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Department of Mathematical Sciences

MATHEMATICS

University of South AfricaPretoria

Only Study guide for MAT3701

Linear Algebra

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MAT3701/1

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iii MAT3701/1

CONTENTS.

Study unit Friedberg Title Page1 Overview 1

PART 1 FAMILIAR LINEAR ALGEBRA RESULTS OVERTHE REAL AND COMPLEX NUMBERS 6

2 Sections 1.1 � 1.6 Vector Spaces over the Real and Complex Numbers 83 Sections 2.1 � 2.5 Linear Transformations over the Real and Complex Numbers 314 Section 4.4 Summary of Important Facts about Determinants over the 52

Real and Complex Numbers

PART 2 DIAGONALISATION 545 Appendix E Polynomials 566 Section 5.1 Eigenvalues and Eigenvectors 607 Section 5.2 Diagonalisabililty 768 Section 5.3 Matrix Limits and Markov Chains 849 Section 5.4 Invariant Subspaces and the Caley�Hamilton Theorem 88

PART 3 INNER�PRODUCT SPACES 9910 Section 6.1 Inner Products and Norms 10111 Section 6.2 The Gram�Schmidt Orthogonalisation Process and Orthogonal 107

Complements12 Section 6.3 The Adjoint of a Linear Operator 11713 Section 6.4 Normal and Self�Adjoint Operators 12614 Section 6.5 Unitary and Orthogonal Operators and Their Matrices 13615 Section 6.6 Orthogonal Projections and the Spectral Theorem 14916 Section 6.8 (4th ed.) Bilinear and Quadratic Forms 161

Section 6.7 (2nd or 3rded.)17 Section 6.10 (4th ed.) Conditioning and the Rayleigh Quotient 165

Section 6.9 (2nd or 3rded.)

1 MAT3701/1

.

STUDY UNIT 1.

OVERVIEW

Page

1.1 About this Module 2

1.2 Prescribed Text 2

1.3 The Syllabus for this Module 2

1.4 Mind Map 3

1.5 How to Use the Study Guide 4

1.6 How to Study for this Module 4

1.7 The Greek Alphabet 5

2

1.1 About this ModuleLinear algebra can be studied from different perspectives. One is the �matrix�oriented� (or �coordinate�based�)perspective, where the emphasis is on matrices and their manipulations. This allows for a concrete introductionto the subject, with a variety of practical applications, which therefore today is very popular as far as introductorycourses are concerned. This, to a large extent, is the type of course we offer on the �rst and second levels. A secondperspective, and the one on which this module is based, is the �coordinate�free� perspective. Here the approach ismore formal and abstract, with the focus on the study of linear transformations. Results derived from this perspectiveoften have equivalent formulations in terms of matrices, obtained by considering the matrix representations of thelinear transformations involved. It is often very convenient and useful to translate between these two perspectives,and therefore at this level it is important for you to be exposed to both perspectives.

1.2 Prescribed TextThe prescribed text for this module is

Stephen H. Friedberg, Arnold J. Insel and Lawrence E. Spence:Linear Algebra, 2nd (1992), 3rd (1997) or 4th (2003) edition,Prentice Hall, New Jersey.

Any of the three editions may be used. In the study guide which accompanies the text, references to all three editionsare given throughout.

1.3 The Syllabus for this ModuleThe sections of the prescribed text on which the module (study guide) will be based are as follows:

� Chapter 1: Vector Spaces (assumed preknowledge, except for SG: Study Unit 2)

Sections 1.1 � 1.6

� Chapter 2: Linear Transformations and Matrices (assumed preknowledge, except for SG: Study Unit 3)

Sections 2.1 � 2.5

� Section 4.4: Summary � Important Facts about Determinants (assumed preknowledge, except for SG:Study Unit 4)

� Chapter 5: Diagonalisation

Sections 5.1 � 5.4, including Appendix E.

� Chapter 6: Inner Product Spaces

Sections 6.1 � 6.6, 6.8 (6.7, 2nd or 3rded.), 6.10 (6.9, 2nd or 3rd ed.)

3 MAT3701/1

1.4 Mind MapThe following is a mind map of this module. It lists, among others, each study unit in the study guide, with thereference to Friedberg in brackets. Some students �nd it useful to draw their own mind map, which could be moredetailed than the one presented here.

4

1.5 How to Use the Study GuideThe study guide is designed to guide you through the prescribed study material. Each study unit, except this one,begins with an overview section in which the signi�cance of the section is explained, and how it �ts in with the restof the material. There is also a section on what to study in each study unit, and a list of worked examples, takenfrom exercises in the prescribed text, previous assignment questions, etc.

Referencing in the Study GuideThe prescribed text is referred to as Friedberg (for example, Friedberg: Section 1.1, Example 1) when it is notnecessary to distinguish between the editions. When it becomes necessary to do so, we use the notations F2, F3and F4 to distinguish between the second, third and fourth editions, respectively � for example: (F2: Section 5.2,Exercise 19 / F3: Section 5.2, Exercise 20 / F4: Section 5.2, Exercise 21) or (F2 and F3: Section 2.1, Exercise 17 /F4: Section 2.1, Exercise 18), when the reference to the second and third editions are the same. References withinthe study guide will be indicated by SG (for example, SG: Section 1.7).

1.6 How to Study for this ModuleThe prescribed text is probably more theoretical than any other text in linear algebra you have encountered before.If you wish to major in mathematics you are expected to be able to read and analyse proofs such as those appearingin the prescribed text, and also to reproduce a selection of these proofs for examination purposes.

The prescribed text and study guide contain a large selection of worked examples of the type you will be expected tosolve. A useful exercise is to �rst cover our solution while attempting your own. This technique can also be appliedto learn the theorems you need to know for examination purposes.

Some students are inclined to read over de�nitions too quickly. It should be realised that de�nitions contain theimportant concepts of the subject � that is why they receive special prominence. They also tell us how to dothings, that is, the de�nition of a linear transformation tells us how to check whether or not a mapping is a lineartransformation. The same applies to the theorems, propositions, lemmas and corollaries proved in the prescribedtext. We should understand and know these statements, since they contain important additional information aboutthe subject, which will be needed later on. Also, by also working through their proofs and by analysing them indetail, one learns the special techniques of the subject. Very often, these special techniques are needed for thetheoretical exercises appearing in the text.

5 MAT3701/1

1.7 The Greek Alphabet

Upper Case Lower Case NameA � alphaB � beta0 gamma4 � deltaE " epsilonZ � zetaH � eta2 � thetaI � iotaK � kappa3 � lambdaM � muN � nu4 � xiO o omicron5 � piP � rho6 � sigmaT � tau7 � upsilon8 � phiX � chi9 psi� ! omega

6

.

PART 1.

FAMILIAR LINEAR ALGEBRA RESULTSOVERTHEREALANDCOMPLEXNUMERS

Page

Study unit 2 (Friedberg: Sections 1.1 � 1.6): Vector Spaces over the Realand Complex Numbers

8

Study unit 3 (Friedberg: Sections 2.1 � 2.5): Linear Transformations over theReal and Complex Numbers

31

Study unit 4 (Friedberg: Section 4.4): Summary of Important Facts about Determinantsover the Real and Complex Numbers

52

7 MAT3701/1

Part 1 OverviewThe basic properties of a general vector space, linear transformations, and determinants are revised. Unlike in mostprevious courses in linear algebra, the scalars may come from either the real or the complex �eld, instead of justthe real �eld, as is usually the case. The advantage of this will become evident when we investigate the structure oflinear transformations, for example, since there the fact that the characteristic polynomial of an operator splits overthe complex �eld plays an important role. Most of the material, except that which is treated in the study units, isassumed to be known from a previous course in linear algebra over the real scalar �eld, since generally the proofsare also valid over the complex �eld.

OutcomesAfter studying Part 1, you should be able to

� solve a range of linear algebra problems related to a general vector space, linear transformations, and deter-minants over the real or complex numbers

� solve basic theoretic linear algebra problems over the real or complex numbers

� evaluate determinants over the real or complex numbers.

8

.

STUDY UNIT 2.

VECTOR SPACES OVER THE REAL ANDCOMPLEX NUMERS

(Friedberg: Sections 1.1 � 1.6)

2.1 Overview

The basic properties of a general vector space are revised. Since these properties are assumed to be familiar froma previous course in linear algebra over the real scalar �eld, and since the proofs in the case of the complex scalar�eld are esentially the same, most of the material is assumed to be known except for the topics treated in this studyunit, which should be studied for examination pruposes. Many of the examples in this study unit involve complexnumbers in order to familiarize you with working with vector spaces over the complex numbers.

2.2 What to Study

Read through Friedberg: Sections 1.1 � 1.6 to refresh your memory. Most of the material should be familiar to youexcept that Friedberg allows the scalars to come from a general �eld F instead of only the real numbers as you areprobably used to. For the purposes of this module, you may assume F to denote either the real numbers R or thecomplex numbers C: This study unit is for exam purposes and should be studied in detail.

2.3 Vector Spaces over R or C

When linear algebra is introduced for the �rst time it is customany to begin with geometric vectors in 2� or 3�space,like in Section 1.1 of Friedberg. Vectors are added geometrically by means of the parallelogram law, and scalarmultiplication by a real number r is done by stretching or contracting the vector by a factor of jr j ; depending onwhether jr j > 1 or jr j < 1; respectively, and by also reversing the direction if r < 0: The next step is to generalizeto the vector spaces Rn; n � 1; where the vectors consist of n�tuples of real numbers and the scalars consist of realnumbers. Addition of vectors is done component wise, as is scalar multiplication by a real number.

The next step in the generalization process is general vector spaces over the real numbers. Here the vectors maycome from an arbitrary nonempty set V on which an addition and a scalar multiplication with real numbers arede�ned, and which satisfy properties V S1 � V S8 in Section 1.2 of Friedberg, with F representing the set of real

9 MAT3701/1

numbers. With each of these generalizations it is possible to apply vector space principles to a larger class ofproblems. In this module, we take the generalizations one step further by allowing the scalars to come from any�eld. Thus far, the scalars have always been real numbers. Now they may come from any set F in which we canadd, subtract, multiply, and divide � things we normally do with numbers. Formally, F should satisfy the arithmeticrules F1 � F5 stated in Friedberg: Appendix C. Such an algebraic structure F is called a �eld. The rationalnumbers Q; real numbers R; and complex numbers C are all �elds. For the purposes of this module we restrictourselves to F being R or C (unless explicitly stated otherwise, F may be either of the two). Thus, in our context,a vector space V over F is a set V of vectors on which an addition is de�ned as well as a scalar multiplication withscalars from F; where F may be the real or complex numbers, such that properties V S1 � V S8 in Section 1.2 ofFriedberg are satis�ed.

Why allowing complex numbers as scalars? One of the reasons you have actually already encountered, and has todo with �nding the eigenvalues of a matrix (or linear operator).

For example,

A D

"0 1�1 0

#is a real matrix, yet its eigenvalues are complex, namely � D �i: This means that A cannot be diagonalised over

R; but, indeed, over C it can be; since for X D 1p2

"1 1i �i

#;

X�1AX D1p2

"1 �i1 i

#"0 1�1 0

#1p2

"1 1i �i

#D

"i 00 �i

#

And being diagonalizable is often very useful. One of the reasons why this works overC is because the characteristicequation of A can be solved over C: This is true for any polynomial equation over C; and for this reason C is calledalgebraically closed. This example demonstrates that R; on the other hand, is not algebraically closed. It can beproved that this property of C causes every matrix over C to be similar to a triangular matrix, which is another veryuseful property. Again, the above example demonstrates that this is not the case over R, since then the eigenvaluesof R would have been real:

The following example provides some practice with vector spaces over R and C:

2.3.1 Example

In each of the following cases, determine whether the set V is a vector space over the indicated �eld F: Unlessstated otherwise, operations are the usual ones.

(a) V D C; F D R

(b) V D R; F D C

(c) V D C2; F D R

(d) V D C2; F D C

10

(e) V D C2; F D C; with scalar multiplication de�ned by z .z1; z2/ D .zz1; zz2/ where z denotes the complexconjugate of z:

(f) V D C2; F D C; with scalar multiplication de�ned by z .z1; z2/ D .zz1; zz2/

SOLUTION

(a) Yes. Axioms V S1� V S8 follow from the properties of complex arithmetic.

(b) No. Scalar multiplication is not closed. For example, take v D 1 and z D i; then zv D i =2 V :

(c) and (d) Yes. Axioms V S1� V S8 follow from the properties of complex arithmetic.

(e) Yes. V S1� V S5 are easy to check, since the �rst four axioms do not involve scalar muliplication.

V S6 : .z1z2/ .z3; z4/ D .z1z2z3; z1z2z4/ D .Nz1 Nz2z3; Nz1 Nz2z4/

D z1 .Nz2z3; Nz2z4/ D z1 .z2 .z3; z4// :

V S7 : z�.z1; z2/C

�z3;z4

��D z .z1 C z3; z2 C z4/ D .Nz .z1 C z3/ ; Nz .z2 C z4//

D .Nzz1 C Nzz3; Nzz2 C Nzz4/ D .Nzz1; Nzz2/C .Nzz3; Nzz4/

D z .z1; z2/C z .z3; z4/

V S8 : .z1 C z2/ .z3; z4/ D�z1 C z2 z3; z1 C z2 z4

�D ..Nz1 C Nz2/ z3; .Nz1 C Nz2/ z4/

D .Nz1z3 C Nz2z3; Nz1z4 C Nz2z4/ D .Nz1z3; Nz1z4/C .Nz2z3; Nz2z4/

D z1 .z3; z4/C z2 .z3; z4/

(f) No. V S5 does not hold. E.g.,1 .i; i/ D .�i;�i/ 6D .i; i/

2.3.2 Remark

Whenever we refer to the vector spaces Fn; Mm�n .F/ ; F .S; F/ ; P .F/ ; or Pn .F/ we always assume F to bethe scalar �eld, unless explicitly stated otherwise.

2.4 Subspaces

The familiar Subspace Test to determine whether a subset W of a vector space V is a subspace of V (and hence avector space in its own right) appears in Friedberg: Theorem 1.3:

Subspace Test

A subset W of a vector space V is a subspace of V iff(ST 1) 0 2 W(ST 2) x C y 2 W whenever x 2 W and y 2 W(ST 3) cx 2 W whenever c 2 F and x 2 W

11 MAT3701/1

Examples 1�5 in Friedberg: Section 1.3 should be mostly familiar to you. Please read through them, keeping inmind that F may be the �eld of complex numbers.

2.4.1 Example

Determine whether or not each of the following is a subspace of V over F:

(a) W D�.z1; z2/ 2 C2 : z1z2 D 0

; V D C2; F D C

(b) W D fa C ib 2 C : a and b are rational numbersg ; V D C; F D C

(c) W D

(A 2 M2�2 .C/ : A

"11

#D

"00

#); V D M2�2 .C/ ; F D C:

SOLUTION

(a) No. Not closed under addition, e.g., .1; 0/ 2 W and .0; 1/ 2 W but .1; 0/C .0; 1/ D .1; 1/ =2 W:

(b) No. Not closed under scalar multiplication, e.g., 1C i 2 W andp2 2 C but

p2 .1C i/ D

p2C

p2i =2 W:

(c) Yes. ST 1:

"0 00 0

#2 W since

"0 00 0

#"11

#D

"00

#:

ST 2: Suppose A1; A2 2 W; hence A1

"11

#D A2

"11

#D

"00

#:

Then

.A1 C A2/

"11

#D A1

"11

#C A2

"11

#D

"00

#C

"00

#D

"00

#

so that A1 C A2 2 W:

ST 3: Suppose A 2 W and z 2 C; hence A

"11

#D

"00

#: Then

.zA/

"11

#D z

A

"11

#!D z

"00

#D

"00

#

so that zA 2 W:

2.5 Linear Dependence, Linear Independence, and Bases

2.5.1 Selecting a Basis from a Generating Set in Fn

In Example 6 of Friedberg: Section 1.6, a method is explained for selecting a basis from a generating set based onF2 and F3: Theorem 1.8 / F4: Theorem 1.7. The following alternative method is similar to F3 and F4: Section 3.4,Example 2, and is assumed to be familiar from a previous course.

12

Method for Selecting a Basis from a Generating Set in Fn

Write the generators as the columns of a matrix A; and reduce toechelon form R: The columns of A corresponding to the leadingcolumns of R will be a basis.

To illustrate this, we redo the example mentioned at the beginning according to this method:ExampleFriedberg: Section 1.6, Example 6 (alternative method)Given that

S D f.2;�3; 5/ ; .8;�12; 20/ ; .1; 0;�2/ ; .0; 2;�1/ ; .7; 2; 0/g

is a generating set of R3; select a basis for R3 from among the vectors of S:

SOLUTIONWrite the vectors in S as the columns of a matrix and reduce to echelon form.264 2 8 1 0 7

�3 �12 0 2 25 20 �2 �1 0

375!264 2 8 1 0 7�1 �4 1 2 91 4 �4 �1 �14

375 R2 C R1R3 � 2R1

!

264 0 0 3 4 25�1 �4 1 2 90 0 �3 1 �5

375 R1 C 2R2

R3 C R2

!

264 0 0 3 4 251 4 �1 �2 �90 0 0 5 20

375 �R2R3 C R1

!

264 1 4 �1 �2 �90 0 1 4

3253

0 0 0 1 4

375 R213 R115 R3

The last matrix is in echelon form. Its leading columns are 1; 3 and 4; since the leading 1's appear in these columns.Thus, columns 1; 3 and 4 of the original matrix form a basis, namely,

f.2;�3; 5/ ; .1; 0;�2/ ; .0; 2;�1/g :

The following provides some practice in linear dependence and independence over R and C:

2.5.2 Examples

1. Suppose � D fv1; v2; v3g is a linearly independent subset of R3 (considered as a vector space over R/:

(a) Show that R3 is a subset of C3:

(b) Is R3 a subspace of C3; considered as a vector space over C? Give reasons your answer.

