

unique to be assured that these postulates are consistent and sufficient (we do 

not need any more postulates). That is, we’ve got an intuitive justification to 

interpret det(�v1,...,�vn) as the size of the box formed by the vectors. (Comment. 

An even more basic approach, which also leads to the definition below, 

is [Weston].) 

1.1 Example The volume of this parallelepiped, which can be found by the 

usual formula from high school geometry, is 12. 

� � 2 

0 

2 

� � 0 

3 

1 

�e1 

� � −1 

0 

1 

�v 

� � 

� 

�2 

0 −1� 

� 

� 

�0 

3 0 � 

� 

�2 

1 1 � =12


1.2 Remark Although property (2) of the definition of determinants is redundant, 

it raises an important point. Consider these two. 

�v 

�u 

� � 

� 

�4 

1� 

� 

�2 3� 

=10 

� � 

� 

�1 

4� 

� 

�3 2� 

= −10 

The only difference between them is in the order in which the vectors are taken. 

If we take �u first and then go to �v, follow the counterclockwise arc shown, then 

the sign is positive. Following a clockwise arc gives a negative sign. The sign 

returned by the size function reflects the ‘orientation’ or ‘sense’ of the box. (We 

see the same thing if we picture the effect of scalar multiplication by a negative 

scalar.) 

Although it is both interesting and important, the idea of orientation turns 

out to be tricky. It is not needed for the development below, and so we will pass 

it by. (See Exercise 27.) 

1.3 Definition The box (or parallelepiped) formed by 〈�v1,...,�vn〉 (where each 

vector is from Rn � 

) includes all of the set {t1�v1 + ···+ tn�vn � t1,... ,tn ∈ [0..1]}. 

The volume of a box is the absolute value of the determinant of the matrix with 

those vectors as columns. 

1.4 Example Volume, because it is an absolute value, does not depend on 

the order in which the vectors are given. The volume of the parallelepiped in 

Exercise 1.1, can also be computed as the absolute value of this determinant. 

� � 

� 

�0 

2 0� 

� 

� 

�3 

0 3� 

� = −12 

�1 

2 1� 

The definition of volume gives a geometric interpretation to something in 

the space, boxes made from vectors. The next result relates the geometry to 

the functions that operate on spaces. 

1.5 Theorem A transformation t: R n → R n changes the size of all boxes by 

the same factor, namely the size of the image of a box |t(S)| is |T | times the 

size of the box |S|, where T is the matrix representing t with respect to the 

standard basis. That is, for all n×n matrices, the determinant of a product is 

the product of the determinants |TS| = |T |·|S|. 

The two sentences state the same idea, first in map terms and then in matrix 

terms. Although we tend to prefer a map point of view, the second sentence, 

the matrix version, is more convienent for the proof and is also the way that 

we shall use this result later. (Alternate proofs are given as Exercise 23 and 

Exercise 28.) 

�v 

�u


Proof. The two statements are equivalent because |t(S)| = |TS|, as both give 

the size of the box that is the image of the unit box En under the composition 

t ◦ s (where s is the map represented by S with respect to the standard basis). 

First consider the case that |T | = 0. A matrix has a zero determinant if and 

only if it is not invertible. Observe that if TS is invertible, so that there is an 

M such that (TS)M = I, then the associative property of matrix multiplication 

T (SM) =I shows that T is also invertible (with inverse SM). Therefore, if T 

is not invertible then neither is TS —if|T | = 0 then |TS| = 0, and the result 

holds in this case. 

Now consider the case that |T |�= 0, that T is nonsingular. Recall that any 

nonsingular matrix can be factored into a product of elementary matrices, so 

that TS = E1E2 ···ErS. In the rest of this argument, we will verify that if E 

is an elementary matrix then |ES| = |E| ·|S|. The result will follow because 

then |TS| = |E1 ···ErS| = |E1|···|Er|·|S| = |E1 ···Er|·|S| = |T |·|S|. 

If the elementary matrix E is Mi(k) then Mi(k)S equals S except that row i 

has been multiplied by k. The third property of determinant functions then 

gives that |Mi(k)S| = k ·|S|. But |Mi(k)| = k, again by the third property 

because Mi(k) is derived from the identity by multiplication of row i by k, and 

so |ES| = |E|·|S| holds for E = Mi(k). The E = Pi,j = −1 andE = Ci,j(k) 

checks are similar. QED 

1.6 Example Application of the map t represented with respect to the standard 

bases by 

� � 

1 1 

−2 0 

will double sizes of boxes, e.g., from this 

to this 

�v 

t( �w) 

�w 

� � 

� 

�2 

1� 

� 

�1 2� 

=3 

t(�v) � � 

�� 3 3 � 

� 

−4 −2� 

=6 

1.7 Corollary If a matrix is invertible then the determinant of its inverse is 

the inverse of its determinant |T −1 | =1/|T |. 

Proof. 1=|I| = |TT −1 | = |T |·|T −1 | QED 

Recall that determinants are not additive homomorphisms, det(A + B) need 

not equal det(A) + det(B). The above theorem says, in contrast, that determinants 

are multiplicative homomorphisms: det(AB) does equal det(A) · det(B).


Exercises 

1.8 Find � the � �volume � of the region formed. 

1 −1 

(a) 〈 , 〉 

3 4 

� � � � � � 

2 3 8 

(b) 〈 1 , −2 , −3 〉 

0 

⎛ ⎞ 

1 

4 

⎛ ⎞ 

2 

8 

⎛ ⎞ 

−1 

⎛ ⎞ 

0 

⎜2⎟ 

⎜2⎟ 

⎜ 3 ⎟ ⎜1⎟ 

(c) 〈 ⎝ 

0 

⎠ , ⎝ 

2 

⎠ , ⎝ 

0 

⎠ , ⎝ 

0 

⎠〉 

1 2 5 7 

� 1.9 Is 

� � 

4 

1 

2 

inside of the box formed by these three? 

� � � � 

3 2 

� � 

1 

3 6 0 

1 1 5 

� 1.10 Find the volume of this region. 

� 1.11 Suppose that |A| = 3. By what factor do these change volumes? 

(a) A (b) A 2 

(c) A −2 

� 1.12 By what factor does each transformation change the size of boxes? 

� � � � � � � � � � � � 

x x − y 

x 2x 

x 3x − y 

(a) ↦→ (b) ↦→ 

(c) y ↦→ x + y + z 

y 3y 

y −2x + y 

z y − 2z 

1.13 What is the area of the image of the rectangle [2..4] × [2..5] under the action 

of this matrix? 

� 

2 

� 

3 

4 −1 

1.14 If t: R 3 → R 3 changes volumes by a factor of 7 and s: R 3 → R 3 changes volumes 

by a factor of 3/2 then by what factor will their composition changes volumes? 

1.15 In what way does the definition of a box differ from the defintion of a span? 

� 1.16 Why doesn’t this picture contradict Theorem 1.5? 

� � 

21 

01 

−→ 

area is 2 determinant is 2 area is 5 

� 1.17 Does |TS| = |ST|? |T (SP)| = |(TS)P |? 

1.18 (a) Suppose that |A| =3andthat|B| =2. Find|A 2 · B trans · B −2 · A trans |. 

(b) Assume that |A| =0. Provethat|6A 3 +5A 2 +2A| =0.


� 1.19 Let T be the matrix representing (with respect to the standard bases) the 

map that rotates plane vectors counterclockwise thru θ radians. By what factor 

does T change sizes? 

� 1.20 Must a transformation t: R 2 → R 2 that preserves areas also preserve lengths? 

� 1.21 What is the volume of a parallelepiped in R 3 bounded by a linearly dependent 

set? 

� 1.22 Find the area of the triangle in R 3 with endpoints (1, 2, 1), (3, −1, 4), and 

(2, 2, 2). (Area, not volume. The triangle defines a plane—what is the area of the 

triangle in that plane?) 

� 1.23 An alternate proof of Theorem 1.5 uses the definition of determinant functions. 

(a) Note that the vectors forming S make a linearly dependent set if and only if 

|S| = 0, and check that the result holds in this case. 

(b) For the |S| �= 0 case, to show that |TS|/|S| = |T | for all transformations, 

consider the function d: Mn×n → R given by T ↦→ |TS|/|S|. Show that d has 

the first property of a determinant. 

(c) Show that d has the remaining three properties of a determinant function. 

(d) Conclude that |TS| = |T |·|S|. 

1.24 Give a non-identity matrix with the property that A trans = A −1 . Show that 

if A trans = A −1 then |A| = ±1. Does the converse hold? 

1.25 The algebraic property of determinants that factoring a scalar out of a single 

row will multiply the determinant by that scalar shows that where H is 3×3, the 

determinant of cH is c 3 times the determinant of H. Explain this geometrically, 

that is, using Theorem 1.5, 

� 1.26 Matrices H and G are said to be similar if there is a nonsingular matrix P 

such that H = P −1 GP (we will study this relation in Chapter Five). Show that 

similar matrices have the same determinant. 

1.27 We usually represent vectors in R 2 with respect to the standard basis so 

vectors in the first quadrant have both coordinates positive. 

� � 

�v 

+3 

RepE2 (�v) = 

+2 

Moving counterclockwise around the origin, we cycle thru four regions: 

� � � � � � � � 

+ − − + 

··· −→ −→ −→ −→ −→ ··· . 

+ + − − 

Using this basis 

� � 

0 

B = 〈 , 

1 

� � 

−1 

〉 

0 

β2 � 

gives the same counterclockwise cycle. We say these two bases have the same 

orientation. 

(a) Whydotheygivethesamecycle? 

(b) What other configurations of unit vectors on the axes give the same cycle? 

(c) Find the determinants of the matrices formed from those (ordered) bases. 

(d) What other counterclockwise cycles are possible, and what are the associated 

determinants? 

(e) What happens in R 1 ? 

(f) What happens in R 3 ? 

A fascinating general-audience discussion of orientations is in [Gardner]. 

�β 1


1.28 This question uses material from the optional Determinant Functions Exist 

subsection. Prove Theorem 1.5 by using the permutation expansion formula for 

the determinant. 

� 1.29 (a) Show that this gives the equation of a line in R 2 thru (x2,y2)and(x3,y3). 

� � 

�x 

x2 x3� 

� � 

�y 

y2 y3� 

� 

1 1 1 

� =0 

(b) [Petersen] Prove that the area of a triangle with vertices (x1,y1), (x2,y2), 

and (x3,y3) is 

� 

�x1 

1 � 

�y1 

2 � 

1 

x2 

y2 

1 

� 

x3� 

� 

y3� 

1 

� . 

(c) [Math. Mag., Jan. 1973] Prove that the area of a triangle with vertices at 

(x1,y1), (x2,y2), and (x3,y3) whose coordinates are integers has an area of N 

or N/2 for some positive integer N.


4.III Other Formulas 

(This section is optional. Later sections do not depend on this material.) 

Determinants are a fount of interesting and amusing formulas. Here is one 

that is often seen in calculus classes and used to compute determinants by hand. 

4.III.1 Laplace’s Expansion 

1.1 Example In this permutation expansion 

� 

�t1,1 

� 

�t2,1 

� 

�t3,1 

t1,2 

t2,2 

t3,2 

� 

� � 

� � 

t1,3� 

� 

� 

�1 

0 0� 

� 

� 

�1 

0 0� 

� 

t2,3� 

� = t1,1t2,2t3,3 � 

�0 

1 0� 

� + t1,1t2,3t3,2 � 

�0 

0 1� 

� 

t3,3� 

�0 

0 1� 

�0 

1 0� 

� � 

� 

� 

�0 

1 0� 

� 

� 

�0 

1 

+ t1,2t2,1t3,3 � 

�1 

0 0� 

� + t1,2t2,3t3,1 � 

�0 

0 

�0 

0 1� 

�1 

0 

� � 

� 

� 

�0 

0 1� 

� 

� 

�0 

0 

+ t1,3t2,1t3,2 

� 

�1 

0 0� 

� + t1,3t2,2t3,1 

� 

�0 

1 

�0 

1 0� 

�1 

0 

� 

0� 

� 

1� 

� 

0� 

� 

1� 

� 

0� 

� 

0� 

we can, for instance, factor out the entries from the first row 

⎡ � � � � ⎤ 

� 

�1 

0 0� 

� 

� �1 

0 0� 

� 

= t1,1 · ⎣t2,2t3,3 

� 

�0 

1 0� 

� + t2,3t3,2 

� 

�0 

0 1�⎦ 

� 

�0 

0 1� 

�0 

1 0� 

⎡ � � � � ⎤ 

� 

�0 

1 0� 

� 

� �0 

1 0� 

� 

+ t1,2 · ⎣t2,1t3,3 

� 

�1 

0 0� 

� + t2,3t3,1 

� 

�0 

0 1�⎦ 

� 

�0 

0 1� 

�1 

0 0� 

⎡ � � � � ⎤ 

� 

�0 

0 1� 

� 

� �0 

0 1� 

� 

+ t1,3 · ⎣t2,1t3,2 � 

�1 

0 0� 

� + t2,2t3,1 � 

�0 

1 0�⎦ 

� 

�0 

1 0� 

�1 

0 0� 

and swap rows in the permutation matrices to get this. 

⎡ � � � � ⎤ 

� 

�1 

0 0� 

� 

� �1 

0 0� 

� 

= t1,1 · ⎣t2,2t3,3 

� 

�0 

1 0� 

� + t2,3t3,2 

� 

�0 

0 1�⎦ 

� 

�0 

0 1� 

�0 

1 0� 

⎡ � � � � ⎤ 

� 

�1 

0 0� 

� 

� �1 

0 0� 

� 

− t1,2 · ⎣t2,1t3,3 

� 

�0 

1 0� 

� + t2,3t3,1 

� 

�0 

0 1�⎦ 

� 

�0 

0 1� 

�0 

1 0� 

⎡ � � � � ⎤ 

� 

�1 

0 0� 

� 

� �1 

0 0� 

� 

+ t1,3 · ⎣t2,1t3,2 

� 

�0 

1 0� 

� + t2,2t3,1 

� 

�0 

0 1�⎦ 

� 

�0 

0 1� 

�0 

1 0�

Section III. Other Formulas 327 

The point of the swapping (one swap to each of the permutation matrices on 

the second line and two swaps to each on the third line) is that the three lines 

simplify to three terms. 

� 

� 

= t1,1 · � 

� t2,2 

� 

t2,3� 

� 

� − t1,2 

� 

� 

· � 

� t2,1 

� 

t2,3� 

� 

� + t1,3 

� 

� 

· � 

� t2,1 

� 

t2,2� 

� 

� 

t3,2 t3,3 

t3,1 t3,3 

t3,1 t3,2 

The formula given in Theorem 1.5, which generalizes this example, is a recurrence 

— the determinant is expressed as a combination of determinants. This 

formula isn’t circular because, as here, the determinant is expressed in terms of 

determinants of matrices of smaller size. 

1.2 Definition For any n×n matrix T , the (n − 1)×(n − 1) matrix formed by 

deleting row i and column j of T is the i, j minor of T .Thei, j cofactor Ti,j of 

T is (−1) i+j times the determinant of the i, j minor of T . 

1.3 Example The 1, 2 cofactor of the matrix from Example 1.1 is the negative 

of the second 2×2 determinant. 

� 

� 

T1,2 = −1 · � 

� t2,1 

� 

t2,3� 

� 

� 

1.4 Example Where 

t3,1 t3,3 

⎛ 

1 2 

⎞ 

3 

T = ⎝4 

5 6⎠ 

7 8 9 

these are the 1, 2 and 2, 2 cofactors. 

T1,2 =(−1) 1+2 � 

� 

· �4 

�7 � 

6� 

� 

9� 

=6 T2,2 =(−1) 2+2 � 

� 

· �1 

�7 � 

3� 

� 

9� 

= −12 

1.5 Theorem (Laplace Expansion of Determinants) Where T is an n×n 

matrix, the determinant can be found by expanding by cofactors on row i or 

column j. 

|T | = ti,1 · Ti,1 + ti,2 · Ti,2 + ···+ ti,n · Ti,n 

= t1,j · T1,j + t2,j · T2,j + ···+ tn,j · Tn,j 

Proof. Exercise 27. QED 

1.6 Example We can compute the determinant 

� 

� 

�1 

|T | = � 

�4 

�7 

2 

5 

8 

� 

3� 

� 

6� 

� 

9�


by expanding along the first row, as in Example 1.1. 

� 

� 

|T | =1· (+1) �5 

�8 � � 

6� 

� 

� 

9� 

+2· (−1) �4 

�7 � � 

6� 

� 

� 

9� 

+3· (+1) �4 

�7 � 

5� 

� 

8� 

= −3+12−9=0 Alternatively, we can expand down the second column. 

� 

� 

|T | =2· (−1) �4 

�7 � � 

6� 

� 

� 

9� 

+5· (+1) �1 

�7 � � 

3� 

� 

� 

9� 

+8· (−1) �1 

�4 � 

3� 

� 

6� 

=12−60 + 48 = 0 

1.7 Example A row or column with many zeroes suggests a Laplace expansion. 

�� 

1 

2 

3 

5 

1 

−1 

� 

0� 

� 

� � 

1� 

� =0· (+1) �2 

� 

0� 

3 

� � 

1 � � 

� 

−1� 

+1· (−1) �1 

�3 � � 

5 � � 

� 

−1� 

+0· (+1) �1 

�2 � 

5� 

� 

1� 

=16 

We finish by applying this result to derive a new formula for the inverse 

of a matrix. With Theorem 1.5, the determinant of an n × n matrix T can 

be calculated by taking linear combinations of entries from a row and their 

associated cofactors. 

ti,1 · Ti,1 + ti,2 · Ti,2 + ···+ ti,n · Ti,n = |T | (∗) 

Recall that a matrix with two identical rows has a zero determinant. Thus, for 

any matrix T , weighing the cofactors by entries from the “wrong” row — row k 

with k �= i — gives zero 

ti,1 · Tk,1 + ti,2 · Tk,2 + ···+ ti,n · Tk,n =0 (∗∗) 

because it represents the expansion along the row k of a matrix with row i equal 

to row k. This equation summarizes (∗) and (∗∗). 

⎛ 

⎜ 

⎝ 

t1,1 

t2,1 

t1,2 

t2,2 

. 

... 

... 

t1,n 

t2,n 

⎞ ⎛ 

T1,1 

⎟ ⎜T1,2 

⎟ ⎜ 

⎟ ⎜ 

⎠ ⎝ 

T2,1 

T2,2 

. 

... 

... 

⎞ 

Tn,1 

Tn,2 ⎟ 

⎠ = 

⎛ 

|T | 

⎜ 0 

⎜ 

⎝ 

0 

|T | 

. 

... 

... 

⎞ 

0 

0 ⎟ 

⎠ 

0 0 ... |T | 

tn,1 tn,2 ... tn,n 

T1,n T2,n ... Tn,n 

Note that the order of the subscripts in the matrix of cofactors is opposite to 

the order of subscripts in the other matrix; e.g., along the first row of the matrix 

of cofactors the subscripts are 1, 1 then 2, 1, etc. 

1.8 Definition The matrix adjoint to the square matrix T is 

⎛ 

T1,1 

⎜T1,2 

⎜ 

adj(T )= ⎜ 

⎝ 

T2,1 

T2,2 

. 

... 

... 

⎞ 

Tn,1 

Tn,2 ⎟ 

⎠ 

where Tj,i is the j, i cofactor. 

T1,n T2,n ... Tn,n

Section III. Other Formulas 329 

1.9 Theorem Where T is a square matrix, T · adj(T )=adj(T ) · T = |T |·I. 

Proof. Equations (∗) and (∗∗). QED 


⎛ 

1 

T = ⎝2 0 

1 

⎞ 

4 

−1⎠ 

1 0 1 

then the adjoint adj(T )is 

⎛ 

T1,1 

⎝T1,2 

T1,3 

T2,1 

T2,2 

T2,3 

⎛ � 

� 

�1 

⎞ ⎜ � 

⎜ 0 

T3,1 ⎜ 

T3,2⎠= 

⎜ 

T3,3 ⎜ 

⎝ 

� 

−1� 

� 

1 � − 

� 

� 

− �2 

�1 � 

−1� 

� 

1 � 

� 

� 

�0 

�0 � 

� 

�1 

�1 � 

4� 

� 

1� 

� 

4� 

� 

1� 

� 

� 

�0 

�1 � 

4 � 

� 

−1� 

− 

� 

� 

�2 

�1 � 

1� 

� 

0� 

� 

� 

− �1 

�1 � 

0� 

� 

0� 

� 

� 

�1 

�2 � 

� 

�1 

�2 ⎞ 

⎟ 

�⎟ 

⎛ 

4 �⎟ 

1 

�⎟ 

−1�⎟= 

⎝−3 

⎟ 

� ⎟ −1 

0�⎠ 

� 

1� 

0 

−3 

0 

⎞ 

−4 

9 ⎠ 

1 

and taking the product with T gives the diagonal matrix |T |·I. 

⎛ 

1 

⎝2 

0 

1 

⎞ ⎛ 

4 1 

−1⎠ 

⎝−3 

0 

−3 

⎞ ⎛ 

−4 −3 

9 ⎠ = ⎝ 0 

0 

−3 

⎞ 

0 

0 ⎠ 

1 0 1 −1 0 1 0 0 −3 

1.11 Corollary If |T |�= 0 then T −1 =(1/|T |) · adj(T ). 

1.12 Example The inverse of the matrix from Example 1.10 is (1/−3)·adj(T ). 

T −1 ⎛ 

1/−3 

= ⎝−3/−3 

0/−3 

−3/−3 

⎞ ⎛ 

−4/−3 −1/3 

9/−3⎠ 

= ⎝ 1 

0 

1 

⎞ 

4/3 

−3 ⎠ 

−1/−3 0/−3 1/−3 1/3 0 −1/3 

The formulas from this section are often used for by-hand calculation and 

are sometimes useful with special types of matrices. However, they are not the 

best choice for computation with arbitrary matrices because they require more 

arithmetic than, for instance, the Gauss-Jordan method. 

Exercises 

� 1.13 Find the cofactor. 

T = 

� 

1 0 

� 

2 

−1 1 3 

0 2 −1 

(a) T2,3 (b) T3,2 (c) T1,3 

� 1.14 Find the determinant by expanding 

� 

� 3 0 

� 

� 1 2 

� 

−1 3 

� 

1� 

� 

2� 

0 

�


(a) on the first row (b) on the second row (c) on the third column. 

1.15 Find the adjoint of the matrix in Example 1.6. 

� 1.16 Find the matrix adjoint to each. 

� � 

2 1 4 � � 

3 −1 

(a) −1 0 2 (b) 

2 4 

1 0 1 

(c) 

� 

1 

5 

� 

1 

0 

(d) 

� 

1 

−1 

1 

4 

0 

8 

� 

3 

3 

9 

� 1.17 Find the inverse of each matrix in the prior question with Theorem 1.9. 

1.18 Find the matrix adjoint to this one. 

⎛ 

2 1 0 

⎞ 

0 

⎜1 

⎝ 

0 

2 

1 

1 

2 

0⎟ 

1 

⎠ 

0 0 1 2 

� 1.19 Expand across the first row to derive the formula for the determinant of a 2×2 

matrix. 

� 1.20 Expand across the first row to derive the formula for the determinant of a 3×3 

matrix. 

� 1.21 (a) Give a formula for the adjoint of a 2×2 matrix. 

(b) Use it to derive the formula for the inverse. 

� 1.22 Can we compute a determinant by expanding down the diagonal? 

1.23 Give a formula for the adjoint of a diagonal matrix. 

� 1.24 Prove that the transpose of the adjoint is the adjoint of the transpose. 

1.25 Prove or disprove: adj(adj(T )) = T . 

1.26 A square matrix is upper triangular if each i, j entry is zero in the part above 

the diagonal, that is, when i>j. 

(a) Must the adjoint of an upper triangular matrix be upper triangular? Lower 

triangular? 

(b) Prove that the inverse of a upper triangular matrix is upper triangular, if an 

inverse exists. 

1.27 This question requires material from the optional Determinants Exist subsection. 

Prove Theorem 1.5 by using the permutation expansion. 

1.28 Prove that the determinant of a matrix equals the determinant of its transpose 

using Laplace’s expansion and induction on the size of the matrix. 

1.29 [Am. Math. Mon., Jun. 1949] Show that 

� 

� 

�1 

−1 1 −1 1 −1 ... � 

� 

� 

�1 

1 0 1 0 1 ... � 

� 

� 

Fn = �0 

1 1 0 1 0 ... � 

� 

�0 

0 1 1 0 1 ... 

� 

� 

�. 

. . . . . ... � 

where Fn is the n-thtermof1, 1, 2, 3, 5,...,x,y,x+y,..., the Fibonacci sequence, 

and the determinant is of order n − 1.

Topic: Cramer’s Rule 331 

Topic: Cramer’s Rule 

We introduced determinant functions algebraically, looking for a formula to 

decide whether a matrix is nonsingular, that is, whether a linear system has a 

unique solution. Then we saw a geometric interpretation, that the determinant 

function gives the size of the box with sides formed by the columns of the matrix. 

Here we will see a nice formula that connects the two views. 

Consider this system 

Rewriting in vector form 

x1 · 

and picturing with parallelograms 

� � 1 

3 

x1 +2x2 =6 

3x1 + x2 =8 

� � 

1 

+ x2 · 

3 

� � 2 

1 

� � 

2 

= 

1 

x 1 · 

� � 

6 

8 

� � � � 

1 

2 

+ x2 · 

3 

1 

gives a geometric interpretation of solving the linear system: by what factor x1 

must we dilate the first vector, and by what factor x2 must we dilate the second 

vector, to expand the small parallegram to fill the larger one? 

Of course, we routinely find the answer with the algebraic manipulations of 

Gauss’ method. Nonetheless, the geometry can give us some insights — compare 

the sizes of these three shaded boxes. 

� � 1 

3 

� � 2 

1 

x 1 · 

� � 1 

3 

� � 2 

1 

� � � � 1 2 

The second box is formed from x1 3 and 1 , and one of the properties of the 

size function (that is, the determinant function) is that the size of the second 

box is therefore x1 � � �times�the � size of the first box. Since the third box is formed 

1 

2 

from x1 + x2 and , and sizes are unchanged by side operations (that 

3 

� 2 

1 

1 

� � 2 

1 

� � 6 

8


is, the determinant is unchanged by adding x2 times the second column to the 

first column), the size of the third box equals the size of the second box. 

� � 

� 

�6 

2� 

� 

�8 1� 

= x1 

� � 

� 

· �1 

2� 

� 

�3 1� 

Solving gives the value of one of the variables. 

� � 

� 

�6 

2� 

� 

�8 1� 

x1 = � � 

� 

�1 

2� 

= 

� 

�3 1� 

−10 

−5 =2 

The theorem that generalizes this example, Cramer’s Rule, is:if|A| �=0 

then the system A�x = �b has the unique solution xi = |Bi|/|A| where the matrix 

Bi is formed from A by replacing column i with the vector �b. Exercise 3 asks 

for a proof. 

For instance, to solve this system for x2 

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 

1 0 4 x1 2 

⎝2 

1 −1⎠ 

⎝x2⎠ 

= ⎝ 1 ⎠ 

1 0 1 

−1 

we do this computation. 

x3 

� 

� 

� 

�1 

2 4 � 

� 

� 

�2 

1 −1� 

� 

�1 

−1 1 � 

x2 = � � 

� 

�1 

0 4 � 

� 

� 

�2 

1 −1� 

� 

�1 

0 1 � 

= −18 

−3 

Cramer’s Rule, with practice, allows us to solve two equations/two unknown 

systems by eye. It is also sometimes used for three equations/three unknowns 

systems. But computing large determinants takes a long time so that solving 

large systems by Cramer’s Rule is impractical. 

Exercises 

1 Use Cramer’s Rule to solve each for each of the variables. 

x − y = 4 −2x + y = −2 

(a) 

(b) 

−x +2y = −7 

x − 2y = −2 

2 Use Cramer’s Rule to solve this system for z. 

2x + y + z =1 

3x + z =4 

x − y − z =2 

3 Prove Cramer’s Rule.

Topic: Cramer’s Rule 333 

4 Suppose that a linear system with as many equations as unknowns, and with 

integer coefficients and constants, has a matrix of coefficients with determinant 1. 

Prove that the entries in the solution are all integers. (Remark. This is often used 

to invent linear systems for exercises. If an instructor makes the linear system with 

this property then the solution is not some disagreeable fraction.) 

5 Use Cramer’s Rule to give a formula for the solution of a two equation/two 

unknown linear system. 

6 Can Cramer’s Rule tell the difference between a system with no solutions and 

one with infinitely many?


Topic: Speed of Calculating Determinants 

The permutation expansion formula for computing determinants is useful for 

proving theorems, but the method of using row operations is a much better for 

finding the determinants of a large matrix. We can make this statement precise 

by considering, as computer algorithm designers do, the number of arithmetic 

operations that each method uses. 

The speed of an algorithm is measured by finding how the time taken by 

the computer grows as the size of its input data set grows. For instance, how 

much longer will the algorithm take if we increase the size of the input data by 

a factor of ten, say from a 1000 row matrix to a 10, 000 row matrix, or from 

10, 000 to 100, 000? Does the time taken grow by a factor of ten or by a factor 

of a hundred, or by a factor of a thousand? That is, is the time taken by the 

algorithm proportional to the size of the data set, or to the square of that size, 

or to the cube of that size, etc.? 

Recall the permutation expansion formula for determinants. 

� 

� 

�t1,1 

t1,2 � 

... t1,n� 

� 

�t2,1 

t2,2 � 

... t2,n� 

� 

� . 

� . 

� = 

� . 

� 

� 

� 

� 

� 

t1,φ(1)t2,φ(2) ···tn,φ(n) |Pφ| 


tn,1 tn,2 ... tn,n 

= t1,φ1(1) · t2,φ1(2) ···tn,φ1(n) |Pφ1 | 

+ t1,φ2(1) · t2,φ2(2) ···tn,φ2(n) |Pφ2 | 

. 