(c) Considering � as a subset of C3; is it linearly independent over C? If yes, supply a proof. If no, supplya counter example.

SOLUTION

13 MAT3701/1

(a) Every element v 2 R3 is of the form

v D .r1; r2; r3/ ; where r1; r2; r3 are real numbers.

Since each real number automatically is also a complex number, it follows that v also lies in C3: HenceR3 is a subset of C3:

(b) No. For example, .1; 0; 0/ 2 R3; but for i 2 C; i � .1; 0; 0/ D .i; 0; 0/ =2 R3; since i is not a real number.Hence, R3 is not closed under scalar multiplication with complex numbers.

(c) Here the question is whether

c1v1 C c2v2 C c3v3 D 0; with c1; c2; c3 complex, ... (i)

will imply that c1 D c2 D c3 D 0; given that this is the case when c1; c2; c3 are real.

Letc1 D a1 C ib1; c2 D c2 C ib2 and c3 D c3 C ib3; ... (ii)

where a1; a2; a3 and b1; b2; b3 are real, then (i) becomes

.a1 C ib1/ v1 C .a2 C ib2/ v2 C .a3 C ib3/ v3 D 0

i.e..a1v1 C a2v2 C a3v3/C i .b1v1 C b2v2 C b3v3/ D 0: ... (iii)

Since v1; v2; v3 2 R3 and a1; a2; a3 are real numbers, it follows that a1v1 C a2v2 C a3v3 2 R3 � say

a1v1 C a2v2 C a3v3 D .r1; r2; r3/ ; where r1; r2; r3 2 R: ... (iv)

Similarly

b1v1 C b2v2 C b3v3 D�r 01; r

02; r

03�; where r 01; r

02; r

03 2 R: ... (v)

Thus, (iii) becomes

.r1; r2; r3/C i�r 01; r

02; r

03�D�r1 C ir 01; r2 C ir

02; r3 C ir

03�D 0;

hence,r1 C ir 01 D r2 C ir

02 D r3 C ir

03 D 0;

and consequently, from the de�nition of equality of complex numbers,

r1 D r2 D r3 D r 01 D r02 D r

03 D 0;

since ri ; r 0i .1 � i � 3/ are all real numbers.

Substitution in (iv) and (v) yieldsa1v1 C a2v2 C a3v3 D 0

andb1v1 C b2v2 C b3v3 D 0:

14

Since fv1; v2; v3g is linearly independent over R and ai ; bi .1 � i � 3/ are real,

a1 D a2 D a3 D 0 and b2 D b2 D b3 D 0:

Thus, from (ii) it follows that c1 D c2 D c3 D 0, which shows that � D fv1; v2; v3g is also linearlyindependent over C:

2. This example illustrates the effect that different scalar �elds can have on a vector space. Let V denotethe vector space C2 over the complex �eld C; and let W denote the vector space C2 over the real �eld R:(According to Friedberg: Section 1.2, Example 1, V is a vector space over C: It can be similarly shown thatW is a vector space over R: Note that W is not a subspace of V; since they have different scalar �elds.)

(a) Find a basis for V :

(b) Find a basis for W:

(c) Show that a D .0; 2/ is a linear combination of S D f.1; i/ ; .i; 1/g in V :

(d) Show that (c) is false in W:

SOLUTION

(a) The set� D f.1; 0/ ; .0; 1/g

is a basis of V; since for .x; y/ 2 V;

.x; y/ D .x; 0/C .0; y/

D x .1; 0/C y .0; 1/ ; ... (i)

where x; y 2 C: Thus, � generates V : It is also linearly independent, since

x .1; 0/C y .0; 1/ D 0

iff .x; 0/C .0; y/ D 0

iff .x; y/ D 0

iff x D y D 0

(b) The set � in (a) is not a basis of W; since in (i), the scalars x and y are complex. To express .x; y/ 2 Was a linear combination of �xed vectors in W with real scalars, let

x D x1 C i x2 and y D y1 C iy2;

where x1; x2; y1; y2 are real. Hence,

.x; y/ D .x1 C i x2; y1 C iy2/

D .x1 C i x2; 0/C .0; y1 C iy2/

D .x1; 0/C .i x2; 0/C .0; y1/C .0; iy2/

D x1 .1; 0/C x2 .i; 0/C y1 .0; 1/C y2 .0; i/ :

15 MAT3701/1

Since x1; x2; y1; y2 are real, D f.1; 0/ ; .i; 0/ ; .0; 1/ ; .0; i/g

generates W:

To check whether is linearly independent over R; suppose

r1 .1; 0/C r2 .i; 0/C r3 .0; 1/C r4 .0; i/ D 0;

where r1; r2; r3; r4 are real. Writing the left side as a single 2�tuple, we �nd that

.r1 C ir2; r3 C ir4/ D 0;

so thatr1 C ir2 D 0 and r3 C ir4 D 0;

and thereforer1 D r2 D 0 and r3 D r4 D 0;

since r1 C ir2 and r3 C ir4 represent complex numbers because r1; r2; r3; r4 are real. Thus, is linearlyindependent over R; and therefore it is a basis of W: It follows that W is a four�dimensional vectorspace while, according to (a), V is a two�dimensional vector space.

(c) For a D .0; 2/ to be a linear combination of S in V; we must �nd complex numbers x and y such that

.0; 2/ D x .1; i/C y .i; 1/ : ... (ii)

Equating corresponding coordinates on both sides, we have that

0 D x C iy2 D i x C y

)

Solve by Gaussian elimination:"1 i : 0i 1 : 2

#!

"1 i : 00 2 : 2

#R2 � i R1

!

"1 i : 00 1 : 1

#12 R2

!

"1 0 : �i0 1 : 1

#R1 � i R2

Thus, x D �i; y D 1 is a solution; and the result follows.

(d) For the result to be true in W; we must �nd real solutions for x and y in (ii). But from the explanationabove it follows that (ii) has only one solution, which turns out to be complex. Thus, the result is falsein W .

16

2.6 Lagrange Interpolation Formula

This topic appears in Friedberg: Section 1.6.

2.6.1 Examples

(1) Friedberg: Section 1.6, remark at the end of "The Lagrange Interpolation Formula"

If f 2 Pn .F/ and f .ci / D 0 for n C 1 distinct elements c0; c1; : : : ; cn in F; then f is the zero function.

SOLUTION

f 2 Pn .F/ can be uniquely expressed in terms of its values at c0; c1; : : : ; cn as

f .x/ D f .c0/ f0 .x/C f .c1/ f1 .x/C : : :C f .cn/ fn .x/ :

Since each f .ci / D 0; the equation becomes

f .x/ D 0 f0 .x/C 0 f1 .x/C : : :C 0 fn .x/ D 0I

hence, f is the zero polynomial.

(2) (a) Calculate the Lagrange polynomials f0 .x/ ; f1 .x/ and f2 .x/ associated with �1; 0 and 1 respectively.

(b) Express a C bx C cx2 2 P2 .C/ as a linear combination of f0 .x/ ; f1 .x/ and f2 .x/ calculated in (a).

(c) Is it possible to construct a polynomial g 2 P2 .C/ such that g .�1/ D g .0/ D g .1/ D 0 and g .2/ D 1?

SOLUTION

(a) Let c0 D �1; c1 D 0 and c2 D 1: According to the de�nition of the Lagrange polynomials,

f0 D.x � c1/ .x � c2/.c0 � c1/ .c0 � c2/

Dx .x � 1/.�1/ .�2/

D12x .x � 1/ I

f1 .x/ D.x � c0/ .x � c2/.c1 � c0/ .c1 � c2/

D.x C 1/ .x � 1/.1/ .�1/

D � .x C 1/ .x � 1/ I

17 MAT3701/1

f2 .x/ D.x � c0/ .x � c1/.c2 � c0/ .c2 � c1/

D.x C 1/ x2 � 1

D12x .x C 1/ :

(b) According to the Lagrange interpolation formula, g .x/ D a C bx C cx2 can be expressed as

g .x/ D g .c0/ f0 .x/C g .c1/ f1 .x/C g .c2/ f2 .x/

D g .�1/ f0 .x/C g .0/ f1 .x/C g .1/ f2 .x/

.c0 D �1I c1 D 0I c2 D 1/

D .a � b C c/ f0 .x/C a f1 .x/C .a C b C c/ f0 .x/

D.a � b C c/

2x .x � 1/� a .x C 1/ .x � 1/C

.a C b C c/2

x .x C 1/

A direct calculation of the right hand side veri�es this representation of g .x/ :

(c) According to the Lagrange interpolation formula each polynomial g .x/ in P2 .C/ can be uniquely ex-pressed as

g .x/ D g .c0/ f0 .x/C g .c1/ f1 .x/C g .c2/ f2 .x/ ; ... (i)

where c0; c1 and c2 are distinct elements in C: We may choose, for example, c0 D �1; c1 D 0 andc2 D 1; then (i) becomes

g .x/ D g .�1/ f0 .x/C g .0/ f1 .x/C g .1/ f2 .x/

Hence, for g .x/ to be such that g .�1/ D g .0/ D g .1/ D 0; we must have that

g .x/ D 0 � f0 .x/C 0 � f1 .x/C 0 � f2 .x/ D 0;

that is, g .x/ must be the zero polynomial. Thus g .2/ D 0; and so it is impossible to have g .2/ D 1; asstated in the question. Thus, a polynomial g .x/ satisfying the conditions as stated does not exist.

2.7 Sum and Direct Sum of Subspaces

Friedberg has delegated this topic to the exercises. Because of this, and since it is important to us later on, we treatit in more detail here.

Let W1 and W2 be subspaces of a vector space V over F: Can we �nd a �smallest� subspace of V containing bothW1 and W2? If we look at W1 and W2 simply as sets, then W1 [W2 is the smallest subset of V containing both W1and W2, but in general this is not a subspace, as the following result shows.

18

2.7.1 Example

F2: Section 1.3, Exercise 18 / F3 and F4: Section 1.3, Exercise 19Let W1 and W2 be subspaces of a vector space V : Prove that W1 [W2 is a subspace of V if and only if W1 � W2 orW2 � W1:

SOLUTIONIf W1 � W2; then W1 [ W2 D W2, which we already know is a subspace of V : Similarly, if W2 � W1; thenW1 [W2 D W1 is a subspace of V :

Conversely, suppose W1 [W2 is a subspace of V :

Let us proceed by reductio ad absurdum, that is, by assuming the conclusion we wish to reach is false, and bydeducing a contradiction. So, in this case we assume neither W1 � W2 nor W2 � W1 holds. Then there must exist avector w1 2 W1 such that w1 =2 W2 and a vector w2 2 W2 such that w2 =2 W1: Since we assume that W1 [ W2 is asubspace, w1 C w2 must lie in W1 [W2; that is, w1 C w2 must lie in W1 or in W2: But, if w D w1 C w2 lies in W1;then w2 D w�w1 2 W1 (since W1 is a subspace and w;w1 2 W1/: This is impossible, since we assumed w2 =2 W1:Similarly, if w 2 W2; then w1 D w � w2 2 W2; which is also impossible. We have reached a contradiction, andtherefore either W1 � W2 or W2 � W1 holds. This completes the proof.

As proved in the following example, it turns out that the smallest subspace of V containing both W1 and W2 is thesum of W1 and W2; namely,

W1 CW2 D fw1 C w2 : w1 2 W1; w2 2 W2g

(see the de�nition before F2: Section 1.3, Exercise 21 / F3 and F4: Section 1.3, Exercise 23).

2.7.2 Example

F2: Section 1.3, Exercise 21 / F3 and F4: Section 1.3, Exercise 23Let W1 and W2 be subspaces of a vector space V :

(a) Prove that W1 CW2 is a subspace of V that contains both W1 and W2:

(b) Prove that any subspace of V that contains both W1 and W2 must also contain W1 CW2:

SOLUTIONLet

W D W1 CW2 D fw1 C w2 : w1 2 W1 and w2 2 W2g

(a) We show that W satis�es the three subspace conditions of Friedberg: Theorem 1.3.

1.3 (a): Since 0 lies in W1 and in W2 (both are subspaces), and 0 D 0C 0; it follows that 0 2 W:

1.3 (b): Suppose x; y 2 W: Hence, x D w1Cw2 and y D w01Cw02 for somew1; w01 2 W1 and w2; w02 2 W2:Then,

x C y D .w1 C w2/C�w01 C w

02�

D�w1 C w

01�C�w2 C w

02�;

which lies in W; since w1 C w01 2 W1 and w2 C w02 2 W2:

19 MAT3701/1

1.3 (c): Suppose c 2 F and x 2 W: Hence, x D w1 C w2 for some w1 2 W1 and w2 2 W2: Then,

cx D c .w1 C w2/ D .cw1/C .cw2/ 2 W;

since cw1 2 W1 and cw2 2 W2:Since W satis�es all three conditions, it is a subspace of V by Friedberg: Theorem 1.3.To show that W1 lies in W; consider any w1 2 W1: Since w1 can be expressed as w1 D w1 C 0; with0 2 W2; it follows that w1 2 W: Hence,

W1 � W D W1 CW2

Similarly, W2 � W:

(b) LetU be any subspace of V that contains both W1 and W2. We must show that W D W1CW2 � U: Take anyw 2 W: Hence, w D w1 Cw2 for some w1 2 W1 and w2 2 W2: Since U contains both W1 and W2; it followsthat w1; w2 2 U; and therefore w D w1 C w2 2 U; since U is a subspace. Since w was chosen arbitrarily, itfollows that

W D W1 CW2 � U

If, in addition, W1 \W2 D f0g for subspaces W1 and W2 of V; than we refer to the sum of W1 and W2 as the directsum of W1 and W2; and write W1 � W2: Thus, for a subspace W of V; W D W1 � W2 if W D W1 C W2 andW1 \W2 D f0g :

2.7.3 Examples

(1) F2: Section 1.3, Exercise 22 / F3 and F4: Section 1.3, Exercise 24

Show that Fn is the direct sum of the subspaces

W1 D�.a1; a2; : : : ; an/ 2 Fn : an D 0

and

W2 D�.a1; a2; : : : ; an/ 2 Fn : a1 D a2 D : : : D an�1 D 0

SOLUTION

Clearly, W1 C W2 � Fn; since W1 and W2 are subspaces of Fn: But also Fn � W1 C W2; since anyx D .a1; a2; : : : ; an/ in Fn can be expressed as

x D .a1; a2; : : : ; an�1; 0/C .0; 0; : : : ; 0; an/ ;

where.a1; a2; : : : ; an�1; 0/ 2 W1 and .0; 0; : : : ; 0; an/ 2 W2

Hence, Fn D W1 C W2: Turning to the second condition for a direct sum, let x 2 W1 \ W2: Since x 2 W1; itcan be expressed as

x D .a1; a2; : : : ; an�1; 0/ for some a1; : : : ; an�1 2 F; ... (i)

20

and since x 2 W2; it can be expressed as

x D .0; 0; : : : ; 0; an/ for some an 2 F ... (ii)

Comparing the corresponding entries of x in (i) and (ii), it follows that

a1 D a2 D : : : D an D 0

Hence, x D 0; and therefore W1 \W2 D f0g : Thus, we have that

Fn D W1 �W2

(2) Let V denote the vector space of all complex numbers over the �eld F D R of real numbers. Let

W1 D fx 2 V : x is realg ;

andW2 D fi x 2 V : x is realg

(a) Show that W1 and W2 are subspaces of V :

(b) Show that V D W1 �W2:

SOLUTION

Note that the vectors of V consist of all complex numbers, that is, numbers of the form a C ib with a; breal, while the scalars (i.e., the elements of F/ are restricted to being real numbers only. The additionof vectors is the ordinary addition of complex numbers, while scalar multiplication is the ordinarymultiplication of complex numbers, except that the scalar on the left must be real.

(a) In both cases we check whether the conditions of Friedberg: Theorem 1.3 are satis�ed, beginning withW1: Firstly, since 0 is a real number, 0 2 W1: Secondly, suppose x; y 2 W1; that is, x and y are real.Since the sum of two real numbers is again real, x C y 2 W1: Thirdly, suppose x 2 W1 and c 2 F:According to the de�nitions of W1 and F; both x and c are real, so that cx is real, since the product oftwo real numbers is again real. Thus, cx 2 W1; and we conclude that W1 is a subspace of V :

Turning to W2; note, �rstly, that 0 2 W2; since it can be expressed as 0 D i0; and 0 is real. Secondly,suppose i x; iy 2 W2; with x and y real. Then i xC iy D i .x C y/ ; which clearly lies in W2; since xC yis real, being the sum of two real numbers. Thirdly, suppose c 2 F and i x 2 W2 ; with x real. Thenc .i x/ D i .cx/ 2 W2; since cx is real, being the product of two real numbers. Thus, we conclude byFriedberg: Theorem 1.3 that W2 is a subspace of V :

(b) According to the de�nition, V D W1 � W2 if V D W1 C W2 and W1 \ W2 D f0g (from (a), we alreadyknow that W1 and W2 are subspaces of V /: Since any v 2 V is a complex number, it can be expressedas

v D a C ib; with a; b real numbers

Thus a 2 W1 and ib 2 W2; so that v 2 W1 CW2: Since v was chosen arbitrarily in V; we conclude thatV � W1 CW2, and since, obviously, W1 CW2 � V; we have that V D W1 CW2:

21 MAT3701/1

Finally, suppose w 2 W1 \ W2: Then, since w 2 W1; it is real, and since w 2 W2; it can be expressedas w D ib; with b real. Writing the last equation as wC i .�b/ D 0 D 0C i0; and keeping in mind thatw and �b are real, it follows from the equality of complex numbers that w D �b D 0: Since w waschosen arbitrarily in W1 \W2; we conclude that W1 \W2 D f0g : Thus, V D W1 �W2; as required.