+ t1,φk(1) · t2,φk(2) ···tn,φk(n) |Pφk | 

There are n! =n · (n − 1) · (n − 2) ···2 · 1 different n-permutations. For numbers 

n of any size at all, this is a quite large number; for instance, even if n is only 

10 then the expansion has 10! = 3, 628, 800 terms, all of which are obtained by 

multiplying n entries together. This is a very large number of multiplications 

(for instance, [Knuth] suggests 10! steps as a rough boundary for the limit 

of practical calculation). The factorial function grows faster than the square 

function. It grows faster than the cube function, the fourth power function, 

or any polynomial function. (One way to see that the factorial function grows 

faster than the square is to note that multiplying the first two factors in n! 

gives n · (n − 1), which for large n is approximately n 2 , and then multiplying 

in more factors will make it even larger. The same argument works for the 

cube function, etc.) So a computer that is programmed to use the permutation 

expansion formula, and thus to perform a number of operations that is greater 

than or equal to the factorial of the number of rows, would take times that grow 

very quickly as the input data set grows. 

In contrast, the time taken by the row reduction method does not grow so 

fast. This fragment of row-reduction code is in the computer language FOR- 

TRAN. The matrix is stored in the N ×N array A. ForeachROW between 1 

and N parts of the program not shown here have already found the pivot entry

Topic: Speed of Calculating Determinants 335 

A(ROW, COL). Now the program pivots. 

−PIVINV · ρROW + ρi 

(This code fragment is for illustration only, and is incomplete. Nonetheless, 

analysis of a finished versions, including all of the tests and subcases, is messier 

but gives essentially the same result.) 

PIVINV=1.0/A(ROW,COL) 

DO 10 I=ROW+1, N 

DO 20 J=I, N 

A(I,J)=A(I,J)-PIVINV*A(ROW,J) 

20 CONTINUE 

10 CONTINUE 

The outermost loop (not shown) runs through N − 1 rows. For each row, the 

nested I and J loops shown perform arithmetic on the entries in A that are 

below and to the right of the pivot entry. Assume that the pivot is found in 

the expected place, that is, that COL = ROW . Then there are (N − ROW ) 2 

entries below and to the right of the pivot. On average, ROW will be N/2. 

Thus, we estimate that the arithmetic will be performed about (N/2) 2 times, 

that is, will run in a time proportional to the square of the number of equations. 

Taking into account the outer loop that is not shown, we get the estimate that 

the running time of the algorithm is proportional to the cube of the number of 

equations. 

Finding the fastest algorithm to compute the determinant is a topic of current 

research. Algorithms are known that run in time between the second and 

third power. 

Speed estimates like these help us to understand how quickly or slowly an 

algorithm will run. Algorithms that run in time proportional to the size of the 

data set are fast, algorithms that run in time proportional to the square of the 

size of the data set are less fast, but typically quite usable, and algorithms that 

run in time proportional to the cube of the size of the data set are still reasonable 

in speed. However, algorithms that run in time (greater than or equal to) the 

factorial of the size of the data set are not practical. 

There are other methods besides the two discussed here that are also used 

for computation of determinants. Those lie outside of our scope. Nonetheless, 

this contrast of the two methods for computing determinants makes the point 

that although in principle they give the same answer, in practice the idea is to 

select the one that is fast. 

Exercises 

Most of these problems presume access to a computer. 

1 Computer systems generate random numbers (of course, these are only pseudorandom, 

in that they are generated by an algorithm, but they pass a number of 

reasonable statistical tests for randomness). 

(a) Fill a 5×5 array with random numbers (say, in the range [0..1)). See if it is 

singular. Repeat that experiment a few times. Are singular matrices frequent 

or rare (in this sense)?


(b) Time your computer algebra system at finding the determinant of ten 5×5 

arrays of random numbers. Find the average time per array. Repeat the prior 

item for 15×15 arrays, 25×25 arrays, and 35×35 arrays. (Notice that, when an 

array is singular, it can sometimes be found to be so quite quickly, for instance 

if the first row equals the second. In the light of your answer to the first part, 

do you expect that singular systems play a large role in your average?) 

(c) Graph the input size versus the average time. 

2 Compute the determinant of each of these by hand using the two methods discussed 

above. 

� 

� 

� � � � �2 

1 0 0� 

� 

(a) �2 

1 � � 3 1 1 � � 

� 

� � � �1 

3 2 0� 

�5 −3� 

(b) �−1 

0 5 � (c) � 

� 

� 

−1 2 −2 

� �0 

−1 −2 1� 

�0 

0 −2 1� 

Count the number of multiplications and divisions used in each case, for each of 

the methods. (On a computer, multiplications and divisions take much longer than 

additions and subtractions, so algorithm designers worry about them more.) 

3 What 10×10 array can you invent that takes your computer system the longest 

to reduce? The shortest? 

4 Write the rest of the FORTRAN program to do a straightforward implementation 

of calculating determinants via Gauss’ method. (Don’t test for a zero pivot.) 

Compare the speed of your code to that used in your computer algebra system. 

5 The FORTRAN language specification requires that arrays be stored “by column”, 

that is, the entire first column is stored contiguously, then the second column, 

etc. Does the code fragment given take advantage of this, or can it be 

rewritten to make it faster, by taking advantage of the fact that computer fetches 

are faster from contiguous locations?

Topic: Projective Geometry 337 

Topic: Projective Geometry 

There are geometries other than the familiar Euclidean one. One such geometry 

arose in art, where it was observed that what a viewer sees is not necessarily 

what is there. This is Leonardo da Vinci’s masterpiece The Last Supper. 

What is there in the room, for instance where the ceiling meets the left and 

right walls, are lines that are parallel. However, what a viewer sees is lines 

that, if extended, would intersect. The intersection point is called the vanishing 

point. This aspect of perspective is also familiar as the image of a long stretch 

of railroad tracks that appear to converge at the horizon. 

To depict the room, da Vinci has adopted a model of how we see, of how we 

project the three dimensional scene to a two dimensional image. This model is 

only a first approximation — it does not take into account that our retina is 

curved and our lens bends the light, that we have binocular vision, or that our 

brain’s processing greatly affects what we see — but nonetheless it is interesting, 

both artistically and mathematically. 

The projection is not orthogonal, it is a central projection from a single 

point, to the plane of the canvas. 

A 

B 

C 

(It is not an orthogonal projection since the line from the viewer to C is not 

orthogonal to the image plane.) As the picture suggests, the operation of central 

projection preserves some geometric properties — lines project to lines. 

However, it fails to preserve some others — equal length segments can project 

to segments of unequal length; the length of AB is greater than the length of 

BC because the segment projected to AB is closer to the viewer and closer 

things look bigger. The study of the effects of central projections is projective 

geometry. We will see how linear algebra can be used in this study.


There are three cases of central projection. The first is the projection done 

by a movie projector. 

projector P 

source S 

image I 

We can think that each source point is “pushed” from the domain plane outward 

to the image point in the codomain plane. This case of projection has a 

somewhat different character than the second case, that of the artist “pulling” 

the source back to the canvas. 

painter P 

image I 

source S 

In the first case S is in the middle while in the second case I is in the middle. 

One more configuration is possible, with P in the middle. An example of this 

is when we use a pinhole to shine the image of a solar eclipse onto a piece of 

paper. 

image I 

pinhole P 

source S 

We shall take each of the three to be a central projection by P of S to I. 

To illustrate some of the geometric effects of these projections, consider again 

the effect of railroad tracks that appear to converge to a point. We model this 

with parallel lines in a domain plane S and a projection via a P to a codomain 

plane I. (The dotted lines are parallel to S and I.) 

P 

All three projection cases appear here. The first picture below shows P acting 

like a movie projector by pushing points from part of S out to image points on 

the lower half of I. The middle picture shows P acting like the artist by pulling 

points from another part of S back to image points in I. In the third picture, P 

acts like the pinhole. This picture is the trickiest—the points that are projected 

near to the vanishing point are the ones that are far out to the bottom left of 

S. Points in S that are near to the vertical dotted line are sent high up on I. 

I 

S


P P P 

There are two awkward things about this situation. The first is that neither of 

the two points in the domain nearest to the vertical dotted line (see below) has 

an image because a projection from those two is along the dotted line that is 

parallel to the codomain plane (we sometimes say that these two are projected 

“to infinity”). The second awkward thing is that the vanishing point in I isn’t 

the image of any point from S because a projection to this point would be along 

the dotted line that is parallel to the domain plane (we sometimes say that the 

vanishing point is the image of a projection “from infinity”). 

For a better model, put the projector P at the origin. Imagine that P is 

covered by a glass hemispheric dome. As P looks outward, anything in the line 

of vision is projected to the same spot on the dome. This includes things on 

the line between P and the dome, as in the case of projection by the movie 

projector. It includes things on the line further from P than the dome, as in 

the case of projection by the painter. It also includes things on the line that lie 

behind P , as in the case of projection by a pinhole. 

ℓ = {k · 

� � 

1 �� 

2 k ∈ R} 

3 

From this perspective P , all of the spots on the line are seen as the same point. 

Accordingly, for any nonzero vector �v ∈ R3 , we define the associated point v 

in the projective plane to be the set {k�v � � k ∈ R and k �= 0} of nonzero vectors 

lying on the same line through the origin as �v. To describe a projective point 

we can give any representative member of the line, so that the projective point 

shown above can be represented in any of these three ways. 

⎛ 

⎝ 1 

⎞ ⎛ 

2⎠ 

⎝ 

3 

1/3 

⎞ ⎛ 

2/3⎠ 

⎝ 

1 

−2 

⎞ 

−4⎠ 

−6


Each of these is a homogeneous coordinate vector for v. 

This picture, and the above definition that arises from it, clarifies the description 

of central projection but there is something awkward about the dome 

model: what if the viewer looks down? If we draw P ’s line of sight so that 

the part coming toward us, out of the page, goes down below the dome then 

we can trace the line of sight backward, up past P and toward the part of the 

hemisphere that is behind the page. So in the dome model, looking down gives 

a projective point that is behind the viewer. Therefore, if the viewer in the 

picture above drops the line of sight toward the bottom of the dome then the 

projective point drops also and as the line of sight continues down past the 

equator, the projective point suddenly shifts from the front of the dome to the 

back of the dome. This discontinuity in the drawing means that we often have 

to treat equatorial points as a separate case. That is, while the railroad track 

discussion of central projection has three cases, the dome model has two. 

We can do better than this. Consider a sphere centered at the origin. Any 

line through the origin intersects the sphere in two spots, which are said to be 

antipodal. Because we associate each line through the origin with a point in the 

projective plane, we can draw such a point as a pair of antipodal spots on the 

sphere. Below, the two antipodal spots are shown connected by a dotted line 

to emphasize that they are not two different points, the pair of spots together 

make one projective point. 

While drawing a point as a pair of antipodal spots is not as natural as the onespot-per-point 

dome mode, on the other hand the awkwardness of the dome 

model is gone, in that if as a line of view slides from north to south, no sudden 

changes happen on the picture. This model of central projection is uniform — 

the three cases are reduced to one. 

So far we have described points in projective geometry. What about lines? 

What a viewer at the origin sees as a line is shown below as a great circle, the 

intersection of the model sphere with a plane through the origin. 

(One of the projective points on this line is shown to bring out a subtlety. 

Because two antipodal spots together make up a single projective point, the 

great circle’s behind-the-paper part is the same set of projective points as its 

in-front-of-the-paper part.) Just as we did with each projective point, we will 

also describe a projective line with a triple of reals. For instance, the members


of this plane through the origin in R3 ⎛ 

{ ⎝ x 

⎞ 

y⎠ 

z 

� � x + y − z =0} 

project to a line that we can described with the triple � 1 1 −1 � (we use row 

vectors to typographically distinguish lines from points). In general, for any 

nonzero three-wide row vector � L we define the associated line in the projective 

plane, tobethesetL = {k� L � � k ∈ R and k �= 0} of nonzero multiples of L. � 

The reason that this description of a line as a triple is convienent is that 

in the projective plane, a point v and a line L are incident — the point lies 

on the line, the line passes throught the point — if and only if a dot product 

of their representatives v1L1 + v2L2 + v3L3 is zero (Exercise 4 shows that this 

is independent of the choice of representatives �v and � L). For instance, the 

projective point described above by the column vector with components 1, 2, 

and 3 lies in the projective line described by � 1 1 −1 � , simply because any 

vector in R3 whose components are in ratio 1 : 2 : 3 lies in the plane through the 

origin whose equation is of the form 1k · x +1k · y − 1k · z = 0 for any nonzero k. 

That is, the incidence formula is inherited from the three-space lines and planes 

of which v and L are projections. 

Thus, we can do analytic projective geometry. For instance, the projective 

line L = � 1 1 −1 � has the equation 1v1 +1v2 − 1v3 = 0, because points 

incident on the line are characterized by having the property that their representatives 

satisfy this equation. One difference from familiar Euclidean anlaytic 

geometry is that in projective geometry we talk about the equation of a point. 

For a fixed point like 

⎛ ⎞ 

1 

v = ⎝2⎠ 

3 

the property that characterizes lines through this point (that is, lines incident 

on this point) is that the components of any representatives satisfy 1L1 +2L2 + 

3L3 = 0 and so this is the equation of v. 

This symmetry of the statements about lines and points brings up the Duality 

Principle of projective geometry: in any true statement, interchanging ‘point’ 

with ‘line’ results in another true statement. For example, just as two distinct 

points determine one and only one line, in the projective plane, two distinct 

lines determine one and only one point. Here is a picture showing two lines that 

cross in antipodal spots and thus cross at one projective point. 

Contrast this with Euclidean geometry, where two distinct lines may have a 

unique intersection or may be parallel. In this way, projective geometry is 

simpler, more uniform, than Euclidean geometry. 

(∗)


That simplicity is relevant because there is a relationship between the two 

spaces: the projective plane can be viewed as an extension of the Euclidean 

plane. Take the sphere model of the projective plane to be the unit sphere in 

R 3 and take Euclidean space to be the plane z = 1. This gives us a way of 

viewing some points in projective space as corresponding to points in Euclidean 

space, because all of the points on the plane are projections of antipodal spots 

from the sphere. 

(∗∗) 

Note though that projective points on the equator don’t project up to the plane. 

Instead, these project ‘out to infinity’. We can thus think of projective space 

as consisting of the Euclidean plane with some extra points adjoined — the 

Euclidean plane is embedded in the projective plane. These extra points, the 

equatorial points, are the ideal points or points at infinity and the equator is the 

ideal line or line at infinity (note that it is not a Euclidean line, it is a projective 

line). 

The advantage of the extension to the projective plane is that some of the 

awkwardness of Euclidean geometry disappears. For instance, the projective 

lines shown above in (∗) cross at antipodal spots, a single projective point, on 

the sphere’s equator. If we put those lines into (∗∗) then they correspond to 

Euclidean lines that are parallel. That is, in moving from the Euclidean plane to 

the projective plane, we move from having two cases, that lines either intersect 

or are parallel, to having only one case, that lines intersect (possibly at a point 

at infinity). 

The projective case is nicer in many ways than the Euclidean case but has 

the problem that we don’t have the same experience or intuitions with it. That’s 

one advantage of doing analytic geometry, where the equations can lead us to 

the right conclusions. Analytic projective geometry uses linear algebra. For 

instance, for three points of the projective plane t, u, andv, setting up the 

equations for those points by fixing vectors representing each, shows that the 

three are collinear — incident in a single line — if and only if the resulting threeequation 

system has infinitely many row vector solutions representing that line. 

That, in turn, holds if and only if this determinant is zero. 

� � 

�t1 

u1 v1� 

� � 

�t2 

u2 v2� 

� � 

� � 

t3 u3 v3 

Thus, three points in the projective plane are collinear if and only if any three 

representative column vectors are linearly dependent. Similarly (and illustrating 

the Duality Principle), three lines in the projective plane are incident on a 

single point if and only if any three row vectors representing them are linearly 

dependent.


The following result is more evidence of the ‘niceness’ of the geometry of the 

projective plane, compared to the Euclidean case. These two triangles are said 

to be in perspective from P because their corresponding vertices are collinear. 

O 

T1 

U1 V1 

T2 

U2 

V2 

Desargue’s Theorem is that when the three pairs of corresponding sides — T1U1 

and T2U2, T1V1 and T2V2, U1V1 and U2V2 — are extended, they intersect 

and further, those three intersection points are collinear. 

We will prove this theorem, using projective geometry. (These are drawn as 

Euclidean figures because it is the more familiar image. To consider them as 

projective figures, we can imagine that, although the line segments shown are 

parts of great circles and so are curved, the model has such a large radius 

compared to the size of the figures that the sides appear in this sketch to be 

straight.) 

For this proof, we need a preliminary lemma [Coxeter]: if W , X, Y , Z are 

four points in the projective plane (no three of which are collinear) then there 

is a basis B for R3 such that 

⎛ ⎞ 

⎛ ⎞ 

⎛ ⎞ 

⎛ ⎞ 

1 

0 

0 

1 

RepB( �w) = ⎝0⎠ 

RepB(�x) = ⎝1⎠ 

RepB(�y) = ⎝0⎠ 

RepB(�z) = ⎝1⎠ 

0 

0 

1 

1 

where �w, �x, �y, �z are homogeneous coordinate vectors for the projective points. 

The proof is straightforward. Because W, X, Y are not on the same projective 

line, any homogeneous coordinate vectors �w0, �x0, �y0 do not line on the same 

plane through the origin in R 3 and so form a spanning set for R 3 . Thus any 

homogeneous coordinate vector for Z can be written as a combination �z0 = 

a · �w0 + b · �x0 + c · �y0. Then �w = a · �w0, �x = b · �x0, �y = c · �y0, and�z = �z0 will do, 

for B = 〈 �w, �x, �y〉. 

Now, to prove of Desargue’s Theorem, use the lemma to fix homogeneous 

coordinate vectors and a basis. 

⎛ 

RepB(�t1) = ⎝ 1 

⎞ 

⎛ 

0⎠ 

RepB(�u1) = ⎝ 

0 

0 

⎞ 

⎛ 

1⎠ 

RepB(�v1) = ⎝ 

0 

0 

⎞ 

⎛ 

0⎠ 

RepB(�o) = ⎝ 

1 

1 

⎞ 

1⎠ 

1


Because the projective point T2 is incident on the projective line OT1, any 

homogeneous coordinate vector for T2 lies in the plane through the origin in R3 that is spanned by homogeneous coordinate vectors of O and T1: 

⎛ 

RepB(�t2) =a ⎝ 1 

⎞ ⎛ 

1⎠ 

+ b ⎝ 

1 

1 

⎞ 

0⎠ 

0 

for some scalars a and b. That is, the homogenous coordinate vectors of members 

T2 of the line OT1 are of the form on the left below, and the forms for U2 and 

V2 are similar. 

⎛ 

RepB(�t2) = ⎝ t2 

⎞ 

⎛ 

1 ⎠ RepB(�u2) = ⎝ 

1 

1 

⎞ 

⎛ 

u2⎠ 

RepB(�v2) = ⎝ 

1 

1 

⎞ 

1 ⎠ 

The projective line T1U1 is the image of a plane through the origin in R 3 . A 

quick way to get its equation is to note that any vector in it is linearly dependent 

on the vectors for T1 and U1 and so this determinant is zero. 

� � 

� 

�1 

0 x� 

� 

� 

�0 

1 y� 

� =0 =⇒ z =0 

�0 

0 z� 

The equation of the plane in R3 whose image is the projective line T2U2 is this. 

� 

�t2 

� 

� 

� 1 

� 1 

1 

u2 

1 

� 

x� 

� 

y� 

� 

z� 

=0 =⇒ (1 − u2) · x +(1− t2) · y +(t2u2 − 1) · z =0 

Finding the intersection of the two is routine. 

⎛ ⎞ 

t2 − 1 

T1U1 ∩ T2U2 = ⎝1 

− u2⎠ 

0 

(This is, of course, the homogeneous coordinate vector of a projective point.) 

The other two intersections are similar. 

⎛ ⎞ 

⎛ 

1 − t2 

T1V1 ∩ T2V2 = ⎝ 0 ⎠ U1V1 ∩ U2V2 = ⎝ 

v2 − 1 

0 

⎞ 

u2 − 1⎠ 

1 − v2 

The proof is finished by noting that these projective points are on one projective 

line because the sum of the three homogeneous coordinate vectors is zero. 

Every projective theorem has a translation to a Euclidean version, although 

the Euclidean result is often messier to state and prove. Desargue’s theorem 

illustrates this. In the translation to Euclidean space, the case where O lies on 

v2


the ideal line must be treated separately for then the lines T1T2, U1U2, andV1V2 

are parallel. 

The parenthetical remark following the statement of Desargue’s Theorem 

suggests thinking of the Euclidean pictures as figures from projective geometry 

for a model of very large radius. That is, just as a small area of the earth appears 

flat to people living there, the projective plane is also ‘locally Euclidean’. 

Although its local properties are the familiar Euclidean ones, there is a global 

property of the projective plane that is quite different. The picture below shows 

a projective point. At that point is drawn an xy-axis. There is something 

interesting about the way this axis appears at the antipodal ends of the sphere. 

In the northern hemisphere, where the axis are drawn in black, a right hand put 

down with fingers on the x-axis will have the thumb point along the y-axis. But 

the antipodal axis, drawn in gray, has just the opposite: a right hand placed with 

its fingers on the x-axis will have the thumb point in the wrong way, instead, 

a left hand comes out correct. Briefly, the projective plane is not orientable: in 

this geometry, left and right handedness are not fixed properties of figures. 

The sequence of pictures below dramatizes this non-orientability. They sketch a 

trip around this space in the direction of the y part of the xy-axis. (This trip is 

not halfway around, it is a circuit, because antipodal spots are not two points, 

they are one point, and so the antipodal spots in the third picture below form 

the same projective point as the antipodal spots in the picture above.) 

=⇒ =⇒ 

At the end of the circuit, the x arrow from the xy-axis sticks out in the other 

direction. Another example of the same effect is that a clockwise spiral, on taking 

the same trip, would switch to counterclockwise. (This orientation reversal 

appeared earlier, in the pinhole/eclipse picture.) 

This exhibition of the existence of a non-orientable space raises the question 

of whether our space is orientable: is is possible for a right handed astronaut to 

take a trip to Mars and return left handed? An excellent nontechnical reference 

is [Gardner]. An intriguing science fiction story about orientation reversal is 

[Clarke]. 

So projective geometry is mathematically interesting and rewarding, in addition 

to the natural way in which it arises in art. It is more than just a technical 

device to shorten some proofs. For an overview, see [Courant & Robbins]. The 

approach we’ve taken here, the analytic approach, leads to quick theorems and 

— most importantly for us — illustrates the power of linear algebra (see [Hanes],


[Ryan], and [Eggar]). But, note that the other possible approach, the synthetic 

approach of deriving the results from an axiom system, is both extraordinarily 

beautiful and is also the historical route of development. Two fine sources for 

this approach are [Coxeter] or[Seidenberg]. An interesting, easy, application is 

[Davies] 

Exercises 

1 What is the equation of this point? 

�1� 

2 (a) Find the line incident on these points in the projective plane. 

� � � � 

1 4 

2 , 5 

3 6 

(b) Find the point incident on both of these projective lines. 

� 1 2 3 � , � 4 5 6 � 

0 

0 

3 Find the formula for the line incident on two projective points. Find the formula 

for the point incident on two projective lines. 

4 Prove that the definition of incidence is independent of the choice of the representatives 

of p and L. That is, if p1, p2, p3, andq1, q2, q3 are two triples of 

homogeneous coordinates for p, andL1, L2, L3, andM1, M2, M3 are two triples 

of homogeneous coordinates for L, provethatp1L1 + p2L2 + p3L3 =0ifandonly 

if q1M1 + q2M2 + q3M3 =0. 

5 Give a drawing to show that central projection does not preserve circles, that a 

circle may project to an ellipse. Can a (non-circular) ellipse project to a circle? 

6 Give the formula for the correspondence between the non-equatorial part of the 

antipodal modal of the projective plane, and the plane z =1. 

7 (Pappus’s Theorem) Assume that T0, U0, andV0 are collinear and that T1, U1, 

and V1 are collinear. Consider these three points: (i) the intersection V2 of the lines 

T0U1 and T1U0, (ii) the intersection U2 of the lines T0V1 and T1V0, and (iii) the 

intersection T2 of U0V1 and U1V0. 

(a) Draw a (Euclidean) picture. 

(b) Apply the lemma used in Desargue’s Theorem to get simple homogeneous 

coordinate vectors for the T ’s and V0. 

(c) Find the resulting homogeneous coordinate vectors for U’s (these must each 

involve a parameter as, e.g., U0 couldbeanywhereontheT0V0 line). 

(d) Find the resulting homogeneous coordinate vectors for V1. (Hint: it involves 

two parameters.) 

(e) Find the resulting homogeneous coordinate vectors for V2. (It also involves 

two parameters.) 

(f) Show that the product of the three parameters is 1. 

(g) Verify that V2 is on the T2U2 line..

Chapter 5 

Similarity 

While studying matrix equivalence, we have shown that for any any homomorphism 

there are bases B and D such that the representation matrix has a block 

partial-identity form. 

Rep B,D(h) = 

� 

Identity 

� 

Zero 

Zero Zero 

This representation lets us think of the map as sending c1 � β1 + ···+ cn � βn to 

c1 � δ1 + ···+ ck � δk +�0+···+�0, where n is the dimension of the domain and k is 

the dimension of the range. So, under this representation the action of the map 

is easy to understand because most of the matrix entries are zero. 

This chapter considers the special case where the domain and the codomain 

are equal, that is, where the homomorphism is a transformation. In this case 

we naturally ask to find a single basis B so that Rep B,B(t) is as simple as 

possible (we will take ‘simple’ to mean that it has many zeroes). A matrix 

having the above block partial-identity form is not always possible here. But, 

we will develop a form that comes close — the representation will be nearly 

diagonal. 

5.I Complex Vector Spaces 

This chapter will require that we factor polynomials. Of course, many polynomials 

do not factor over the real numbers. For instance, x 2 + 1 does not factor 

into the product of two linear polynomials with real coefficients. For that reason, 

we shall from now on take our scalars from the complex numbers. In this 

chapter the c’s in c1�v1 + c2�v2 + ···+ cn�vn will be complex numbers. 

So we are shifting from studying vector spaces over the real numbers to 

vector spaces over the complex numbers. As a consequence, in this chapter 

vector and matrix entries are complex. (The real numbers are a subset of the 

complex numbers, and a quick glance through this chapter shows that most of 

347

348 Chapter 5. Similarity 

the examples use only pure-real numbers. Nonetheless, the critical theorems 

require the use of the complex number system to go through.) Therefore, the 

first section of this chapter is a quick review of complex numbers. 

The idea of taking scalars from a structure other than the real numbers is 

an interesting one. However, in this book we are moving to this more general 

context only for the pragmatic reason that we must do so in order to develop 

the representation. We will not go into using other sets of scalars in more 

detail because it could distract from our task. For the more general approach, 

delightful presentations are in [Halmos] or[Hoffman & Kunze]. 

5.I.1 Factoring and Complex Numbers; A Review 

This subsection is a review only and we take the main results as known. 

For proofs, see [Birkhoff & MacLane] or[Ebbinghaus]. 

Just as integers have a division operation — e.g., ‘4 goes 5 times into 21 

with remainder 1’ — so do polynomials. 

1.1 Theorem (Division Theorem for Polynomials) Let c(x) be a polynomial. 

If m(x) is a non-zero polynomial then there are quotient and remainder 

polynomials q(x) andr(x) such that 

c(x) =m(x) · q(x)+r(x) 

where the degree of r(x) is strictly less than the degree of m(x). 

In this book constant polynomials, including the zero polynomial, are said to 

have degree 0. (This is not the standard definition, but it is convienent here.) 

The point of the integer division statement ‘4 goes 5 times into 21 with 

remainder 1’ is that the remainder is less than 4 — while 4 goes 5 times, it does 

not go 6 times. In the same way, the point of the polynomial division statement 

is its final clause. 

1.2 Example If c(x) =2x 3 − 3x 2 +4x and m(x) =x 2 + 1 then q(x) =2x − 3 

and r(x) =2x + 3. Note that r(x) has a lower degree than m(x). 

1.3 Corollary The remainder when c(x) is divided by x − λ is the constant 

polynomial r(x) =c(λ). 

Proof. The remainder must be a constant polynomial. because it is of degree 

less than the divisor x − λ, To determine the constant, taking m(x) fromthe 

theorem to be x − λ and substituting λ for x yields c(λ) =(λ − λ) · q(λ) + 

r(x). QED 

If a divisor m(x) goes into a dividend c(x) evenly, meaning that r(x) isthe 

zero polynomial, then m(x) isafactor of c(x). Any root of the factor (any 

λ ∈ R such that m(λ) = 0) is a root of c(x) since c(λ) =m(λ) · q(λ) = 0. The 

prior corollary immediately yields the following converse.

Section I. Complex Vector Spaces 349 

1.4 Corollary If λ is a root of the polynomial c(x) then x − λ divides c(x) 

evenly, that is, x − λ is a factor of c(x). 

Finding the roots and factors of a high-degree polynomial can be hard. But 

for second-degree polynomials we have the quadratic formula: the roots of ax 2 + 

bx + c are 

λ1 = −b + √ b 2 − 4ac 

2a 

λ2 = −b − √ b 2 − 4ac 

2a 

(if the discriminant b 2 − 4ac is negative then the polynomial has no real number 

roots). A polynomial that cannot be factored into two lower-degree polynomials 

with real number coefficients is irreducible over the reals. 

1.5 Theorem Any constant or linear polynomial is irreducible over the reals. 

A quadratic polynomial is irreducible over the reals if and only if its discriminant 

is negative. No cubic or higher-degree polynomial is irreducible over the reals. 

1.6 Corollary Any polynomial with real coefficients can be factored into linear 

and irreducible quadratic polynomials. That factorization is unique; any two 

factorizations have the same powers of the same factors. 

Note the analogy with the prime factorization of integers. In both cases, the 

uniqueness clause is very useful. 

1.7 Example Because of uniqueness we know, without multiplying them out, 

that (x +3) 2 (x 2 +1) 3 does not equal (x +3) 4 (x 2 + x +1) 2 . 

1.8 Example By uniqueness, if c(x) =m(x) · q(x) then where c(x) =(x − 

3) 2 (x +2) 3 and m(x) =(x − 3)(x +2) 2 , we know that q(x) =(x − 3)(x + 2). 

While x 2 + 1 has no real roots and so doesn’t factor over the real numbers, 

if we imagine a root — traditionally denoted i so that i 2 + 1 = 0 — then x 2 +1 

factors into a product of linears (x − i)(x + i). 

So we adjoin this root i to the reals and close the new system with respect 

to addition, multiplication, etc. (i.e., we also add 3 + i, and 2i, and 3 + 2i, etc., 

putting in all linear combinations of 1 and i). We then get a new structure, the 

complex numbers, denoted C. 

In C we can factor (obviously, at least some) quadratics that would be irreducible 

if we were to stick to the real numbers. Surprisingly, in C we can 

not only factor x 2 + 1 and its close relatives, we can factor any quadratic. Any 

quadratic polynomial factors over the complex numbers. 

ax 2 + bx + c = a · � x − −b + √ b 2 − 4ac 

2a 

� � −b − 

· x − √ b2 − 4ac� 

2a 

1.9 Example The second degree polynomial x2 +x+1 factors over the complex 

numbers into the product of two first degree polynomials. 

� −1+ 

x − √ −3�� 

−1 − 

x − 

2 

√ −3� 

� 1 

= x − (− 

2 

2 + 

√ 

3 

2 i)��x − (− 1 

2 − 

√ 

3 

2 i)�


1.10 Corollary (Fundamental Theorem of Algebra) Polynomials with 

complex coefficients factor into linear polynomials with complex coefficients. 

The factorization is unique. 

5.I.2 Complex Representations 

Recall the definitions of the complex number operations. 

(a + bi) +(c + di) =(a + c)+(b + d)i 

(a + bi)(c + di) =ac + adi + bci + bd(−1) = (ac − bd)+(ad + bc)i 

2.1 Example For instance, (1−2i) +(5+4i) =6+2i and (2−3i)(4 − 0.5i) = 

6.5 − 13i. 

Handling scalar operations with those rules, all of the operations that we’ve 

covered for real vector spaces carry over unchanged. 

2.2 Example Matrix multiplication is the same, although the scalar arithmetic 

involves more bookkeeping. 

� 

1+1i 

�� 

2 − 0i 1+0i 

� 

1 − 0i 

i −2+3i 3i −i 

� 

(1 + 1i) · (1 + 0i)+(2−0i) · (3i) 

= 

(i) · (1 + 0i)+(−2+3i) · (3i) 

� � 

1+7i 1 − 1i 

= 

−9 − 5i 3+3i 

� 

(1+1i) · (1 − 0i)+(2−0i) · (−i) 

(i) · (1 − 0i)+(−2+3i) · (−i) 

Everything else from prior chapters that we can, we shall also carry over 

unchanged. For instance, we shall call this 

⎛ ⎞ 

1+0i 

⎜ 

⎜0+0i 

⎟ 

〈 ⎜ 

⎝ . 

⎟ 

. ⎠ 

0+0i 

,..., 

⎛ ⎞ 

0+0i 

⎜ 

⎜0+0i 

⎟ 

⎜ 

⎝ . 

⎟ 

. ⎠ 

1+0i 

〉 

the standard basis for C n as a vector space over C and again denote it En.

Section II. Similarity 351 

5.II Similarity 

5.II.1 Definition and Examples 

We’ve defined H and ˆ H to be matrix-equivalent if there are nonsingular 

matrices P and Q such that ˆ H = PHQ. That definition is motivated by this 

diagram 

Vw.r.t. B 

h 

−−−−→ 

H 

Ww.r.t. D 

⏐ 

⏐ 

⏐ 

⏐ 

id� 

id� 

V w.r.t. ˆ B 

h 

−−−−→ 

ˆH 

W w.r.t. ˆ D 

showing that H and ˆ H both represent h but with respect to different pairs of 

bases. We now specialize that setup to the case where the codomain equals the 

domain, and where the codomain’s basis equals the domain’s basis. 

Vw.r.t. B 

t 

−−−−→ Vw.r.t. B 

⏐ 

⏐ 

⏐ 

⏐ 

id� 

id� 

Vw.r.t. D 

t 

−−−−→ Vw.r.t. D 

To move from the lower left to the lower right we can either go straight over, or 

up, over, and then down. In matrix terms, 

Rep D,D(t) =Rep B,D(id) Rep B,B(t) � Rep B,D(id) � −1 

(recall that a representation of composition like this one reads right to left). 

1.1 Definition The matrices T and S are similar if there is a nonsingular P 

such that T = PSP −1 . 

Since nonsingular matrices are square, the similar matrices T and S must be 

square and of the same size. 

1.2 Example With these two, 

� � 

2 1 

P = 

1 1 

S = 

calculation gives that S is similar to this matrix. 

� � 

0 −1 

T = 

1 1 

� � 

2 −3 

1 −1 

1.3 Example The only matrix similar to the zero matrix is itself: PZP −1 = 

PZ = Z. The only matrix similar to the identity matrix is itself: PIP −1 = 

PP −1 = I.


Since matrix similarity is a special case of matrix equivalence, if two matrices 

are similar then they are equivalent. What about the converse: must 

matrix equivalent square matrices be similar? The answer is no. The prior 

example shows that the similarity classes are different from the matrix equivalence 

classes, because the matrix equivalence class of the identity consists of 

all nonsingular matrices of that size. Thus, for instance, these two are matrix 

equivalent but not similar. 

T = 

� � 

1 0 

0 1 

S = 

� � 

1 2 

0 3 

So some matrix equivalence classes split into two or more similarity classes — 

similarity gives a finer partition than does equivalence. This picture shows some 

matrix equivalence classes subdivided into similarity classes. 

All square matrices: 

✥ 

✪ 

✩ 

✦ ✜ 

✢ 

.A 

... 

B. 

A matrix-equivalent to, 

but not similar to, B. 

To understand the similarity relation we shall study the similarity classes. 

We approach this question in the same way that we’ve studied both the row 

equivalence and matrix equivalence relations, by finding a canonical form for 

representatives ∗ of the similarity classes, called Jordan form. With this canonical 

form, we can decide if two matrices are similar by checking whether they 

reduce to the same representative. We’ve also seen with both row equivalence 

and matrix equivalence that a canonical form gives us insight into the ways in 

which members of the same class are alike (e.g., two identically-sized matrices 

are matrix equivalent if and only if they have the same rank). 

Exercises 

1.4 For 

S = 

� 

1 

� 

3 

−2 −6 

T = 

� 

0 

� 

0 

−11/2 −5 

P = 

� 

4 

� 

2 

−3 2 

check that T = PSP −1 . 

� 1.5 Example 1.3 shows that the only matrix similar to a zero matrix is itself and 

that the only matrix similar to the identity is itself. 

(a) Show that the 1×1 matrix (2), also, is similar only to itself. 

(b) Is a matrix of the form cI for some scalar c similar only to itself? 

(c) Is a diagonal matrix similar only to itself? 

1.6 Show that these matrices are not similar. 

� � � � 

1 0 4 1 0 1 

1 1 3 0 1 1 

2 1 7 3 1 2 

∗ More information on representatives is in the appendix.


1.7 Consider the transformation t: P2 →P2 described by x 2 ↦→ x +1, x ↦→ x 2 − 1, 

and 1 ↦→ 3. 

(a) Find T =Rep B,B(t) whereB = 〈x 2 ,x,1〉. 

(b) Find S =Rep D,D(t) whereD = 〈1, 1+x, 1+x + x 2 〉. 

(c) Find the matrix P such that T = PSP −1 . 

� 1.8 Exhibit an nontrivial similarity relationship in this way: let t: C 2 → C 2 act by 

� � 

1 

2 

↦→ 

� � � � 

3 −1 

0 1 

↦→ 

� � 

−1 

2 

and pick two bases, and represent t with respect to then T = Rep B,B(t) and 

S =Rep D,D(t). Then compute the P and P −1 to change bases from B to D and 

back again. 

1.9 Explain Example 1.3 in terms of maps. 

� 1.10 Are there two matrices A and B that are similar while A 2 and B 2 are not 

similar? 

� 1.11 Prove that if two matrices are similar and one is invertible then so is the 

other. 

� 1.12 Show that similarity is an equivalence relation. 

1.13 Consider a matrix representing, with respect to some B,B, reflection across 

the x-axis in R 2 . Consider also a matrix representing, with respect to some D, D, 

reflection across the y-axis. Must they be similar? 

1.14 Prove that similarity preserves determinants and rank. Does the converse 

hold? 

1.15 Is there a matrix equivalence class with only one matrix similarity class inside? 

One with infinitely many similarity classes? 

1.16 Can two different diagonal matrices be in the same similarity class? 

� 1.17 Prove that if two matrices are similar then their k-th powers are similar when 

k>0. What if k ≤ 0? 

� 1.18 Let p(x) be the polynomial cnx n + ···+ c1x + c0. Show that if T is similar to 

S then p(T )=cnT n + ···+ c1T + c0I is similar to p(S) =cnS n + ···+ c1S + c0I. 

1.19 List all of the matrix equivalence classes of 1×1 matrices. Also list the similarity 

classes, and describe which similarity classes are contained inside of each 

matrix equivalence class. 

1.20 Does similarity preserve sums? 

1.21 Show that if T − λI and N are similar matrices then T and N + λI are also 

similar. 

5.II.2 Diagonalizability 

The prior subsection defines the relation of similarity and shows that, although 

similar matrices are necessarily matrix equivalent, the converse does not 

hold. Some matrix-equivalence classes break into two or more similarity classes 

(the nonsingular n × n matrices, for instance). This means that the canonical 

form for matrix equivalence, a block partial-identity, cannot be used as a


canonical form for matrix similarity because the partial-identities cannot be in 

more than one similarity class, so there are similarity classes without one. This 

picture illustrates. As earlier in this book, class representatives are shown with 

stars. 

Equivalence classes 

subdivided into 

similarity classes. 

⋆ ⋆ 

⋆ 

⋆ 

✪ 

✥ 

✩ 

✦ ✜ 

⋆ 

⋆ ✢ 

... 

⋆ 

⋆ ⋆ 

⋆ ⋆ 

This finer partition needs 

more representatives. 

We are developing a canonical form for representatives of the similarity classes. 

We naturally try to build on our previous work, meaning first that the partial 

identity matrices should represent the similarity classes into which they fall, 

and beyond that, that the representatives should be as simple as possible. The 

simplest extension of the partial-identity form is a diagonal form. 

2.1 Definition A transformation is diagonalizable if it has a diagonal representation 

with respect to the same basis for the codomain as for the domain. A 

diagonalizable matrix is one that is similar to a diagonal matrix: T is diagonalizable 

if there is a nonsingular P such that PTP −1 is diagonal. 


� � 

4 −2 

1 1 

is diagonalizable. 

� 

2 

0 

� � 

0 −1 

= 

3 1 

�� 

2 4 

−1 1 

�� 

−2 −1 

1 1 

�−1 2 

−1 

2.3 Example Not every matrix is diagonalizable. The square of 

� � 

0 0 

N = 

1 0 

is the zero matrix. Thus, for any map n that N represents (with respect to the 

same basis for the domain as for the codomain), the composition n ◦ n is the 

zero map. This implies that no such map n can be diagonally represented (with 

respect to any B,B) because no power of a nonzero diagonal matrix is zero. 

That is, there is no diagonal matrix in N’s similarity class. 

That example shows that a diagonal form will not do for a canonical form — 

we cannot find a diagonal matrix in each matrix similarity class. However, the 

canonical form that we are developing has the property that if a matrix can 

be diagonalized then the diagonal matrix is the canonical representative of the 

similarity class. The next result characterizes which maps can be diagonalized.


2.4 Corollary A transformation t is diagonalizable if and only if there is a 

basis B = 〈 � β1,... , � βn〉 and scalars λ1,... ,λn such that t( � βi) =λi � βi for each i. 

Proof. This follows from the definition by considering a diagonal representation 

matrix. 

⎛ 

. 

. 

⎜ 

. 

. 

RepB,B(t) = ⎜ 

⎝RepB(t( 

� β1)) ··· RepB(t( � ⎞ 

⎟ 

βn)) ⎟ 

⎠ 

. 

. 

. 

. 

= 

⎛ 

⎞ 

λ1 0 

⎜ . 

⎝ . . 

. 

.. 

. 

⎟ 

. ⎠ 

0 λn 

This representation is equivalent to the existence of a basis satisfying the stated 

conditions simply by the definition of matrix representation. QED 

2.5 Example To diagonalize 

T = 

� � 

3 2 

0 1 

we take it as the representation of a transformation with respect to the standard 

basis T =RepE2,E2 (t) and we look for a basis B = 〈 � β1, � β2〉 such that 

� � 

λ1 0 

RepB,B(t) = 

0 λ2 

that is, such that t( � β1) =λ1 and t( � β2) =λ2. 

� 

3 

0 

� 

2 �β1 = λ1 · 

1 

� β1 

� 

3 

0 

� 

2 

1 

We are looking for scalars x such that this equation 

� �� 

3 2 b1 b1 

= x · 

0 1 

b2 

b2 

�β2 = λ2 · � β2 

has solutions b1 and b2, which are not both zero. Rewrite that as a linear system. 

(3 − x) · b1 + 2· b2 =0 

(1 − x) · b2 =0 

In the bottom equation the two numbers multiply to give zero only if at least 

one of them is zero so there are two possibilities, b2 =0andx =1. Intheb2 =0 

possibility, the first equation gives that either b1 =0orx = 3. Since the case 

of both b1 =0andb2 = 0 is disallowed, we are left looking at the possibility of 

x = 3. With it, the first equation in (∗) is0· b1 +2· b2 = 0 and so associated 

with 3 are vectors with a second component of zero and a first component that 

is free. 

� �� 

3 2 b1 

=3· 

0 1 0 

� � 

b1 

0 

(∗)


That is, one solution to (∗) isλ1 = 3, and we have a first basis vector. 

�β1 = 

� � 

1 

0 

In the x = 1 possibility, the first equation in (∗) is2· b1 +2· b2 =0,andso 

associated with 1 are vectors whose second component is the negative of their 

first component. 

� �� 

3 2 b1 b1 

=1· 

0 1 −b1 −b1 

Thus, another solution is λ2 = 1 and a second basis vector is this. 

� � 

�β2 

1 

= 

−1 

To finish, drawing the similarity diagram 

t 

−−−−→ 

T 

R2 w.r.t. R E2 

2 w.r.t. E2 

⏐ 

⏐ 

⏐ 

⏐ 

id� 

id� 

R 2 w.r.t. B 

t 

−−−−→ 

D 

R 2 w.r.t. B 

and noting that the matrix Rep B,E2 (id) is easy leads to this diagonalization. 

� � � �−1 � �� 

3 0 1 1 3 2 1 1 

= 

0 1 0 −1 0 1 0 −1 

In the next subsection, we will expand on that example by considering more 

closely the property of Corollary 2.4. This includes seeing another way, the way 

that we will routinely use, to find the λ’s. 

Exercises 

� 2.6 Repeat Example 2.5 for the matrix from Example 2.2. 

2.7 Diagonalize � �these 

upper � triangular � matrices. 

−2 1 

5 4 

(a) 

(b) 

0 2 

0 1 

� 2.8 What form do the powers of a diagonal matrix have? 

2.9 Give two same-sized diagonal matrices that are not similar. Must any two 

different diagonal matrices come from different similarity classes? 

2.10 Give a nonsingular diagonal matrix. Can a diagonal matrix ever be singular? 

� 2.11 Show that the inverse of a diagonal matrix is the diagonal of the the inverses, 

if no element on that diagonal is zero. What happens when a diagonal entry is 

zero?


2.12 The equation ending Example 2.5 

� �−1 � �� 

1 1 3 2 1 

0 −1 0 1 0 

� � 

1 3 

= 

−1 0 

� 

0 

1 

is a bit jarring because for P we must take the first matrix, which is shown as an 

inverse, and for P −1 we take the inverse of the first matrix, so that the two −1 

powers cancel and this matrix is shown without a superscript −1. 

(a) Check that this nicer-appearing equation holds. 

� � � �� 

3 0 1 1 3 2 1 

= 

0 1 0 −1 0 1 0 

�−1 1 

−1 

(b) Is the previous item a coincidence? Or can we always switch the P and the 

P −1 ? 

2.13 Show that the P used to diagonalize in Example 2.5 is not unique. 

2.14 Find a formula for the powers of this matrix Hint: see Exercise 8. 

� � 

−3 1 

−4 2 

� 2.15 Diagonalize � � these. � � 

1 1 

0 1 

(a) 

(b) 

0 0 

1 0 

2.16 We can ask how diagonalization interacts with the matrix operations. Assume 

that t, s: V → V are each diagonalizable. Is ct diagonalizable for all scalars c? 

What about t + s? t ◦ s? 

� 2.17 Show that matrices of this form are not diagonalizable. 

� � 

1 c 

c �= 0 

0 1 

2.18 Show � that � each of these � is� diagonalizable. 

1 2 

x y 

(a) 

(b) 

x, y, z scalars 

2 1 

y z 

5.II.3 Eigenvalues and Eigenvectors 

In this subsection we will focus on the property of Corollary 2.4. 

3.1 Definition A transformation t: V → V has a scalar eigenvalue λ if there 

is a nonzero eigenvector � ζ ∈ V such that t( � ζ)=λ · � ζ. 

(“Eigen” is German for “characteristic of” or “peculiar to”; some authors call 

these characteristic values and vectors. No authors call them “peculiar”.) 

3.2 Example The projection map 

⎛ 

⎝ x 

⎞ 

y⎠ 

z 

π 

⎛ 

↦−→ ⎝ x 

⎞ 

y⎠ 

x, y, z ∈ C 

0


has an eigenvalue of 1 associated with any eigenvector of the form 

⎛ 

⎝ x 

⎞ 

y⎠ 

0 

where x and y are non-0 scalars. On the other hand, 2 is not an eigenvalue of 

π since no non-�0 vector is doubled. 

That example shows why the ‘non-�0’ appears in the definition. Disallowing 

�0 as an eigenvector eliminates trivial eigenvalues. 

3.3 Example The only transformation on the trivial space {�0 } is �0 ↦→ �0. This 

map has no eigenvalues because there are no non-�0 vectors �v mapped to a scalar 

multiple λ · �v of themselves. 

3.4 Example Consider the homomorphism t: P1 →P1 given by c0 + c1x ↦→ 

(c0 + c1)+(c0 + c1)x. The range of t is one-dimensional. Thus an application of 

t to a vector in the range will simply rescale that vector: c + cx ↦→ (2c)+(2c)x. 

That is, t has an eigenvalue of 2 associated with eigenvectors of the form c + cx 

where c �= 0. 

This map also has an eigenvalue of 0 associated with eigenvectors of the form 

c − cx where c �= 0. 

3.5 Definition A square matrix T has a scalar eigenvalue λ associated with 

the non-�0 eigenvector � ζ if T � ζ = λ · � ζ. 

3.6 Remark Although this extension from maps to matrices is obvious, there 

is a point that must be made. Eigenvalues of a map are also the eigenvalues of 

matrices representing that map, and so similar matrices have the same eigenvalues. 

But the eigenvectors are different — similar matrices need not have the 

same eigenvectors. 

For instance, consider again the transformation t: P1 →P1 given by c0 + 

c1x ↦→ (c0+c1)+(c0+c1)x. It has an eigenvalue of 2 associated with eigenvectors 

of the form c + cx where c �= 0. If we represent t with respect to B = 〈1 + 

1x, 1 − 1x〉 

� � 

2 0 

T =RepB,B(t) = 

0 0 

then 2 is an eigenvalue of T , associated with these eigenvectors. 

� � � �� 

c0 �� 2 0 c0 2c0 c0 �� 

{ 

= } = { c0 ∈ C, c0�= 0} 

c1 0 0 c1 2c1 0 

On the other hand, representing t with respect to D = 〈2+1x, 1+0x〉 gives 

� � 

3 0 

S =RepD,D(t) = 

−3 2


and the eigenvectors of S associated with the eigenvalue 2 are these. 

� � � �� 

c0 �� 3 0 c0 2c0 0 �� 

{ 

= } = { c1 ∈ C, c1�= 0} 

c1 −3 2 c1 2c1 c1 

Thus similar matrices can have different eigenvectors. 

Here is an informal description of what’s happening. The underlying transformation 

doubles the eigenvectors �v ↦→ 2 ·�v. But when the matrix representing 

the transformation is T =Rep B,B(t) then it “assumes” that column vectors are 

representations with respect to B. In contrast, S =Rep D,D(t) “assumes” that 

column vectors are representations with respect to D. So the vectors that get 

doubled by each matrix look different. 

The next example illustrates the basic tool for finding eigenvectors and eigenvalues. 

3.7 Example What are the eigenvalues and eigenvectors of this matrix? 

⎛ 

1 

T = ⎝ 2 

2 

0 

⎞ 

1 

−2⎠ 

−1 2 3 

To find the scalars x such that T � ζ = x� ζ for non-�0 eigenvectors � ζ, bring everything 

to the left-hand side 

⎛ 

1 

⎝ 2 

2 

0 

⎞ ⎛ ⎞ ⎛ ⎞ 

1 z1 z1 

−2⎠ 

⎝z2⎠ 

− x ⎝z2⎠ 

= �0 

−1 2 3 

z3 

and factor (T −xI) � ζ = �0. (Note that it says T −xI; the expression T −x doesn’t 

make sense because T is a matrix while x is a scalar.) This homogeneous linear 

system 

⎛ 

1 − x 

⎝ 2 

2 

0−x 1 

−2 

⎞ ⎛ ⎞ ⎛ ⎞ 

z1 0 

⎠ ⎝z2⎠ 

= ⎝0⎠ 

−1 2 3−x 0 

has a non-�0 solution if and only if the matrix is nonsingular. We can determine 

when that happens. 

z3 

z3 

0=|T − xI| 

� 

� 

� 

�1 

− x 2 1 � 

� 

= � 

� 2 0−x −2 � 

� 

� −1 2 3−x� = x 3 − 4x 2 +4x 

= x(x − 2) 2


The eigenvalues are λ1 =0andλ2 = 2. To find the associated eigenvectors, 

plug in each eigenvalue: 

⎛ 

1 − 0 

⎝ 2 

2 

0−0 ⎞ ⎛ 

1 

−2 ⎠ ⎝ 

−1 2 3−0 z1 

⎞ ⎛ 

z2⎠ 

= ⎝ 0 

⎞ 

0⎠ 

=⇒ 

⎛ 

⎝ 

0 

z1 

⎞ ⎛ 

z2⎠ 

= ⎝ a 

⎞ 

0⎠ 

a 

z3 

for a scalar parameter a �= 0(a is non-0 because eigenvectors must be non-�0). 

In the same way, 

⎛ 

2 − 1 

⎝ 2 

2 

2−0 ⎞ ⎛ 

1 

−2 ⎠ ⎝ 

−1 2 2−3 z1 

⎞ ⎛ 

z2⎠ 

= ⎝ 0 

⎞ 

0⎠ 

=⇒ 

⎛ 

⎝ 

0 

z1 

⎞ ⎛ 

z2⎠ 

= ⎝ b 

⎞ 

−b⎠ 

b 

with b �= 0. 


z3 

S = 

� � 

π 1 

0 3 

(here π is not a projection map, it is the number 3.14 ...) then 

�� 

�� 

� 

� π − x 1 �� 

� 

=(x− π)(x − 3) 

0 3−x so S has eigenvalues of λ1 = π and λ2 = 3. To find associated eigenvectors, first 

plug in λ1 for x: 

� �� 

� � � � 

π − π 1 z1 0 

z1 a 

= =⇒ = 

0 3−π 0 

0 

for a scalar a �= 0, and then plug in λ2: 

� �� 

π − 3 1 z1 0 

= 

0 3−3 0 

where b �= 0. 

z2 

z2 

=⇒ 

z2 

z3 

z3 

� � � � 

z1 −(1/π)b 

= 

b 

3.9 Definition The characteristic polynomial of a square matrix T is the determinant 

of the matrix T − xI, where x is a variable. The characteristic 

equation is |T − xI| = 0. The characteristic polynomial of a transformation t is 

the polynomial of any Rep B,B(t). 

Exercise 30 checks that the characteristic polynomial of a transformation is 

well-defined, that is, any choice of basis yields the same polynomial. 

3.10 Lemma A linear transformation on a nontrivial vector space has at least 

one eigenvalue. 

z2


Proof. Any root of the characteristic polynomial is an eigenvalue. Over the 

complex numbers, any polynomial of degree one or greater has a root. (This is 

the reason that in this chapter we’ve gone to scalars that are complex.) QED 

Notice the familiar form of the sets of eigenvectors in the above examples. 

3.11 Definition The eigenspace of a transformation t associated with the 

eigenvalue λ is Vλ = { � ζ � � t( � ζ )=λ � ζ }∪{�0 }. The eigenspace of a matrix is 

defined analogously. 

3.12 Lemma An eigenspace is a subspace. 

Proof. An eigenspace must be nonempty — for one thing it contains the zero 

vector — and so we need only check closure. Take vectors � ζ1,... , � ζn from Vλ, 

to show that any linear combination is in Vλ 

t(c1 � ζ1 + c2 � ζ2 + ···+ cn � ζn) =c1t( � ζ1)+···+ cnt( � ζn) 

= c1λ� ζ1 + ···+ cnλ� ζn 

= λ(c1 � ζ1 + ···+ cn � ζn) 

(the second equality holds even if any � ζi is �0 since t(�0) = λ · �0 =�0). QED 

3.13 Example In Example 3.8 the eigenspace associated with the eigenvalue 

π and the eigenspace associated with the eigenvalue 3 are these. 

� � � � 

a �� −b/π �� 

Vπ = { a ∈ R} V3 = { b ∈ R} 

0 

b 

3.14 Example In Example 3.7, these are the eigenspaces associated with the 

eigenvalues 0 and 2. 

⎛ ⎞ 

a 

V0 = { ⎝0⎠ 

a 

� ⎛ ⎞ 

b 

� a ∈ R}, V2 = { ⎝−b⎠ 

b 

� � b ∈ R}. 

3.15 Remark The characteristic equation is 0 = x(x − 2) 2 so in some sense 2 

is an eigenvalue “twice”. However there are not “twice” as many eigenvectors, 

in that the dimension of the eigenspace is one, not two. The next example shows 

a case where a number, 1, is a double root of the characteristic equation and 

the dimension of the associated eigenspace is two. 

3.16 Example With respect to the standard bases, this matrix 

⎛ 

1 

⎝0 0 

1 

⎞ 

0 

0⎠ 

0 0 0


represents projection. 

⎛ 

⎝ x 

⎞ 

y⎠ 

z 

π 

⎛ 

↦−→ ⎝ x 

⎞ 

y⎠ 

x, y, z ∈ C 

0 

Its eigenspace associated with the eigenvalue 0 and its eigenspace associated 

with the eigenvalue 1 are easy to find. 

⎛ 

V0 = { ⎝ 0 

⎞ 

0 ⎠ � ⎛ 

� c3 ∈ C} V1 = { ⎝ c1 

⎞ 

c2⎠ 

0 

� � c1,c2 ∈ C} 

c3 

By the lemma, if two eigenvectors �v1 and �v2 are associated with the same 

eigenvalue then any linear combination of those two is also an eigenvector associated 

with that same eigenvalue. But, if two eigenvectors �v1 and �v2 are 

associated with different eigenvalues then the sum �v1 + �v2 need not be related 

to the eigenvalue of either one. In fact, just the opposite. If the eigenvalues are 

different then the eigenvectors are not linearly related. 

3.17 Theorem For any set of distinct eigenvalues of a map or matrix, a set of 

associated eigenvectors, one per eigenvalue, is linearly independent. 

Proof. We will use induction on the number of eigenvalues. If there is no eigenvalue 

or only one eigenvalue then the set of associated eigenvectors is empty or is 

a singleton set with a non-�0 member, and in either case is linearly independent. 

For induction, assume that the theorem is true for any set of k distinct eigenvalues, 

suppose that λ1,...,λk+1 are distinct eigenvalues, and let �v1,...,�vk+1 

be associated eigenvectors. If c1�v1 + ···+ ck�vk + ck+1�vk+1 = �0 then after multiplying 

both sides of the displayed equation by λk+1, applying the map or matrix 

to both sides of the displayed equation, and subtracting the first result from the 

second, we have this. 

c1(λk+1 − λ1)�v1 + ···+ ck(λk+1 − λk)�vk + ck+1(λk+1 − λk+1)�vk+1 = �0 

The induction hypothesis now applies: c1(λk+1−λ1) =0,...,ck(λk+1−λk) =0. 

Thus, as all the eigenvalues are distinct, c1, ..., ck are all 0. Finally, now ck+1 

must be 0 because we are left with the equation �vk+1 �= �0. QED 

3.18 Example The eigenvalues of 

⎛ 

2 

⎝ 0 

−2 

1 

⎞ 

2 

1⎠ 

−4 8 3 

are distinct: λ1 =1,λ2 =2,andλ3 = 3. A set of associated eigenvectors like 

⎛ 

{ ⎝ 2 

⎞ ⎛ 

1⎠ 

, ⎝ 

0 

9 

⎞ ⎛ 

4⎠ 

, ⎝ 

4 

2 

⎞ 

1⎠} 

2 

is linearly independent.