(3) Let

W1 D

(A 2 M2�2 .C/ : A

"11

#D

"00

#)

and

W2 D

(A 2 M2�2 .C/ : A

"1

�1

#D

"00

#)

(a) Show that W1 is a subspace of M2�2 .C/ :

(b) Assuming that W2 is also a subspace of M2�2 .C/ ; express"a bc d

#2 M2�2 .C/

as the sum of a vector in W1 and a vector in W2:

(c) Show thatM2�2 .C/ D W1 �W2

SOLUTION

(a) Note that the vectors of W1 are complex 2 � 2 matrices. Apply the Subspace Test given by Friedberg:Theorem 1.3. The �rst condition states that W1 must contain the zero vector, which in this case is"

0 00 0

#

Since, obviously, "0 00 0

#"11

#D

"0 � 1C 0 � 10 � 1C 0 � 1

#D

"00

#;

it follows that "0 00 0

#2 W1

To check the second condition, suppose A; B 2 W1: Hence,

A

"11

#D

"00

#and B

"11

#D

"00

#... (i)

For A C B to also lie in W1; it must satisfy the de�ning property of W1; namely,

.A C B/

"11

#D

"00

#... (ii)

22

Since

.A C B/

"11

#D A

"11

#C B

"11

#;

D

"00

#C

"00

#; from (i)

D

"00

#;

it follows that property (ii) is satis�ed: hence A C B 2 W1:

For the third subspace condition, we must show that zA 2 W1 whenever z 2 C and A 2 W1: Now,

.zA/

"11

#D z

A

"11

#!

D z

"00

#; since A 2 W1

D

"00

#I

hence zA 2 W1:

Since W1 satis�es all three conditions of the Subspace Test (Friedberg: Theorem 1.3), we conclude thatit is a subspace of M2�2 .C/ :

(b) First we need deeper insight into the structure of the matrices contained in W1 and W2: If we writeA 2 M2�2 .C/ as A D [A1 A2] where A1; A2 denote the �rst and second columns of A respectively,then

A D [A1 A2] 2 W1 , A

"11

#D [A1 A2]

"11

#D A1 C A2 D 0

, A D [A1 � A1]

Show, similarly, that A 2 W2 iff A D [A1 A1] : Hence, a matrix in W1 is of the form [X � X ]where X 2 C2; and a matrix in W2 is of the form [Y Y ] where Y 2 C2: Thus,"

a bc d

#D [X � X ]C [Y Y ] ... (iii)

D [X C Y � X C Y ]

,

"ac

#D X C Y and

"bd

#D �X C Y ... (iv)

23 MAT3701/1

To solve for X and Y in terms of a; b, c and d; add the last two equations:

2Y D

"ac

#C

"bd

#

D

"a C bc C d

#;

thus

Y D12

"a C bc C d

#;

and, from the �rst equation in (iv),

X D

"ac

#� Y

D

"ac

#�12

"a C bc C d

#

D12

"a � bc � d

#

Substituting these values for X and Y in (iii) yields"a bc d

#D12

"a � b � .a � b/c � d � .c � d/

#C12

"a C b a C bc C d c C d

#;

where the �rst matrix on the right belongs to W1 and the second to W2:

(c) By (b) we already have that M2�2 .C/ D W1 CW2. All that remains to show is that

W1 \W2 D

("0 00 0

#)... (v)

Let A 2 W1 \ W2: In (b) we showed that A 2 W1 iff A D [X � X ] for some X 2 C2; and A 2 W2iff A D [Y Y ] for some Y 2 C2: Thus,

A D [X � X ] D [Y Y ] ;

which implies thatX D Y and � X D Y

Adding these two equations yields"00

#D 2Y; that is, Y D

"00

#

Hence

A D [Y Y ] D

"0 00 0

#;

which proves (v). Since, also, M22 .C/ D W1 CW2; we conclude that M22 .C/ D W1 �W2:

24

(4) F2: Section 1.3, Exercise 28 / F3 and F4: Section 1.3, Exercise 30

Let W1 and W2 be subspaces of a vector space V : Prove that V D W1 � W2 iff each vector in V can beuniquely written as x1 C x2; where x1 2 W1 and x2 2 W2:

SOLUTION

Suppose V D W1 � W2: Let v 2 V be arbitrary. From the de�nition of direct sum, V D W1 C W2; so thatv D w1 C w2 for some w1 2 W1 and w2 2 W2: To show the expression is unique, suppose that also

v D w01 C w02; with w

01 2 W1 and w02 2 W2

Then,v D w1 C w2 D w

01 C w

02

implies thatw1 � w

01 D w

02 � w2 D x (say)

According to the left side, x lies in W1; and according to the right side, x 2 W2; so that x 2 W1 \ W2 Df0g : Thus w1 D w01 and w2 D w02; which completes the �rst part.

Conversely, suppose that each v 2 V can be uniquely expressed as v D w1 C w2: Then, clearly, V D

W1 CW2: Further, if x 2 W1 \W2; then

0 D x C .�x/ D 0C 0

are two expressions of 0 as the sum of a vector fromW1 and a vector fromW2: From the uniqueness condition,it follows that x D 0; so that W1 \W2 D f0g : This completes the second part, and hence the proof.

(5) Show that C3 is the direct sum of the subspaces

W1 D span .f.1; 0; 1/g/

andW2 D span .f.1; 1; 0/ ; .0; 1; 1/g/

SOLUTION

To show that C3 D W1CW2; we need to establish that any vector x D .a; b; c/ in C3 can be expressed in theform x D w1 C w2; where w1 2 W1 and w2 2 W2: According to the de�nition of span we may write

W1 D fa1 .1; 0; 1/ : a1 2 Cg

andW2 D fa2 .1; 1; 0/C a3 .0; 1; 1/ : a2; a3 2 Cg

25 MAT3701/1

Hence we must show that

x D .a; b; c/ D a1 .1; 0; 1/C a2 .1; 1; 0/C a3 .0; 1; 1/ for some a1; a2; a3 2 C ... (i)

Writing this as a system of linear equations yields

a1 C a2 D aa2 C a3 D b

a1 C a3 D c

9>=>;In matrix form: 264 1 1 0 : a

0 1 1 : b1 0 1 : c

375!264 1 1 0 : a0 1 1 : b0 �1 1 : �a C c

375R3 � R1

!

264 1 1 0 : a0 1 1 : b0 0 2 : �a C b C c

375R3 C R2

!

264 1 0 �1 : a � b0 1 1 : b0 0 1 : 1

2 .�a C b C c/

375 R1 � R2

12 R3

!

264 1 0 0 : 12 .a � b C c/

0 1 0 : 12 .a C b C c/

0 0 1 : 12 .�a C b C c/

375 R1 C R3R2 � R3

Hence, system (i) has a unique solution, namely,

a1 D 12 .a � b C c/

a2 D 12 .a C b � c/

a3 D 12 .�a C b C c/

9>=>;It follows that C3 D W1 C W2: Since we have that the solution to equation (i) is unique, it follows from theprecious example that C3 D W1 �W2:

2.8 The Direct-Sum TestsSG: Examples 2.7.3.1�2.7.3.3 illustrate how a direct sum decomposition of a vector space can be established from�rst principles (that is, directly from the de�nition of the direct sum).

The following tests are also useful, as some of the examples in the following sections will demonstrate.

2.8.1

First Direct-Sum TestLet �1 and �2 be disjoint bases for subspaces W1 and W2; respectively, of a �nite�dimensional vector space V :If �1 [ �2 is a basis for V; then V D W1 �W2:

26

PROOF (F2: Section 1.6, Exercise 23(b) / F3: Section 1.6, Exercise 31(b) / F4: Section 1.6, Exercise 33(b))Since �1 [ �2 is a basis for V;

V D span��1 [ �2

�D span

��1�C span

��2�

(show)

D W1 CW2; ... (i)

since �1 and �2 are bases for W1 and W2 respectively.Also, if we denote by jSj the number of elements in a �nite set S; then

dim .V / D��1 [ �2�� ; since �1 [ �2 is a basis for V

D��1��C ��2�� ; since �2 and �2 are disjoint

D dim .W1/C dim .W2/ ; ... (ii)

since �1 and �2 are bases for W1 and W2 respectively.By the dimension formula for the sum of subspaces,

dim .V / D dim .W1 CW2/ ; from (i)

D dim .W1/C dim .W2/� dim .W1 \W2/ ... (iii)

A comparison of (ii) and (iii) yieldsdim .W1 \W2/ D 0;

hence,W1 \W2 D f0g ... (iv)

From (i) and (iv) we conclude thatV D W1 �W2;

as required.

2.8.2Second Direct-Sum Test

Let W1 and W2 be �nite�dimensional subspaces of a vector space V; and let V D W1 CW2:Then V is the direct sum of W1 and W2 if and only if dim .V / D dim .W1/C dim .W2/ :

SOLUTIONF2: Section 1.6, Exercise 20(b) / F3: Section 1.6, Exercise 27(b) / F4: Section 1.6, Exercise 19(b)

Since it is given that V D W1 CW2;

V D W1 �W2 iff W1 \W2 D f0g ... (i)

According to the Dimension Formula for the Sum of Subspaces,

dim .V / D dim .W1 CW2/

D dim .W1/C dim .W2/� dim .W2 \W2/

Hence, from (i),V D W1 �W2 iff dim .V / D dim .W1/C dim .W2/

27 MAT3701/1

2.8.3 Examples

(1) F2: Section 1.6, Exercise 23(a) / F3: Section 1.6, Exercise 31(a) / F4: Section 1.6, Exercise 3(a)

Let W1 and W2 be subspaces of a �nite�dimensional vector space V such that V D W1 � W2: If �1 and �2are bases for W1 and W2; respectively, show that �1 \ �2 D � and �1 [ �2 is a basis for V :

SOLUTION

Since �1 � W1 and �2 � W2; it follows that

�1 \ �2 � W1 \W2 D f0g

Since a basis cannot contain the zero vector,

�1 \ �2 D � ... (i)

Also,

V D W1 CW2D span

��1�C span

��2�

D span��1 [ �2

�; ... (ii)

and

dim .V / D dim .W1/C dim .W2/ ; previous example

D��1��C ��2�� .jSj denotes the number of vectors in S/

D��1 [ �2�� ; by (i).

Hence, it follows from (ii) that �1 [ �2 is a basis for V :

(2) F2: Section 1.6, Exercise 24 / F3: Section 1.6, Exercise 32 / F4: Section 1.6, Exercise 34(a)

Prove that if W1 is any subspace of a �nite�dimensional vector space V; then there exists a subspace W2 of Vsuch that V D W1 �W2:

SOLUTION

Let �1 be basis of W1: Since V is �nite�dimensional, �1 is �nite and can be extended to a basis �1 [�2 of V;where �1 and �2 are disjoint. Let W2 D span

��2�: Then �2 is a basis of W2; and by the First Direct-SumTest,

V D W1 �W2:

(3) Let A D

0 11 0

!; and consider the following two subsets of M2�2 .C/ ; the vector space of 2� 2 matrices

over the complex numbers C :S D fX 2 M2�2 .C/ j AX D Xg

andS0 D fX 2 M2�2 .C/ j AX D �Xg

28

(a) Show that S is a subspace of M2�2 .C/ :

(b) Determine a basis of S:

(c) Given that S0 is a subspace of M2�2 .C/ ; determine a basis of S0:

(d) Determine whether M2�2 .C/ D S � S0:

SOLUTION

(a) Since AO2�2 D O2�2; where O2�2 denotes the 2 � 2 zero matrix, it follows that O2�2 2 S: SupposeX; Y 2 S; that is, X and Y are 2� 2 complex matrices, and AX D X and AY D Y: Then

A .X C Y / D AX C AY (matrix multiplication distributes over addition)

D X C Y; by the above assumptions.

Thus, X C Y 2 S: Finally, suppose � 2 C and X 2 S: Then

A .�X/ D �AX

D �X; since AX D X:

Thus, �X 2 S:Hence, S satis�es the three requirements for a subspace, and the solution is complete.

(b) Let X D

"a bc d

#: Then

AX D X ,

"0 11 0

#"a bc d

#D

"a bc d

#

,

"c da b

#D

"a bc d

#

, a D c and b D d

Thus,

X 2 S , X D

"a ba b

#; where a; b 2 C

Since

X D

"a 0a 0

#C

"0 b0 b

#

D a

"1 01 0

#C b

"0 10 1

#;

it follows that S is generated by

� D

("1 01 0

#;

"0 10 1

#)I

29 MAT3701/1

which is also linearly independent, since

a

"1 01 0

#C b

"0 10 1

#D

"0 00 0

#

,

"a ba b

#D

"0 00 0

#, a D b D 0

Thus, � is a basis of S:

(c) Following the same procedure for S0; let

X D

"a bc d

#

Then

X 2 S0 , AX D �X

,

"0 11 0

#"a bc d

#D �

"a bc d

#

,

"c da b

#D

"�a �b�c �d

#

, a D �c and b D �d

, X D

"a b

�a �b

#

Since

X D

"a 0

�a 0

#C

"0 b0 �b

#

D a

"1 0

�1 0

#C b

"0 10 �1

#;

it follows that S0 is generated by

� 0 D

("1 0

�1 0

#;

"0 10 �1

#);

which also turns out to be a basis (the veri�cation of which is left to you).

(d) Apply the First Direct-Sum test.

We need to show that

� [ � 0 D

("1 01 0

#;

"0 10 1

#;

"1 0

�1 0

#;

"0 10 �1

#)

30

is a basis of M2�2 .C/ ; with � and � 0 as obtained in (b) and (c) respectively. Since � [ � 0 has fourmatrices, and we know that dim .M2�2 .C// D 4; it suf�ces to show that � [ � 0 is linearly independent.Now,

a

"1 01 0

#C b

"0 10 1

#C c

"1 0

�1 0

#C d

"0 10 �1

#D

"0 00 0

#

,

"a C c b C da � c b � d

#D

"0 00 0

#

,

(a C c D 0a � c D 0

and

(b C d D 0b � d D 0

:

Adding the two equations in the �rst system yields 2a D 0; that is, a D 0; and hence c D 0 from thesecond equation. Thus a D c D 0I and similarly b D d D 0:

It follows that � [ � 0 is a basis of M2�2 .C/ ; and hence by the Direct Sum Test, M2�2 .C/ D S � S0:

31 MAT3701/1

.

STUDY UNIT 3.

LINEAR TRANSFORMATIONS OVER THEREAL AND COMPLEX NUMBERS

(Friedberg: Sections 2.1 � 2.5)

3.1 Overview

The basic properties of linear transformations are revised. Since these properties are familiar from a previouscourse in linear algebra over the real scalar �eld, and since the proofs in the case of the complex scalar �eld areessentially the same, most of the material is assumed to be known except for the topics treated in this study unit,which should be studied for exam purposes. Many of the examples in this study unit involve complex numbers inorder to familiarize you with working with linear transformations over the complex numbers.

3.2 What to Study

Read through Friedberg: Sections 2.1�2.5 and study this unit in detail � the section on projections may not be sofamiliar to you.

3.3 Linear Transformations

Take note of the examples in Friedberg concerning rotation, re�ection and projection (F2: Section 2.1, Examples5-7/F3 and F4: Section 2.1, Examples 2�4) as these types of operators will play a special role later on.

3.3.1 Examples

(1) (a) Show thatT : C2 ! C de�ned by T .x; y/ D x C iy

is a linear transformation.

(b) Show thatT : C2 ! C de�ned by T .x; y/ D xy

is not a linear transformation.

32

SOLUTIONNote: When a vector space of the form Fn is under discussion, the scalar �eld is assumed to be F;unless explicitly stated otherwise. Thus in each of the above cases, the scalar �eld will be C:

(a) Let v D .x; y/ and w D�x 0; y0

�be vectors in C2; and let c be a complex number. Then

T .v C w/ D T�x C x 0; y C y0

�D x C x 0 C i

�y C y0

�D x C iy C x 0 C iy0

D T .v/C T .w/

and

T .cv/ D T .cx; cy/

D cx C i .cy/

D c .x C iy/

D cT .v/

Thus, T is a linear transformation.

(b) We need to show that one of the conditions for a linear transformation fails. The simplest would bescalar multiplication. Take, for example, i 2 C and v D .x; y/ 2 C2: Then

T .iv/ D T .i x; iy/

D .i x/ .iy/

D �xy

and

iT .v/ D iT .x; y/

D i xy

Hence,T .iv/ 6D iT .v/ ;

so T is not a linear transformation.

(2) Let V denote the vector space C2 with scalar multiplication over the real numbers R: De�ne T : V ! V by

T .x; y/ D .x � x; y � y/ ; where .x; y/ 2 V D C2

(a) Show that T is a linear transformation over R:

(b) Find a basis for N .T / :

(c) Find a basis for R .T / :

(d) Verify the Dimension Theorem for Linear Transformations in this case, namely,

nullity .T /C rank .T / D dim .V /

33 MAT3701/1

(e) Determine whether V D N .T /� R .T / :

SOLUTION

(a) Let v1 D .x1; y1/ and v2 D .x2; y2/ be vectors in V; and let r 2 R:

T .v1 C v2/ D T .x1 C x2; y1 C y2/

D�x1 C x2 � x1 C x2; y1 C y2 � y1 C y2

�; de�nition of T

D .x1 C x2 � x1 � x2; y1 C y2 � y1 � y2/ ; property of complex conjugation

D .x1 � x1; y1 � y1/C .x2 � x2; y2 � y2/

D T .v1/C T .v2/ ;

and

T .rv1/ D T .r x1; r y1/

D�r x1 � r x1; r y1 � r y1

�; de�nition of T

D�r x1 � r x1; r y1 � r y1

�; since r is real

D r�x1 � x1; y1 � y1

�D rT .v1/

Thus, T is a linear transformation, since its satis�es both properties above.