3.19 Corollary An n×n matrix with n distinct eigenvalues is diagonalizable. 

Proof. Form a basis of eigenvectors. Apply Corollary 2.4. QED 

Exercises 

3.20 For �each, find � the characteristic � � polynomial � and � the eigenvalues. � � 

10 −9 

1 2 

0 3 

0 0 

(a) 

(b) 

(c) 

(d) 

4 −2 

4 3 

7 0 

0 0 

� � 

1 0 

(e) 

0 1 

� 3.21 For each matrix, find the characteristic equation, and the eigenvalues and 

associated � eigenvectors. � � � 

3 0 

3 2 

(a) 

(b) 

8 −1 

−1 0 

3.22 Find the characteristic equation, and the eigenvalues and associated eigenvectors 

for this matrix. Hint. The eigenvalues are complex. 

� � 

−2 −1 

5 2 

3.23 Find the characteristic polynomial, the eigenvalues, and the associated eigenvectorsofthismatrix. 

� 

1 1 

� 

1 

0 0 1 

0 0 1 

� 3.24 For each matrix, find the characteristic equation, and the eigenvalues and 

associated eigenvectors. 

� � 

3 −2 0 

� 

0 1 

� 

0 

(a) −2 3 0 (b) 0 0 1 

0 0 5 

4 −17 8 

� 3.25 Let t: P2 →P2 be 

a0 + a1x + a2x 2 ↦→ (5a0 +6a1 +2a2) − (a1 +8a2)x +(a0 − 2a2)x 2 . 

Find its eigenvalues and the associated eigenvectors. 

3.26 Find the eigenvalues and eigenvectors of this map t: M2 →M2. 

� � � � 

a b 2c a+ c 

↦→ 

c d b − 2c d 

� 3.27 Find the eigenvalues and associated eigenvectors of the differentiation operator 

d/dx: P3 →P3. 

3.28 Prove that the eigenvalues of a triangular matrix (upper or lower triangular) 

are the entries on the diagonal. 

� 3.29 Find the formula for the characteristic polynomial of a 2×2 matrix. 

3.30 Prove that the characteristic polynomial of a transformation is well-defined. 

� 3.31 (a) Can any non-�0 vector in any nontrivial vector space be a eigenvector? 

That is, given a �v �= �0 from a nontrivial V , is there a transformation t: V → V 

and a scalar λ ∈ R such that t(�v) =λ�v? 

(b) Given a scalar λ, can any non-�0 vector in any nontrivial vector space be an 

eigenvector associated with the eigenvalue λ?


� 3.32 Suppose that t: V → V and T =RepB,B(t). Prove that the eigenvectors of T 

associated with λ are the non-�0 vectors in the kernel of the map represented (with 

respect to the same bases) by T − λI. 

3.33 Prove that if a,... , d are all integers and a + b = c + d then 

� � 

a b 

c d 

has integral eigenvalues, namely a + b and a − c. 

� 3.34 Prove that if T is nonsingular and has eigenvalues λ1,...,λn then T −1 has 

eigenvalues 1/λ1,...,1/λn. Is the converse true? 

� 3.35 Suppose that T is n×n and c, d are scalars. 

(a) Prove that if T has the eigenvalue λ with an associated eigenvector �v then �v 

is an eigenvector of cT + dI associated with eigenvalue cλ + d. 

(b) Prove that if T is diagonalizable then so is cT + dI. 

� 3.36 Show that λ is an eigenvalue of T if and only if the map represented by T −λI 

is not an isomorphism. 

3.37 [Strang 80] 

(a) Show that if λ is an eigenvalue of A then λ k is an eigenvalue of A k . 

(b) What is wrong with this proof generalizing that? “If λ is an eigenvalue of A 

and µ is an eigenvalue for B, thenλµ is an eigenvalue for AB, for, if A�x = λ�x 

and B�x = µ�x then AB�x = Aµ�x = µA�xµλ�x”? 

3.38 Do matrix-equivalent matrices have the same eigenvalues? 

3.39 Show that a square matrix with real entries and an odd number of rows has 

at least one real eigenvalue. 

3.40 Diagonalize. 

� 

−1 2 

� 

2 

2 2 2 

−3 −6 −6 

3.41 Suppose that P is a nonsingular n×n matrix. Show that the similarity transformation 

map tP : Mn×n →Mn×n sending T ↦→ PTP −1 is an isomorphism. 

3.42 [Math. Mag., Nov. 1967] Show that if A is an n square matrix and each row 

(column) sums to c then c is a characteristic root of A.

Section III. Nilpotence 365 

5.III Nilpotence 

The goal of this chapter is to show that every square matrix is similar to one 

that is a sum of two kinds of simple matrices. The prior section focused on the 

first kind, diagonal matrices. We now consider the other kind. 

5.III.1 Self-Composition 

This subsection is optional, although it is necessary for later material in this 

section and in the next one. 

A linear transformations t: V → V , because it has the same domain and 

codomain, can be iterated. ∗ That is, compositions of t with itself such as t 2 = t◦t 

and t 3 = t ◦ t ◦ t are defined. 

�v 

t(�v ) 

t 2 (�v ) 

Note that this power notation for the linear transformation functions dovetails 

with the notation that we’ve used earlier for their square matrix representations 

because if Rep B,B(t) =T then Rep B,B(t j )=T j . 

1.1 Example For the derivative map d/dx: P3 →P3 given by 

a + bx + cx 2 3 d/dx 

+ dx ↦−→ b +2cx +3dx 2 

the second power is the second derivative 

the third power is the third derivative 

and any higher power is the zero map. 

a + bx + cx 2 + dx 3 d2 /dx 2 

↦−→ 2c +6dx 

a + bx + cx 2 + dx 3 d3 /dx 3 

↦−→ 6d 

1.2 Example This transformation of the space of 2×2 matrices 

� 

a 

c 

� � 

b t b 

↦−→ 

d d 

� 

a 

0 

∗ More information on function interation is in the appendix.


has this second power 

and this third power. 

� � 

2 a b t 

↦−→ 

c d 

� � 

3 a b t 

↦−→ 

c d 

After that, t 4 = t 2 and t 5 = t 3 ,etc. 

� � 

a b 

0 0 

� � 

b a 

0 0 

These examples suggest that on iteration more and more zeros appear until 

there is a settling down. The next result makes this precise. 

1.3 Lemma For any transformation t: V → V , the rangespaces of the powers 

form a descending chain 

V ⊇ R(t) ⊇ R(t 2 ) ⊇··· 

and the nullspaces form an ascending chain. 

{�0 }⊆N (t) ⊆ N (t 2 ) ⊆··· 

Further, there is a k such that for powers less than k the subsets are proper (if 

j


and this chain of nullspaces. 

{�0 }⊂P0 ⊂P1 ⊂P2 ⊂P3 = P3 = ··· 

1.5 Example The transformation π : C 3 → C 3 projecting onto the first two 

coordinates 

⎛ 

⎝ c1 

⎞ 

c2⎠ 

π 

⎛ 

↦−→ ⎝ c1 

⎞ 

c2⎠ 

0 

has C 3 ⊃ R(π) =R(π 2 )=··· and {�0 }⊂N (π) =N (π 2 )=···. 

c3 

1.6 Example Let t: P2 →P2 be the map c0 + c1x + c2x 2 ↦→ 2c0 + c2x. As the 

lemma describes, on iteration the rangespace shrinks 

R(t 0 )=P2 R(t) ={a + bx � � a, b ∈ C} R(t 2 )={a � � a ∈ C} 

and then stabilizes R(t 2 )=R(t 3 )=···, while the nullspace grows 

N (t 0 )={0} N (t) ={cx � � c ∈ C} N (t 2 )={cx + d � � c, d ∈ C} 

and then stabilizes N (t 2 )=N (t 3 )=···. 

This graph illustrates Lemma 1.3. The horizontal axis gives the power j 

of a transformation. The vertical axis gives the dimension of the rangespace 

of t j as the distance above zero — and thus also shows the dimension of the 

nullspace as the distance below the gray horizontal line, because the two add to 

the dimension n of the domain. 

n 

rank(t j ) 

0 1 2 j 

As sketched, on iteration the rank falls and with it the nullity grows until the 

two reach a steady state. This state must be reached by the n-th iterate. The 

steady state’s distance above zero is the dimension of the generalized rangespace 

and its distance below n is the dimension of the generalized nullspace. 

1.7 Definition Let t be a transformation on an n-dimensional space. The 

generalized rangespace (or the closure of the rangespace) isR∞(t) =R(t n )The 

generalized nullspace (or the closure of the nullspace) isN∞(t) =N (t n ). 

n


Exercises 

1.8 Give the chains of rangespaces and nullspaces for the zero and identity transformations. 

1.9 For each map, give the chain of rangespaces and the chain of nullspaces, and 

the generalized rangespace and the generalized nullspace. 

(a) t0 : P2 →P2, a + bx + cx 2 ↦→ b + cx 2 

(b) t1 : R 2 → R 2 , 

� � 

a 

↦→ 

b 

� � 

0 

a 

(c) t2 : P2 →P2, a + bx + cx 2 ↦→ b + cx + ax 2 

(d) t3 : R 3 → R 3 , 

� � 

a 

b ↦→ 

c 

� � 

a 

a 

b 

1.10 Prove that function composition is associative (t ◦ t) ◦ t = t ◦ (t ◦ t) andsowe 

can write t 3 without specifying a grouping. 

1.11 Check that a subspace must be of dimension less than or equal to the dimension 

of its superspace. Check that if the subspace is proper (the subspace does not 

equal the superspace) then the dimension is strictly less. (This is used in the proof 

of Lemma 1.3.) 

1.12 Prove that the generalized rangespace R∞(t) is the entire space, and the 

generalized nullspace N∞(t) is trivial, if the transformation t is nonsingular. Is 

this ‘only if’ also? 

1.13 Verify the nullspace half of Lemma 1.3. 

1.14 Give an example of a transformation on a three dimensional space whose 

range has dimension two. What is its nullspace? Iterate your example until the 

rangespace and nullspace stabilize. 

1.15 Show that the rangespace and nullspace of a linear transformation need not 

be disjoint. Are they ever disjoint? 

5.III.2 Strings 

This subsection is optional, and requires material from the optional Direct Sum 

subsection. 

The prior subsection shows that as j increases, the dimensions of the R(t j )’s 

fall while the dimensions of the N (t j )’s rise, in such a way that this rank and 

nullity split the dimension of V . Can we say more; do the two split a basis — 

is V = R(t j ) ⊕ N (t j )? 

The answer is yes for the smallest power j = 0 since V = R(t 0 ) ⊕ N (t 0 )= 

V ⊕{�0}. The answer is also yes at the other extreme. 

2.1 Lemma Where t: V → V is a linear transformation, the space is the direct 

sum V = R∞(t) ⊕ N∞(t). That is, both dim(V )=dim(R∞(t)) + dim(N∞(t)) 

and R∞(t) ∩ N∞(t) ={�0 }.


Proof. We will verify the second sentence, which is equivalent to the first. The 

first clause, that the dimension n of the domain of t n equals the rank of t n plus 

the nullity of t n , holds for any transformation and so we need only verify the 

second clause. 

Assume that �v ∈ R∞(t) ∩ N∞(t) =R(t n ) ∩ N (t n ), to prove that �v is �0. 

Because �v is in the nullspace, t n (�v) =�0. On the other hand, because R(t n )= 

R(t n+1 ), the map t: R∞(t) → R∞(t) is a dimension-preserving homomorphism 

and therefore is one-to-one. A composition of one-to-one maps is one-to-one, 

and so t n : R∞(t) → R∞(t) is one-to-one. But now — because only �0 issentby 

a one-to-one linear map to �0 — the fact that t n (�v) =�0 implies that �v = �0. QED 

2.2 Note Technically we should distinguish the map t: V → V from the map 

t: R∞(t) → R∞(t) because the domains or codomains might differ. The second 

one is said to be the restriction ∗ of t to R(t k ). We shall use later a point from 

that proof about the restriction map, namely that it is nonsingular. 

In contrast to the j =0andj = n cases, for intermediate powers the space 

V might not be the direct sum of R(t j )andN (t j ). The next example shows 

that the two can have a nontrivial intersection. 

2.3 Example Consider the transformation of C2 defined by this action on the 

elements of the standard basis. 

� � � � � � � � 

� � 

1 n 0 0 n 0 

0 0 

↦−→ 

↦−→ N =RepE2,E2 (n) = 

0 1 1 0 

1 0 

The vector 

�e2 = 

� � 

0 

1 

is in both the rangespace and nullspace. Another way to depict this map’s 

action is with a string. 

�e1 ↦→ �e2 ↦→ �0 

2.4 Example A map ˆn: C 4 → C 4 whose action on E4 is given by the string 

�e1 ↦→ �e2 ↦→ �e3 ↦→ �e4 ↦→ �0 

has R(ˆn) ∩ N (ˆn) equal to the span [{�e4}], has R(ˆn 2 ) ∩ N (ˆn 2 )=[{�e3,�e4}], and 

has R(ˆn 3 ) ∩ N (ˆn 3 )=[{�e4}]. The matrix representation is all zeros except for 

some subdiagonal ones. 

⎛ 

0 

⎜ 

ˆN =RepE4,E4 (ˆn) = ⎜1 

⎝0 

0 

0 

1 

0 

0 

0 

⎞ 

0 

0 ⎟ 

0⎠ 

0 0 1 0 

∗ More information on map restrictions is in the appendix.


2.5 Example Transformations can act via more than one string. A transformation 

t acting on a basis B = 〈 � β1,..., � β5〉 by 

�β1 ↦→ � β2 ↦→ � β3 ↦→ �0 

�β4 ↦→ � β5 ↦→ �0 

is represented by a matrix that is all zeros except for blocks of subdiagonal ones 

⎛ 

0 

⎜ 

⎜1 


⎜0 

⎝0 

0 

0 

1 

0 

0 

0 

0 

0 

0 

0 

0 

0 

⎞ 

0 

0 ⎟ 

0 ⎟ 

0⎠ 

0 0 0 1 0 

(the lines just visually organize the blocks). 

In those three examples all vectors are eventually transformed to zero. 

2.6 Definition A nilpotent transformation is one with a power that is the zero 

map. A nilpotent matrix is one with a power that is the zero matrix. In either 

case, the least such power is the index of nilpotency. 

2.7 Example In Example 2.3 the index of nilpotency is two. In Example 2.4 

it is four. In Example 2.5 it is three. 

2.8 Example The differentiation map d/dx: P2 →P2 is nilpotent of index 

three since the third derivative of any quadratic polynomial is zero. This map’s 

action is described by the string x2 ↦→ 2x ↦→ 2 ↦→ 0 and taking the basis 

B = 〈x2 , 2x, 2〉 gives this representation. 

⎛ 

0 0 

⎞ 

0 

RepB,B(d/dx) = ⎝1 

0 0⎠ 

0 1 0 

Not all nilpotent matrices are all zeros except for blocks of subdiagonal ones. 

2.9 Example With the matrix ˆ N from Example 2.4, and this four-vector basis 

⎛ ⎞ 

1 

⎜ 

D = 〈 ⎜0 

⎟ 

⎝1⎠ 

0 

, 

⎛ ⎞ 

0 

⎜ 

⎜2 

⎟ 

⎝1⎠ 

0 

, 

⎛ ⎞ 

1 

⎜ 

⎜1 

⎟ 

⎝1⎠ 

0 

, 

⎛ ⎞ 

0 

⎜ 

⎜0 

⎟ 

⎝0⎠ 

1 

〉 

a change of basis operation produces this representation with respect to D, D. 

⎛ 

1 

⎜ 

⎜0 

⎝1 

0 

2 

1 

1 

1 

1 

⎞ ⎛ 

0 0 

0⎟⎜ 

⎟ ⎜1 

0⎠⎝0 

0 

0 

1 

0 

0 

0 

⎞ ⎛ 

0 1 

0⎟⎜ 

⎟ ⎜0 

0⎠⎝1 

0 

2 

1 

1 

1 

1 

⎞−1 

⎛ 

0 −1 

0⎟ 

⎜ 

⎟ 

0⎠ 

= ⎜−3 

⎝−2 

0 

−2 

−1 

1 

5 

3 

⎞ 

0 

0 ⎟ 

0⎠ 

0 0 0 1 0 0 1 0 0 0 0 1 2 1 −2 0


The new matrix is nilpotent; it’s fourth power is the zero matrix since 

(P ˆ NP −1 ) 4 = P ˆ NP −1 · P ˆ NP −1 · P ˆ NP −1 · P ˆ NP −1 = P ˆ N 4 P −1 

and ˆ N 4 is the zero matrix. 

The goal of this subsection is Theorem 2.13, which shows that the prior 

example is prototypical in that every nilpotent matrix is similar to one that is 

all zeros except for blocks of subdiagonal ones. 

2.10 Definition Let t be a nilpotent transformation on V . A t-string generated 

by �v ∈ V is a sequence 〈�v, t(�v),... ,t k−1 (�v)〉. This sequence has length k. 

A t-string basis is a basis that is a concatenation of t-strings. 

2.11 Example In Example 2.5, thet-strings 〈 � β1, � β2, � β3〉 and 〈 � β4, � β5〉, of length 

three and two, can be concatenated to make a basis for the domain of t. 

2.12 Lemma If a space has a t-string basis then the longest string in it has 

length equal to the index of nilpotency of t. 

Proof. Suppose not. Those strings cannot be longer; if the index is k then 

t k sends any vector — including those starting the string — to �0. So suppose 

instead that there is a transformation t of index k on some space, such that 

the space has a t-string basis where all of the strings are shorter than length 

k. Because t has index k, there is a vector �v such that t k−1 (�v) �= �0. Represent 

�v as a linear combination of basis elements and apply t k−1 . We are supposing 

that t k−1 sends each basis element to �0 but that it does not send �v to �0. That 

is impossible. QED 

We shall show that every nilpotent map has an associated string basis. Then 

our goal theorem, that every nilpotent matrix is similar to one that is all zeros 

except for blocks of subdiagonal ones, is immediate, as in Example 2.5. 

Looking for a counterexample — a nilpotent map without an associated 

string basis that is disjoint — will suggest the idea for the proof. Consider the 

map t: C 5 → C 5 with this action. 

�e1 

�e2 

↦→ 

↦→ �e3 ↦→ �0 

�e4 ↦→ �e5 ↦→ �0 

⎛ 

0 

⎜ 

⎜0 

RepE5,E5 (t) = ⎜ 

⎜1 

⎝0 

0 

0 

1 

0 

0 

0 

0 

0 

0 

0 

0 

0 

⎞ 

0 

0 ⎟ 

0 ⎟ 

0⎠ 

0 0 0 1 0 

Even after ommitting the zero vector, these three strings aren’t disjoint, but 

that doesn’t end hope of finding a t-string basis. It only means that E5 will not 

do for the string basis. 

To find a basis that will do, we first find the number and lengths of its 

strings. Since t’s index of nilpotency is two, Lemma 2.12 says that at least one


string in the basis has length two. Thus the map must act on a string basis in 

one of these two ways. 

�β1 ↦→ � β2 ↦→ �0 

�β3 ↦→ � β4 ↦→ �0 

�β5 ↦→ �0 

�β1 ↦→ � β2 ↦→ �0 

�β3 ↦→ �0 

�β4 ↦→ �0 

�β5 ↦→ �0 

Now, the key point. A transformation with the left-hand action has a nullspace 

of dimension three since that’s how many basis vectors are sent to zero. A 

transformation with the right-hand action has a nullspace of dimension four. 

Using the matrix representation above, calculation of t’s nullspace 

⎛ ⎞ 

x 

⎜ 

⎜−x 

⎟ � 

N (t) ={ ⎜ z ⎟ � 

⎟ x, z, r ∈ C} 

⎝ 0 ⎠ 

r 

shows that it is three-dimensional, meaning that we want the left-hand action. 

To produce a string basis, first pick � β2 and � β4 from R(t) ∩ N (t) 

⎛ ⎞ 

0 

⎜ 

⎜0 

⎟ 

�β2 = ⎜ 

⎜1 

⎟ 

⎝0⎠ 

0 

⎛ ⎞ 

0 

⎜ 

⎜0 

⎟ 

�β4 = ⎜ 

⎜0 

⎟ 

⎝0⎠ 

1 

(other choices are possible, just be sure that { � β2, � β4} is linearly independent). 

For � β5 pick a vector from N (t) that is not in the span of { � β2, � β4}. 

⎛ ⎞ 

1 

⎜ 

⎜−1 

⎟ 

�β5 = ⎜ 0 ⎟ 

⎝ 0 ⎠ 

0 

Finally, take � β1 and � β3 such that t( � β1) = � β2 and t( � β3) = � β4. 

⎛ ⎞ 

0 

⎜ 

⎜1 

⎟ 

�β1 = ⎜ 

⎜0 

⎟ 

⎝0⎠ 

0 

⎛ ⎞ 

0 

⎜ 

⎜0 

⎟ 

�β3 = ⎜ 

⎜0 

⎟ 

⎝1⎠ 

0


Now, with respect to B = 〈 � β1,... , � β5〉, the matrix of t is as desired. 

⎛ 

0 

⎜ 

1 


⎜0 

⎝0 

0 

0 

0 

0 

0 

0 

0 

1 

0 

0 

0 

0 

⎞ 

0 

0 ⎟ 

0 ⎟ 

0⎠ 

0 0 0 0 0 

2.13 Theorem Any nilpotent transformation t is associated with a t-string 

basis. While the basis is not unique, the number and the length of the strings 

is determined by t. 

This illustrates the proof. Basis vectors are categorized into kind 1, kind 2, and 

kind 3. They are also shown as squares or circles, according to whether they 

are in the nullspace or not. 

3 ↦→ 1 ↦→ ··· ··· ↦→ 1 ↦→ 1 ↦→ �0 

3 ↦→ 1 ↦→ ··· ··· ↦→ 1 ↦→ 1 ↦→ �0 

. 

3 ↦→ 1 ↦→ · · · ↦→ 1 ↦→ 1 ↦→ �0 

2 ↦→ 

. 

�0 

2 ↦→ �0 

Proof. Fix a vector space V ; we will argue by induction on the index of nilpotency 

of t: V → V . If that index is 1 then t is the zero map and any basis is 

a string basis � β1 ↦→ �0, ... , � βn ↦→ �0. For the inductive step, assume that the 

theorem holds for any transformation with an index of nilpotency between 1 

and k − 1 and consider the index k case. 

First observe that the restriction to the rangespace t: R(t) → R(t) is also 

nilpotent, of index k − 1. Apply the inductive hypothesis to get a string basis 

for R(t), where the number and length of the strings is determined by t. 

B = 〈 � β1,t( � β1),...,t h1 ( � β1)〉 ⌢ 〈 � β2,... ,t h2 ( � β2)〉 ⌢ ··· ⌢ 〈 � βi,... ,t hi ( � βi)〉 

(In the illustration these are the basis vectors of kind 1, so there are i strings 

shown with this kind of basis vector.) 

Second, note that taking the final nonzero vector in each string gives a basis 

C = 〈t h1 ( � β1),...,t hi ( � βi)〉 for R(t) ∩ N (t). (These are illustrated with 1’s in 

squares.) For, a member of R(t) is mapped to zero if and only if it is a linear 

combination of those basis vectors that are mapped to zero. Extend C to a 

basis for all of N (t). 

Ĉ = C ⌢ 〈 � ξ1,..., � ξp〉


(The � ξ’s are the vectors of kind 2 so that Ĉ is the set of squares.) While many 

choices are possible for the � ξ’s, their number p is determined by the map t as it 

is the dimension of N (t) minus the dimension of R(t) ∩ N (t). 

Finally, B ⌢ Ĉ is a basis for R(t)+N (t) because any sum of something in the 

rangespace with something in the nullspace can be represented using elements 

of B for the rangespace part and elements of Ĉ for the part from the nullspace. 

Note that 

dim � R(t)+N (t) � =dim(R(t)) + dim(N (t)) − dim(R(t) ∩ N (t)) 

=rank(t) + nullity(t) − i 

=dim(V ) − i 

and so B ⌢ Ĉ can be extended to a basis for all of V by the addition of i more 

vectors. Specifically, remember that each of � β1,..., � βi is in R(t), and extend 

B ⌢ Ĉ with vectors �v1,...,�vi such that t(�v1) = � β1,...,t(�vi) = � βi. (In the 

illustration, these are the 3’s.) The check that linear independence is preserved 

by this extension is Exercise 29. QED 

2.14 Corollary Every nilpotent matrix is similar to a matrix that is all zeros 

except for blocks of subdiagonal ones. That is, every nilpotent map is represented 

with respect to some basis by such a matrix. 

This form is unique in the sense that if a nilpotent matrix is similar to two 

such matrices then those two simply have their blocks ordered differently. Thus 

this is a canonical form for the similarity classes of nilpotent matrices provided 

that we order the blocks, say, from longest to shortest. 


M = 

� � 

1 −1 

1 −1 

has an index of nilpotency of two, as this calculation shows. 

p M p N (M p � � � � 

) 

1 −1 x �� 

1 M = 

{ x ∈ C} 

1 −1 x 

2 M 2 � � 

0 0 

= 

C 

0 0 

2 

The calculation also describes how a map m represented by M must act on any 

string basis. With one map application the nullspace has dimension one and so 

one vector of the basis is sent to zero. On a second application, the nullspace 

has dimension two and so the other basis vector is sent to zero. Thus, the action 

of the map is � β1 ↦→ � β2 ↦→ �0 and the canonical form of the matrix is this. 

� � 

0 0 

1 0


We can exhibit such a m-string basis and the change of basis matrices witnessing 

the matrix similarity. For the basis, take M to represent m with respect 

to the standard bases, pick a � β2 ∈ N (m) and also pick a � β1 so that m( � β1) = � β2. 

�β2 = 

� � 

1 

1 

�β1 = 

� � 

1 

0 

(If we take M to be a representative with respect to some nonstandard bases 

then this picking step is just more messy.) Recall the similarity diagram. 

C2 w.r.t. E2 

⏐ 

id 

m 

−−−−→ 

M 

C2 w.r.t. E2 

⏐ 

�P id�P 

C 2 w.r.t. B 

m 

−−−−→ C 2 w.r.t. B 

The canonical form equals Rep B,B(m) =PMP −1 , where 

P −1 =Rep B,E2 (id) = 

� � 

1 1 

0 1 

P =(P −1 ) −1 = 

and the verification of the matrix calculation is routine. 

� 

1 

0 

�� 

−1 1 

1 1 

�� 

−1 1 

−1 0 

� � 

1 0 

= 

1 1 

� 

0 

0 


⎛ 

0 

⎜ 1 

⎜ 

⎜−1 

⎝ 0 

0 

0 

1 

1 

0 

0 

1 

0 

0 

0 

−1 

0 

⎞ 

0 

0 ⎟ 

1 ⎟ 

0 ⎠ 

1 0 −1 1 −1 

is nilpotent. These calculations show the nullspaces growing. 

� � 

1 −1 

0 1 

p N p N (N p 1 

⎛ 

0 

⎜ 1 

⎜ 

⎜−1 

⎝ 0 

0 

0 

1 

1 

0 

0 

1 

0 

0 

0 

−1 

0 

⎞ 

0 

0 ⎟ 

1 ⎟ 

0 ⎠ 

) 

1 0 −1 1 −1 

{ 

⎛ ⎞ 

0 

⎜ 0 ⎟ � 

⎜ 

⎜u 

− v⎟ 

� 

⎟ u, v ∈ C} 

⎝ u ⎠ 

2 

⎛ 

0 0 0 

⎜ 

⎜0 

0 0 

⎜ 

⎜1 

0 0 

⎝1 

0 0 

0 

0 

0 

0 

⎞ 

0 

0 ⎟ 

0 ⎟ 

0⎠ 

⎛ ⎞v 

0 

⎜ 

⎜y 

⎟ � 

{ ⎜ 

⎜z⎟ 

� 

⎟ y, z, u, v ∈ C} 

⎝u⎠ 

0 0 0 0 0 

v 

3 –zero matrix– C5


That table shows that any string basis must satisfy: the nullspace after one map 

application has dimension two so two basis vectors are sent directly to zero, 

the nullspace after the second application has dimension four so two additional 

basis vectors are sent to zero by the second iteration, and the nullspace after 

three applications is of dimension five so the final basis vector is sent to zero in 

three hops. 

�β1 ↦→ � β2 ↦→ � β3 ↦→ �0 

�β4 ↦→ � β5 ↦→ �0 

To produce such a basis, first pick two independent vectors from N (n) 

⎛ ⎞ ⎛ ⎞ 

0 

0 

⎜ 

⎜0⎟ 

⎜ 

⎟ ⎜0 

⎟ 

�β3 = ⎜ 

⎜1⎟ 

�β5 ⎟ = ⎜ 

⎜0 

⎟ 

⎝1⎠ 

⎝1⎠ 

0 

1 

then add � β2, � β4 ∈ N (n2 ) such that n( � β2) = � β3 and n( � β4) = � β5 

⎛ ⎞ ⎛ ⎞ 

0 

0 

⎜ 

⎜1⎟ 

⎜ 

⎟ ⎜1 

⎟ 

�β2 = ⎜ 

⎜0⎟ 

�β4 ⎟ = ⎜ 

⎜0 

⎟ 

⎝0⎠ 

⎝1⎠ 

0 

0 

and finish by adding � β1 ∈ N (n3 )=C5 ) such that n( � β1) = � β2. 

⎛ ⎞ 

1 

⎜ 

⎜0 

⎟ 

�β1 = ⎜ 

⎜1 

⎟ 

⎝0⎠ 

0 

Exercises 

� 2.17 What is the index of nilpotency of the left-shift operator, here acting on the 

space of triples of reals? 

(x, y, z) ↦→ (0,x,y) 

� 2.18 For each string basis state the index of nilpotency and give the dimension of 

the rangespace and nullspace of each iteration of the nilpotent map. 

(a) � β1 ↦→ � �β3 

β2 

↦→ 

↦→ �0 

� β4 ↦→ �0 

(b) � β1 ↦→ � β2 ↦→ � �β4 

�β5 

�β6 

↦→ �0 

↦→ �0 

↦→ �0 

β3 ↦→ �0 

(c) � β1 ↦→ � β2 ↦→ � β3 ↦→ �0 

Also give the canonical form of the matrix. 

2.19 Decide which of these matrices are nilpotent.


� 

−2 

(a) 

−1 

� 

45 

� 

4 

2 

−22 

� 

3 

(b) 

1 

� 

−19 

� 

1 

3 

(c) 

� 

−3 

−3 

−3 

2 

2 

2 

� 

1 

1 

1 

(e) 33 −16 −14 

69 −34 −29 

� 2.20 Find the canonical form of 

⎛ 

this matrix. 

0 1 1 0 

⎞ 

1 

⎜0 

⎜ 

⎜0 

⎝0 

0 

0 

0 

1 

0 

0 

1 

0 

0 

1 ⎟ 

0⎟ 

0⎠ 

0 0 0 0 0 

(d) 

� 

1 1 

� 

4 

3 0 −1 

5 2 7 

� 2.21 Consider the matrix from Example 2.16. 

(a) Use the action of the map on the string basis to give the canonical form. 

(b) Find the change of basis matrices that bring the matrix to canonical form. 

(c) Use the answer in the prior item to check the answer in the first item. 

� 2.22 Each of these matrices is nilpotent. 

� � 

1/2 −1/2 

(a) 

(b) 

1/2 −1/2 

� 0 0 0 

0 −1 1 

0 −1 1 

� 

(c) 

� 

−1 1 

� 

−1 

1 0 1 

1 −1 1 

Put each in canonical form. 

2.23 Describe the effect of left or right multiplication by a matrix that is in the 

canonical form for nilpotent matrices. 

2.24 Is nilpotence invariant under similarity? That is, must a matrix similar to a 

nilpotent matrix also be nilpotent? If so, with the same index? 

� 2.25 Show that the only eigenvalue of a nilpotent matrix is zero. 

2.26 Is there a nilpotent transformation of index three on a two-dimensional space? 

2.27 In the proof of Theorem 2.13, why isn’t the proof’s base case that the index 

of nilpotency is zero? 

� 2.28 Let t: V → V be a linear transformation and suppose �v ∈ V is such that 

t k (�v) =�0 but t k−1 (�v) �= �0. Consider the t-string 〈�v, t(�v),...,t k−1 (�v)〉. 

(a) Prove that t is a transformation on the span of the set of vectors in the string, 

that is, prove that t restricted to the span has a range that is a subset of the 

span. We say that the span is a t-invariant subspace. 

(b) Prove that the restriction is nilpotent. 

(c) Prove that the t-string is linearly independent and so is a basis for its span. 

(d) Represent the restriction map with respect to the t-string basis. 

2.29 Finish the proof of Theorem 2.13. 

2.30 Show that the terms ‘nilpotent transformation’ and ‘nilpotent matrix’, as 

given in Definition 2.6, fit with each other: a map is nilpotent if and only if it is 

represented by a nilpotent matrix. (Is it that a transformation is nilpotent if an 

only if there is a basis such that the map’s representation with respect to that 

basis is a nilpotent matrix, or that any representation is a nilpotent matrix?) 

2.31 Let T be nilpotent of index four. How big can the rangespace of T 3 be? 

2.32 Recall that similar matrices have the same eigenvalues. Show that the converse 

does not hold. 

2.33 Prove a nilpotent matrix is similar to one that is all zeros except for blocks of 

super-diagonal ones.


� 2.34 Prove that if a transformation has the same rangespace as nullspace. then the 

dimension of its domain is even. 

2.35 Prove that if two nilpotent matrices commute then their product and sum are 

also nilpotent. 

2.36 Consider the transformation of Mn×n given by tS(T )=ST − TS where S is 

an n×n matrix. Prove that if S is nilpotent then so is tS. 

2.37 Show that if N is nilpotent then I − N is invertible. Is that ‘only if’ also?

Section IV. Jordan Form 379 

5.IV Jordan Form 

This section uses material from three optional subsections: Direct Sum, Determinants 

Exist, and Other Formulas for the Determinant. 

The chapter on linear maps shows that every h: V → W can be represented 

by a partial-identity matrix with respect to some bases B ⊂ V and D ⊂ W . 

This chapter revisits this issue in the special case that the map is a linear 

transformation t: V → V . Of course, the general result still applies but with 

the codomain and domain equal we naturally ask about having the two bases 

also be equal. That is, we want a canonical form to represent transformations 

as Rep B,B(t). 

After a brief review section, we began by noting that a block partial identity 

form matrix is not always obtainable in this B,B case. We therefore considered 

the natural generalization, diagonal matrices, and showed that if its eigenvalues 

are distinct then a map or matrix can be diagonalized. But we also gave an 

example of a matrix that cannot be diagonalized and in the section prior to this 

one we developed that example. We showed that a linear map is nilpotent — 

if we take higher and higher powers of the map or matrix then we eventually 

get the zero map or matrix — if and only if there is a basis on which it acts via 

disjoint strings. That led to a canonical form for nilpotent matrices. 

Now, this section concludes the chapter. We will show that the two cases 

we’ve studied are exhaustive in that for any linear transformation there is a 

basis such that the matrix representation Rep B,B(t) is the sum of a diagonal 

matrix and a nilpotent matrix in its canonical form. 

5.IV.1 Polynomials of Maps and Matrices 

Recall that the set of square matrices is a vector space under entry-by-entry 

addition and scalar multiplication and that this space Mn×n has dimension n2 . 

Thus, for any n×n matrix T the n2 +1-member set {I,T,T 2 ,...,Tn2} is linearly 

dependent and so there are scalars c0,...,cn2 such that cn2T n2 +···+c1T +c0I 

is the zero matrix. 

1.1 Remark This observation is small but important. It says that every transformation 

exhibits a generalized nilpotency: the powers of a square matrix cannot 

climb forever without a “repeat”. 

1.2 Example Rotation of plane vectors π/6 radians counterclockwise is represented 

with respect to the standard basis by 

�√ 

3/2 −1/2 

T = 

1/2 √ � 

3/2 

and verifying that 0T 4 +0T 3 +1T 2 − 2T − 1I equals the zero matrix is easy.


1.3 Definition For any polynomial f(x) =cnx n + ···+ c1x + c0, where t is a 

linear transformation then f(t) is the transformation cnt n + ···+ c1t + c0(id) 

on the same space and where T is a square matrix then f(T ) is the matrix 

cnT n + ···+ c1T + c0I. 

1.4 Remark If, for instance, f(x) =x − 3, then most authors write in the 

identity matrix: f(T )=T − 3I. But most authors don’t write in the identity 

map: f(t) =t − 3. In this book we shall also observe this convention. 

Of course, if T =Rep B,B(t) then f(T )=Rep B,B(f(t)), which follows from 

the relationships T j = Rep B,B(t j ), and cT = Rep B,B(ct), and T1 + T2 = 

Rep B,B(t1 + t2). 

As Example 1.2 shows, there may be polynomials of degree smaller than n 2 

that zero the map or matrix. 

1.5 Definition The minimal polynomial m(x) of a transformation t or a square 

matrix T is the polynomial of least degree and with leading coefficient 1 such 

that m(t) is the zero map or m(T ) is the zero matrix. 

A minimal polynomial always exists by the observation opening this subsection. 

A minimal polynomial is unique by the ‘with leading coefficient 1’ clause. 

This is because if there are two polynomials m(x) and ˆm(x) that are both of the 

minimal degree to make the map or matrix zero (and thus are of equal degree), 

and both have leading 1’s, then their difference m(x) − ˆm(x) has a smaller degree 

than either and still sends the map or matrix to zero. Thus m(x) − ˆm(x) is 

the zero polynomial and the two are equal. (The leading coefficient requirement 

also prevents a minimal polynomial from being the zero polynomial.) 

1.6 Example We can see that m(x) =x2 − 2x − 1 is minimal for the matrix 

of Example 1.2 by computing the powers of T up to the power n2 =4. 

T 2 � √ � 

1/2 − 3/2 

= √ T 

3/2 1/2 

3 � � 

0 −1 

= 

T 

1 0 

4 � √ � 

−1/2 − 3/2 

= √ 

3/2 −1/2 

Next, put c4T 4 + c3T 3 + c2T 2 + c1T + c0I equal to the zero matrix 

and use Gauss’ method. 

−(1/2)c4 + (1/2)c2 +( √ 3/2)c1 + c0 =0 

−( √ 3/2)c4 − c3 − ( √ 3/2)c2 − (1/2)c1 =0 

( √ 3/2)c4 + c3 +( √ 3/2)c2 + (1/2)c1 =0 

−(1/2)c4 + (1/2)c2 +( √ 3/2)c1 + c0 =0 

c4 − c2 − √ 3c1 − 2c0 =0 

c3 + √ 3c2 + 2c1 + √ 3c0 =0 

Setting c4, c3, andc2 to zero forces c1 and c0 to also come out as zero. To get 

a leading one, the most we can do is to set c4 and c3 to zero. Thus the minimal 

polynomial is quadratic.


Using the method of that example to find the minimal polynomial of a 3×3 

matrix would mean doing Gaussian reduction on a system with nine equations 

in ten unknowns. We shall develop an alternative. To begin, note that we can 

break a polynomial of a map or a matrix into its components. 

1.7 Lemma Suppose that the polynomial f(x) =cnx n + ···+ c1x + c0 factors 

as k(x − λ1) q1 ···(x − λℓ) qℓ . If t is a linear transformation then these two are 

equal maps. 

cnt n + ···+ c1t + c0 = k · (t − λ1) q1 ◦···◦(t − λℓ) qℓ 

Consequently, if T is a square matrix then f(T )andk·(T −λ1I) q1 ···(T −λℓI) qℓ 

are equal matrices. 

Proof. This argument is by induction on the degree of the polynomial. The 

cases where the polynomial is of degree 0 and 1 are clear. The full induction 

argument is Exercise 1.7 but the degree two case gives its sense. 

A quadratic polynomial factors into two linear terms f(x) =k(x − λ1) · (x − 

λ2) =k(x2 +(λ1 + λ2)x + λ1λ2) (the roots λ1 and λ2 might be equal). We can 

check that substituting t for x in the factored and unfactored versions gives the 

same map. 

� 

k · (t − λ1) ◦ (t − λ2) � (�v) = � k · (t − λ1) � (t(�v) − λ2�v) 

= k · � t(t(�v)) − t(λ2�v) − λ1t(�v) − λ1λ2�v � 

= k · � t ◦ t (�v) − (λ1 + λ2)t(�v)+λ1λ2�v � 

= k · (t 2 − (λ1 + λ2)t + λ1λ2)(�v) 

The third equality holds because the scalar λ2 comes out of the second term, as 

t is linear. QED 

In particular, if a minimial polynomial m(x) for a transformation t factors 

as m(x) =(x − λ1) q1 ···(x − λℓ) qℓ then m(t) =(t − λ1) q1 ◦···◦(t − λℓ) qℓ is 

the zero map. Since m(t) sends every vector to zero, at least one of the maps 

t − λi sends some nonzero vectors to zero. So, too, in the matrix case — if m is 

minimal for T then m(T )=(T − λ1I) q1 ···(T − λℓI) qℓ is the zero matrix and at 

least one of the matrices T −λiI sends some nonzero vectors to zero. Rewording 

both cases: at least some of the λi are eigenvalues. (See Exercise 29.) 

Recall how we have earlier found eigenvalues. We have looked for λ such that 

T�v = λ�v by considering the equation �0 =T�v−x�v =(T −xI)�v and computing the 

determinant of the matrix T − xI. That determinant is a polynomial in x, the 

characteristic polynomial, whose roots are the eigenvalues. The major result 

of this subsection, the next result, is that there is a connection between this 

characteristic polynomial and the minimal polynomial. This results expands 

on the prior paragraph’s insight that some roots of the minimal polynomial 

are eigenvalues by asserting that every root of the minimal polynomial is an 

eigenvalue and further that every eigenvalue is a root of the minimal polynomial 

(this is because it says ‘1 ≤ qi’ and not just ‘0 ≤ qi’).


1.8 Theorem (Cayley-Hamilton) If the characteristic polynomial of a 

transformation or square matrix factors into 

k · (x − λ1) p1 (x − λ2) p2 ···(x − λℓ) pℓ 

then its minimal polynomial factors into 

(x − λ1) q1 (x − λ2) q2 ···(x − λℓ) qℓ 

where 1 ≤ qi ≤ pi for each i between 1 and ℓ. 

The proof takes up the next three lemmas. Although they are stated only in 

matrix terms, they apply equally well to maps. We give the matrix version only 

because it is convenient for the first proof. 

The first result is the key — some authors call it the Cayley-Hamilton Theorem 

and call Theorem 1.8 above a corollary. For the proof, observe that a matrix 

of polynomials can be thought of as a polynomial with matrix coefficients. 

� 

2 2 2x +3x− 1 x +2 

3x2 +4x +1 4x2 � 

= 

+ x +1 

� � 

2 1 

x 

3 4 

2 + 

� � 

3 0 

x + 

4 1 

� � 

−1 2 

1 1 

1.9 Lemma If T is a square matrix with characteristic polynomial c(x) then 

c(T ) is the zero matrix. 

Proof. Let C be T − xI, the matrix whose determinant is the characteristic 

polynomial c(x) =cnx n + ···+ c1x + c0. 

⎛ 

t1,1 − x t1,2 ... 

⎜ t2,1 t2,2 ⎜ 

− x 

C = ⎜ . 

⎝ 

. 

. 

.. 

⎞ 

⎟ 

⎠ 

tn,n − x 

Recall that the product of the adjoint of a matrix with the matrix itself is the 

determinant of that matrix times the identity. 

c(x) · I = adj(C)C = adj(C)(T − xI) =adj(C)T − adj(C) · x (∗) 

The entries of adj(C) are polynomials, each of degree at most n − 1 since the 

minors of a matrix drop a row and column. Rewrite it, as suggested above, as 

adj(C) =Cn−1x n−1 + ···+ C1x + C0 where each Ci is a matrix of scalars. The 

left and right ends of equation (∗) above give this. 

cnIx n + cn−1Ix n−1 + ···+ c1Ix + c0I =(Cn−1T )x n−1 + ···+(C1T )x + C0T 

− Cn−1x n − Cn−2x n−1 −···−C0x


Equate the coefficients of x n , the coefficients of x n−1 ,etc. 

cnI = −Cn−1 

cn−1I = −Cn−2 + Cn−1T 

. 

c1I = −C0 + C1T 

c0I = C0T 

Multiply (from the right) both sides of the first equation by T n , both sides 

of the second equation by T n−1 , etc. Add. The result on the left is cnT n + 

cn−1T n−1 + ···+ c0I, and the result on the right is the zero matrix. QED 

We sometimes refer to that lemma by saying that a matrix or map satisfies 

its characteristic polynomial. 

1.10 Lemma Where f(x) is a polynomial, if f(T ) is the zero matrix then f(x) 

is divisible by the minimal polynomial of T . That is, any polynomial satisfied 

by T is divisable by T ’s minimal polynomial. 

Proof. Let m(x) be minimal for T . The Division Theorem for Polynomials 

gives f(x) =q(x)m(x) +r(x) where the degree of r is strictly less than the 

degree of m. Plugging T in shows that r(T ) is the zero matrix, because T 

satisfies both f and m. That contradicts the minimality of m unless r is the 

zero polynomial. QED 

Combining the prior two lemmas gives that the minimal polynomial divides 

the characteristic polynomial. Thus, any root of the minimal polynomial is 

also a root of the characteristic polynomial. That is, so far we have that if 

m(x) =(x − λ1) q1 ...(x − λi) qi then c(x) must has the form (x − λ1) p1 ...(x − 

λi) pi (x − λi+1) pi+1 ...(x − λℓ) pℓ where each qj is less than or equal to pj. The 

proof of the Cayley-Hamilton Theorem is finished by showing that in fact the 

characteristic polynomial has no extra roots λi+1, etc. 

1.11 Lemma Each linear factor of the characteristic polynomial of a square 

matrix is also a linear factor of the minimal polynomial. 

Proof. Let T be a square matrix with minimal polynomial m(x) and assume 

that x − λ is a factor of the characteristic polynomial of T , that is, assume that 

λ is an eigenvalue of T . We must show that x − λ is a factor of m, that is, that 

m(λ) =0. 

In general, where λ is associated with the eigenvector �v, for any polynomial 

function f(x), application of the matrix f(T )to�v equals the result of 

multiplying �v by the scalar f(λ). (For instance, if T has eigenvalue λ associated 

with the eigenvector �v and f(x) =x 2 +2x + 3 then (T 2 +2T +3)(�v) = 

T 2 (�v)+2T (�v)+3�v = λ 2 · �v +2λ · �v +3· �v =(λ 2 +2λ +3)· �v.) Now, as m(T )is 

the zero matrix, �0 =m(T )(�v) =m(λ) · �v and therefore m(λ) =0. QED


1.12 Example We can use the Cayley-Hamilton Theorem to help find the 

minimal polynomial of this matrix. 

⎛ 

2 

⎜ 

T = ⎜1 

⎝0 

0 

2 

0 

0 

0 

2 

⎞ 

1 

2 ⎟ 

−1⎠ 

0 0 0 1 

First, its characteristic polynomial c(x) =(x − 1)(x − 2) 3 can be found with the 

usual determinant. Now, the Cayley-Hamilton Theorem says that T ’s minimal 

polynomial is either (x − 1)(x − 2) or (x − 1)(x − 2) 2 or (x − 1)(x − 2) 3 .Wecan 

decide among the choices just by computing: 

⎛ 

1 

⎜ 

(T − 1I)(T − 2I) = ⎜1 

⎝0 

0 

1 

0 

0 

0 

1 

⎞ ⎛ 

1 0 

2 ⎟ ⎜ 

⎟ ⎜1 

−1⎠ 

⎝0 

0 

0 

0 

0 

0 

0 

⎞ 

1 

2 ⎟ 

−1⎠ 

0 0 0 0 0 0 0 −1 

= 

⎛ 

0 

⎜ 

⎜1 

⎝0 

0 

0 

0 

0 

0 

0 

⎞ 

0 

1 ⎟ 

0⎠ 

0 0 0 0 

and 

(T − 1I)(T − 2I) 2 ⎛ 

0 

⎜ 

= ⎜1 

⎝0 

0 

0 

0 

0 

0 

0 

⎞ ⎛ 

0 0 

1⎟⎜ 

⎟ ⎜1 

0⎠⎝0 

0 

0 

0 

0 

0 

0 

⎞ 

1 

2 ⎟ 

−1⎠ 

0 0 0 0 0 0 0 −1 

= 

⎛ 

0 

⎜ 

⎜0 

⎝0 

0 

0 

0 

0 

0 

0 

⎞ 

0 

0 ⎟ 

0⎠ 

0 0 0 0 

and so m(x) =(x − 1)(x − 2) 2 . 

Exercises 

� 1.13 What are the possible minimal polynomials if a matrix has the given characteristic 

polynomial? 

(a) 8 · (x − 3) 4 

(d) 5 · (x +3) 2 (x − 1)(x − 2) 2 

What is the degree of each possibility? 

(b) (1/3) · (x +1) 3 (x − 4) (c) −1 · (x − 2) 2 (x − 5) 2 

� 1.14 Find the minimal polynomial of each matrix. 

� � � � � 

3 0 0 

3 0 0 

3 0 

� 

0 

(a) 1 3 0 (b) 1 3 0 (c) 1 3 0 

(e) 

0 

� 

2 

0 

0 

0 

2 

6 

0 

4 

� 

1 

2 

2 

0 0 3 

⎛ 

−1 4 

⎜ 0 3 

⎜ 

(f) ⎜ 0 −4 

⎝ 3 −9 

0 

0 

−1 

−4 

0 

0 

0 

2 

0 1 

⎞ 

0 

0 ⎟ 

0 ⎟ 

−1⎠ 

3 

1 5 4 1 4 

1.15 Find the minimal polynomial of this matrix. 

� � 

0 1 0 

0 0 1 

1 0 0 

(d) 

� 

2 0 

� 

1 

0 6 2 

0 0 2 

� 1.16 What is the minimal polynomial of the differentiation operator d/dx on Pn?


� 1.17 Find the minimal polynomial of matrices of this form 

⎛ 

⎞ 

λ 0 0 ... 0 

⎜1 

⎜ 

⎜0 

⎜ 

⎝ 

λ 

1 

0 

λ 

. .. 

λ 

0 ⎟ 

0⎠ 

0 0 ... 1 λ 

where the scalar λ is fixed (i.e., is not a variable). 

1.18 What is the minimal polynomial of the transformation of Pn that sends p(x) 

to p(x +1)? 

1.19 What is the minimal polynomial of the map π : C 3 → C 3 projecting onto the 

first two coordinates? 

1.20 Find a 3×3 matrix whose minimal polynomial is x 2 . 

1.21 What is wrong with this claimed proof of Lemma 1.9: “ifc(x) =|T −xI| then 

c(T )=|T − TI| = 0”? 

1.22 Verify Lemma 1.9 for 2×2 matrices by direct calculation. 

� 1.23 Prove that the minimal polynomial of an n × n matrix has degree at most 

n (not n 2 as might be guessed from this subsection’s opening). Verify that this 

maximum, n, can happen. 

� 1.24 The only eigenvalue of a nilpotent map is zero. Show that the converse statement 

holds. 

1.25 What is the minimal polynomial of a zero map or matrix? Of an identity map 

or matrix? 

� 1.26 Interpret the minimal polynomial of Example 1.2 geometrically. 

1.27 What is the minimal polynomial of a diagonal matrix? 

� 1.28 A projection is any transformation t such that t 2 = t. (For instance, the 

transformation of the plane R 2 projecting each vector onto its first coordinate will, 

if done twice, result in the same value as if it is done just once.) What is the 

minimal polynomial of a projection? 

1.29 The first two items of this question are review. 

(a) Prove that the composition of one-to-one maps is one-to-one. 

(b) Prove that if a linear map is not one-to-one then at least one nonzero vector 

from the domain is sent to the zero vector in the codomain. 

(c) Verify the statement, excerpted here, that preceeds Theorem 1.8. 

... if a minimial polynomial m(x) for a transformation t factors as 

m(x) =(x − λ1) q1 ···(x − λℓ) q ℓ then m(t) =(t − λ1) q1 ◦···◦(t − λℓ) q ℓ 

is the zero map. Since m(t) sends every vector to zero, at least one 

of the maps t − λi sends some nonzero vectors to zero. ... Rewording 

... : at least some of the λi are eigenvalues. 

1.30 True or false: for a transformation on an n dimensional space, if the minimal 

polynomial has degree n then the map is diagonalizable. 

1.31 Let f(x) be a polynomial. Prove that if A and B are similar matrices then 

f(A) is similar to f(B). 

(a) Now show that similar matrices have the same characteristic polynomial. 

(b) Show that similar matrices have the same minimal polynomial.


(c) Decide if these are similar. 

�1 � 

3 

� 

4 

� 

−1 

2 3 1 1 

1.32 (a) Show that a matrix is invertible if and only if the constant term in its 

minimal polynomial is not 0. 

(b) Show that if a square matrix T is not invertible then there is a nonzero 

matrix S such that ST and TS both equal the zero matrix. 

� 1.33 (a) Finish the proof of Lemma 1.7. 

(b) Give an example to show that the result does not hold if t is not linear. 

5.IV.2 Jordan Canonical Form 

This subsection moves from the canonical form for nilpotent matrices to the 

one for all matrices. 

We have shown that if a map is nilpotent then all of its eigenvalues are zero. 

We can now prove the converse. 

2.1 Lemma A linear transformation whose only eigenvalue is zero is nilpotent. 

Proof. If a transformation t on an n-dimensional space has only the single 

eigenvalue of zero then its characteristic polynomial is x n . The Cayley-Hamilton 

Theorem says that a map satisfies its characteristic polynimial so t n is the zero 

map. Thus t is nilpotent. QED 

We have a canonical form for nilpotent matrices, that is, for each matrix 

whose single eigenvalue is zero: each such matrix is similar to one that is all 

zeroes except for blocks of subdiagonal ones. (To make this representation 

unique we can fix some arrangement of the blocks, say, from longest to shortest.) 

We next extend this to all single-eigenvalue matrices. 

Observe that if t’s only eigenvalue is λ then t − λ’s only eigenvalue is 0 

because t(�v) =λ�v if and only if (t − λ)(�v) =0· �v. The natural way to extend 

the results for nilpotent matrices is to represent t − λ in the canonical form N, 

and try to use that to get a simple representation T for t. The next result says 

that this try works. 

2.2 Lemma If the matrices T − λI and N are similar then T and N + λI are 

also similar, via the same change of basis matrices. 

Proof. With N = P (T − λI)P −1 = PTP −1 − P (λI)P −1 we have N = 

PTP −1 − PP −1 (λI) since the diagonal matrix λI commutes with anything, 

and so N = PTP −1 − λI. Therefore N + λI = PTP −1 , as required. QED 

2.3 Example The characteristic polynomial of 

� � 

2 −1 

T = 

1 4


is (x − 3) 2 and so T has only the single eigenvalue 3. Thus for 

� � 

−1 −1 

T − 3I = 

1 1 

the only eigenvalue is 0, and T − 3I is nilpotent. The null spaces are routine 

to find; to ease this computation we take T to represent the transformation 

t: C 2 → C 2 with respect to the standard basis (we shall maintain this convention 

for the rest of the chapter). 

� � 

−y �� 

2 2 

N (t − 3) = { y ∈ C} N ((t − 3) )=C 

y 

The dimensions of these null spaces show that the action of an associated map 

t − 3 on a string basis is � β1 ↦→ � β2 ↦→ �0. Thus, the canonical form for t − 3 with 

one choice for a string basis is 

Rep B,B(t − 3) = N = 

� � 

0 0 

1 0 

and by Lemma 2.2, T is similar to this matrix. 

Rep t(B,B) =N +3I = 

� � � � 

1 −2 

B = 〈 , 〉 

1 2 

� � 

3 0 

1 3 

We can produce the similarity computation. Recall from the Nilpotence 

section how to find the change of basis matrices P and P −1 to express N as 

P (T − 3I)P −1 . The similarity diagram 

C2 w.r.t. E2 

⏐ 

id 

t−3 

−−−−→ 

T −3I 

C2 w.r.t. E2 

⏐ 

�P id�P 

C 2 w.r.t. B 

t−3 

−−−−→ 

N 

C 2 w.r.t. B 

describes that to move from the lower left to the upper left we multiply by 

P −1 = � RepE2,B(id) � � � 

−1 1 −2 

=RepB,E2 (id) = 

1 2 

and to move from the upper right to the lower right we multiply by this matrix. 

� 

1 

P = 

1 

�−1 � 

−2 1/2 

= 

2 −1/4 

� 

1/2 

1/4 

So the similarity is expressed by 

� 

3 

1 

� � 

0 1/2 

= 

3 −1/4 

�� 

1/2 2 

1/4 1 

�� 

−1 1 

4 1 

� 

−2 

2 

which is easily checked.


2.4 Example This matrix has characteristic polynomial (x − 4) 4 

⎛ 

4 

⎜ 

T = ⎜0 

⎝0 

1 

3 

0 

0 

0 

4 

⎞ 

−1 

1 ⎟ 

0 ⎠ 

1 0 0 5 

and so has the single eigenvalue 4. The nullities of t − 4 are: the null space of 

t − 4 has dimension two, the null space of (t − 4) 2 has dimension three, and the 

null space of (t − 4) 3 has dimension four. Thus, t − 4 has the action on a string 

basis of � β1 ↦→ � β2 ↦→ � β3 ↦→ �0 and � β4 ↦→ �0. This gives the canonical form N for 

t − 4, which in turn gives the form for t. 

⎛ 

4 

⎜ 

N +4I = ⎜1 

⎝0 

0 

4 

1 

0 

0 

4 

⎞ 

0 

0 ⎟ 

0⎠ 

0 0 0 4 

An array that is all zeroes, except for some number λ down the diagonal 

and blocks of subdiagonal ones, is a Jordan block. We have shown that Jordan 

block matrices are canonical representatives of the similarity classes of singleeigenvalue 

matrices. 

2.5 Example The 3×3 matrices whose only eigenvalue is 1/2 separate into 

three similarity classes. The three classes have these canonical representatives. 

⎛ 

1/2 0 0 

⎞ ⎛ 

1/2 0 0 

⎞ ⎛ 

1/2 0 0 

⎞ 

⎝ 0 1/2 0 ⎠ ⎝ 1 1/2 0 ⎠ ⎝ 1 1/2 0 ⎠ 

0 0 1/2 0 0 1/2 0 1 1/2 

In particular, this matrix 

⎛ 

1/2 0 0 

⎞ 

⎝ 0 1/2 0 ⎠ 

0 1 1/2 

belongs to the similarity class represented by the middle one, because we have 

adopted the convention of ordering the blocks of subdiagonal ones from the 

longest block to the shortest. 

We will now finish the program of this chapter by extending this work to 

cover maps and matrices with multiple eigenvalues. The best possibility for 

general maps and matrices would be if we could break them into a part involving 

their first eigenvalue λ1 (which we represent using its Jordan block), a part with 

λ2, etc. 