(b)

v D .x; y/ 2 N .T / , T .x; y/ D .0; 0/

, .x � x; y � y/ D .0; 0/

, x D x and y D y

, x and y are real

Thus,

N .T / D f.x; y/ 2 V : x; y are realg

Also, v can be expressed as

v D .x; y/ D x .1; 0/C y .0; 1/ ;

where x and y are real, that is, scalars in R:

Thus,

�1 D f.1; 0/ ; .0; 1/g

is a basis of N .T / :

(c) According to SG: Example 2.5.2.2(b),

� D f.1; 0/ ; .0; 1/ ; .i; 0/ ; .0; i/g ... (i)

34

is a basis of V : Thus, by Friedberg: Theorem 2.2,

R .T / D span .fT .1; 0/ ; T .0; 1/ ; T .i; 0/ ; T .0; i/g/

D span .f.0; 0/ ; .0; 0/ ; .2i; 0/ ; .0; 2i/g/

D span .f.i; 0/ ; .0; i/g/

Thus,

�2 D f.i; 0/ ; .0; i/g

is a basis of R .T / :

(d) Since dim .V / D 4; according to (i), dim .N .T // D 2; according to (b), and dim .R .T // D 2; accord-ing to (c), substitution of these values in the given equation veri�es the theorem.

(e) According to the First Direct Sum Test (SG: Section 2.8.1), it suf�ces to show that

�1 [ �2 D f.1; 0/ ; .0; 1/ ; .i; 0/ ; .0; i/g

is a basis of V : But this is exactly the statement in (i); thus the result follows.

3.4 Constructing Linear Transformations

Friedberg: Theorem 2.6 demonstrates that we have much freedom in the construction of linear transformations. Infact, we can take a basis of the domain and arbitrarily specify what their images should be in the codomain, in otherwords:

Constructing Linear TransformationsIf fv1; v2; : : : ; vng is a basis of a vector space V; and w1; w2; : : : ; wnare arbitrary vectors in a vector space W; then there exists a lineartransformation T : V ! W such that

T .vi / D wi for i D 1; 2; : : : ; nAlso, T is unique.

(Note that w1; w2; : : : ; wn need not be a basis for W � they do not even have to be distinct.)

3.4.1 Examples

(1) Let T : C3 ! C3 denote the linear transformation de�ned by

T .x; y; z/ D .x C iy; �i x C y; �z/

(where C3 is considered as a vector space over C/: Find the formula of a linear transformation U : C3 ! C3

such that UT D 0 and U .0; 1; 0/ D .0; 1; 0/ :

35 MAT3701/1

SOLUTION

Let v1 D .1; 0; 0/ ; v2 D .0; 1; 0/ ; v3 D .0; 0; 1/ denote the standard basis vectors of C3:

.UT / .x; y; z/ D U .T .x; y; z//

D U .x C iy; �i x C y; �z/

D U ..x C iy/ v1 C .�i x C y/ v2 � zv3/

D .x C iy/U .v1/C .�i x C y/U .v2/� zU .v3/

D x .U .v1/� iU .v2//C y .iU .v1/CU .v2//� zU .v3/

This is zero for all x; y and z in C iff

U .v1/� iU .v2/ D 0iU .v1/C U .v2/ D 0

U .v2/ D 0

9>=>;Since it is also give that U .v2/ D v2; it follows that

U .v1/ D iv2I

U .v2/ D v2I

U .v3/ D 0

To obtain the formula of U from this, note that

U .x; y; z/ D U .xv1 C yv2 C zv3/

D xU .v1/C yU .v2/C zU .v3/

D xiv2 C yv2D .0; xi C y; 0/

(2) Give an example of a linear transformation T : C2 ! C2 such that N .T / D R .T / :

SOLUTION

Suppose we take the standard basisf.1; 0/ ; .0; 1/g

and letN .T / D R .T / D span f.1; 0/g :

ThenT .1; 0/ D .0; 0/ ;

and by Friedberg: Theorem 2.2,

R .T / D span fT .1; 0/ ; T .0; 1/g

D span f.0; 0/ ; T .0; 1/g

D span fT .0; 1/g

36

Since we also have thatR .T / D span f.1; 0/g ;

chooseT .0; 1/ D .1; 0/

Now T is completely determined. In fact, we can easily derive its formula from the above information:

T .x; y/ D T .x .1; 0/C y .0; 1//

D xT .1; 0/C yT .0; 1/ ; T is linear

D x .0; 0/C y .1; 0/

D .y; 0/

3.5 Projections

Friedberg has delegated this topic to the exercises. Because of this, and since it is important to us later on, we treatit in more detail here. Note that the de�nition of a projection appears in the exercise set of Friedberg: Section 2.1.

3.5.1 Examples

(1) (Not in F2) / F3: Section 2.1, Exercise 23 / F4: Section 2.1, Exercise 25

Let T : R3 ! R3:

(a) If T .a; b; c/ D .a; b; 0/, show that T is the projection on the xy�plane along the z�axis.

(b) Find a formula for T .a; b; c/, where T represents the projection on the z�axis along the xy�plane.

(c) If T .a; b; c/ D .a � c; b; 0/ ; show that T is the projection on the xy�plane along the line L D

f.a; 0; a/ : a 2 Rg :

SOLUTION

(a) Since.a; b; c/ D .a; b; 0/C .0; 0; c/ ; ... (i)

where .a; b; 0/ lies in the xy�plane and .0; 0; c/ on the z�axis, it follows from the de�nition of T that itis the projection on the xy�plane along the z�axis.

(b) From (i) and the de�nition of the projection on the z�axis along the xy�plane, it follows that

T .a; b; c/ D .0; 0; c/

(c) Since.a; b; c/ D .a � c; b; 0/C .c; 0; c/ ;

where .a � c; b; 0/ lies in the xy�plane and .c; 0; c/ on the line L ; it follows that T is the projection onthe xy�plane along the line L :

37 MAT3701/1

(2) F2: Section 2.1, Exercises 22�24 / F3: Section 2.1, Exercise 24 / F4: Section 2.1, Exercise 26

Let T : V ! V be the projection on W1 along W2; where V D W1 �W2:

(a) Prove that T is linear and W1 D fx 2 V : T .x/ D xg :

(b) Prove that W1 D R .T / and W2 D N .T / :

(c) Describe T if W1 D V :

(d) Describe T if W1 is the zero subspace.

SOLUTION

(a) Since V D W1 �W2; the vectors x; y 2 V can be expressed as

x D w1 C w2 and y D w01 C w02; ... (i)

where w1; w01 2 W1 and w2; w02 2 W2: Let a 2 R; then

ax C y D a .w1 C w2/C w01 C w02

D�aw1 C w01

�C�aw2 C w02

�;

where aw1 C w01 2 W1 and aw2 C w02 2 W2; since W1 and W2 are subspaces. So

T .ax C y/ D aw1 C w01D aT .x/C T .y/ according to (i)

Thus, T is linear.

To show the second part, suppose, �rstly, T .x/ D x : From (i), T .x/ D w1; therefore x D w1; so thatx 2 W1: Conversely, if x 2 W1; writing x D x C 0 with x 2 W1 and 0 2 W2; we have T .x/ D x : Thus,

W1 D fx 2 V : T .x/ D xg

(b) Writing w1 D w1 C 0 and w2 D 0 C w2 where w1 2 W1 and w2 2 W2; we have T .w1/ D w1 andT .w1/ D 0: So, W1 � R .T / and W2 � N .T / : Conversely, from (i), T .x/ D w1I so R .T / � W1:Also, if T .x/ D 0; then w1 D 0; so that x 2 W2; and therefore N .T / � W2: We conclude thatW1 D R .T / and W2 D N .T / :

(c) If W1 D V; then for any x 2 V; x D x C 0; with x 2 W1 and 0 2 W2: So T .x/ D x D I .x/ ; andtherefore T D I:

(d) If W1 D f0g ; then from (i), w1 D 0 and w2 D x : So T .x/ D 0 D T0 .x/ ; and therefore T D T0:

38

3.5.2 A Characterisation of Projections

A Characterisation of ProjectionsAn operator T on a �nite�dimensional vector spaceV is a projection if, and only if, T 2 D T :

Proof(See F2: Section 2.3, Example 14 / F3: Section 2.3, Example 16 / F4: Section 2.3, Example 17)Suppose T is a projection. By SG: Example 3.5.1.2,

V D R .T /� N .T / and R .T / D fx 2 V : T .x/ D xg

So T .T .x// D T .x/ ; since T .x/ 2 R .T / ; and therefore T 2 D T :Conversely, suppose T 2 D T :We show that V D R .T /� N .T / : For any x 2 V ,

x D T .x/C .x � T .x// ... (i)

Clearly, T .x/ 2 R .T / : Also,

T .x � T .x// D T .x/� T .T .x//

D T .x/� T .x/ ; since T 2 D T

D 0I

so x � T .x/ 2 N .T / ; and therefore

V D R .T /C N .T / ... (ii)

Let x 2 R .T / \ N .T / : Since x 2 R .T / ; x D T .y/ for some y 2 V : Since x 2 N .T / ;

0 D T .x/ D T .T .y// D T 2 .y/ D T .y/ D xI

so x D 0, and therefore R .T / \ N .T / D f0g : Thus,

V D R .T /� N .T /

Let P denote the projection on R .T / along N .T /. From (i), P .x/ D T .x/ for all x 2 V; that is, P D T; andhence T is a projection.

3.5.3 Example

Let T : C2 ! C2 be de�ned by

T .a; b/ D .�a � b; 2a C 2b/ for all .a; b/ 2 C2

(a) Show that T is a projection on C2:

(b) Find bases for the subspaces W1 and W2 with C2 D W1 �W2; such that T is the projection on W1 along W2.

39 MAT3701/1

SOLUTION

(a)

T 2 .a; b/ D T .�a � b; 2a C 2b/ ; de�nition of T

D .� .�a � b/� .2a C 2b/ ; 2 .�a � b/C 2 .2a C 2b// ; de�nition of T

D .�a � b; 2a C 2b/

D T .a; b/ I

hence T 2 D T; and therefore T is a projection.

(b) By SG: Example 3.5.1.2,W1 D R .T / and W2 D N .T /

Thus,

W1 D f.�a � b; 2a C 2b/ : a; b 2 Cg

D fa .�1; 2/C b .�1; 2/ : a; b 2 Cg

D span f.�1; 2/g ;

so that f.�1; 2/g is a basis for W1:

Also,

W2 D f.a; b/ : �a � b D 2a C 2b D 0g

D f.a; b/ : a C b D 0g

D f.a;�a/ : a 2 Cg

D fa .1;�1/ : a 2 Cg

D span f.1;�1/g ;

and therefore f.1;�1/g is a basis for W2:

3.6 Matrix Representation of a Linear Transformation

From the remark in Friedberg: Section 2.2 following the de�nition of matrix representation, it follows that thematrix representation of T : V ! W in � D fv1; v2; : : : ; vng and is such that its �rst column is the coordinatevector of T .v1/ relative to ; its second column is the coordinate vector of T .v2/ relative to ; etc.

A Formula for the Matrix RepresentationThe matrix representation of T in the ordered bases � D fv1; v2 : : : ; vngand is given by

[T ] � D��T .v1/

�[T .v2/] : : : [T .vn/]

�Note that the j�th column of [T ] � is given by

�T�v j�� (the coordinate vector of T

�v j�relative to /:

40

3.6.1 Examples

(1) Let T : C2 ! M2�2 .C/ denote the mapping de�ned by

T .a; b/ D

"a �bb a

#;

and let

D

("1 00 0

#;

"0 11 0

#;

"0 00 1

#;

"0 01 0

#)

(a) Find the coordinate vector of

v D

"1 �22 1

#

relative to the basis :

(b) Find [T ] � ; where � denotes the standard basis of C2:

SOLUTION

(a) Solve for x1; x2; x3 and x4 in the matrix equation"1 �22 1

#D x1

"1 00 0

#C x2

"0 11 0

#C x3

"0 00 1

#C x4

"0 01 0

#;

i.e., "1 �22 1

#D

"x1 x2x2 C x4 x3

#;

i.e., 8>>>><>>>>:x1 D 1

x2 D �2x2 Cx4 D 2

x3 D 1;

i.e.,

x1 D 1I x2 D �2I x3 D 1 and x4 D 4

Thus,

[v] D

2666641

�214

377775(b) Since � D f.1; 0/ ; .0; 1/g ; the above formula becomes

[T ] � D�[T .0; 1/] [T .0; 1/]

�... (i)

41 MAT3701/1

By the de�nition of T;

T .1; 0/ D

"1 00 1

#

D 1 �

"1 00 0

#C 0 �

"0 11 0

#C 1 �

"0 00 1

#C 0 �

"0 01 0

#;

hence

[T .1; 0/] D

2666641010

377775 ... (ii)

Similarly,

[T .0; 1/] D

"0 �11 0

#

D 0 �

"1 00 0

#�

"0 11 0

#C 0 �

"0 00 1

#C 2 �

"0 01 0

#;

hence

[T .0; 1/] D

2666640

�102

377775 ... (iii)

Substitution of (ii) and (iii) in (i) yields

[T ] � D

2666641 00 �11 00 2

377775

(2) F2 & F3: Section 2.2, Exercise 7 / F4: Section 2.2, Exercise 8

Let V be an n�dimensional vector space with an ordered basis �: De�ne T : V ! Fn by T .x/ D [x]� :Prove that T is linear.

SOLUTION

By the Linear Transformation Test, it suf�ces to verify the single condition

T .cx C y/ D cT .x/C T .y/ for x; y 2 V and c 2 F

Since � D fv1; : : : ; vng is a basis for V; there exist unique scalars ai ; bi 2 F .1 � i � n/ such that

x D a1v1 C : : :C anvn and y D b1v1 C : : :C bnvn

42

Hence,

cx C y D c .a1v1 C : : :C anvn/C .b1v1 C : : :C bnvn/

D .ca1 C b1/ v1 C : : :C .can C bn/ vn

So,

T .cx C y/ D�cx C y

��

D

2664ca1 C b1

:::

can C bn

3775

D c

2664a1:::

an

3775C2664b1:::

bn

3775D c [x]� C

�y��

D cT .x/C T .y/ ;

which completes the proof.

(3) F2 and F3: Section 2.2, Exercise 8 / F4: Section 2.2, Exercise 9

Let V be the vector space of complex numbers over the �eld R: De�ne T : V ! V by T .z/ D z; where z isthe complex conjugate of z: Prove that T is linear, and compute [T ]� , where � D f1; ig : Show that T is notlinear if V is regarded as a vector space over the �eld C:

SOLUTION

The vectors of V are the complex numbers, addition is the usual addition of complex numbers, and scalarmultiplication is the usual multiplication of complex numbers, but where the �rst number is real. To checkwhether T is linear, use the Linear Transformation Test with c in cx C y real and x and y complex.

T .cx C y/ D cx C y

D cx C y

D cx C y

D cx C y; since c is real ... (i)

D cT .x/C T .y/

Hence, T is linear.

We leave it to you to show that V is a two�dimensional vector space over R with basis � D f1; ig : To compute[T ]� ; we have

T .1/ D 1 D 1 D 1 � 1C 0T .i/ D i D �i D 0 � 1� 1 � i

)

43 MAT3701/1

Hence,

[T ]� D

"1 00 �1

#

Finally, if the underlying �eld is complex, then equation (i) does not hold, since c is complex. Hence, T isnot linear in this case.


Find linear transformations U; T : F2 ! F2 such that UT D T0 (the zero transformation), but TU 6D T0:Use your answer to �nd matrices A and B such that AB D 0 but BA 6D 0:

SOLUTION

Let � D fe1 D .1; 0/ ; e2 D .0; 1/g denote the standard basis of F2: By Friedberg: Theorem 2.6, there existlinear transformations U; T : F2 ! F2 such that

T .e1/ D T .e2/ D e1 ... (i)

andU .e1/ D 0; U .e2/ D e2 ... (ii)

(It can be shown that the formulas for T and U are

T .x; y/ D .x C y; 0/ and U .x; y/ D .0; y/ ;

but it will not be needed.)

SinceUT .e1/ D U .T .e1// D U .e1/ D 0UT .e2/ D U .T .e2// D U .e1/ D e1

)... (iii)

andTU .e1/ D T .U .e1// D T .0/ D 0TU .e2/ D T .U .e2// D T .e2/ D e1

); ... (iv)

it follows from Friedberg: Theorem 2.6 that UT D T0; but TU 6D T0: From this, we can derive values for Aand B as follows:

[UT ]� D�[UT .e1/]� [UT .e2/]�

�

D

"0 00 0

#according to (iii)

D [U ]� [T ]� by Friedberg: Theorem 2.11

D

"0 00 1

#"1 10 0

#according to (i) and (ii)

44

and, similarly,

[TU ]� D

"0 10 0

#

D [T ]� [U ]�

D

"1 10 0

#"0 00 1

#:

Thus,

A D [U ]� D

"0 00 1

#and B D [T ]� D

"1 10 0

#

are suitable choices.