This ideal is in fact what happens. For any transformation t: V → V ,we 

shall break the space V into the direct sum of a part on which t−λ1 is nilpotent, 

plus a part on which t − λ2 is nilpotent, etc. More precisely, we shall take three


steps to get to this section’s major theorem and the third step shows that 

V = N∞(t − λ1) ⊕···⊕N∞(t − λℓ) where λ1,... ,λℓ are t’s eigenvalues. 

Suppose that t: V → V is a linear transformation. Note that the restriction ∗ 

of t to a subspace M need not be a linear transformation on M because there may 

be an �m ∈ M with t(�m) �∈ M. To ensure that the restriction of a transformation 

to a ‘part’ of a space is a transformation on the partwe need the next condition. 

2.6 Definition Let t: V → V be a transformation. A subspace M is t invariant 

if whenever �m ∈ M then t(�m) ∈ M (shorter: t(M) ⊆ M). 

Two examples are that the generalized null space N∞(t) and the generalized 

range space R∞(t) of any transformation t are invariant. For the generalized null 

space, if �v ∈ N∞(t) then t n (�v) =�0 where n is the dimension of the underlying 

space and so t(�v) ∈ N∞(t) because t n ( t(�v) ) is zero also. For the generalized 

range space, if �v ∈ R∞(t) then �v = t n ( �w) for some �w and then t(�v) =t n+1 ( �w) = 

t n ( t( �w) ) shows that t(�v) is also a member of R∞(t). 

Thus the spaces N∞(t − λi) andR∞(t − λi) aret − λi invariant. Observe 

also that t − λi is nilpotent on N∞(t − λi) because, simply, if �v has the property 

that some power of t − λi maps it to zero — that is, if it is in the generalized 

null space — then some power of t − λi maps it to zero. The generalized null 

space N∞(t − λi) is a ‘part’ of the space on which the action of t − λi is easy 

to understand. 

The next result is the first of our three steps. It establishes that t−λj leaves 

t − λi’s part unchanged. 

2.7 Lemma A subspace is t invariant if and only if it is t − λ invariant for any 

scalar λ. In particular, where λi is an eigenvalue of a linear transformation t, 

then for any other eigenvalue λj, the spaces N∞(t − λi) andR∞(t − λi) are 

t − λj invariant. 

Proof. For the first sentence we check the two implications of the ‘if and only 

if’ separately. One of them is easy: if the subspace is t − λ invariant for any λ 

then taking λ =0showsthatitist invariant. For the other implication suppose 

that the subspace is t invariant, so that if �m ∈ M then t(�m) ∈ M, and let λ 

be any scalar. The subspace M is closed under linear combinations and so if 

t(�m) ∈ M then t(�m) − λ�m ∈ M. Thus if �m ∈ M then (t − λ)(�m) ∈ M, as 

required. 

The second sentence follows straight from the first. Because the two spaces 

are t − λi invariant, they are therefore t invariant. From this, applying the first 

sentence again, we conclude that they are also t − λj invariant. QED 

The second step of the three that we will take to prove this section’s major 

result makes use of an additional property of N∞(t − λi) andR∞(t − λi), that 

they are complementary. Recall that if a space is the direct sum of two others 

V = N ⊕ R then any vector �v in the space breaks into two parts �v = �n + �r 

where �n ∈ N and �r ∈ R, and recall also that if BN and BR are bases for N 

∗ More information on restrictions of functions is in the appendix.


⌢ BR is linearly independent (and so the two 

and R then the concatenation BN 

parts of �v do not “overlap”). The next result says that for any subspaces N 

and R that are complementary as well as t invariant, the action of t on �v breaks 

into the “non-overlapping” actions of t on �n and on �r. 

2.8 Lemma Let t: V → V be a transformation and let N and R be t invariant 

complementary subspaces of V . Then t can be represented by a matrix with 

blocks of square submatrices T1 and T2 

� 

T1 Z2 

Z1 T2 

where Z1 and Z2 are blocks of zeroes. 

� }dim(N )-many rows 

}dim(R)-many rows 

Proof. Since the two subspaces are complementary, the concatenation of a basis 

for N and a basis for R makes a basis B = 〈�ν1,...,�νp,�µ1,... ,�µq〉 for V .We 

shall show that the matrix 

⎛ 

⎞ 

. 

. 

⎜ . 

. ⎟ 


⎝ 

RepB(t(�ν1)) ··· RepB(t(�µq)) ⎟ 

⎠ 

. 

. 

. 

. 

has the desired form. 

Any vector �v ∈ V is in N if and only if its final q components are zeroes 

when it is represented with respect to B. As N is t invariant, each of the 

vectors Rep B(t(�ν1)), ... ,Rep B(t(�νp)) has that form. Hence the lower left of 

Rep B,B(t) is all zeroes. 

The argument for the upper right is similar. QED 

To see that t has been decomposed into its action on the parts, observe 

that the restrictions of t to the subspaces N and R are represented, with 

respect to the obvious bases, by the matrices T1 and T2. So, with subspaces 

that are invariant and complementary, we can split the problem of examining a 

linear transformation into two lower-dimensional subproblems. The next result 

illustrates this decomposition into blocks. 

2.9 Lemma If T is a matrices with square submatrices T1 and T2 

� � 

T1 Z2 

T = 

Z1 T2 

where the Z’s are blocks of zeroes, then |T | = |T1|·|T2|. 

Proof. Suppose that T is n×n, that T1 is p×p, and that T2 is q ×q. In the 

permutation formula for the determinant 

|T | = 

� 

t1,φ(1)t2,φ(2) ···tn,φ(n) sgn(φ) 

permutations φ


each term comes from a rearrangement of the column numbers 1,...,n into a 

new order φ(1),...,φ(n). The upper right block Z2 is all zeroes, so if a φ has at 

least one of p +1,...,n among its first p column numbers φ(1),...,φ(p) then 

the term arising from φ is zero, e.g., if φ(1) = n then t 1,φ(1)t 2,φ(2) ...t n,φ(n) = 

0 · t 2,φ(2) ...t n,φ(n) =0. 

So the above formula reduces to a sum over all permutations with two 

halves: any significant φ is the composition of a φ1 that rearranges only 1,...,p 

and a φ2 that rearranges only p +1,...,p+ q. Now, the distributive law (and 

the fact that the signum of a composition is the product of the signums) gives 

that this 

� 

� 

|T1|·|T2| = 

perms φ1 

of 1,...,p 

� 

t1,φ1(1) ···tp,φ1(p) sgn(φ1) 

� 

� 

· 

perms φ2 

of p+1,...,p+q 

� 

tp+1,φ2(p+1) ···tp+q,φ2(p+q) sgn(φ2) 

equals |T | = � 

significant φ t 1,φ(1)t 2,φ(2) ···t n,φ(n) sgn(φ). QED 

2.10 Example 

� 

� 

� 

�2 

0 0 0� 

� � � 

� 

�1 

2 0 0� 

� 

� 

� 

�0 

0 3 0�= 

�2 

0� 

� 

� 

� 1 2� 

�0 

0 0 3� 

· 

� � 

� 

�3 

0� 

� 

�0 3� 

=36 

From Lemma 2.9 we conclude that if two subspaces are complementary and 

t invariant then t is nonsingular if and only if its restrictions to both subspaces 

are nonsingular. 

Now for the promised third, final, step to the main result. 

2.11 Lemma If a linear transformation t: V → V has the characteristic polynomial 

(x − λ1) p1 ...(x − λℓ) pℓ then (1) V = N∞(t − λ1) ⊕···⊕N∞(t − λℓ) 

and (2) dim(N∞(t − λi)) = pi. 

Proof. Because dim(V ) is the degree p1 + ···+ pℓ of the characteristic polynomial, 

to establish statement (1) we need only show that statement (2) holds 

and that N∞(t − λi) ∩ N∞(t − λj) is trivial whenever i �= j. 

For the latter, by Lemma 2.7, both N∞(t−λi)andN∞(t−λj)aret invariant. 

Notice that an intersection of t invariant subspaces is t invariant and so the 

restriction of t to N∞(t − λi) ∩ N∞(t − λj) is a linear transformation. But both 

t − λi and t − λj are nilpotent on this subspace and so if t has any eigenvalues 

on the intersection then its “only” eigenvalue is both λi and λj. That cannot 

be, so this restriction has no eigenvalues: N∞(t − λi) ∩ N∞(t − λj) is trivial 

(Lemma 3.10 shows that the only transformation without any eigenvalues is on 

the trivial space).


To prove statement (2), fix the index i. 

R∞(t − λi) and apply Lemma 2.8. 

Decompose V as N∞(t − λi) ⊕ 

� 

T1 

T = 

� 

Z2 }dim( N∞(t − λi) )-many rows 

}dim( R∞(t − λi) )-many rows 

Z1 T2 

By Lemma 2.9, |T − xI| = |T1 − xI|·|T2 − xI|. By the uniqueness clause of the 

Fundamental Theorem of Arithmetic, the determinants of the blocks have the 

same factors as the characteristic polynomial |T1 −xI| =(x−λ1) q1 ...(x−λℓ) qℓ 

and |T2 − xI| =(x − λ1) r1 ...(x − λℓ) rℓ , and the sum of the powers of these 

factors is the power of the factor in the characteristic polynomial: q1 + r1 = p1, 

... , qℓ + rℓ = pℓ. Statement (2) will be proved if we will show that qi = pi and 

that qj = 0 for all j �= i, because then the degree of the polynomial |T1 − xI| — 

which equals the dimension of the generalized null space — is as required. 

For that, first, as the restriction of t − λi to N∞(t − λi) is nilpotent on that 

space, the only eigenvalue of t on it is λi. Thus the characteristic equation of t 

on N∞(t − λi) is|T1 − xI| =(x − λi) qi . And thus qj = 0 for all j �= i. 

Now consider the restriction of t to R∞(t − λi). By Note II.2.2, the map 

t − λi is nonsingular on R∞(t − λi) andsoλi is not an eigenvalue of t on that 

subspace. Therefore, x − λi is not a factor of |T2 − xI|, andsoqi = pi. QED 

Our major result just translates those steps into matrix terms. 

2.12 Theorem Any square matrix is similar to one in Jordan form 

⎛ 

Jλ1 

⎜ 

⎝ 

Jλ2 

–zeroes– 

. .. 

⎞ 

⎟ 

⎠ 

Jλℓ−1 

–zeroes– Jλℓ 

where each Jλ is the Jordan block associated with the eigenvalue λ of the original 

matrix (that is, is all zeroes except for λ’s down the diagonal and some 

subdiagonal ones). 

Proof. Given an n×n matrix T , consider the linear map t: Cn → Cn that it 

represents with respect to the standard bases. Use the prior lemma to write 

Cn = N∞(t − λ1) ⊕···⊕N∞(t − λℓ) where λ1,... ,λℓ are the eigenvalues of t. 

Because each N∞(t − λi) ist invariant, Lemma 2.8 and the prior lemma show 

that t is represented by a matrix that is all zeroes except for square blocks along 

the diagonal. To make those blocks into Jordan blocks, pick each Bλi to be a 

string basis for the action of t − λi on N∞(t − λi). QED 

Jordan form is a canonical form for similarity classes of square matrices, 

provided that we make it unique by arranging the Jordan blocks from least 

eigenvalue to greatest and then arranging the subdiagonal 1 blocks inside each 

Jordan block from longest to shortest.


2.13 Example This matrix has the characteristic polynomial (x − 2) 2 (x − 6). 

⎛ 

2 

T = ⎝0 0 

6 

⎞ 

1 

2⎠ 

0 0 2 

We will handle the eigenvalues 2 and 6 separately. 

Computation of the powers, and the null spaces and nullities, of T − 2I is 

routine. (Recall from Example 2.3 the convention of taking T to represent a 

transformation, here t: C 3 → C 3 , with respect to the standard basis.) 

power p (T − 2I) p N ((t − 2) p 1 

⎛ ⎞ 

0 0 1 

⎜ ⎟ 

⎝0 

4 2⎠ 

) 

⎛ ⎞ 

x 

⎜ ⎟ 

{ ⎝0⎠ 

nullity 

0 0 0 

0 

� � x ∈ C} 1 

2 

⎛ 

0 

⎜ 

⎝0 

0 

16 

⎞ 

0 

⎟ 

8⎠ 

⎛ ⎞ 

x 

⎜ ⎟ 

{ ⎝−z/2⎠ 

0 0 0 z 

� � x, z ∈ C} 2 

3 

⎛ 

0 

⎜ 

⎝0 

0 

64 

⎞ 

0 

⎟ 

32⎠ 

–same– — 

0 0 0 

So the generalized null space N∞(t − 2) has dimension two. We’ve noted that 

the restriction of t − 2 is nilpotent on this subspace. From the way that the 

nullities grow we know that the action of t − 2 on a string basis � β1 ↦→ � β2 ↦→ �0. 

Thus the restriction can be represented in the canonical form 

⎛ ⎞ ⎛ ⎞ 

� � 

1 −2 

0 0 

N2 = =Rep 

1 0 

B,B(t − 2) B2 = 〈 ⎝ 1 ⎠ , ⎝ 0 ⎠〉 

−2 0 

where many choices of basis are possible. Consequently, the action of the restriction 

of t to N∞(t − 2) is represented by this matrix. 

� � 

2 0 

J2 = N2 +2I =RepB2,B2 (t) = 

1 2 

The second eigenvalue’s computations are easier. Because the power of x−6 

in the characteristic polynomial is one, the restriction of t−6 toN∞(t−6) must 

be nilpotent of index one. Its action on a string basis must be � β3 ↦→ �0 and since 

it is the zero map, its canonical form N6 is the 1×1 zero matrix. Consequently, 

the canonical form J6 for the action of t on N∞(t−6) is the 1×1 matrix with the 

single entry 6. For the basis we can use any nonzero vector from the generalized 

null space. 

⎛ 

B6 = 〈 ⎝ 0 

⎞ 

1⎠〉 

0


Taken together, these two give that the Jordan form of T is 

⎛ 

2 

RepB,B(t) = ⎝1 0 

0 

2 

0 

⎞ 

0 

0⎠ 

6 

where B is the concatenation of B2 and B6. 

2.14 Example Contrast the prior example with 

⎛ 

2 

T = ⎝0 2 

6 

⎞ 

1 

2⎠ 

0 0 2 

which has the same characteristic polynomial (x − 2) 2 (x − 6). 

While the characteristic polynomial is the same, 

power p (T − 2I) p N ((t − 2) p 1 

⎛ ⎞ 

0 2 1 

⎜ ⎟ 

⎝0 

4 2⎠ 

) 

⎛ ⎞ 

x 

⎜ ⎟ 

{ ⎝−z/2⎠ 

nullity 

0 0 0 z 

� � x, z ∈ C} 2 

2 

⎛ 

0 

⎜ 

⎝0 

8 

16 

⎞ 

4 

⎟ 

8⎠ 


0 0 0 

here the action of t−2 is stable after only one application — the restriction of of 

t−2 toN∞(t−2) is nilpotent of index only one. (So the contrast with the prior 

example is that while the characteristic polynomial tells us to look at the action 

of the t − 2 on its generalized null space, the characteristic polynomial does not 

describe completely its action and we must do some computations to find, in 

this example, that the minimal polynomial is (x − 2)(x − 6).) The restriction of 

t − 2 to the generalized null space acts on a string basis as � β1 ↦→ �0 and� β2 ↦→ �0, 

and we get this Jordan block associated with the eigenvalue 2. 

� � 

2 0 

J2 = 

0 2 

For the other eigenvalue, the arguments for the second eigenvalue of the 

prior example apply again. The restriction of t − 6toN∞(t− 6) is nilpotent 

of index one (it can’t be of index less than one, and since x − 6 is a factor of 

the characteristic polynomial to the power one it can’t be of index more than 

one either). Thus t − 6’s canonical form N6 is the 1×1 zero matrix, and the 

associated Jordan block J6 is the 1×1 matrix with entry 6. 

Therefore, T is diagonalizable. 

⎛ ⎞ 

⎛ 

2 0 0 

⌢ 

RepB,B(t) = ⎝0 2 0⎠ 

B = B2 B6 = 〈 ⎝ 

0 0 6 

1 

⎞ ⎛ 

0⎠ 

, ⎝ 

0 

0 

⎞ ⎛ 

1 ⎠ , ⎝ 

−2 

3 

⎞ 

4⎠〉 

0 

(Checking that the third vector in B is in the nullspace of t − 6 is routine.)


2.15 Example A bit of computing with 

⎛ 

−1 

⎜ 0 

T = ⎜ 0 

⎝ 3 

4 

3 

−4 

−9 

0 

0 

−1 

−4 

0 

0 

0 

2 

⎞ 

0 

0 ⎟ 

0 ⎟ 

−1⎠ 

1 5 4 1 4 

shows that its characteristic polynomial is (x − 3) 3 (x +1) 2 . This table 

power p (T − 3I) p N ((t − 3) p 1 

⎛ 

−4 

⎜ 0 

⎜ 0 

⎜ 

⎝ 3 

4 0 0 

0 0 0 

−4 −4 0 

−9 −4 −1 

) nullity 

⎞ ⎛ ⎞ 

0 −(u + v)/2 

⎟ ⎜ ⎟ 

0 ⎟ ⎜−(u 

+ v)/2⎟ 

⎟ ⎜ ⎟ � 

0 ⎟ { ⎜ (u + v)/2 ⎟ � 

⎟ u, v ∈ C} 2 

⎟ ⎜ ⎟ 

−1⎠ 

⎝ u ⎠ 

2 

1 

⎛ 

16 

⎜ 0 

⎜ 0 

⎜ 

⎝−16 

5 4 

−16 

0 

16 

32 

1 

0 

0 

16 

16 

0 

0 

0 

0 

1 

⎞ 

0 

⎟ 

0 ⎟ 

0 ⎟ 

0⎠ 

v 

⎛ ⎞ 

−z 

⎜ ⎟ 

⎜−z⎟ 

⎜ ⎟ � 

{ ⎜ z ⎟ � 

⎟ z,u,v ∈ C} 

⎜ ⎟ 

⎝ u ⎠ 

3 

3 

0 

⎛ 

−64 

⎜ 0 

⎜ 0 

⎜ 

⎝ 64 

−16 

64 

0 

−64 

−128 

−16 

0 

0 

−64 

−64 

0 

0 

0 

0 

0 

0 

⎞ 

0 

⎟ 

0 ⎟ 

0 ⎟ 

0⎠ 

v 


0 64 64 0 0 

shows that the restriction of t − 3toN∞(t − 3) acts on a string basis via the 

two strings � β1 ↦→ � β2 ↦→ �0 and � β3 ↦→ �0. 

A similar calculation for the other eigenvalue 

power p (T +1I) p N ((t +1) p 1 

⎛ 

0 

⎜ 

⎜0 

⎜ 

⎜0 

⎜ 

⎝3 

4 0 0 

4 0 0 

−4 0 0 

−9 −4 3 

⎞ 

0 

⎟ 

0 ⎟ 

0 ⎟ 

−1⎠ 

) 

⎛ ⎞ 

−(u + v) 

⎜ ⎟ 

⎜ 0 ⎟ 

⎜ ⎟ � 

{ ⎜ −v ⎟ � 

⎟ u, v ∈ C} 

⎜ ⎟ 

⎝ u ⎠ 

nullity 

2 

2 

1 

⎛ 

0 

⎜ 

⎜0 

⎜ 

⎜0 

⎜ 

⎝8 

5 

16 

16 

−16 

−40 

4 1 

0 0 

0 0 

0 0 

−16 8 

5 

⎞ 

0 

⎟ 

0 ⎟ 

0 ⎟ 

−8⎠ 

v 


8 24 16 8 24


shows that the restriction of t + 1 to its generalized null space acts on a string 

basis via the two separate strings � β4 ↦→ �0 and� β5 ↦→ �0. 

Therefore T is similar to this Jordan form matrix. 

⎛ 

−1 

⎜ 0 

⎜ 0 

⎝ 0 

0 

−1 

0 

0 

0 

0 

3 

1 

0 

0 

0 

3 

⎞ 

0 

0 ⎟ 

0 ⎟ 

0⎠ 

0 0 0 0 3 

We close with the statement that the subjects considered earlier in this 

Chpater are indeed, in this sense, exhaustive. 

2.16 Corollary Every square matrix is similar to the sum of a diagonal matrix 

and a nilpotent matrix. 

Exercises 

2.17 Do the check for Example 2.3. 

2.18 Each matrix is in Jordan form. State its characteristic polynomial and its 

minimal polynomial. 

� � � � 

3 0 

−1 0 

(a) 

(b) 

1 3 

0 −1 

⎛ ⎞ ⎛ 

3 0 0 0 

4 

⎜1 

3 0 0⎟ 

⎜1 

(e) ⎝ 

0 0 3 0 

⎠ (f) ⎝ 

0 

0 0 1 3 

0 

⎛ ⎞ ⎛ 

5 0 0 0 

5 

� 

2 0 

(c) 1 2 

0 0 

⎞ 

0 0 0 

4 0 0 ⎟ 

0 −4 0 

⎠ 

0 1 −4 

⎞ 

0 0 0 

� 

0 

0 

−1/2 

� 

5 

(g) 0 

0 

� 

3 

(d) 1 

0 

� 

0 0 

2 0 

0 3 

0 

3 

1 

� 

0 

0 

3 

⎜0 

(h) ⎝ 

0 

2 

0 

0 

2 

0⎟ 

0 

⎠ 

⎜0 

(i) ⎝ 

0 

2 

1 

0 

2 

0⎟ 

0 

⎠ 

0 0 0 3 

0 0 0 3 

� 2.19 Find the Jordan form from the given data. 

(a) The matrix T is 5×5 with the single eigenvalue 3. The nullities of the powers 

are: T − 3I has nullity two, (T − 3I) 2 has nullity three, (T − 3I) 3 has nullity 

four, and (T − 3I) 4 has nullity five. 

(b) The matrix S is 5×5 with two eigenvalues. For the eigenvalue 2 the nullities 

are: S − 2I has nullity two, and (S − 2I) 2 has nullity four. For the eigenvalue 

−1 the nullities are: S +1I has nullity one. 

2.20 Find the change of basis matrices for each example. 

(a) Example 2.13 (b) Example 2.14 (c) Example 2.15 

� 2.21 Find � the Jordan � form and a Jordan basis for each matrix. 

−10 4 

(a) 

−25 10 

� � 

5 −4 

(b) 

9 −7 

� � 

4 0 0 

(c) 2 1 3 

5 0 4


� 

5 4 

� 

3 

(d) −1 0 −3 

1 

� 

9 

−2 

7 

1 

� 

3 

(e) −9 −7 −4 

4 

� 

2 

4 

2 

4 

� 

−1 

(f) −1 −1 1 

−1 

⎛ 

7 

−2 

1 

2 

2 

⎞ 

2 

⎜ 1 

(g) ⎝ 

−2 

4 

1 

−1 

5 

−1⎟ 

−1 

⎠ 

1 1 2 8 

� 2.22 Find all possible Jordan forms of a transformation with characteristic polynomial 

(x − 1) 2 (x +2) 2 . 

2.23 Find all possible Jordan forms of a transformation with characteristic polynomial 

(x − 1) 3 (x +2). 

� 2.24 Find all possible Jordan forms of a transformation with characteristic polynomial 

(x − 2) 3 (x + 1) and minimal polynomial (x − 2) 2 (x +1). 

2.25 Find all possible Jordan forms of a transformation with characteristic polynomial 

(x − 2) 4 (x + 1) and minimal polynomial (x − 2) 2 (x +1). 

� 2.26 Diagonalize � � these. � 

1 1 

0 

(a) 

(b) 

0 0 

1 

� 

1 

0 

� 2.27 Find the Jordan matrix representing the differentiation operator on P3. 

� 2.28 Decide if these two are similar. 

� � 

1 −1 

� 

−1 

� 

0 

4 −3 1 −1 

2.29 Find the Jordan form of this matrix. 

� � 

0 −1 

1 0 

Also give a Jordan basis. 

2.30 How many similarity classes are there for 3×3 matrices whose only eigenvalues 

are −3 and4? 

� 2.31 Prove that a matrix is diagonalizable if and only if its minimal polynomial 

has only linear factors. 

2.32 Give an example of a linear transformation on a vector space that has no 

non-trivial invariant subspaces. 

2.33 Show that a subspace is t − λ1 invariant if and only if it is t − λ2 invariant. 

2.34 Prove or disprove: two n×n matrices are similar if and only if they have the 

same characteristic and minimal polynomials. 

2.35 The trace of a square matrix is the sum of its diagonal entries. 

(a) Find the formula for the characteristic polynomial of a 2×2 matrix. 

(b) Show that trace is invariant under similarity, and so we can sensibly speak 

of the ‘trace of a map’. (Hint: see the prior item.) 

(c) Is trace invariant under matrix equivalence?


(d) Show that the trace of a map is the sum of its eigenvalues (counting multiplicities). 

(e) Show that the trace of a nilpotent map is zero. Does the converse hold? 

2.36 To use Definition 2.6 to check whether a subspace is t invariant, we seemingly 

have to check all of the infinitely many vectors in a (nontrivial) subspace to see if 

they satisfy the condition. Prove that a subspace is t invariant if and only if its 

subbasis has the property that for all of its elements, t( � β) is in the subspace. 

� 2.37 Is t invariance preserved under intersection? Under union? Complementation? 

Sums of subspaces? 

2.38 Give a way to order the Jordan blocks if some of the eigenvalues are complex 

numbers. That is, suggest a reasonable ordering for the complex numbers. 

2.39 Let Pj(R) be the vector space over the reals of degree j polynomials. Show 

that if j ≤ k then Pj(R) is an invariant subspace of Pk(R) under the differentiation 

operator. In P7(R), does any of P0(R), ... , P6(R) have an invariant complement? 

2.40 In Pn(R), the vector space (over the reals) of degree n polynomials, 

E = {p(x) ∈Pn(R) � � p(−x) =p(x) for all x} 

and 

O = {p(x) ∈Pn(R) � � p(−x) =−p(x) for all x} 

are the even and the odd polynomials; p(x) =x 2 is even while p(x) =x 3 is odd. 

Show that they are subspaces. Are they complementary? Are they invariant under 

the differentiation transformation? 

2.41 Lemma 2.8 says that if M and N are invariant complements then t has a 

representation in the given block form (with respect to the same ending as starting 

basis, of course). Does the implication reverse? 

2.42 AmatrixS is the square root of another T if S 2 = T . Show that any nonsingular 

matrix has a square root.

Topic: Computing Eigenvalues—the Method of Powers 399 

Topic: Computing Eigenvalues—the Method of 

Powers 

In practice, calculating eigenvalues and eigenvectors is a difficult problem. Finding, 

and solving, the characteristic polynomial of the large matrices often encountered 

in applications is too slow and too hard. Other techniques, indirect 

ones that avoid the characteristic polynomial, are used. Here we shall see such 

a method that is suitable for large matrices that are ‘sparse’ (the great majority 

of the entries are zero). 

Suppose that the n×n matrix T has the n distinct eigenvalues λ1, λ2, ... , λn. 

Then R n has a basis that is composed of the associated eigenvectors 〈 � ζ1,..., � ζn〉. 

For any �v ∈ R n , where �v = c1 � ζ1 + ···+ cn � ζn, iterating T on �v gives these. 

T�v = c1λ1 � ζ1 + c2λ2 � ζ2 + ···+ cnλn � ζn 

T 2 �v = c1λ 2 1 � ζ1 + c2λ 2 2 � ζ2 + ···+ cnλ 2 n � ζn 

T 3 �v = c1λ 3 1 � ζ1 + c2λ 3 2 � ζ2 + ···+ cnλ 3 n � ζn 

. 

T k �v = c1λ k 1 � ζ1 + c2λ k 2 � ζ2 + ···+ cnλ k n � ζn 

If one of the eigenvaluse, say, λ1, has a larger absolute value than any of the 

other eigenvalues then its term will dominate the above expression. Put another 

way, dividing through by λ k 1 gives this, 

T k �v 

λ k 1 

= c1 � ζ1 + c2 

λk 2 

λk 1 

�ζ2 + ···+ cn 

and, because λ1 is assumed to have the largest absolute value, as k gets larger 

the fractions go to zero. Thus, the entire expression goes to c1 � ζ1. 

That is (as long as c1 is not zero), as k increases, the vectors T k �v will 

tend toward the direction of the eigenvectors associated with the dominant 

eigenvalue, and, consequently, the ratios of the lengths � T k �v �/� T k−1 �v � will 

tend toward that dominant eigenvalue. 

For example (sample computer code for this follows the exercises), because 

the matrix 

T = 

� � 

3 0 

8 −1 

is triangular, its eigenvalues are just the entries on the diagonal, 3 and −1. 

Arbitrarily taking �v to have the components 1 and 1 gives 

λk n 

λk 1 

�v T�v T 2 �v ··· T 9 �v T 10 �v 

� � 

1 

� � 

3 

� � 

9 

1 7 17 

··· 

�ζn 

� � 

19 683 

� � 

59 049 

39 367 118 097 

and the ratio between the lengths of the last two is 2.999 9.


Two implementation issues must be addressed. The first issue is that, instead 

of finding the powers of T and applying them to �v, we will compute �v1 as T�v and 

then compute �v2 as T�v1, etc. (i.e., we never separately calculate T 2 , T 3 , etc.). 

These matrix-vector products can be done quickly even if T is large, provided 

that it is sparse. The second issue is that, to avoid generating numbers that are 

so large that they overflow our computer’s capability, we can normalize the �vi’s 

at each step. For instance, we can divide each �vi by its length (other possibilities 

are to divide it by its largest component, or simply by its first component). We 

thus implement this method by generating 

�w0 = �v0/��v0� 

�v1 = T�w0 

�w1 = �v1/��v1� 

�v2 = T�w2 

. 

�wk−1 = �vk−1/��vk−1� 

�vk = T�wk 

until we are satisfied. Then the vector �vk is an approximation of an eigenvector, 

and the approximation of the dominant eigenvalue is the ratio ��vk�/� �wk−1� = 

��vk�. 