(5) Let W denote the vector space of all symmetric 2 � 2 matrices

"a bb c

#with entries in C: De�ne the

invertible linear transformation

T : P2 .C/! W by T .p .x// D

"p .1/ p .0/p .0/ p .�1/

#

Let

� D�1; x; x2

and D

("1 00 0

#;

"0 00 1

#;

"0 11 0

#)

denote ordered bases of P2 .C/ and W respectively.

(a) Compute [T ] � :

(b) Compute�T�1

�� :

(c) Use (b) to �nd the formula for T�1:

45 MAT3701/1

SOLUTION

(a) In order to calculate [T ] � ; we need to express T .1/ ; T .x/ and T�x2�in terms of : Now,

T .1/ D

"1 11 1

#; de�nition of T

D

"1 00 0

#C

"0 00 1

#C

"0 11 0

#I

T .x/ D

"1 00 �1

#; de�nition of T

D

"1 00 0

#�

"0 00 1

#C 0

"0 11 0

#I

T�x2�D

"1 00 1

#; de�nition of T

D

"1 00 0

#C

"0 00 1

#C 0

"1 00 1

#

Thus, according to the de�nition of [T ] � ;

[T ] � Dh[T .1/] ; [T .x/] ;

�T�x2��

i

D

264 1 1 11 �1 11 0 0

375

(b) According to F2: Theorem 2.19 / F3 and F4: Theorem 2.18,

�T�1

�� D

�[T ] �

��1

D

264 1 1 11 �1 11 0 0

375�1

D

264 0 0 112 � 1

2 012

12 �1

375 ;

using the Matrix Inversion Algorithm (check the details).

46

(c) From the de�nition of�T�1

�� ; it follows that

"T�1

"1 00 0

##�

D

264 01212

375 ;that is, T�1

"1 00 0

#D 0 � 1C

12� x C

12� x2 ... (i)

Similarly,

T�1"0 00 1

#D �

12x C

12x2 ... (ii)

and

T�1"0 11 0

#D 1� x2: ... (iii)

So,

T�1"a bb c

#

D T�1 "

a 00 0

#C

"0 bb 0

#C

"0 00 c

#!

D T�1"a 00 0

#C T�1

"0 bb 0

#C T�1

"0 00 c

#

D aT�1"1 00 0

#C bT�1

"0 11 0

#C cT�1

"0 00 1

#

D a�12x C

12x2�C b

�1� x2

�C c

��12x C

12x2�

from (i), (ii) and (iii)

D b C12.a � c/ x C

12.a � 2b C c/ x2


Let T : M2�2 .R/! M2x2 .R/ denote the linear transformation de�ned by T .M/ D M t ; and let

� D

(E11 D

1 00 0

!; E12 D

0 10 0

!; E21 D

0 01 0

!; E22 D

0 00 1

!)

be an ordered basis for M2�2 .R/ :

(a) Compute [T ]� :

(b) Verify that L A�� .M/ D ��T .M/ for A D [T ]� and

M D

1 23 4

!

47 MAT3701/1

SOLUTION

(a) T�E11

�D�E11

�tD E11;

T�E12

�D�E12

�tD E21;

T�E21

�D�E21

�tD E12;

T�E22

�D�E22

�tD E22;

so

[T ]� D

2666641 0 0 00 0 1 00 1 0 00 0 0 1

377775(b) M D 1 � E11 C 2 � E12 C 3 � E21 C 4E22;

so

�� .M/ D [M]� D

2666641234

377775and

L A�� .M/ D

2666641 0 0 00 0 1 00 1 0 00 0 0 1

3777752666641234

377775 D2666641324

377775 ... (i)

Since

T .M/ D M t D

1 32 4

!

D 1 � E11 C 3 � E12 C 2 � E21 C 4E22;

��T .M/ D�T .M/�

��D

2666641324

377775 ;

which agrees with (i).

(7) Let T : P2 .C/! P2 .C/ denote the linear transformation with matrix

[T ] D

264 1 0 11 1 10 1 1

375with respect to D

�1C x; 1� x; 1C x2

:

48

(a) Express each vector in � D�1; x; x2

as a linear combination of :

(b) Find the change of coordinate matrix P which changes ��coordinates into �coordinates.

(c) Use P to express p .x/ D 1C x C x2 as a linear combination of 1C x; 1� x and 1C x2:

SOLUTION

(a)

1 D a .1C x/C b .1� x/C c�1C x2

�D .a C b C c/C .a � b/ x C cx2

iff

a C b C c D 1a � b D 0

c D 0

9>=>;Adding the �rst two equations and substituting c D 0 in the result yields 2a D 1; that is, a D 1

2 andb D 1

2 (from the second equation). Thus,

1 D12.1C x/C

12.1� x/C 0

�1C x2

�... (i)

Similarly, show that

x D12.1C x/�

12.1� x/C 0

�1C x2

�... (ii)

and

x2 D �12.1C x/�

12.1� x/C

�1C x2

�... (iii)

(b)

P D�IP2 .R/

� �

Dh[1] ; [x] ;

�x2�

iD

26412

12 � 1

212 � 1

2 � 12

0 0 1

375 ; according to (i), (ii) and (iii)

D12

264 1 1 �11 �1 �10 0 2

375

49 MAT3701/1

(c) According to F2: Theorem 2.23(b) / F3 and F4: Theorem 2.22(b),�p .x/

� D P

�p .x/

��

D P

264 111

375 ; since � is the standard basis of P2 .C/

D12

264 1 1 �11 �1 �10 0 2

375264 111

375

D

26412

� 121

375 ;hence,

p .x/ D12.1C x/�

12.1� x/C

�1C x2

�(8) Let T : P2 .C/! P2 .C/ denote the linear transformation with matrix

[T ] D

264 1 0 11 1 10 1 1

375with respect to D

�1C x; 1� x; 1C x2

:

(a) Express each vector in � D�1; x; x2

as a linear combination of :

(b) Find the change of coordinate matrix P which changes ��coordinates to �coordinates.

(c) Use P to express p .x/ D 1C x C x2 as a linear combination of 1C x; 1� x and 1C x2:

(d) Calculate [T ]� :

(e) Use your matrix in (d) to �nd T�a C bx C cx2

�:

SOLUTION

(a) See Example 7(a) above.

(b) According to Example 7(b) above,

P D12

264 1 1 �11 �1 �10 0 2

375(c) See Example 7(c) above.

(d) According to F2: Theorem 2.24 / F3 and F4: Theorem 2.23,

[T ]� D��IP2.R/

� �

��1[T ]

�IP2.R/

� �

D P�1 [T ] P ...(i)

50

Use the Matrix Inversion Algorithm to calculate P�1 :

264 1 0 0 : 12

12 � 1

20 1 0 : 1

2 � 12 � 1

20 0 1 : 0 0 1

375#264 1 0 1

2 : 12

12 0

0 1 12 : 1

2 � 12 0

0 0 1 : 0 0 1

375 R1 C 12 R3

R2 C 12 R3

#264 1 1 1 : 1 0 00 1 1

2 : 12 � 1

2 00 0 1 : 0 0 1

375 R1 C R2

#264 1 1 1 : 1 0 0� 12

12 0 : 0 � 1

2 00 0 1 : 0 0 1

375 R2 � 12 R1

264 1 1 1 : 1 0 01 �1 0 : 0 1 00 0 1 : 0 0 1

375 �2R2

Thus,

P�1 D

264 1 1 11 �1 00 0 1

375

Substituting the value P and P�1 in (i) yields

[T ]� D12

264 1 1 11 �1 00 0 1

375264 1 0 11 1 10 1 1

375264 1 1 �11 �1 �10 0 2

375

D12

264 2 2 30 �1 00 1 1

375264 1 1 �11 �1 �10 0 2

375

D12

264 4 0 2�1 1 11 �1 1

375

51 MAT3701/1

(e) According to F2: Theorem 2.15 / F3 and F4: Theorem 2.14,�T�a C bx C cx2

��D [T ]�

�a C bx C cx2

�

D12

264 4 0 2�1 1 11 �1 1

375264 abc

375

D12

264 4a C 2c�a C b C ca � b C c

375 ;hence,

T�a C bx C cx2

�D .2a C c/C

12.�a C b C c/ x C

12.a � b C c/ x2

52

.

STUDY UNIT 4.

SUMMARY OF THE IMPORTANT FACTSABOUTDETERMINANTS OVER THEREAL AND COMPLEX NUMBERS(Friedberg: Section 4.4)

4.1 OverviewIn this study unit, the important properties of determinants needed for the remainder of this module, and with whichyou are assumed to be familiar, are summarised. A result on the deteminant of block�triangular matrices with whichyou may not be familiar is discussed further on.

4.2 What to StudyRead through Friedberg: Section 4.4, and make sure that you are familiar with all the properties listed there, andthen study this unit.

4.3 The Determinant of a Block Triangular MatrixF2: Section 4.3, Exercise 9 / F3: Section 4.3, Exercise 20 / F4: Section 4.3, Exercise 21Suppose that M 2 Mn�n .F/ can be written in the form

M D

"A B0 C

#;

where A and C are square matrices and 0 is a zero matrix. Prove that

det .M/ D det .A/ � det .C/

SOLUTIONDenote by eMi j the submatrix of M obtained by deleting the i�th row and j�th column. The proof is done byinduction on the order k of A: For k D 1 (hence A D [A11] is a scalar and B a 1� .n � 1/ matrix), expansion alongthe �rst column yields

det .M/ D M11 � det� eM11�

D A11 � det .C/ ; since M11 D A11 and eM11 D CD det .A/ � det .C/ I

53 MAT3701/1

hence, the result holds for k D 1: Suppose k > 1, and the result holds for k � 1: Expansion along the �rst columnyields

det .M/ DnXiD1

.�1/iC1 Mi1 � det� eMi1�

DkXiD1

.�1/iC1 Ai1 � det� eMi1� ; ... (i)

since M11 D A11; : : : ;Mk1 D Ak1 and Mi1 D 0 for k < i < n:Each eMi1 .1 � i � k/ is of the form eMi1 D " eAi1 Bi

0 C

#;

where Bi is obtained from B by deleting the i�th row. Since eAi1 is of order k � 1, it follows by induction thatdet

� eMi1� D det �eAi1� � det .C/Substitution in (i) yields

det .M/ DkXiD1

.�1/iC1 Ai1 � det�eAi1� � det .C/

D

kXiD1

.�1/iC1 Ai1 � det�eAi1�! � det .C/

D det .A/ � det .C/ :

4.3.1 ExampleCalculate the determinant of

M D

266666642 1 2 3 41 2 5 6 70 0 1 1 80 0 1 4 90 0 0 0 1

37777775SOLUTIONPartition M as

M D

"A B0 C

#;

where

A D

"2 11 2

#and C D

264 1 1 81 4 90 0 1

375By the previous result,

det .M/ D det

"2 11 2

#� det

264 1 1 81 4 90 0 1

375D 3 � 3 D 9

54

.

PART 2

DIAGONALIZATION

Page

Study unit 5 (Friedberg: Appendix E): Polynomials 56

Study unit 6 (Friedberg: Section 5.1): Eigenvalues and Eigenvectors 60

Study unit 7 (Friedberg: Section 5.2): Diagonalisability 76

Study unit 8 (Friedberg: Section 5.3): Matrix Limits and Markov Chains 84

Study unit 9 (Friedberg: Section 5.4): Invariant Subspaces and the Cayley�Hamilton Theorem 88

55 MAT3701/1

Part 2 OverviewThe chapter overview of Friedberg: Chapter 5 serves our purpose well in this case. So, you may wish to refer backto the end of SG: Section 2.3, where diagonalisability is also addressed, and how it relates to the type of �eld chosenas the scalar �eld.


� solve a range of problems in a general real or complex vector space related to eigenvalues and eigenvectors,the characteristic polynomial, diagonalisability, matrix limits, Markov chains, the location of eigenvalues,invariant subspaces, cyclic subspaces, and the Cayley�Hamilton Theorem

� reproduce theoretic arguments, including proofs, and solve basic theoretic problems related to the abovetopics.

56

.

STUDY UNIT 5.

POLYNOMIALS

(Friedberg: Appendix E)

5.1 OverviewProperties of polynomials over an arbitrary �eld are developed which are needed for the remainder of this module.A concept which plays an important role in what follows is that of a polynomial in a linear operator or a squarematrix.

5.2 What to StudyStudy Friedberg: Appendix E to the end of Theorem E.5.

5.3 The Matrix Representation of f .T /(Friedberg: Theorem E.3)This theorem shows that f .T / is again an operator, and expresses its matrix representation in terms of the matrixrepresentation of T :

Friedberg: Theorem E.3Let f .x/ be a polynomial with coef�cients from a �eld F; and let T be a linear operator on a vector space V overF: Then

(a) f .T / is a linear operator on V

(b) if � is a �nite ordered basis for V and A D [T ]� ; then�f .T /

��D f .A/

PROOFSuppose

f .x/ D a0 C a1x C : : :C anxn

(a) Since T is a linear operator on V; by Friedberg: Theorem 2.9 so are T T; .T T / T; ..T T / T / T; : : : ; whichwe write as T 2; T 3; T 4; : : : ; respectively. If we let T 0 D I; the identity operator on V; then it follows that T n

is a linear operator on V for any n � 0: By Friedberg: Theorem 2.7,

a0 I; a0 I C a1T; .a0 I C a1T /C a2T 2;�.a0 I C a1T /C a2T 2

�C a3T 3; : : :

57 MAT3701/1

are linear operators on V; which we write simply as

a0 I; a0 I C a1T; a0 I C a1T C a2T 2; a0 I C a1T C a2T 2 C a3T 3; : : : ;

since addition of linear transformations are associative by Friedberg: Theorem 2.7(b), which implies that thebrackets are unnecessary. Thus,

f .T / D a0 I C a1T C : : :C anT n

is a linear operator on V :

(b) �f .T /

��D

�a0 I C a1T C : : :C anT n

��

D [a0 I ]� C [a1T ]� C : : :C�anT n

��; Friedberg: Theorem 2.8(a)

D a0 [I ]� C a1 [T ]� C : : :C an�T n��; Friedberg: Theorem 2.8(b)

D a0 [I ]� C a1 [T ]� C : : :C an�[T ]�

�n; Friedberg: Theorem 2.11

D a0 I C a1 [T ]� C : : :C an�[T ]�

�n; since [I ]� D I

D f�[T ]�

�D f .A/ ; where A D [T ]�

We may summarise part (b) as follows.

Matrix Representation of f .T /The matrix representation of f .T / is f of the matrixrepresentation of T :�

f .T /��D f

�[T ]�

�(Friedberg: Theorem E.3)

ExampleLet T : F2 ! F2 be de�ned by

T .a; b/ D .a C b; a � b/ for all .a; b/ 2 F2;

let f .t/ D t2 C 1; and let � denote the standard basis for F2:

(a) Find [T ]� :

(b) Find the formula for f .T / :

(c) Find�f .T /

��:

(d) Verify that�f .T /

��D f

�[T ]�

�:

58

SOLUTION

(a) [T ]� D

"1 11 �1

#(Check the details.)

(b)

f .T / .a; b/ D�T 2 C I

�.a; b/ ; de�nition of f .T /

D T 2 .a; b/C I .a; b/

D T .a C b; a � b/C .a; b/

D .2a; 2b/C .a; b/

D .3a; 3b/

(c)�f .T /

��D

"3 00 3

#(Check.)

(d)

f�[T ]�

�D

�[T ]�

�2C I

D

"1 11 �1

#2C

"1 00 1

#

D

"2 00 2

#C

"1 00 1

#

D

"3 00 3

#;

which agrees with the answer in (c).

5.4 Polynomials in a given Operator (or Square Matrix) Commute(Friedberg: Theorem E.4)The statement of this theorem reads as follows:Let T be a linear operator on a vector space V over a �eld F; and let A be a square matrix with entries from F:Then, for any polynomials f1 .x/ and f2 .x/ with coef�cients from F;

(a) f1 .T / f2 .T / D f2 .T / f1 .T /

(b) f1 .A/ f2 .A/ D f2 .A/ f1 .A/

We illustrate it by means of the following example:

59 MAT3701/1

ExampleVerify part (b) of the above theorem for

f1 .t/ D t2 C t C 1; f2 .t/ D t C 2; and A D

"1 11 �1

#

SOLUTION

f1 .A/ D A2 C A C I2

D

"1 11 �1

#2C

"1 11 �1

#C

"1 00 1

#

D

"2 00 2

#C

"1 11 �1

#C

"1 00 1

#

D

"4 11 2

#I

f2 .A/ D A C 2I2

D

"1 11 �1

#C

"2 00 2

#

D

"3 11 1

#

Now

f1 .A/ f2 .A/ D

"4 11 2

#"3 11 1

#D

"13 55 3

#and

f2 .A/ f1 .A/ D

"3 11 1

#"4 11 2

#D

"13 55 3

#Hence

f1 .A/ f2 .A/ D f2 .A/ f1 .A/

60

.

STUDY UNIT 6.

EIGENVALUES AND EIGENVECTORS

(Friedberg: Section 5.1)

6.1 OverviewThe concepts of eigenvalue and eigenvector are introduced. To compute these entities for a general operator, it isnecessary to revert back to matrices via the matrix representation. Although the material in this section of Friedbergis familiar to you, the treatment here and in the next section will be at a deeper level.

6.2 What to StudyStudy Friedberg: Section 5.1, and try as many of the exercises as possible. Pay special attention to the section onsimilarity which we discuss further on, as it contains some additional information.