One way we could be ‘satisfied’ is to iterate until our approximation of the 

eigenvalue settles down. We could decide, for instance, to stop the iteration 

process not after some fixed number of steps, but instead when ��vk� differs 

from ��vk−1� by less than one percent, or when they agree up to the second 

significant digit. 

The rate of convergence is determined by the rate at which the powers of 

�λ2/λ1� go to zero, where λ2 is the eigenvalue of second largest norm. If that 

ratio is much less than one then convergence is fast, but if it is only slightly 

less than one then convergence can be quite slow. Consequently, the method of 

powers is not the most commonly used way of finding eigenvalues (although it 

is the simplest one, which is why it is here as the illustration of the possibility of 

computing eigenvalues without solving the characteristic polynomial). Instead, 

there are a variety of methods that generally work by first replacing the given 

matrix T with another that is similar to it and so has the same eigenvalues, but 

is in some reduced form such as tridiagonal form: the only nonzero entries are 

on the diagonal, or just above or below it. Then special techniques can be used 

to find the eigenvalues. Once the eigenvalues are known, the eigenvectors of T 

can be easily computed. These other methods are outside of our scope. A good 

reference is [Goult, et al.] 

Exercises 

1 Use ten iterations to estimate the largest eigenvalue of these matrices, starting 

from the vector with components 1 and 2. Compare the answer with the one 

obtained by solving the characteristic equation.

Topic: Computing Eigenvalues—the Method of Powers 401 

(a) 

� � 

1 5 

0 4 

(b) 

� 

3 

� 

2 

−1 0 

2 Redo the prior exercise by iterating until ��vk� −��vk−1� has absolute value less 

than 0.01 At each step, normalize by dividing each vector by its length. How many 

iterations are required? Are the answers significantly different? 

3 Use ten iterations to estimate the largest eigenvalue of these matrices, starting 

from the vector with components 1, 2, and 3. Compare the answer with the one 

obtained by solving the characteristic equation. 

� � � � 

4 0 1 

−1 2 2 

(a) −2 1 0 (b) 2 2 2 

−2 0 1 

−3 −6 −6 

4 Redo the prior exercise by iterating until ��vk� −��vk−1� has absolute value less 

than 0.01. At each step, normalize by dividing each vector by its length. How 

many iterations does it take? Are the answers significantly different? 

5 What happens if c1 = 0? That is, what happens if the initial vector does not to 

have any component in the direction of the relevant eigenvector? 

6 How can the method of powers be adopted to find the smallest eigenvalue? 

Computer Code 

This is the code for the computer algebra system Octave that was used 

to do the calculation above. 

>T=[3, 0; 

8, -1] 

T= 

3 0 

8 -1 

>v0=[1; 2] 

v0= 

1 

1 

>v1=T*v0 

v1= 

3 

7 

>v2=T*v1 

v2= 

9 

17 

>T9=T**9 

T9= 

19683 0 

39368 -1 

>T10=T**10 

T10= 

59049 0 

118096 1 

>v9=T9*v0 

v9= 

19683


39367 

>v10=T10*v0 

v10= 

59049 

118096 

>norm(v10)/norm(v9) 

ans=2.9999 

(It has been lightly edited to remove blank lines, etc. Remark: we are ignoring 

the power of Octave here; there are built-in functions to automatically 

apply quite sophisticated methods to find eigenvalues and eigenvectors. Instead, 

we are using just the system as a calculator.)

Topic: Stable Populations 403 

Topic: Stable Populations 

Imagine a reserve park with animals from a species that we are trying to protect. 

The park doesn’t have a fence and so animals cross the boundary, both from 

the inside out and in the other direction. Every year, 10% of the animals from 

inside of the park leave, and 1% of the animals from the outside find their way 

in. We can ask if we can find a stable level of population for this park: is there a 

population that, once established, will stay constant over time, with the number 

of animals leaving equal to the number of animals entering? 

To answer that question, we must first establish the equations. Let the year 

n population in the park be pn and in the rest of the world be rn. 

pn+1 = .90pn + .01rn 

rn+1 = .10pn + .99rn 

We can set this system up as a matrix equation (see the Markov Chain topic). 

� � � �� 

pn+1 .90 .01 pn 

= 

.10 .99 

rn+1 

Now, “stable level” means that pn+1 = pn and rn+1 = rn, so that the matrix 

equation �vn+1 = T�vn becomes �v = T�v. We are therefore looking for eigenvectors 

for T that are associated with the eigenvalue 1. The equation (I − T )�v = �0 is 

� �� 

.10 .01 p 0 

= 

.10 .01 r 0 

which gives the eigenspace: vectors with the restriction that p = .1r. Coupled 

with additional information, that the total world population of this species is is 

p + r = 110 000, we find that the stable state is p =10, 000 and r = 100, 000. 

If we start with a park population of ten thousand animals, so that the rest of 

the world has one hundred thousand, then every year ten percent (a thousand 

animals) of those inside will leave the park, and every year one percent (a 

thousand) of those from the rest of the world will enter the park. It is stable, 

self-sustaining. 

Now imagine that we are trying to gradually build up the total world population 

of this species. We can try, for instance, to have the world population 

grow at a rate of 1% per year. In this case, we can take a “stable” state for 

the park’s population to be that it also grows at 1% per year. The equation 

�vn+1 =1.01 · �vn = T�vn leads to ((1.01 · I) − T )�v = �0, which gives this system. 

� �� 

.11 .01 p 

= 

.10 .02 r 

rn 

� � 

0 

0 

The matrix is nonsingular, and so the only solution is p =0andr =0. Thus, 

there is no (usable) initial population that we can establish at the park and 

expect that it will grow at the same rate as the rest of the world.


Knowing that an annual world population growth rate of 1% forces an unstable 

park population, we can ask which growth rates there are that would 

allow an initial population for the park that will be self-sustaining. We consider 

λ�v = T�v and solve for λ. 

� 

� 

� 

0= �λ 

− .9 .01 � 

� 

� .10 λ − .99� 

=(λ− .9)(λ − .99) − (.10)(.01) = λ2 − 1.89λ + .89 

A shortcut to factoring that quadratic is our knowledge that λ = 1 is an eigenvalue 

of T , so the other eigenvalue is .89. Thus there are two ways to have a 

stable park population (a population that grows at the same rate as the population 

of the rest of the world, despite the leaky park boundaries): have a world 

population that is does not grow or shrink, and have a world population that 

shrinks by 11% every year. 

So this is one meaning of eigenvalues and eigenvectors—they give a stable 

state for a system. If the eigenvalue is 1 then the system is static. If 

the eigenvalue isn’t 1 then the system is either growing or shrinking, but in a 

dynamically-stable way. 

Exercises 

1 What initial population for the park discussed above should be set up in the case 

where world populations are allowed to decline by 11% every year? 

2 What will happen to the population of the park in the event of a growth in world 

population of 1% per year? Will it lag the world growth, or lead it? Assume 

that the inital park population is ten thousand, and the world population is one 

hunderd thousand, and calculate over a ten year span. 

3 The park discussed above is partially fenced so that now, every year, only 5% of 

the animals from inside of the park leave (still, about 1% of the animals from the 

outside find their way in). Under what conditions can the park maintain a stable 

population now? 

4 Suppose that a species of bird only lives in Canada, the United States, or in 

Mexico. Every year, 4% of the Canadian birds travel to the US, and 1% of them 

travel to Mexico. Every year, 6% of the US birds travel to Canada, and 4% 

go to Mexico. From Mexico, every year 10% travel to the US, and 0% go to 

Canada. 

(a) Give the transition matrix. 

(b) Is there a way for the three countries to have constant populations? 

(c) Find all stable situations.

Topic: Linear Recurrences 405 

Topic: Linear Recurrences 

In 1202 Leonardo of Pisa, also known as Fibonacci, posed this problem. 

A certain man put a pair of rabbits in a place surrounded on all 

sides by a wall. How many pairs of rabbits can be produced from 

that pair in a year if it is supposed that every month each pair begets 

a new pair which from the second month on becomes productive? 

This moves past an elementary exponential growth model for population increase 

to include the fact that there is an initial period where newborns are not 

fertile. However, it retains other simplyfing assumptions, such as that there is 

no gestation period and no mortality. 

The number of newborn pairs that will appear in the upcoming month is 

simply the number of pairs that were alive last month, since those will all be 

fertile, having been alive for two months. The number of pairs alive next month 

is the sum of the number alive last month and the number of newborns. 

f(n +1)=f(n)+f(n − 1) where f(0) = 1, f(1) = 1 

The is an example of a recurrence relation (it is called that because the values 

of f are calculated by looking at other, prior, values of f). From it, we can 

easily answer Fibonacci’s twelve-month question. 

month 0 1 2 3 4 5 6 7 8 9 10 11 12 

pairs 1 1 2 3 5 8 13 21 34 55 89 144 233 

The sequence of numbers defined by the above equation (of which the first few 

are listed) is the Fibonacci sequence. The material of this chapter can be used 

to give a formula with which we can can calculate f(n + 1) without having to 

first find f(n), f(n − 1), etc. 

For that, observe that the recurrence is a linear relationship and so we can 

give a suitable matrix formulation of it. 

� �� 

1 1 f(n) 

= 

1 0 f(n − 1) 

� � 

f(n +1) 

f(n) 

where 

� � 

f(1) 

= 

f(0) 

� � 

1 

1 

Then, where we write T for the matrix and �vn for the vector with components 

f(n+1) and f(n), we have that �vn = T n �v0. The advantage of this matrix formulation 

is that by diagonalizing T we get a fast way to compute its powers: where 

T = PDP −1 we have T n = PD n P −1 , and the n-th power of the diagonal 

matrix D is the diagonal matrix whose entries that are the n-th powers of the 

entries of D. 

The characteristic equation of T is λ 2 − λ − 1. The quadratic formula gives 

its roots as (1 + √ 5)/2 and(1− √ 5)/2. Diagonalizing gives this. 

� � � √ 

1 1 1+ 5 

= 2 

1 0 

1− √ 5 

2 

1 1 

� � 1+ √ 5 

2 

0 

0 

1− √ 5 

2 

�� 1 

√5 − 1−√5 2 √ 5 

√−1 1+ 

5 

√ 5 

2 √ 5 

�


Introducing the vectors and taking the n-th power, we have 

� � � �n � � 

f(n +1) 1 1 f(1) 

= 

f(n) 1 0 f(0) 

� √ 

1+ 5 1− 

= 2 

√ � 

5 

2 

1 1 

� 1+ √ n 

5 

2 

0 

0 

�� 

1 

n 

1− √ 5 

2 

√5 − 1−√5 2 √ 5 

√−1 1+ 

5 

√ 5 

2 √ 5 

We can compute f(n) from the second component of that equation. 

f(n) = 1 

�� 

1+ 

√ 

5 

√ �n � 

5 1 − 

− 

2 

√ �n� 5 

2 

� �f(1) � 

f(0) 

Notice that f is dominated by its first term because (1 − √ 5)/2 is less than 

one, so its powers go to zero. Although we have extended the elementary model 

of population growth by adding a delay period before the onset of fertility, we 

nonetheless still get an (asmyptotically) exponential function. 

In general, a linear recurrence relation has the form 

f(n +1)=anf(n)+an−1f(n − 1) + ···+ an−kf(n − k) 

(it is also called a difference equation). This recurrence relation is homogeneous 

because there is no constant term; i.e, it can be put into the form 0 = −f(n + 

1) + anf(n)+an−1f(n − 1) + ···+ an−kf(n − k). This is said to be a relation 

of order k. The relation, along with the initial conditions f(0), ... , f(k) 

completely determine a sequence. For instance, the Fibonacci relation is of 

order 2 and it, along with the two initial conditions f(0) = 1 and f(1) = 1, 

determines the Fibonacci sequence simply because we can compute any f(n) by 

first computing f(2), f(3), etc. In this Topic, we shall see how linear algebra 

can be used to solve linear recurrence relations. 

First, we define the vector space in which we are working. Let V be the set 

of functions f from the natural numbers N = {0, 1, 2,...} to the real numbers. 

(Below we shall have functions with domain {1, 2,...}, that is, without 0, but 

it is not an important distinction.) 

Putting the initial conditions aside for a moment, for any recurrence, we can 

consider the subset S of V of solutions. For example, without initial conditions, 

in addition to the function f given above, the Fibonacci relation is also solved by 

the function g whose first few values are g(0) = 1, g(1) = 1, g(2) = 3, g(3) = 4, 

and g(4) = 7. 

The subset S is a subspace of V . It is nonempty because the zero function 

is a solution. It is closed under addition since if f1 and f2 are solutions, then 

an+1(f1 + f2)(n +1)+···+ an−k(f1 + f2)(n − k) 

=(an+1f1(n +1)+···+ an−kf1(n − k)) 

+(an+1f2(n +1)+···+ an−kf2(n − k)) 

=0.


And, it is closed under scalar multiplication since 

an+1(rf1)(n +1)+···+ an−k(rf1)(n − k) 

= r(an+1f1(n +1)+···+ an−kf1(n − k)) 

= r · 0 

=0. 

We can give the dimension of S. Consider this map from the set of functions S 

to the set of vectors Rk . 

⎛ ⎞ 

f(0) 

⎜ 

⎜f(1) 

⎟ 

f ↦→ ⎜ . 

⎝ . 

⎟ 

. ⎠ 

f(k) 

Exercise 3 shows that this map is linear. Because, as noted above, any solution 

of the recurrence is uniquely determined by the k initial conditions, this map is 

one-to-one and onto. Thus it is an isomorphism, and thus S has dimension k, 

the order of the recurrence. 

So (again, without any initial conditions), we can describe the set of solutions 

of any linear homogeneous recurrence relation of degree k by taking linear 

combinations of only k linearly independent functions. It remains to produce 

those functions. 

For that, we express the recurrence f(n +1)=anf(n)+···+ an−kf(n − k) 

with a matrix equation. 

⎛ 

⎞ 

an an−1 an−2 ... an−k+1 an−k ⎛ ⎞ 

⎜ 1 0 0 ... 0 0 ⎟ f(n) 

⎜ 0 1 0 

⎟ ⎜ 

⎟ ⎜f(n 

− 1) ⎟ 

⎜ 0 0 1 

⎟ ⎜ . 

⎟ ⎝ . 

⎟ 

⎜ . . 

⎝ 

. 

. . 

.. 

. ⎟ . ⎠ 

. ⎠ f(n − k) 

0 0 0 ... 1 0 

= 

⎛ 

⎞ 

f(n +1) 

⎜ f(n) ⎟ 

⎜ . 

⎝ . 

⎟ 

. ⎠ 

f(n − k +1) 

In trying to find the characteristic function of the matrix, we can see the pattern 

in the 2×2 case 

� � 

an − λ an−1 

= λ 

1 −λ 

2 − anλ − an−1 

and 3×3 case. 

⎛ 

⎝ an − λ 

1 

an−1 

−λ 

⎞ 

an−2 

0 

0 1 −λ 

⎠ = −λ 3 + anλ 2 + an−1λ + an−2


Exercise 4 shows that the characteristic equation is this. 

� 

�an 

� − λ 

� 

� 1 

� 

� 0 

� 

� 0 

� . 

� . 

� . 

� 0 

an−1 

−λ 

1 

0 

. 

0 

an−2 

0 

−λ 

1 

0 

... 

... 

. .. 

... 

an−k+1 

0 

1 

� 

an−k� 

� 

0 � 

� 

� 

� 

� 

� 

. 

� 

. � 

� 

−λ � 

= ±(−λ k + anλ k−1 + an−1λ k−2 + ···+ an−k+1λ + an−k) 

We call that the polynomial ‘associated’ with the recurrence relation. (We will 

be finding the roots of this polynomial and so we can drop the ± as irrelevant.) 

If −λ k + anλ k−1 + an−1λ k−2 + ···+ an−k+1λ + an−k has no repeated roots 

then the matrix is diagonalizable and we can, in theory, get a formula for f(n) 

as in the Fibonacci case. But, because we know that the subspace of solutions 

has dimension k, we do not need to do the diagonalization calculation, provided 

that we can exhibit k linearly independent functions satisfying the relation. 

Where r1, r2, ... , rk are the distinct roots, consider the functions fr1 (n) = 

of powers of those roots. Exercise 5 shows that each is 

rn 1 through frk (n) =rn k 

a solution of the recurrence and that the k of them form a linearly independent 

set. So, given the homogeneous linear recurrence f(n +1) = anf(n) +···+ 

an−kf(n−k) (that is, 0 = −f(n+1)+anf(n)+···+an−kf(n−k)) we consider 

the associated equation 0 = −λk + anλk−1 + ···+ an−k+1λ + an−k. We find its 

roots r1, ... , rk, and if those roots are distinct then any solution of the relation 

has the form f(n) =c1rn 1 + c2rn 2 + ···+ ckrn k for c1,...,cn ∈ R. (The case of 

repeated roots is also easily done, but we won’t cover it here—see any text on 

Discrete Mathematics.) 

Now, given some initial conditions, so that we are interested in a particular 

solution, we can solve for c1, ... , cn. For instance, the polynomial associated 

with the Fibonacci relation is −λ2 + λ + 1, whose roots are (1 ± √ 5)/2 andso 

any solution of the Fibonacci equation has the form f(n) =c1((1 + √ 5)/2) n + 

c2((1 − √ 5)/2) n . Including the initial conditions for the cases n =0andn =1 

gives 

c1 + c2 =1 

(1 + √ 5/2)c1 +(1− √ 5/2)c2 =1 

which yields c1 =1/ √ 5andc2 = −1/ √ 5, as was calculated above. 

We close by considering the nonhomogeneous case, where the relation has the 

form f(n+1) = anf(n)+an−1f(n−1)+···+an−kf(n−k)+b for some nonzero 

b. As in the first chapter of this book, only a small adjustment is needed to make 

the transition from the homogeneous case. This classic example illustrates. 

In 1883, Edouard Lucas posed the following problem. 

In the great temple at Benares, beneath the dome which marks 

the center of the world, rests a brass plate in which are fixed three


diamond needles, each a cubit high and as thick as the body of a 

bee. On one of these needles, at the creation, God placed sixty four 

disks of pure gold, the largest disk resting on the brass plate, and 

the others getting smaller and smaller up to the top one. This is the 

Tower of Bramah. Day and night unceasingly the priests transfer 

the disks from one diamond needle to another according to the fixed 

and immutable laws of Bramah, which require that the priest on 

duty must not move more than one disk at a time and that he must 

place this disk on a needle so that there is no smaller disk below 

it. When the sixty-four disks shall have been thus transferred from 

the needle on which at the creation God placed them to one of the 

other needles, tower, temple, and Brahmins alike will crumble into 

dusk, and with a thunderclap the world will vanish. (Translation of 

[De Parville] from[Ball & Coxeter].) 

How many disk moves will it take? Instead of tackling the sixty four disk 

problem right away, we will consider the problem for smaller numbers of disks, 

starting with three. 

To begin, all three disks are on the same needle. 

After moving the small disk to the far needle, the mid-sized disk to the middle 

needle, and then moving the small disk to the middle needle we have this. 

Now we can move the big disk over. Then, to finish, we repeat the process of 

moving the smaller disks, this time so that they end up on the third needle, on 

top of the big disk. 

So the thing to see is that to move the very largest disk, the bottom disk, 

at a minimum we must: first move the smaller disks to the middle needle, then 

move the big one, and then move all the smaller ones from the middle needle to 

the ending needle. Those three steps give us this recurence. 

T (n +1)=T (n)+1+T (n) =2T (n) + 1 where T (1) = 1 

We can easily get the first few values of T . 

n 1 2 3 4 5 6 7 8 9 10 

T (n) 1 3 7 15 31 63 127 255 511 1023


We recognize those as being simply one less than a power of two. 

To derive this equation instead of just guessing at it, we write the original 

relation as −1 =−T (n +1)+2T (n), consider the homogeneous relation 0 = 

−T (n)+2T (n − 1), get its associated polynomial −λ + 2, which obviously has 

the single, unique, root of r1 = 2, and conclude that functions satisfying the 

homogeneous relation take the form T (n) =c12 n . 

That’s the homogeneous solution. Now we need a particular solution. 

Because the nonhomogeneous relation −1 =−T (n +1)+2T (n) is so simple, 

in a few minutes (or by remembering the table) we can spot the particular 

solution T (n) =−1 (there are other particular solutions, but this one is easily 

spotted). So we have that—without yet considering the initial condition—any 

solution of T (n +1)=2T (n) + 1 is the sum of the homogeneous solution and 

this particular solution: T (n) =c12 n − 1. 

The initial condition T (1) = 1 now gives that c1 = 1, and we’ve gotten the 

formula that generates the table: the n-disk Tower of Hanoi problem requires a 

minimum of 2 n − 1 moves. 

Finding a particular solution in more complicated cases is, naturally, more 

complicated. A delightful and rewarding, but challenging, source on recurrence 

relations is [Graham, Knuth, Patashnik]., For more on the Tower of Hanoi, 

[Ball & Coxeter] or[Gardner 1957] are good starting points. So is [Hofstadter]. 

Some computer code for trying some recurrence relations follows the exercises. 

Exercises 

1 Solve each homogeneous linear recurrence relations. 

(a) f(n +1)=5f(n) − 6f(n − 1) 

(b) f(n +1)=4f(n− 1) 

(c) f(n +1)=6f(n)+7f(n − 1) + 6f(n − 2) 

2 Give a formula for the relations of the prior exercise, with these initial conditions. 

(a) f(0) = 1, f(1) = 1 

(b) f(0) = 0, f(1) = 1 

(c) f(0) = 1, f(1) = 1, f(2) = 3. 

3 Check that the isomorphism given betwween S and R k is a linear map. It is 

argued above that this map is one-to-one. What is its inverse? 

4 Show that the characteristic equation of the matrix is as stated, that is, is the 

polynomial associated with the relation. (Hint: expanding down the final column, 

and using induction will work.) 

5 Given a homogeneous linear recurrence relation f(n +1) = anf(n) +··· + 

an−kf(n − k), let r1, ... , rk be the roots of the associated polynomial. 

(a) Prove that each function fr i (n) =r n k satisfies the recurrence (without initial 

conditions). 

(b) Prove that no ri is 0. 

(c) Prove that the set {fr1,...,fr k } is linearly independent. 

6 (This refers to the value T (64) = 18, 446, 744, 073, 709, 551, 615 given in the computer 

code below.) Transferring one disk per second, how many years would it 

take the priests at the Tower of Hanoi to finish the job?


Computer Code 

This code allows the generation of the first few values of a function defined 

by a recurrence and initial conditions. It is in the Scheme dialect of 

LISP (specifically, it was written for A. Jaffer’s free scheme interpreter SCM, 

although it should run in any Scheme implementation). 

First, the Tower of Hanoi code is a straightforward implementation of 

the recurrence. 

(define (tower-of-hanoi-moves n) 

(if (= n 1) 

1 

(+ (* (tower-of-hanoi-moves (- n 1)) 

2) 

1) ) ) 

(Note for readers unused to recursive code: to compute T (64), the computer 

is told to compute 2 ∗ T (63) − 1, which requires, of course, computing T (63). 

The computer puts the ‘times 2’ and the ‘plus 1’ aside for a moment to 

do that. It computes T (63) by using this same piece of code (that’s what 

‘recursive’ means), and to do that is told to compute 2 ∗ T (62) − 1. This 

keeps up (the next step is to try to do T (62) while the other arithmetic is 

held in waiting), until, after 63 steps, the computer tries to compute T (1). 

It then returns T (1) = 1, which now means that the computation of T (2) 

can proceed, etc., up until the original computation of T (64) finishes.) 

The next routine calculates a table of the first few values. (Some language 

notes: ’() is the empty list, that is, the empty sequence, and cons pushes 

something onto the start of a list. Note that, in the last line, the procedure 

proc is called on argument n.) 

(define (first-few-outputs proc n) 

(first-few-outputs-helper proc n ’()) ) 

; 

(define (first-few-outputs-aux proc n lst) 

(if (< n 1) 

lst 

(first-few-outputs-aux proc (- n 1) (cons (proc n) lst)) ) ) 

The session at the SCM prompt went like this. 

>(first-few-outputs tower-of-hanoi-moves 64) 

Evaluation took 120 mSec 

(1 3 7 15 31 63 127 255 511 1023 2047 4095 8191 16383 32767 

65535 131071 262143 524287 1048575 2097151 4194303 8388607 

16777215 33554431 67108863 134217727 268435455 536870911 

1073741823 2147483647 4294967295 8589934591 17179869183 

34359738367 68719476735 137438953471 274877906943 549755813887 

1099511627775 2199023255551 4398046511103 8796093022207 

17592186044415 35184372088831 70368744177663 140737488355327 

281474976710655 562949953421311 1125899906842623 

2251799813685247 4503599627370495 9007199254740991 

18014398509481983 36028797018963967 72057594037927935 

144115188075855871 288230376151711743 576460752303423487


1152921504606846975 2305843009213693951 4611686018427387903 

9223372036854775807 18446744073709551615) 

This is a list of T (1) through T (64). (The 120 mSec came on a 50 mHz ’486 

running in an XTerm of XWindow under Linux. The session was edited to 

put line breaks between numbers.)

Appendix 

Introduction 

Mathematics is made of arguments (reasoned discourse that is, not pottery 

throwing). This section is a reference to the most used techniques. A reader 

having trouble with, say, proof by contradiction, can turn here for an outline of 

that method. 

But this section gives only a sketch. For more, these are classics: Propositional 

Logic by Copi, Induction and Analogy in Mathematics by Pólya, and 

Naive Set Theory by Halmos. 

Propositions 

Thepointatissueinanargumentistheproposition. Mathematicians usually 

write the point in full before the proof and label it either Theorem for major 

points, Lemma for results chiefly used to prove others, or Corollary for points 

that follow immediately from a prior result. 

Propositions can be complex, with many subparts. The truth or falsity of 

the entire proposition depends both on the truth value of the parts, and on the 

words used to assemble the statement from its parts. 

Not. For example, where P is a proposition, ‘it is not the case that P ’istrue 

provided P is false. Thus ‘n is not prime’ is true only when n is the product of 

smaller integers. 

We can picture the ‘not’ operation with a Venn diagram: 

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Where the box encloses all natural numbers, and inside the circle are the primes, 

the dots are numbers satisfying ‘not P ’. 

To prove a ‘not P ’ statement holds, show P is false. 

A-1

A-2 

And. Consider the statement form ‘P and Q’. For the statement to be true 

both halves must hold: ‘7 is prime and so is 3’ is true, while ‘7 is prime and 3 

is not’ is false. 

Here is the Venn diagram for ‘P and Q’. 

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To prove ‘P and Q’, prove each half holds. 

Or. A‘P or Q’ is true when either half holds: ‘7 is prime or 4 is prime’ is 

true, while ‘7 is not prime or 4 is prime’ is false. We take ‘or’ inclusively so that 

if both halves are true ‘7 is prime or 4 is not’ then the statement as a whole 

is true. (In everyday speech sometimes ‘or’ is meant in an exclusive way: “Eat 

your vegetables or no dessert” does not intend both halves to hold.) 

The Venn diagram for ‘or’ includes all of both circles. 

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To prove ‘P or Q’ show that at all times at least one half holds (perhaps 

sometimes one half and sometimes the other, but at all times at least one). 

Implication. A‘P implies Q’ statement (perhaps instead phrased ‘if P then 

Q’ or‘P =⇒ Q’) is true unless P is true while Q is false. Thus ‘7 is prime 

implies 4 is not’ is true (contrary to its use in casual speech, in mathematics ‘if 

P then Q’ does not connote that P precedes Q or causes Q) while ‘7 is prime 

implies 4 is also prime’ is false. 

More subtly, ‘P =⇒ Q’ is always true when P is false: ‘4 is prime implies 7 

is prime’ and ‘4 is prime implies 7 is not’ are both true statements, sometimes 

said to be vacuously true. (We adopt this convention because we want ‘if a 

number is a perfect square then it is not prime’ to be always true, for instance 

when the number is 5 or when the number is 6.) 

The diagram 

✬✬✩✩ 

P Q 

✫✫✪✪ 

shows Q holds whenever P does (another phrasing is ‘P is sufficient to give Q’). 

Notice again that if P does not hold, Q may or may not be in force. 

There are two main ways to establish an implication. The first way is direct: 

assume P is true and, using that assumption, prove Q. For instance, to show

A-3 

‘if a number is divisible by 5 then twice that number is divisible by 10’, assume 

the number is 5n and deduce that 2(5n) =10n. The second way is indirect: 

prove the contrapositive statement: ‘if Q is false then P is false’ (rephrased, ‘Q 

can only be false when P is also false’). As an example, to show ‘if a number is 

prime then it is not a perfect square’, argue that if it were a square p = n 2 then 

it could be factored p = n · n where n

A-4 

Venn diagrams aren’t very helpful with quantifiers, but in a sense the box we 

draw to border the diagram represents the universal quantifier since it dilineates 

the universe of possible members. 

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To show a statement holds in all cases, we must show it holds in each case. 

Thus, to prove ‘every number divisible by p has its square divisible by p 2 ’, take a 

single number of the form pn and square it: (pn) 2 = p 2 n 2 . Hence the statement 

holds for each number divisible by p. (This is a “typical element” proof. This 

kind of argument requires that no properties are assumed for that element other 

than those in the hypothesis—for instance, this is a wrong argument: “if n is 

divisible by a prime, say 2, so that n =2k then n 2 =(2k) 2 =4k 2 and the square 

of the number is divisible by the square of the prime”. That is an argument 

about the case p = 2, but it isn’t a proof for general p.) 

There exists. The other common quantifier is ‘there exists’, symbolized ∃. 