6.3 Computing the Eigenvalues and Eigenvectors of a Square Matrix(F2 and F3: Theorem 5.7, Corollary 1, and Theorem 5.9 / F4: Theorems 5.2 and 5.4)We summarise the process as follows:

Computing the Eigenvalues and Eigenvectors of a Matrix A 2 Mn�n .F/The eigenvalues of A are given by the zeros of det .A � t In/ ; which represents apolynomial of degree n in t:The eigenvectors of A corresponding to an eigenvalue � 2 F are given by the non�zero solutions of .A � �In/ x D 0; where x 2 Fn: This represents a homogeneouslinear system in n variables in matrix form.

ExampleFriedberg: Section 5.1, Exercise 3(d)Let

A D

264 2 0 �14 1 �42 0 �1

375 over F D R

(a) Determine all the eigenvalues of A:

(b) For each eigenvalue � of A; �nd the set of eigenvectors corresponding to �:

(c) If possible, �nd a basis for F3 consisting of eigenvectors of A:

61 MAT3701/1

SOLUTION

(a)

det .A � t I / D det

264 2� t 0 �14 1� t �42 0 �1� t

375 ... (i)

D .1� t/ det

"2� t �1

2 �1� t

#; expanding along the 2nd column

D .1� t/ [.t C 1/ .t � 2/C 2]

D �t .t � 1/2

Hence, the eigenvalues of A are 0 and 1:

(b) The eigenvectors corresponding to the eigenvalue � are given by the nonzero vectors in N .A � �I / ; that is,all the nonzero solutions of .A � �I / x D 0: So for � D 0; the coef�cient matrix of the homogeneous systemto solve is simply A: In the row reduction that follows, we do not augment the coef�cient matrix, since forhomogeneous systems the column of constant terms consists of zeros only.264 2 0 �1

4 1 �42 0 �1

375!264 2 0 �10 1 �20 0 0

375 R2 � 2R1R3 � R1

Hence

2x1 � x3 D 0

x2 � 2x3 D 0

Letting x3 D 2t (in order to avoid fractions in the �rst equation), we get x1 D t and x2 D 3t: Hence,

N .A/ D ft .1; 4; 2/ : t 2 Rg D span f.1; 4; 2/g ; ... (ii)

and so the eigenvectors corresponding to 0 consist of the nonzero vectors in this set.

For the coef�cient matrix in the case of � D 1; substitute � D 1 in matrix (i).264 1 0 �14 0 �42 0 �2

375!264 1 0 �10 0 00 0 0

375 R2 � 4R1R3 � 2R1

Solving this system, we get

N .A � I / D span f.1; 0; 1/ ; .0; 1; 0/g ; ... (iii)

hence the eigenvectors corresponding to 1 consist of the nonzero vectors in this set.

62

(c) Together, the spanning vectors in (ii) and (iii), namely,

� D f.1; 4; 2/ ; .1; 0; 1/ ; .0; 1; 0/g ;

is a basis for R3 (verify).

6.4 Diagonalizing a Square Matrix(See the discussion following F2 and F3: Section 5.1, Example 3 / F4: Section 5.1, Example 6.)Suppose that � is a basis for Fn consisting of eigenvectors of A 2 Mn�n .F/ : Then [L A]� D D is the diagonalmatrix consisting of the eigenvalues of A; arranged in the same order as the corresponding eigenvectors in �: Now

[L A]� D [IFn ]� [L A] [IFn ] � ;

where denotes the standard basis for FnI hence,

D D Q�1AQ;

where Q is the matrix whose j�th column is the j�th vector of �:

In summary:

Diagonalizing a Square MatrixA 2 Mn�n .F/ can be diagonalised only if Fn has a basis � consistingof eigenvectors of A: If this is the case,

Q�1AQ D D; a diagonal matrix,where the j�th column of Q is the j�th vector of � correspondingto the eigenvalue of A appearing in j�th diagonal entry of D:

ExampleFriedberg: Section 5.1, Exercise 3(d)Determine an invertible matrix Q and a diagonal matrix D such that Q�1AQ D D; where

A D

264 2 0 �14 1 �42 0 �1

375 over F D R

SOLUTIONBy SG: Example 6.3(c) above,

� D f.1; 4; 2/ ; .1; 0; 1/ ; .0; 1; 0/g

is a basis of R3 consisting of eigenvectors of A: Hence,

Q D

264 1 1 04 0 12 1 0

375

63 MAT3701/1

and

D D

264 0 0 00 1 00 0 1

375 ;since the �rst column of Q consists of an eigenvector of A with eigenvalue 0; and the second and third columns ofA consist of eigenvectors of A with eigenvalue 1:

6.5 Similarity of MatricesIt is now possible to derive some general properties of similar matrices.

6.5.1 Properties of Similar MatricesSuppose the n � n matrices A and B (over F/ are similar. Then

(a) rank.A/ D rank.B/

(b) tr.A/ D tr.B/

(c) det .A/ D det .B/

(d) A and B have the same characteristic polynomial (Friedberg: Section 5.1, Exercise 12(a))

(e) A and B have the same eigenvalues

PROOFSince A and B are similar, there exists a nonsingular matrix Q such that

B D Q�1AQ

(a) The following property from a previous course is used:

rank .M/ D rank .RM/ D rank .MS/ ;

where M; R and S are matrices such that RM and MS are de�ned and R and S are nonsingular.

Now

rank .B/ D rank�Q�1AQ

�D rank .AQ/ since Q�1 is nonsingular

D rank .A/ since Q is nonsingular.

(b)

tr .B/ D tr�Q�1AQ

�D tr

�QQ�1A

�; since tr .XY / D tr .Y X/

D tr .A/

64

(c)

det .B/ D det�Q�1AQ

�D det

�Q�1

�det .A/ det .Q/

D .det .Q//�1 det .A/ det .Q/

D det .A/

(d)

det .B � t I / D det�Q�1AQ � t I

�D det

�Q�1 .A � t I / Q

�since t I D Q�1t I Q

D det .A � t I / by the argument in (c)

(e) Since the eigenvalues are the roots of the characteristic polynomial, the result follows from (d).

Note: These properties are not suf�cient to guarantee similarity, as the next example demonstrates.

ExampleThe matrices

A D

"1 00 1

#and B D

"1 10 1

#satisfy all the above properties, but are not similar.

SOLUTIONVerify that for both these matrices the rank is 2; the trace is 2; the determinant is 1; the characteristic polynomial is.t � 1/2 ; and the only eigenvalue is 1:However, they are not similar, since for any nonsingular 2� 2 matrix Q;

Q�1AQ D Q�1 I2Q

D Q�1Q

D I26D B

6.5.2 A Test for NonsimilaritySometimes, these conditions are useful to check for nonsimilarity (but remember, if none of these conditions fails itdoes not imply similarity, as demonstrated in the previous example).

A Test for NonsimilarityIf any of the following conditions fail for two n � n matricesA and B; they are not similar:� rank.A/ D rank.B/� tr.A/ D tr.B/� det .A/ D det .B/� A and B have the same characteristic polynomial� A and B have the same eigenvalues

65 MAT3701/1

ExampleDetermine whether 264 0 0 0

1 0 10 1 0

375 and

264 0 0 01 0 40 1 0

375are similar.SOLUTIONLet

A D

264 0 0 01 0 10 1 0

375 and B D

264 0 0 01 0 40 1 0

375Check whether or not the above �ve conditions hold.

rank .A/ D 2 D rank .B/ � condition (1) holds

tr .A/ D 0 D tr .B/ � condition (2) holds

det .A/ D 0 D det .B/ � condition (3) holds

det .A � t I / D

��t 0 01 �t 10 1 �t

��D �t3 C t (check),

and

det .B � t I / D

��t 0 01 �t 40 1 �t

��D �t3 C 4t (check)

Since det .A � t I / 6D det .B � t I / ; condition (4) fails; hence A and B are not similar.

6.6 Computing the Eigenvalues and Eigenvectors of an OperatorThe eigenvalues and eigenvectors of an operator T are derived from the eigenvalues and eigenvectors of a matrixrepresentation of T; with which we are now familiar.Suppose T : V ! V is a linear operator on an n�dimensional vector space V with ordered basis �:

The Eigenvalues of TTo �nd the eigenvalues of T we compute the roots of the polynomial det

�[T ]� � t In

�, where � is a basis for V : But

what if we choose � differently, may we not get different roots?

Friedberg: Section 5.1, Exercise 12(b)Show that the de�nition of the characteristic polynomial of a linear operator on a �nite�dimensional vector space Vis independent of the choice of basis for V :

66

SOLUTIONLet T : V ! V be linear, and let � and � 0 be ordered bases for V : Since [T ]� and [T ]� 0 are similar by F2: Theorem2.24 / F3 and F4: Theorem 2.23, it follows from SG: Property 6.5.1(d) thaT

det�[T ]� � t I

�D det

�[T ]� 0 � t I

�I

hence the characteristic polynomial of T is independent of the basis.

The Eigenvectors of TTo �nd the eigenvectors of T corresponding to a given eigenvalue �; we compute the eigenvectors of A D [T ]�corresponding to �I that is, we solve for the nonzero solutions in the homogeneous linear system .A � �In/ x D 0in Fn: If

N .A � �In/ D span .fx1; x2; : : : ; xkg/ ;

thenN .T � �I / D span

��1� .x1/ ; ��1� .x2/ ; : : : ; ��1� .xk/

�;

where �� : V ! Fn is the standard representation of V with respect to � (that is, �� .v/ D [v]�/:This procedure is based on the following:

Friedberg: Section 5.1, Exercise 13(In F2, replace P2 .R/ by V and R3 by Fn throughout.)Let T be a linear operator on a �nite�dimensional vector space V over a �eld F; let � be an ordered basis for V;and let A D [T ]� :With reference to Friedberg: Figure 5.1, prove the following.

(a) If v 2 V and �� .v/ is an eigenvector of A corresponding to the eigenvalue �; then v is an eigenvector of Tcorresponding to �:

(b) If � is an eigenvalue of A (and hence of T /; then a vector y 2 Fn is an eigenvector of A corresponding to �if and only if ��1� .y/ is an eigenvector of T corresponding to �:

SOLUTION

(a) Since A�� .v/ D �� .v/ ;

A�� .v/ D L A�� .v/ D ��T .v/ and �� .v/ D �� .�v/ I

hence�� .T .v// D �� .�v/ ;

soT .v/ D �v;

since �� is an isomorphism. Since �� .v/ 6D 0 (being an eigenvector) and �� is an isomorphism, v 6D 0: Thusv is an eigenvector of T corresponding to �:

67 MAT3701/1

(b) Suppose y 2 Fn is an eigenvector of A corresponding to �: Since v D ��1� .y/ 2 V and �� .v/ D y; it followsfrom (a) that ��1� .y/ is an eigenvector of T corresponding to �:

Conversely, suppose ��1� .y/ is an eigenvector of T corresponding to �: Then

T��1� .y/

�D ��1� .y/ I

hence

L A .y/ D L A��1� .y/

�D ��T

��1� .y/

�since L A�� D ��T by Friedberg: Figure 5.1

D ��1� .y/

�D �y:

Thus y is an eigenvector of A corresponding to �:

In Summary:

Computing the Eigenvalues and Eigenvectors of an OperatorSuppose T : V ! V is linear, where V is n�dimensional with basis �: For theeigenvalues of T; compute the roots of det

�[T ]� � t In

�: For the eigenvectors

of T corresponding to an eigenvalue �; �rst solve�[T ]� D �In

�x D 0; that is,

compute N�[T ]� � �In

�(the eigenvectors of [T ]� corresponding to �; together

with 0/: IfN .T � �I / D span.fx1; x2; : : : ; xkg/ ;

then the nonzero vectors ofN .T � �I / D span

��1� .x1/ ; ��1� .x2/ ; : : : ; ��1� .xk/

�are the eigenvectors of T corresponding to �:

ExampleLet

A D

"�1 01 2

#and de�ne T : M2�2 .C/! M2�2 .C/ by T .X/ D AX; where X 2 M2�2 .C/ :

(a) Show that T is a linear operator on M2�2 .C/ :

(b) Calculate the characteristic polynomial of T; and show that it splits (i.e., factors into linear polynomials).

(c) Find a basis (consisting of 2� 2 matrices over C/ for each eigenspace N .T � �I / of T :

(d) Find a basis � of M2�2 .C/ consisting of eigenvectors of T :

(e) Calculate [T ]� with � as in (d), and show that it is diagonal.

68

SOLUTION

(a) Let X; Y 2 M2�2 .C/ and c 2 C:

T .cX C Y / D A .cX C Y / ; de�nition of T

D A .cX/C AY; matrix arithmetic property

D cAX C AY; matrix arithmetic property

D cT .X/C T .Y / ; de�nition of T

Thus, T is a linear operator on M2�2 .C/ :

(b) First we calculate [T ] with respect to the (standard) basis

D

(v1 D

"1 00 0

#I v2 D

"0 10 0

#I v3 D

"0 01 0

#I v4 D

"0 00 1

#)

of M2�2 .C/ :

T .v1/ D A

"1 00 0

#; de�nition of T

D

"�1 01 2

#"1 00 0

#

D

"�1 01 0

#

D �v1 C v3;

hence,

[T .v1/] D

266664�1010

377775Similarly,

T .v2/ D

"�1 01 2

#"0 10 0

#

D

"0 �10 1

#

D �v2 C v4;

hence,

[T .v2/] D

2666640

�101

377775 I

69 MAT3701/1

and

T .v3/ D

"�1 01 2

#"0 01 0

#

D

"0 02 0

#

D 2v3;

hence,

[T .v3/] D

2666640020

377775 I

and

T .v4/ D

"0 00 2

#

D 2v4;

hence,

[T .v4/] D

2666640002

377775

Therefore,

[T ] D�[T .v1/] [T .v2] [T .v3/] [T .v4/]

�

D

266664�1 0 0 00 �1 0 01 0 2 00 1 0 2

377775

70

The characteristic polynomial f .t/ of T is given by

f .t/ D det�[T ] � t I4

�

D

��1� t 0 0 0

0 �1� t 0 01 0 2� t 00 1 0 2� t

��... (i)

D .�1� t/ .�1� t/ .2� t/ .2� t/ ; determinant of a lower triangular matrix.

D .t C 1/2 .t � 2/2 ; ... (ii)

showing that f .t/ splits.

(c) First we calculate the eigenspaces of [T ] ; and then deduce the eigenspaces of T from them: According to(b), the eigenvalues of [T ] (and of T / are �1 and 2:

For t D �1; the corresponding homogeneous linear system to solve according to (i) is (in matrix form)2666640 0 0 00 0 0 01 0 3 00 1 0 3

377775 ;with solution space

N�[T ] C I4

�D

8>>>><>>>>:

266664�3s�3tst

377775 : s; t 2 C9>>>>=>>>>;

D span

8>>>><>>>>:

266664�3010

377775 ;266664

0�301

3777759>>>>=>>>>;

Since these column vectors represent coordinate vectors with respect to ; it follows that a basis of N .T C I /is given by

�3v1 C v3 D

"�3 01 0

#; from the �rst column vector,

and

�3v2 C v4 D

"0 �30 1

#; from the second column vector

Thus,

N .T C I / D span

("�3 01 0

#;

"0 �30 1

#)

71 MAT3701/1

(As a check, show that T .X/ D AX D �X for each X in the spanning set of N .T C I / :/

Repeat the process for t D 2: The corresponding homogeneous linear system to solve according to (ii) is, inmatrix form, 266664

�3 0 0 00 �3 0 01 0 0 00 1 0 0

377775 ,with solution space

N�[T ] � 2I4

�D

8>>>><>>>>:

26666400st

377775 : s; t 2 R9>>>>=>>>>;

D span

8>>>><>>>>:

2666640010

377775 ;2666640001

3777759>>>>=>>>>;

Since these columns vectors represent coordinate vectors with respect to ; it follows that a basis of N .T � 2I /is given by

v3 D

"0 01 0

#; from the �rst column vector,

and

v4 D

"0 00 1

#; from the second column vector

Thus,

N .T � 2I / D span

("0 01 0

#;

"0 00 1

#)

(As a check, show that T .X/ D AX D 2X for each X in the spanning set of N .T � 2I / :/

(d) � is simply the union of the bases obtained for N .T C I / and N .T � 2I / ; that is,

� D

(u1 D

"�3 01 0

#I u2 D

"0 �30 1

#I u3 D

"0 01 0

#I u4 D

"0 01 0

#)

72

(e) [T ]� D�[T .u1/]� [T .u2/]� [T .u3/]� [T .u4/]�

�:

For [T .u1/]� ; note that

T .u1/ D A

"�3 01 0

#; de�nition of T

D

"�1 01 2

#"�3 01 0

#

D

"3 0

�1 0

#

D �

"�3 01 0

#

D �u1

Show, similarly, thatT .u2/ D �u2; T .u3/ D 2u3 and T .u4/ D 2u4

Hence,

[T ]� D

266664�1 0 0 00 �1 0 00 0 2 00 0 0 2

3777756.7 Results Needed for Later Use

(1) Friedberg: Section 5.1, Exercise 8

(a) Prove that a linear operator T on a �nite�dimensional vector space is invertible if and only if zero is notan eigenvalue of T :

(b) Let T be an invertible linear operator. Prove that a scalar � is an eigenvalue of T if and only if ��1 is aneigenvalue of T�1:

SOLUTION

(a) T is invertible

iff N .T / D f0g (Friedberg: Theorems 2.4 and 2.5)

iff .T v D 0 D 0 � v ) v D 0/

iff zero is not an eigenvalue of T

(b) � is an eigenvalue of T

iff T .v/ D �v; for some vector v 6D 0

iff T�1 .T .v// D T�1 .�v/ for some v 6D 0

73 MAT3701/1

iff v D �T�1 .v/ for some v 6D 0 .T�1 is an operator)

iff ��1v D T�1 .v/ for some v 6D 0 (� 6D 0 according to (a))

iff ��1 is an eigenvalue of T�1


For any square matrix A; prove that A and At have the same characteristic polynomial (and hence the sameeigenvalues).

SOLUTION

det .A � �I / D det .A � �I /t ; F2: Section 4.4, Property (7) / F3 and F4: Section 4.4, Property (8)

D det�At � �I t

�D det

�At � �I

�(3) Friedberg: Section 5.1, Exercise 15

(a) Let T be a linear operator on a vector space V; and let x be an eigenvector of T corresponding to theeigenvalue �: For any positive integer m; prove that x is an eigenvector of T m corresponding to theeigenvalue �m :

(b) State and prove the analogous result for matrices.

SOLUTION

(a) T .x/ D �x; henceT 2 .x/ D T .T .x// D T .�x/ D �T .x/ D �2x;

hence,T 3 .x/ D T

�T 2 .x/

�D T

��2x

�D �2T .x/ D �3x; etc.

Continuing in this way, the result follows.

(b) Let x 2 Fn be an eigenvector of an n � n matrix A corresponding to the eigenvalue �: For any positiveinteger m; x is an eigenvector of Am corresponding to the eigenvalue �m :

To verify (b), note that x is an eigenvector of L A corresponding to �:Hence by (a), x is an eigenvector of.L A/m corresponding to �m : Since .L A/m D L Am (F2: Theorem 2.16(e) / F3 and F4: Theorem 2.15(e)),it follows that x is an eigenvector of Am corresponding to �m :


Let A and B be similar n � n matrices. Prove that there exists an n�dimensional vector space V; a linearoperator T on V; and ordered bases � and for V such that A D [T ]� and B D [T ] :

74

SOLUTION

Let V D Fn; T D L A and � be the standard basis of Fn: Then

[T ]� D [L A]� D A

Since B is similar to A; there exists a nonsingular matrix Q such that

B D Q�1AQ

Since Q is nonsingular, its columns form a basis of V such that

Q D [IV ]�

Hence,

B D�[IV ]�

��1 [L A]� [IV ]� D [L A] D [T ]


(a) Let T be a linear operator on a vector space V over the �eld F; and let g .t/ be a polynomial withcoef�cients from F: Prove that if x is an eigenvector of T with corresponding eigenvalue �; then

g .T / .x/ D g .�/ x

That is, x is an eigenvector of g .T / with corresponding eigenvalue g .�/ :

(b) State and prove a comparable result for matrices.

(c) Verify (b) for the matrix A in Friedberg: Section 5.1, Exercise 3(a) with polynomial

g .t/ D 2t2 � t C 1;

eigenvector

x D

"23

#;

and corresponding eigenvalue � D 4:

SOLUTION

Let

g .t/ D a0 C a1t C : : :C antn

75 MAT3701/1

(a)

g .T / x D�a0 I C a1T C : : :C anT n

�.x/

D .a0 I / .x/C .a1T / .x/C : : :C�anT n

�.x/

D a0 .I .x//C a1 .T .x//C : : :C an�T n .x/

�D a0x C a1 .�x/C : : :C an

��nx

�; SG: Example 6.7.3 above

D .a0 C a1�C : : :C an�n/x

D g .�/ x

(b) If x 2 Fn is an eigenvector of an n � n matrix A with corresponding eigenvalue �; then

g .A/ x D g .�/ x

That is, x is an eigenvector of g .A/ with corresponding eigenvalue g .�/ :To verify (b), note that x is an eigenvector of L A with corresponding eigenvalue �: Hence by (a), x is aneigenvector of g .L A/ with corresponding eigenvalue g .�/ : But

g .L A/ D a0 I C a1 .L A/C a2 .L A/2 C : : :C an .L A/n

D La0 In C La1A C La2A2 C : : :C Lan An ; F2: Theorem 2.16(c,e,f)/F3 and F4: Theorem 2.15(c,e,f)

D L .a0 InCa1AC:::Can An/; F2: Theorem 2.16(c) / F3 and F4: Theorem 2.15(c)

D Lg.A/;

hence, x is an eigenvector of g .A/ with corresponding eigenvalue g .�/ :

(c)

g .A/ D 2A2 � A C I

D 2

"1 23 2

#2�

"1 23 2

#C

"1 00 1

#

D 2

"7 69 10

#�

"1 23 2

#C

"1 00 1

#

D

"14 1015 19

#

g .A/ x D

"14 1015 19

#"23

#

D

"5887

#

D 29

"23

#Hence, x is an eigenvector of g .A/ with corresponding eigenvalue 29: Since g .�/ D g .A/ D 29; thisveri�es (b) in this case.

76

.

STUDY UNIT 7.

DIAGONALISABILITY


7.1 OverviewIn this study unit, we focus on developing a simple test for diagonalisability, and on a procedure for constructing abasis of eigenvectors in cases where the test is positive. To accomplish these goals, diagonalisability should �rst beinvestigated at a deeper level than in the previous study unit.

7.2 What to StudyStudy Friedberg: Section 5.2, and try as many exercises as possible. The exercises on simultaneous diagonalisationneed not to be studied. The subsection on direct sums is part of the syllabus, and is included in the discussion whichfollows.

7.3 A Test for Diagonalisability(See the discussion in Friedberg following F2 and F3: Theorem 5.14 / F4: Theorem 5.9.)

A Test for the Diagonalisability of an OperatorT : V ! V is diagonalisable if and only if the following two conditions hold for anybasis � of V :(a) The characteristic polynomial of [T ]� (and hence of T / splits.(b) n�rank

�[T ]� � �I

�D multiplicity of �; for each repeated eigenvalue of T :

An equivalent formulation in the case of square matrices is

A Test for the Diagonalisability of a Square MatrixA 2 Mn�n .F/ is diagonalisable if and only if the following two conditions hold.(a) The characteristic polynomial of A splits.(b) n�rank.A � �I / D multiplicity of �; for each repeated eigenvalue � of A:

77 MAT3701/1

ExampleLet T : P2 .R/! P2 .R/ be the linear transformation de�ned by

T�a C bx C cx2

�D .2a C b/C .a C 2b/ x C .a C b C c/ x2

(a) Calculate [T ]� ; where � D�1; x C x2; x2

:

(b) Use (a) and the Test for Diagonalisability to determine whether T is diagonalisable.

SOLUTION

(a)[T ]� D

h[T .1/]�

�T�x C x2

��

�T�x2��

iNow

T .1/ D 2C x C x2; according to the formula for T

D 2:1C�x C x2

�;

so

[T .1/]� D

264 210

375 Iand

T�x C x2

�D 1C 2x C 2x2; formula for T

D 1C 2�x C x2

�;

so �T�x C x2

��D

264 120

375 Iand

T�x2�D x2 according to the formula for T;

so �T�x2��D

264 001

375Hence

[T ]� D

264 2 1 01 2 00 0 1

375

78

(b) The characteristic polynomial of T is

det�[T ]� � t I

�D

��2� t 1 0

1 2� t 00 0 1� t

��D .1� t/

�� 2� t 11 2� t

��D .1� t/

�.2� t/2 � 1

�D .1� t/ .2� t � 1/ .2� t C 1/

D .1� t/2 .3� t/

Thus, the �rst condition is met, namely, the characteristic polynomial splits. We still need to check that foreach eigenvalue �;

n�rank�[T ]� � �I

�D algebraic multiplicity of �; ... (i)

where n denotes the dimension of the domain space of T (3 in this case, the dimension of P2 .R/):

Since the algebraic multiplicity of � D 3 is 1; condition (i) will automatically hold for � D 3:

For � D 1;

[T ]� � I D

264 1 1 01 1 00 0 0

375 ;

which is of rank 1I hence

3�rank�[T ]� � I

�D 2

Since the algebraic multiplicity of � D 1 is 2; condition (i) also holds for � D 1: Thus [T ]� ; and thereforealso T; is diagonalisable.

7.4 Diagonalising an Operator or Square Matrix

(See the discussion in Friedberg following F2 and F3: Theorem 5.14 / F4: Theorem 5.9)

On the basis of this theorem, the procedure described in SG: Section 6.4 can now be extended as follows:

79 MAT3701/1

Diagonalising a Square MatrixIf A 2 Mn�n .F/ satis�es the test for diagonalisability in SG: Section 7.3,then it can be diagonalised as follows:(a) Compute the distinct eigenvalues �1; �2; : : : ; �k of A:(b) Compute an ordered basis � i for each eigenspace E�i :(c) The ordered union � D �1 [ �2 [ : : : [ �k will be a basis for Fn

consisting of eigenvectors of A:(d) If Q is the matrix whose i�th column is the i�th vector of �, and if

D is the diagonal matrix whose i�th diagonal entry is the eigenvaluecorresponding to the i�th vector of �; then

D D Q�1AQ:

The formulation in the case of operators, using the procedure described in SG: Section 6.6, is as follows:

Diagonalising an OperatorIf T : V ! V satis�es the test for diagonalisability in SG: Section 7.3,then it can be diagonalised as follows:(a) Choose a convenient basis � for V :(b) Compute the distinct eigenvalues �1; �2; : : : ; �k of [T ]� (and

hence of T /:(c) Compute an ordered basis i for each E�i .[T ]�/ ; the eigenspace

of [T ]� associated with �i :(d) � i D �

�1�

� i�will be an ordered basis for E�i .T / ; the eigenspace

of T associated with �i :(If D fx1; : : : ; xr g then ��1� . / D

��1� .x1/ ; : : : ; ��1� .xr /

:/

(e) The union � D �1 [ �2 [ : : : [ �k will be an ordered basis for Vconsisting of eigenvectors of T I hence [T ]� is diagonal.

ExampleLet T : P2 .R/! P2 .R/ be the linear transformation de�ned by

T�a C bx C cx2

�D .2a C b/C .a C 2b/ x C .a C b C c/ x2:

(a) Calculate the eigenvalues and eigenspaces of T :

(b) Find a basis of P2 .R/ such that [T ] is diagonal.

SOLUTION

(a) Let � D�1; x C x2; x2

: According to (the previous) SG: Example 7.3,

[T ]� D

264 2 1 01 2 00 0 1

375 ;

80

and the eigenvalues of T are � D 1 and � D 3: For E1�[T ]�

�; the corresponding homogeneous system to

solve is

[T ]� � I D

264 1 1 01 1 00 0 0

375The solution set of this is

x D �; y D �� and z D �; where �; � 2 R

Thus,.x; y; z/ D � .1;�1; 0/C � .0; 0; 1/ ; �; � 2 R;

so thatE1�[T ]�

�D span f.1;�1; 0/ ; .0; 0; 1/g

From this we obtain E1 .T / ; namely,

E1 .T / D span�1�

�x C x2

�; x2

... (i)

Follow the same procedure for E3 .T /:

[T ]� � 3I D

264 �1 1 01 �1 00 0 �2

375 ;hence the solution set is

.x; y; z/ D � .1; 1; 0/ ; � 2 R;

so thatE3�[T ]�

�D span f.1; 1; 0/g

and thereforeE3 .T / D span

�1C x C x2

... (ii)

(b) A suitable is the union of the bases in (i) and (ii), namely,

D�1� x � x2; x2; 1C x C x2

7.5 Direct Sums with Two or More SummandsStudy the subsection on direct sums in Friedberg: Section 5.2.

7.5.1 First Direct-Sum Test for Two or More SummandsThe test in SG: Section 2.8.1 can now be generalised to more than two summands by F2 and F3: Theorem 5.15 /F4: Theorem 5.10.

81 MAT3701/1

First Direct-Sum TestLet �1; �2; : : : ; �k be ordered bases for the subspaces W1;W2; : : : ;Wk;respectively, of a �nite�dimensional vector space V : ThenV D W1 �W2 � : : :�Wk if and only if �1 [ �2 [ : : : [ �k is an ordered basis for V :

ExampleDetermine whether C4 D W1 �W2 �W3; where

W1 D�.x1; x2; x3; x4/ 2 C4 : x1 C x2 D 0; x3 D 0; x4 D 0

;

W2 D�.x1; x2; x3; x4/ 2 C4 : x1 C x3 D 0; x2 C x4 D 0

;

W3 D span f.0; 1; 0; 0/g

SOLUTIONWe leave it to you to show that

�1 D f.1;�1; 0; 0/g

is a basis for W1;�2 D f.1; 0;�1; 0/ ; .0; 1; 0;�1/g

is a basis for W2; and�3 D f.0; 1; 0; 0/g

is a basis for W3: Let

� D �1 [ �2 [ �3 D fv1 D .1;�1; 0; 0/ ; v2 D .1; 0;�1; 0/ ; v3 D .0; 1; 0;�1; / ; v4 D .0; 1; 0; 0/g :

Since

e1 D .1; 0; 0; 0/ D v1 C v4;

e2 D .0; 1; 0; 0/ D v4;

e3 D .0; 0; 1; 0/ D v1 � v2 C v4;

e4 D .0; 0; 0; 1/ D �v3 C v4;

it follows thatC4 D span fe1; e2; e3; e4g � span .�/

Hence � generates C4; and since j�j D 4 D dim�C4�; � is a basis for C4: By the above test,

C4 D W1 �W2 �W3

7.5.2 Second Direct-Sum Test for Two or More SummandsThe test in SG: Section 2.8.2 can be generalised as follows:

Second Direct-Sum TestLet W1;W2; : : : ;Wk be subspaces of a �nite�dimensional vector space V such that

V D W1 CW2 C : : :CWkThenV D W1 �W2 � : : :�Wk iff dim .V / D dim .W1/C dim .W2/C : : :C dim .Wk/

82

Proof (F2: Section 5.2, Exercise 18 / F3: Section 5.2, Exercise 19 / F4: Section 5.2, Exercise 20)For each i , let � i denote an ordered basis of Wi ; and let � D �1 [ �2 [ : : : [ �k :We denote the number of vectorsin � i by

�� i �� (the cardinality of � i /:Suppose V D W1 � : : :�Wk : By F2 and F3: Theorem 5.15(d) / F4: Theorem 5.10(d), � is a basis of V : Hence,

dim .V / D j�j D��1��C ��2��C : : :C ��k��

D dim .W1/C dim .W2/C : : :C dim .Wk/

DkXiD1dim .Wi /

Conversely, suppose

dim .V / DkXiD1dim .Wi /

Since

V DkXiD1Wi D

kXiD1span

�� i�D span

��1 [ : : : [ �k

�D span .�/ ;

� is a spanning set of V : Hence,

dim .V / � j�j ��1��C : : :C ��k�� D kX

iD1dim .Wi / D dim .V / ;

so j�j D dim .V / ; and therefore � is a basis of V :By F2 and F3: Theorem 5.15(e) /F4: Theorem 5.10(e), V D W1 � : : :�Wk; completing the proof.

ExampleDetermine whether V D M2�2 .C/ is the direct sum of the subspaces W1; W2 and W3 de�ned by

W1 D

("a b0 a

#: a; b 2 C

);

W2 D

("a 0b �a

#: a; b 2 C

);

W3 D

("0 00 b

#: b 2 C

)

SOLUTIONSince "

a bc d

#D

"a2 b0 a

2

#C

"a2 0c � a

2

#C

"0 00 d

#;

where the matrix on the left represents an arbitrary vector of M2�2 .C/ ; and the �rst, second and third matrices onthe right belong to W1; W2 and W3; respectively, it follows that

V D W1 CW2 CW3

83 MAT3701/1

Now

dim .W1/ D dim .W2/ D 2 (show),

dim .W3/ D 1 and

dim .V / D 4I

hencedim .V / 6D dim .W1/C dim .W2/C dim .W3/ ;

and therefore V is not the direct sum of W1, W2 and W3:

84

.

STUDY UNIT 8.

MATRIX LIMITS AND MARKOV CHAINS


8.1 Overview

Powers of matrices and their limits are studied, including necessary and suf�cient conditions for the limtis to exist.The results are then applied to Markov Chains, especially for the case where the associated transition matrix A isregular, in which case useful results are developed to �nd lim

m!1Am and the �xed probability vector. In the process,

the Gerschgorin Disk Theorem is presented for the location of the eigenvalues of a complex square matrix in thecomplex plane.

8.2 What to Study

Study Friedberg: Section 5.3, and try as many exercises as possible. The application at the end of the section ongenotypes may be left out.

8.3 Matrix Power Limits

If we consider the sequence f�mg where � is a complex number, then it is known that limm!1

�m exists iff

� 2 S D f� 2 C : j�j < 1 or � D 1g

In fact, it follows that under these conditions,

limm!1

�m D

(0 if j�j < 11 if � D 1

The set S also features when � is replaced by a complex square matrix: limm!1

Am exists iff

(a) the eigenvalues of A all lie in S; and

(b) dim .E1/ D multiplicity of .� D 1/ ; if 1 is an eigenvalue of A:

85 MAT3701/1

(F2 and F3: Theorem 5.18/F4: Theorem 5.13)

Note that, unlike the case for limm!1

�m; the limit of Am; if it exists, need not by 0 or I; e.g. if A D

"12 00 1

#then

limm!1

Am D

"0 00 1

#:

8.3.1 Example

Let A D

"12 112 0

#:

(a) Show that A satis�es the two conditions listed above for limm!1

Am to exist.

(b) Find limm!1

Am :

SOLUTION

(a) Show that the eigenvalues of A are 1 and � 12 : Since the algebraic multiplicity of � D 1 is one, the geometric

multiplicity must also be one. Thus both conditions are satis�ed.