This quantifier is in some sense the opposite of ‘for all’. For instance, contrast 

these two definitions of primality of an integer p: (1) for all n, ifn is not 1 and 

n is not p then n does not divide p, and (2) it is not the case that there exists 

an n (with n �= 1andn �= p) such that n divides p. 

As noted above, Venn diagrams are not much help with quantifiers, but a 

picture of ‘there is a number such that P ’ would show both that there can be 

more than one and that not all numbers need satisfy P . 

. 

✬✩ 

P . 

. 

. 

✫✪ 

An existence proposition can be proved by producing something satisfying 

the property: one time, to settle the question of primality of 225 + 1, Euler 

produced the divisor 641. But there are proofs that show that something exists 

without saying how to find it; Euclid’s argument, given in the next subsection, 

shows there are infinitely many primes without naming them. In general, while 

demonstrating existence is better than nothing, giving an example is better than 

that, and an exhaustive list of all instances is great. Still, mathematicians take 

what they can get. 

Finally, along with “Are there any?” we often ask “How many?” That 

is why the issue of uniqueness often arises in conjunction with questions of 

existence. Many times the two arguments are simpler if separated, so note that 

just as proving something exists does not show it is unique, neither does proving 

something is unique show it exists.

Techniques of Proof 

A-5 

Induction. Many proofs are iterative, “Here’s why it’s true for 1, it then 

follows for 2, from there to 3, and so on ... ”. These are called proofs by 

induction. Such a proof has two steps. In the base step the proposition is 

established for some first number, often 0 or 1. Then in the inductive step we 

assume the proposition holds for numbers up to some k then it holds for the 

next number. 

Examples explain it best. 

We prove that 1 + 2 + 3 + ···+ n = n(n +1)/2. 

For the base step we must show the formula holds when n =1. That’s 

easy, the sum of the first 1 numbers equals 1(1 + 1)/2. 

For the inductive step, assume the formula holds for 1, 2,... ,k 

1=1(1+1)/2 

and 1+2=2(2+1)/2 

and1+2+3=3(3+1)/2 

. 

and1 + ···+ k = k(k +1)/2 

and then show it therefore holds in the k + 1 case. That’s just algebra: 

1+2+···+ k +(k +1)= 

k(k +1) 

2 

+(k +1)= 

(k +1)(k +2) 

. 

2 

The idea of induction is simple. We’ve shown the proposition holds for 1, 

and we’ve shown that if it holds for 1 then it holds for 2, and if it holds for 1 

and 2 then it holds for 3, etc. Thus it holds for any natural number greater 

than or equal to 1. 

We prove that every integer greater than 1 is a product of primes. 

The base step is easy, 2 is the product of a single prime. 

For the inductive step assume each of 2, 3,... ,k is a product of primes, 

aiming to show k + 1 is also. There are two cases: (1) if k +1 isnot divisible by a number smaller than itself then it is a prime and so the 

product of primes, and (2) if k + 1 factors then each factor can be written 

as a product of primes by the inductive hypothesis and so k +1 canbe rewritten as a product of primes. That ends the proof. 

(Remark. The Prime Factorization Theorem of Number Theory says that 

not only does a factorization exist, but that it is unique. We’ve shown the 

easy half.) 

Two remarks about ‘next number’. 

For one thing, while induction works on the integers, it’s no good on the 

reals. There is no ‘next’ real.

A-6 

The other thing is that we sometimes use induction to go down, say, from 

10 to 9 to 8, etc., down to 0. This is OK—‘next number’ could mean ‘next 

lowest number’. Of course, at the end we have not shown the fact for all natural 

numbers, only for those less than or equal to 10. 

Contradiction. Another technique of proof is to show something is true by 

showing it can’t be false. 

The classic example is Euclid’s, that there are infinitely many primes. 

Suppose there are only finitely many primes p1,...,pk. Consider p1 · 

p2 ...pk +1. None of the primes on this supposedly exhaustive list divides 

that number evenly, each leaves a remainder of 1. But every number is 

a product of primes so this can’t be. Thus there cannot be only finitely 

many primes. 

Every proof by contradiction has the same form—assume the proposition is 

false and derive some contradiction to known facts. 

Another example is this proof that √ 2 is not a rational number. 

Suppose √ 2=m/n. Then 

2n 2 = m 2 . 

Factor out the 2’s: n =2 kn · ˆn and m =2 km · ˆm. Rewrite: 

2 · (2 kn · ˆn) 2 =(2 km · ˆm) 2 . 

The Prime Factorization Theorem says there must be the same number 

of factors of 2 on both sides, but there are an odd number (1 + 2kn) on 

the left and an even number (2km) on the right. That’s a contradiction 

so a rational with a square of 2 cannot be. 

Both these examples aimed to prove something doesn’t exist. A negative 

proposition often suggests a proof by contradiction. 

Sets, Functions, and Relations 

Sets. The perfect squares less than 20, the roots of x 5 − 3x 3 + 2, the primes— 

all are collections. Mathematicians work with sets, collections that satisfy the 

Principle of Extensionality stated below. 

A set can be given as a listing between curly braces: {1, 4, 9, 16}, or, if that’s 

unwieldy, by using set-builder notation: {x � � x 5 − 3x 3 +2=0} (read “the set 

of all x such that ... ”). We name sets with capital roman letters: P = 

{2, 3, 5, 7, 11,...} except for the set of real numbers, written R, and the set 

of complex numbers, written C. To denote that something is an element (or 

member) of a set we use ‘ ∈ ’, so 7 ∈{3, 5, 7} while 8 �∈ {3, 5, 7}.

A-7 

The Principle of Extensionality is that two sets with the same elements are 

equal. Hence repeats collapse {7, 7} = {7} and order doesn’t matter {2,π} = 

{π, 2}. 

We use ‘⊂’ for the subset relationship: {2,π}⊂{2,π,7} and ‘⊆’ for subset or 

equality (if A is a subset of B but A �= B then A is a proper subset of B). These 

symbols may be flipped to signify the reverse relationship: {2,π,5} ⊃{2, 5}. 

Because of Extensionality, to prove A = B just show they have the same 

members. Usually we show mutual inclusion: both A ⊆ B and A ⊇ B. 

Set operations. Venn diagrams are handy here. For instance, ‘x ∈ P ’canbe 

pictured 

and ‘P ⊆ Q’ looks like this. 

✬✩ 

P 

.x 

✫✪ 

✬✬✩✩ 

P Q 

✫✫✪✪ 

Note this is also the diagram for implication. That’s because ‘P ⊆ Q’ means 

x ∈ P =⇒ x ∈ Q. 

In general, for every propositional logic operator there is an associated set 

operator. For instance, the complement of P is P comp = {x � � not(x ∈ P )} 

. . 

. ✬✩ 

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the union is P ∪ Q = {x � � (x ∈ P )or(x ∈ Q)} 

✬✩ 

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and the intersection is P ∩ Q = {x � � (x ∈ P ) and (x ∈ Q)}. 

✬✩ 

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P . . . . . . 

Q 

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✫✪

A-8 

When two sets share no members their intersection is the empty set {}, 

symbolized ∅. Any set has the empty set for a subset by the ‘vacuously true’ 

property of the definition of implication. 

Sequences. We shall also use collections where order does matter and where 

repeats do not collapse. These are sequences, denoted with angle brackets: 

〈2, 3, 7〉 �= 〈2, 7, 3〉. A sequence of length 2 is sometimes called an ordered pair 

and written with parentheses: (π, 3). We also sometimes say ‘ordered triple’, 

‘ordered 4-tuple’, etc. The set of ordered n-tuples of elements of a set A is 

denoted A n . Thus the set of pairs of reals is R 2 . 

Functions. Functions are used to express relationships; for instance, Galelio 

rolled balls down inclined planes to find the function relating elapsed time to 

distance covered. 

We first see functions in elementary Algebra, where they are presented as 

formulas (e.g., f(x) =16x 2 − 100), but progressing to more advanced Mathematics 

reveals more general functions—trigonometric ones, exponential and 

logarithmic ones, and even constructs like absolute value that involve piecing 

together parts—and we see that functions aren’t formulas, instead the key idea 

is that a function associates with each input x a single output f(x). 

Consequently, a function (or map) is defined to be a set of ordered pairs 

(x, f(x) ) such that x suffices to determine f(x), that is x1 = x2 =⇒ f(x1) = 

f(x2) (this requirement is referred to by saying a function is well-defined). ∗ 

Each input x is one of the function’s arguments and each output f(x) isa 

value. The set of all arguments is f’s domain and the set of output values is its 

range. Often we don’t need to produce exactly the range, and we instead work 

with a superset of the range, the codomain. The notation for a function f with 

domain X and codomain Y is f : X → Y . 

X✬ 

✩ 

f 

✲ 

✫✪ 

✬Y 

✩ 

✛ ✘ � 

✫✪ 

range of f 

We sometimes instead use the notation x f 

↦−→ 16x 2 − 100, read ‘x maps under 

f to 16x 2 − 100’, or ‘16x 2 − 100 is the image of x’. 

Complicated maps, like x ↦→ sin(1/x), can be thought of as combinations 

of simple maps, for instance here applying the function g(y) =sin(y) tothe 

image of f(x) =1/x. The composition of g : Y → Z with f : X → Y , denoted 

g ◦ f : X → Z, is the map sending x ∈ X to g( f(x)) ∈ Z. This definition only 

makes sense if the range of f is a subset of the domain of g. 

Observe that the identity map id: Y → Y, defined by id(y) =y, has the 

property that for any f : X → Y , the composition id ◦ f is equal to the map f. 

∗ More on this is in the section on isomorphisms

A-9 

So an identity map plays the same role with respect to function composition 

that the number 0 plays in real number addition, or that the number 1 plays in 

multiplication. 

In line with that analogy, define a left inverse of a map f : X → Y to be a 

function g : range(f) → X such that g ◦ f is the identity map on X. Ofcourse, 

a right inverse of f is a h: Y → X such that f ◦ h is the identity. 

A map that is both a left and right inverse of f is called simply an inverse. 

An inverse, if one exists, is unique because if both g1 and g2 are inverses of f 

then g1(x) =g1 ◦ (f ◦ g2)(x) =(g1 ◦ f) ◦ g2(x) =g2(x) (the middle equality 

comes from the associativity of function composition), so we often call it “the” 

inverse, written f −1 . For instance, the inverse of the function f : R → R given 

by f(x) =2x − 3 is the function f −1 : R → R given by f −1 (x) =(x +3)/2. 

The superscript ‘f −1 ’ notation for function inverse can be confusing — it 

doesn’t mean 1/f(x). It is used because it fits into a larger scheme. Functions 

that have the same codomain as domain can be iterated, so that where 

f : X → X, we can consider the composition of f with itself: f ◦f, andf ◦f ◦f, 

etc. Naturally enough, we write f ◦ f as f 2 and f ◦ f ◦ f as f 3 , etc. Note 

that the familiar exponent rules for real numbers obviously hold: f i ◦ f j = f i+j 

and (f i ) j = f i·j . The relationship with the prior paragraph is that, where f is 

invertible, writing f −1 for the inverse and f −2 for the inverse of f 2 , etc., gives 

that these familiar exponent rules continue to hold, once f 0 is defined to be the 

identity map. 

If the codomain Y equals the range of f then we say that the function is onto. 

A function has a right inverse if and only if it is onto (this is not hard to check). 

If no two arguments share an image, if x1 �= x2 implies that f(x1) �= f(x2), 

then the function is one-to-one. A function has a left inverse if and only if it is 

one-to-one (this is also not hard to check). 

By the prior paragraph, a map has an inverse if and only if it is both onto 

and one-to-one; such a function is a correspondence. It associates one and only 

one element of the domain with each element of the range (for example, finite 

sets must have the same number of elements to be matched up in this way). 

Because a composition of one-to-one maps is one-to-one, and a composition of 

onto maps is onto, a composition of correspondences is a correspondence. 

We sometimes want to shrink the domain of a function. For instance, we 

may take the function f : R → R given by f(x) =x 2 and, in order to have an 

inverse, limit input arguments to nonnegative reals ˆ f : R + → R. Technically, 

ˆf is a different function than f; we call it the restriction of f to the smaller 

domain. 

A final point on functions: neither x nor f(x) need be a number. As an 

example, we can think of f(x, y) =x + y as a function that takes the ordered 

pair (x, y) as its argument. 

Relations. Some familiar operations are obviously functions: addition maps 

(5, 3) to 8. But what of ‘

A-10 

the set {(a, b) � � a

A-11 

Similarly, the equivalence relation ‘=’ partitions the integers into one-element 

sets. 

... 

✄ ✄✄✄✄✄ 

.−1 

.0 

.1 

.2 

✄ ✄✄✄✄✄ 

✄ ✄✄✄✄✄ 

✄ ✄✄✄✄✄ 

✄ ✄✄✄✄✄ 

Before we show that equivalence relations always give rise to partitions, 

we first illustrate the argument. Consider the relationship between two integers 

of ‘same parity’, the set {(−1, 3), (2, 4), (0, 0),...} (i.e., ‘give the same 

remainder when divided by 2’). We want to say that the natural numbers 

split into two pieces, the evens and the odds, and inside a piece each member 

has the same parity as each other. So for each x we define the set of 

numbers associated with it: Sx = {y � � (x, y) ∈ ‘same parity’}. Some examples 

are S1 = {... ,−3, −1, 1, 3,...}, andS4 = {... ,−2, 0, 2, 4,...}, andS−1 = 

{... ,−3, −1, 1, 3,...}. These are the parts, e.g., S1 is the odds. 

Theorem. An equivalence relation induces a partition on the underlying set. 

Proof. Call the set S and the relation R. 

For each x ∈ S define Sx = {y � � (x, y) ∈ R}. Observe that, as x is in Sx, the 

union of all these sets is S. 

All that remains is to show that distinct parts are disjoint: if Sx �= Sy then 

Sx ∩ Sy = ∅. To argue the contrapositive, assume Sx ∩ Sy �= ∅, aiming to show 

Sx = Sy. Let p be an element of the intersection, so that each of (x, p), (p, x), 

(y, p), and (p, y) isinR. To show that Sx = Sy we show each is a subset of the 

other. 

Assume q is in Sx so (q, x) isinR. Use transitivity along with (x, p) ∈ R to 

conclude that (q, p) is also an element of R. But (p, y) isinR, so another use 

of transitivity gives that (q, y) isinR. Thus q is in Sy. Hence q ∈ Sx implies 

q ∈ Sy, andsoSx ⊆ Sy. 

The same argument in the other direction gives the other inclusion, and so 

equality holds, completing the contrapositive. QED 

We call each part of a partition an equivalence class (for our purposes ‘class’ 

means the same as ‘set’). 

A last remark about classification. We often pick a single element to be the 

representative. 


✥ 

from each class: ✪ 

✩ 

✦ ✜ 

⋆ 

⋆ 

✢ 

⋆ 

... 

⋆ ⋆ 

...

A-12 

Usually when we pick representatives we have some natural scheme in mind. In 

that case we call them the canonical class representatives. 

For example, when considering the even and odd natural numbers, 

... 

.−3 

.−1 

.1 

.3 

✄ ✄✄✄✄✄ 

.2 

.0 

.−2 

we may pick 0 and 1 as canonical representatives because each is the smallest 

nonnegative member if its class. 

... 

.−3 

.−1 

⋆1 

.3 

✄ 

✄ 

✄ 

✄ 

✄ 

✄ 

.2 

⋆0 

.−2 

Another example is the simplest form of a fraction. We consider 3/5 and 

9/15 to be equivalent fractions. That is, we partition symbols of the form ‘d/n’ 

where d and n are integers and n �= 0 according to the relationship that ‘d1/n1’ 

is equivalent to ‘d2/n2’ if and only if ‘d1n2 = d2n1’. 

All d/n with n �= 0: 

✥ 

✪ 

✩ 

✦ ✜ 

✢ 

. 

... 

3 

5 

9 

15 . 

We usually use the reduced form symbols as representatives. 

... 

... 


✥ 

from each class: ✪ 

✩ 

✦ ✜ 

⋆ 

✢ 

... 

1 

2 

⋆ 3 

5 

⋆ 2 

⋆ −8 

7 

⋆ 2 

3 

1 

3 

9 

equivalent to 5 15 .

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Index 

accuracy 

of Gauss’ method, 67–71 

rounding error, 68 

addition 

vector, 80 

additive inverse, 80 

adjoint matrix, 328 

angle, 42 

antipodal, 340 

antisymmetric matrix, 139 

arrow diagram, 216, 232, 238, 242, 351 

augmented matrix, 14 

automorphism, 163 

dilation, 164 

reflection, 164 

rotation, 164 

back-substitution, 5 

basis, 113–124 

change of, 238 

definition, 113 

orthogonal, 256 

orthogonalization, 257 

orthonormal, 258 

standard, 114, 350 

standard over the complex numbers, 

350 

string, 371 

best fit line, 269 

block matrix, 311 

box, 321 

orientation, 321 

sense, 321 

volume, 321 

C language, 67 

canonical form 

for matrix equivalence, 245 

for nilpotent matrices, 374 

for row equivalence, 57 

for similarity, 392 

canonical representative, A-12 

Cauchy-Schwartz Inequality, 41 

Cayley-Hamilton theorem, 382 

central projection, 337 

change of basis, 238–249 

characteristic 

vectors, values, 357 

characteristic equation, 360 

characteristic polynomial, 360 

characterized, 172 

characterizes, 246 

Chemistry problem, 1, 9 

chemistry problem, 22 

circuits 

parallel, 73 

series, 73 

series-parallel, 74 

closure, 95 

of nullspace, 367 

of rangespace, 367 

codomain, A-8 

cofactor, 327 

column, 13 

rank, 126 

vector, 15 

column rank 

full, 131 

column space, 126 

complementary subspaces, 136 


complex numbers 

vector space over, 91 

component, 15 

composition 

self, 365 

computer algebra systems, 61–62 

concatenation, 134 

conditioning number, 70 

congruent figures, 286

congruent plane figures, 286 

contradiction, A-6 

convex set, 183 

coordinates 

homogeneous, 340 

with respect to a basis, 116 

correspondence, 161, A-9 

coset, 193 

Cramer’s rule, 331–333 

cross product, 298 

crystals, 143–146 

diamond, 144 

graphite, 144 

salt, 143 

unit cell, 144 

da Vinci, Leonardo, 337 

determinant, 294, 299–318 

cofactor, 327 

Cramer’s rule, 332 


exists, 309, 315 

Laplace expansion, 327 

minor, 327 

permutation expansion, 308, 312, 

334 

diagonal matrix, 209, 225 

diagonalizable, 354–357 

difference equation, 406 


dilation, 164, 275 

representing, 203 

dimension, 121 

direct sum, 131 

), 140 


external, 168 

of two subspaces, 136 

direction vector, 35 

distance-preserving, 286 

division theorem, 348 

dot product, 39 

dual space, 193 

echelon form, 5 

free variable, 12 

leading variable, 5 

reduced, 46 

eigenspace, 361 

eigenvalue, eigenvector 

of a matrix, 358 

of a transformation, 357 

elementary 

matrix, 226, 275 

elementary reduction operations, 4 

pivoting, 4 

rescaling, 4 

swapping, 4 

elementary row operations, 4 

entry, 13 

equivalence 

class, A-11 

canonical representative, A-12 

representative, A-11 

equivalence relation, A-10, A-11 

isomorphism, 169 

matrix equivalence, 244 

matrix similarity, 351 

row equivalence, 50 

equivalent statements, A-3 

Erlanger Program, 286 

Euclid, 286 

even functions, 99, 138 

even polynomials, 398 

external direct sum, 168 

Fibonacci sequence, 405 

field, 141–142 


finite-dimensional vector space, 119 

flat, 36 

form, 55 

free variable, 12 

full column rank, 131 

full row rank, 131 

function, A-8 

inverse image, 185 

codomain, A-8 

composition, 215, A-8 

correspondence, A-9 

domain, A-8 

even, 99 

identity, A-8 

inverse, 231, A-9 

left inverse, 231 

multilinear, 305 

odd, 99 

one-to-one, A-9 

onto, A-9 

range, A-8

estriction, A-9 

right inverse, 231 

structure preserving, 161, 165 

seehomomorphism, 176 

two-sided inverse, 231 

well-defined, A-8 

zero, 177 

Fundamental Theorem 

of Linear Algebra, 268 

Gauss’ method, 2 

accuracy, 67–71 

back-substitution, 5 

elementary operations, 4 

Gauss-Jordan, 46 


generalized nullspace, 367 

generalized rangespace, 367 

Geometry of Linear Maps, 274–279 

Gram-Schmidt process, 255–260 

homogeneous coordinate vector, 340 

homogeneous coordinates, 291 

homogeneous equation, 21 

homomorphism, 176 

composition, 215 

matrix representing, 194–204 

nonsingular, 190, 207 

nullity, 188 

nullspace, 188 

rangespace, 184 

rank, 206 

zero, 177 

ideal line, 342 

ideal point, 342 

identity 

function, A-8 

matrix, 224 

ill-conditioned, 68 

implication, A-2 

improper subspace, 92 

incidence matrix, 228 

index 

of nilpotency, 370 

induced map, 275 

induction, 23, A-5 

inner product, 39 

Input-Output Analysis, 63–66 

internal direct sum, 135 

invariant 

subspace, 377 

invariant subspace 


inverse, 231, A-9 

additive, 80 

exists, 231 

left, 231, A-9 

matrix, 329 

right, 231, A-9 

two-sided, A-9 

inverse function, 231 

inverse image, 185 

inversion, 313 

isometry, 286 

isomorphism, 159–175 

characterized by dimension, 172 


of a space with itself, 163 

Jordan block, 388 

Jordan form, 379–398 

represents similarity classes, 392 

kernel, 188 

Kirchhoff’s Laws, 73 

Klein, F., 286 

Laplace expansion, 326–330 

computes determinant, 327 

leading variable, 5 

least squares, 269–273 

length, 39 

Leontief, W., 63 

line 

best fit, 269 

in projective plane, 341 

line at infinity, 342 

line of best fit, 269–273 

linear 

transpose operation, 131 

linear combination, 52 

Linear Combination Lemma, 52 

linear equation, 2 

coefficients, 2 

constant, 2 


inconsistent systems, 269 

satisfied by a vector, 15 

solution of, 2



solutions of 

Cramer’s rule, 332 

system of, 2 

linear map 

dilation, 275 


rotation, 274, 288 

seehomomorphism, 176 

skew, 276 

trace, 397 

linear recurrence, 406 

linear recurrences, 405–412 

linear relationship, 103 

linear surface, 36 

linear transformation 

seetransformation, 180 

linearly dependent, 103 

linearly independent, 103 

LINPACK, 61 

map 

distance-preserving, 286 

extended linearly, 173 

induced, 275 

self composition, 365 

Maple, 61 

Markov chains, 280–285 

Markov matrix, 284 

Mathematica, 61 

mathematical induction, 23 

MATLAB, 61 

matrix, 13 

adjoint, 328 

antisymmetric, 139 

augmented, 14 

block, 246, 311 

change of basis, 238 


cofactor, 327 

column, 13 

column space, 126 

conditioning number, 70 

determinant, 294, 299 

diagonal, 209, 225 

diagonalizable, 354 

diagonalized, 244 

elementary reduction, 226, 275 

entry, 13 

equivalent, 244 

identity,220,224 

incidence, 228 

induced map, 275 

inverse, 329 

main diagonal, 224 

Markov, 229, 284 

matrix-vector product, 197 

minimal polynomial, 220, 380 

minor, 327 

multiplication, 215 

nilpotent, 370 



orthonormal, 286–291 

permutation, 225 

rank, 206 

representation, 196 

row, 13 


row rank, 124 

row space, 124 

scalar multiple, 212 

similar, 324 

similarity, 351 

singular, 27 

skew-symmetric, 311 

submatrix, 303 

sum, 212 

symmetric, 118, 139, 213, 220, 228, 

268 

trace, 213, 229, 397 

transpose, 19, 126, 213 

triangular, 204, 229, 330 

unit, 222 

Vandermonde, 311 

matrix equivalence, 242–249 

canonical form, 245 


matrix:form, 55 

mean 

arithmetic, 44 

geometric, 44 

method of powers, 399–402 

minimal polynomial, 220, 380 

minor, 327 

morphism, 161 

multilinear, 305 

multiplication 

matrix-matrix, 215

matrix-vector, 197 

mutual inclusion, A-7 

natural representative, A-12 

networks, 72–78 

Kirchhoff’s Laws, 73 

nilpotent, 368–378 

canonical form for, 374 


matrix, 370 

transformation, 370 

nilpotentcy 

index, 370 



matrix, 27 

normalize, 258 

nullity, 188 

nullspace, 188 

closure of, 367 

generalized, 367 

Octave, 61 

odd functions, 99, 138 

order 

of a recurrence, 406 

orientation, 321, 324 


basis, 256 

complement, 263 

mutually, 255 

projection, 263 

orthogonal matrix, 288 

orthogonalization, 257 

orthonormal basis, 258 

orthonormal matrix, 286–291 

parallelepiped, 321 

parallelogram rule, 35 

parameter, 13 

partial pivoting, 69 

partition, A-10–A-12 

matrix equivalence classes, 244, 247 

row equivalence classes, 50 

partitions 

into isomorphism classes, 170 

permutation, 308 

inversions, 313 

matrix, 225 

signum, 314 

permutation expansion, 308, 312, 334 

perp, 263 

perpendicular, 42 

perspective 

triangles, 343 

Physics problem, 1 

pivoting 

full, 69 

pivoting on rows, 4 

plane figure, 286 

congruence, 286 

point 

at infinity, 342 

in projective plane, 339 

polynomial 

even, 398 

minimal, 380 

of map, matrix, 379 

polynomials 

division theorem, 348 

populations, stable, 403–404 

powers, method of, 399–402 

preserves structure, 176 

projection, 176, 185, 250, 268, 385 

along a subspace, 260 

central, 337 

vanishing point, 337 

into a line, 251 

into a subspace, 260 

orthogonal, 251, 263 

Projective Geometry, 337–346 

projective geometry 

Duality Principle, 341 

projective plane 

ideal line, 342 

idealpoint,342 

lines, 341 

proof techniques 

induction, 23 

proper subspace, 92 

rangespace, 184 

closure of, 367 

generalized, 367 

rank, 128, 206 

column, 126 

of a homomorphism, 184, 188 

recurrence, 327, 406 


initial conditions, 406

educed echelon form, 46 


glide, 289 

reflection (or flip) about a line, 164 

relation, A-9 

equivalence, A-10 

relationship 

linear, 103 

representation 

of a matrix, 196 

of a vector, 116 

representative, A-11 

canonical, A-12 

for row equivalence classes, 57 

of matrix equivalence classes, 245 

of similarity classes, 392 

rescaling rows, 4 

restriction, A-9 

rigid motion, 286 

rotation, 274, 288 

rotation (or turning), 164 

represented, 199 

row, 13 

rank, 124 

vector, 15 


row rank 

full, 131 

row space, 124 

scalar, 80 

scalar multiple 

matrix, 212 

vector, 15, 34, 80 

scalar product, 39 

Schwartz Inequality, 41 

SciLab, 61 

self composition 

of maps, 365 

sense, 321 

sequence, A-8 

concatenation, 134 

sets, A-6 

dependent, independent, 103 

empty, 105 

mutual inclusion, A-7 

proper subset, A-7 

span of, 95 

subset, A-7 

sgn 

seesignum, 314 

signum, 314 

similar, 298, 324 

canonical form, 392 

similar matrices, 351 

similarity, 351–364 

similarity transformation, 364 

singular 

matrix, 27 

size, 319, 321 

skew, 276 

skew-symmetric, 311 

span, 95 

of a singleton, 99 

spin, 149 

square root, 398 

stable populations, 403–404 

standard basis, 114 

Statics problem, 5 

string, 371 

basis, 371 

of basis vectors, 369 

structure 

preservation, 176 

submatrix, 303 

subspace, 91–101 

closed, 93 

complementary, 136 


direct sum, 135 

improper, 92 

independence, 135 

invariant, 389 

orthocomplement, 139 

proper, 92 

sum, 132 

sum 

of matrices, 212 

of subspaces, 132 

vector, 15, 34, 80 

summation notation 

for permutation expansion, 308 

swapping rows, 4 

symmetric matrix, 118, 139, 213, 220 

system of linear equations, 2 


solving, 2 

trace, 213, 229, 397 

transformation


composed with itself, 365 

diagonalizable, 354 

eigenspace, 361 

eigenvalue, eigenvector, 357 

Jordan form for, 392 

minimal polynomial, 380 

nilpotent, 370 

canonical representative, 374 

projection, 385 

size change, 321 

transpose, 19, 126 

determinant, 309, 317 

interaction with sum and scalar 

multiplication, 213 

Triangle Inequality, 40 

triangular matrix, 229 

Triangularization, 204 

trivial space, 84, 114 

turning map, 164 

unit matrix, 222 

Vandermonde matrix, 311 

vanishing point, 337 

vector, 15, 33 

angle, 42 

canonical position, 33 

column, 15 

component, 15 

cross product, 298 

direction, 35 

dot product, 39 

free, 33 

homogeneous coordinate, 340 

length, 39 


representation of, 116, 238 

row, 15 

satisfies an equation, 15 

scalar multiple, 15, 34, 80 

sum, 15, 34, 35, 80 

unit, 43 

zero, 22, 80 

vector space, 80–101 

basis, 113 

closure, 80 

complex scalars, 91 


dimension, 121 

dual, 193 

finite dimensional, 119 


isomorphism, 161 

map, 176 

over complex numbers, 347 

subspace, 91 

trivial, 84, 114 

volume, 321 

voting paradox, 147 

majority cycle, 147 

rational preference order, 147 

voting paradoxes, 147–151 

spin, 149 

well-defined, A-8 

Wheatstone bridge, 75 

zero 

divisor, 220 

zero divison, 237 

zero divisor, 220 

zero homomorphism, 177 

zero vector, 22, 80

Linear Algebra

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