(b) Show that A D QDQ�1; where

Q D

"1 2

�1 1

#and D D

"� 12 00 1

#

Thus,

limm!1

Am D limm!1

�QDQ�1

�mD lim

m!1

�QDmQ�1

�D Q

�limm!1

Dm�Q�1

D Q

"0 00 1

#Q�1

D13

"2 21 1

#

8.4 Markov Chains

The discussion starts after the example following F2 and F3: Theorem 5.19/F4: Theorem 5.14.

86

8.4.1 Example

Let A D

"12 112 0

#(as in the previous example).

(a) Show that A is a regular transition matrix.

(b) Use F2 and F3: Theorem 5.25/F4: Theorem 5.20 to �nd limm!1

Am :

(c) Find the �xed probability vector of A:

SOLUTION

(a) A is a transition matrix since all its entries are nonnegative and all its column sums are 1. Further,

A2 D

"34

12

14

12

#;

which has only positive entries, so that A is regular too.

(b) It follows from the solution to part (b) of the previous example that

E1 .A/ D span

("21

#);

thus13

"21

#is the (unique) probabillity vector belonging to E1 .A/ : Thus, from part (e) of the stated

theorem, limm!1

Am D13

"2 21 1

#:

(c)13

"21

#(from (b)).

Friedberg presents four examples on Markov chains (Examples 4 and 5, and two earlier non�numbered ones) whichshould all be studied.

8.5 Gerschgorin's Disk Theorem

The theorem appears in F2 and F3: Theorem 5.21 / F4: Theorem 5.16.

8.5.1 Example

Let A D

"12 112 0

#(as in the previous examples).

(a) Write down the Gerschgorin disks associated with A; and deduce that all its eigenvalues lie in

G D�z 2 C :

��z � 12

�� 1 :

87 MAT3701/1

(b) Verify directly that the eigenvalues of A all lie in G:

SOLUTION

(a) The disks are

C1 D�z 2 C :

��z � 12�� 1� and C2 D

�z 2 C : jzj �

12

�Since C2 � C1; the result follows.

(b) The eigenvalues of A are 1 and � 12 ; which both lie in G:

88

.

STUDY UNIT 9.

INVARIANT SUBSPACES AND THECAYLEY�HAMILTON THEOREM


9.1 OverviewAn invariant subspace enables us to investigate the properties of a linear operator on a part of a vector space.Properties derived there can sometimes be used to infer properties of the linear operator on the whole space; forexample, the characteristic polynomial of an operator on an invariant subspace divides the characteristic polynomialof the operator on the whole space (F2 and F3: Theorem 5.26 / F4: Theorem 5.21). This particular property is usedto provide a simple proof of the Cayley�Hamilton Theorem (F2 and F3: Theorem 5.28 / F4: Theorem 5.23).

9.2 What to StudyStudy Friedberg: Section 5.4, and do as many of the exercises as possible. The subsection titled Invariant Subspacesand Direct Sums is part of the syllabus, and should be studied for examination purposes.

9.3 Examples on Invariant Subspaces

1. Let T : C3 ! C3 be the linear operator de�ned by

T .a; b; c/ D .a; a C b; a � b C 2c/ ;

and let W be the subspace of C3 de�ned by

W D f.a; b; c/ jb D cg

Show that W is T �invariant.

SOLUTION

Let c D b in the formula for T; then

T .a; b; b/ D .a; a C b; a C b/ ;

which lies in W since the last two components of T .a; b; b/ are both equal to a C b: Hence, T .W / � W;showing that W is T �invariant.

89 MAT3701/1

2. Consider the vector space C2 over the �eld C; and the linear operator T on C2 with matrix

A D

"2 �45 �2

#

Determine all the invariant subspaces of T :

SOLUTION

C2 and f0g are invariant subspaces of T of dimensions 2 and 0; respectively. All the other invariant subspaceswill be of dimension one. Let W D span.fvg/ ; with 0 6D v 2 C2; denote an invariant subspace of dimensionone. Since Av must lie in W; we have Av D �v for some � 2 CI hence, W is generated by an eigenvector ofA: The eigenvalues of A are �1 D 4i and �2 D �4i ; so the one�dimensional invariant subspace of A are

E�1 D span .f.2; 1� 2i/g/ and E�2 D span .f.2; 1C 2i/g/

(verify).

3. Friedberg: Section 5.4, Exercise 3

Let T be a linear operator on a �nite�dimensional vector space V : Prove that the following subspaces areT �invariant:

(a) f0gand V

(b) N .T / and R .T /

(c) E�; for any eigenvalue � of T

SOLUTION

(a) Let v 2 f0g : Then v D 0; hence T .v/ D T .0/ D 0 since T is linear. So T .v/ 2 f0g ; and therefore f0gis T �invariant.Let v 2 V : Then, obviously, T .v/ 2 V; since T maps V into itself. Hence V is T �invariant.

(b) Let v 2 N .T / : Then T .v/ D 0; according to the de�nition of null space. Hence T .v/ 2 N .T / ; since0 2 N .T / ; being a subspace. Thus, N .T / is T �invariant.Let v 2 R .T / : Then, obviously, v 2 V; and therefore T .v/ 2 R .T / ; according to the de�nition of therange of T : Thus R .T / is T �invariant.

(c) Let v 2 E�: Then T .v/ D �v (also if v D 0; since T .0/ D 0 D � �0/I hence T .v/ 2 E� since �v 2 E�;being a subspace. Thus E� is T �invariant.


Prove that the restriction of a linear operator T to a T �invariant subspace is a linear operator on that subspace.

SOLUTION

Suppose T : V ! V is linear, and W is a T �invariant subspace of V : Clearly, TW : W ! V is linear, sinceT already preserves the vector-space operations on all of V : Also, since T .W / � W; TW maps W to W Ihence TW is a linear operator on W:

90

9.4 An Invariant Subspace TestWhen a subspace is speci�ed by means of a generating set, the following test may be used to determine whether it isinvariant. It shows that it is enough only to check whether the generators are being mapped back into the subspace.

Invariant Subspace TestLet T : V ! V be linear, and W D span.fw1; w2; : : : ; wng/ ; a subspace of V :W is T �invariant iff T .wi / 2 W for i D 1; 2; : : : ; n:

PROOFIf W if T �invariant, then, obviously, T .wi / 2 W for i D 1; 2; : : : ; n; since T .W / � W:Conversely, suppose T .wi / 2 W for i D 1; 2; : : : ; n: Any w 2 W can be expressed as

w D a1w1 C a2w2 C : : :C anwn for some ai 2 F .1 � i � n/ I

hence,

T .w/ D T .a1w1 C a2w2 C : : :C anwn/

D a1T .w1/C a2T .w2/C : : :C anT .wn/ since T is linear

2 W since each T .wi / 2 W

Thus, W is T �invariant.

ExampleLet T : P2 .C/! P2 .C/ be the linear transformation de�ned by

T�a C bx C cx2

�D .2a C b/C .a C 2b/ x C .a C b C c/ x2

Show that U D span�1; x C x2

is T �invariant.

SOLUTION

T .1/ D 2C x C x2 according to the formula for T

D 2 � 1C�x C x2

�2 U

and

T�x C x2

�D 1C 2x C 2x2; formula for T

D 1C 2�x C x2

�2 U

By the above test, U is T �invariant.

9.5 Cyclic SubspacesSuppose v is a nonzero vector of V and T : V ! V is linear. The smallest subspace of V containing v is span.fvg/ ;but it is not necessarily T �invariant. To enlarge fvg to the smallest T �invariant subspace of V; we need to include

v; T .v/ ; T 2 .v/ ; : : :

91 MAT3701/1

We refer toW D span

�v; T .v/ ; T 2 .v/ ; : : :

as the T�cyclic subspace of V generated by v:

9.5.1 Friedberg: Section 5.4, Exercise 11Let T be a linear operator on a vector space V; let v be a nonzero vector in V; and let W be the T �cyclic subspaceof V generated by v: Prove that

(a) W is T �invariant

(b) Any T �invariant subspace of V containing v also contains W

SOLUTION

(a) Let x 2 W: According to the de�nition of span,

x D a0v C a1T .v/C : : :C anT n .v/

for some n � 0 and scalars a0; : : : ; an in F: Hence,

T .x/ D T�a0v C a1T .v/C : : :C anT n .v/

�D T .a0v/C T .a1T .v//C : : :C T

�anT n .v/

�D a0T .v/C a1T 2 .v/C : : :C anT nC1 .v/ ;

which again lies in W: Thus W is T �invariant.

(b) Suppose U is T �invariant and v 2 U: Then T .v/ 2 U; and hence T .T .v// D T 2 .v/ 2 U; T�T 2 .v/

�D

T 3 .v/ 2 U; etc. Continuing this way, it follows that T n .v/ 2 U for n � 0: Since W is the smallest subspaceof V containing v; T .v/ ; : : : ; it must be contained in U:

9.5.2Determining a Basis for a Cyclic Subspace

Suppose T : V ! V is linear and v 2 V is nonzero. To �nd a basis for theT �cyclic subspace W generated by v; successively compute v; T .v/ ;T 2 .v/ ; : : : ; each time checking for linear dependence on the preceding vectors.

If k � 1 is the �rst number for which T k .v/ is linearly dependent onv; T .v/ ; : : : ; T k�1 .v/ ; then a basis for W is given by

�v; T .v/ ; : : : ; T k�1 .v/

:

This procedure is based on the proof of F2 and F3: Theorem 5.27(a) / F4: Theorem 5.22(a).ExampleLet T : P2 .C/! P2 .C/ denote the linear operator de�ned by

T�a C bx C cx2

�D c C ax C bx2;

and let W denote the T �cyclic subspace generated by v D x � 1: Find a basis for W:

92

SOLUTION

T .v/ D T .�1C x/ D T��1C 1 � x C 0 � x2

�D 0C .�1/ � x C 1 � x2; by the de�nition of T

D �x C x2

Observe that T v is not a scalar multiple of v:

T 2 .v/ D T .T v/ D T��x C x2

�D T

�0C .�1/ � x C 1 � x2

�D 1C 0 � x C .�1/ � x2

D 1� x2

Now

T 2 .v/ D �v C �T .v/

iff 1� x2 D � .x � 1/C ��x2 � x

�iff 1� x2 D �� C .� � �/ x C �x2

Equating corresponding coef�cients yields 8><>:� D �1

� � b D 0� D �1;

which shows that T 2 .v/ is linearly dependent on fv; T .v/g: A basis for W is, therefore,

� D fv; T .v/g D�x � 1;�x C x2

9.5.3

The Characteristic Polynomial of T restricted to a T �cyclic SubspaceSuppose T : V ! V is linear and W is the T �cyclic subspace of V with basis�v; T .v/ ; : : : ; T k�1 .v/

obtained above.

Ifa0v C a1T .v/C : : :C ak�1T k�1 .v/C T k .v/ D 0;

then the characteristic polynomial of TW isf .t/ D .�1/k

�a0 C a1t C : : :C ak�1tk�1 C tk

�This procedure is based on F2 and F3: Theorem 5.27(b) / F4: Theorem 5.22(b).

Examples

1. Let T : P2 .C/! P2 .C/ denote the linear operator de�ned by

T�a C bx C cx2

�D c C ax C bx2;

and let W denote the T �cyclic subspace generated by v D x � 1:

93 MAT3701/1

(a) Find a basis for W:

(b) Determine the characteristic polynomial f .t/ of TW :

(c) Show that P2 .C/ itself is T �cyclic.

SOLUTION

(a) According to SG: Example 9.5.2 above,

� D fv; T .v/g D�x � 1;�x C x2

is a basis for W:

(b) First MethodSince

T .v/ D 0v C 1 � T .v/

andT 2 .v/ D .�1/ � v C .�1/ � T .v/ ; ...(i)

according to (i) in SG: Example 9.5.2,

[TW ]� D

"0 �11 �1

#I

hence,

f .t/ D

�� 0� t �11 �1� t

�� D t .1C t/C 1 D 1C t C t2Second MethodFrom (i),

v C T .v/C T 2 .v/ D 0I

hencef .t/ D .�1/2

�1C t C t2

�D 1C t C t2;

according to the procedure above.

(c) To show that P2 .C/ is T �cyclic, we must �nd a vector w such that the T �cyclic subspace generated byw is all of P2 .C/ : Since P2 .C/ is of dimension 3 (with basis

�1; x; x2

/; we must �nd a w such that

w; Tw; T 2w are linearly independent. For w D 1;

Tw D T�1C 0 � x C 0 � x2

�D 0C 1 � x C 0 � x2 D x;

T 2w D T .Tw/ D T .x/

D T�0C 1 � x C 0 � x2

�D 0C 0 � x C 1 � x2 D x2

Thus, P2 .C/ is the T �cyclic subspace generated by w D 1:

94

2. Let V denote the vector space of all symmetric 2 � 2 matrices over C and let T : V ! V be the linearoperator de�ned by

T

"a bb c

#D

"0 cc �a C c

#

Denote by W the T �cyclic subspace of V generated by v D

"1 00 1

#:

(a) Find a basis of W:

(b) Determine the characteristic polynoial g .t/ of TW :

(c) Verify F2 and F3: Theorem 5.26 / F4: Theorem 5.21 by showing that g .t/ divides f .t/ ; the character-istic polynomial of T on V :

SOLUTION

(a)

v D

"1 00 1

#; and T .v/ D

"0 11 0

#according to the formula for T : Since fv; T .v/g is linearly independent, we proceed as follows:

T 2 .v/ D T

"0 11 0

#D

"0 00 0

#

Since�v; T .v/ ; T 2 .v/

is linearly dependent, the required basis for W is

fv; T .v/g D

("1 00 1

#;

"0 11 0

#)(b) Since

T 2 .v/C 0T .v/C 0v D 0;

the characteristic polynomial of TW is

.�1/2�t2 C 0t C 0

�D t2

(c) Choose as a basis for V;

� D

(v1 D

"1 00 0

#; v2 D

"0 11 0

#; v3 D

"0 00 1

#)Now

T .v1/ D

"0 00 �1

#D 0 � v1 C 0 � v2 � v3;

T .v2/ D

"0 00 0

#D 0 � v1 C 0 � v2 C 0 � v3;

T .v3/ D

"0 11 1

#D 0 � v1 C v2 C v3;

95 MAT3701/1

so

[T ]� D

264 0 0 00 0 1

�1 0 1

375 ;and therefore

f .t/ D

��t 0 00 �t 1

�1 0 1� t

�� D �t�� t 10 1� t

��D �t2 .t � 1/

D �g .t/ � .t � 1/

9.6 The Cayley�Hamilton Theorem (F2 and F3: Theorem 5.28 / F4: Theorem 5.23)The theorem states that an operator "satis�es" its characteristic equation.In other word: f .T / D T0; where f .t/ denotes the characteristic polynomial of an operator T : V ! V on a�nite�dimensional vector space V :

ExampleVerify the Cayley�Hamilton Theorem for the linear operator

T : V ! V de�ned by T

"a bb c

#D

"0 cc �a C c

#;

where V denotes the vector space of symmetric 2� 2 matrices over C:

SOLUTIONAccording to SG: Example 9.5.3.1.2 above, f .t/ D �t2 .t � 1/ :Hence,

f .T /

"a bb c

#!D �T 2 .T � I /

"a bb c

#!

D �T 2 "

0 cc �a C c

#�

"a bb c

#!

D �T 2 "

�a c � bc � b �a

#!

D �T

"0 �a

�a 0

#!

D �

"0 00 0

#

96

9.7Matrix Representations of an Operator in Convenient Block-Matrix FormsLet T : V ! V be a linear operator, and suppose V D W1 �W2:If �1 and �2 are bases for W1 and W2; respectively, then � D �1 [ �2 is a basisfor V according to SG: Example 2.8.3.1.

(a) If W1 is T �invariant, then A D [T ]� D

"�1 CO �2

#is in block-triangular

form, where B1 D�TW1

��1:

(b) If both W1 and W2 are T �invariant, then A D [T ]� D

"B1 OO B2

#is in

block-diagonal form, where B1 D�TW1

��1and B2 D

�TW2

��2:

(See also F2 and F3: Theorem 5.30 / F4: Theorem 5.15.)

PROOFLet �1 D fv1; v2; : : : ; vkg and �2 D fvkC1; : : : ; vng :

(a) (F2: Section 5.4, Exercise 10 / F3 and F4: Section 5.4, Exercise 12)

According to the de�nition of B1;

T�v j�D TW1

�v j�D .B1/i j � v1 C : : :C .B1/k j � vk C 0 � vkC1 C : : :C 0 � vn

for 1 � j � k. Hence, the �rst k columns of A yield the matrix

"B1O

#; where O is the zero matrix of order

.n � k/� k: Partitioning A at the k�th row and column completes the proof.

(b) If W2 is also T �invariant, then it follows by a similar argument to (a) that the last .n � k/ columns of A are

of the form

"OB2

#; where B2 D

�TW2

��2and O is the zero matrix of order k � .n � k/ :

ExampleLet T : C4 ! C4 be the linear operator de�ned by

T .x1; x2; x3; x4/ D .�x2; x1;�x1 C x3 � x4;�x2 C x3 C x4/

Denote by U the T �cyclic subspace of C4 generated by u D .1; 0; 1; 0/ ; and let

W D span f.0; 0; 1; 0/ ; .0; 0; 0; 1/g

(a) Determine a basis �1 for U:

(b) Show that W is T �invariant, and write down a basis �2 for W:

(c) Show that A D [T ]� is a block-diagonal matrix, where � D �1 [ �2:

(d) Calculate the characteristic polynomial of T :

97 MAT3701/1

SOLUTION

(a) From the formula for T;

T .u/ D T .1; 0; 1; 0/

D .0; 1; 0; 1/

Since T .u/ is not linearly dependent on u; we proceed as follows:

T 2 .u/ D T .T .u//

D T .0; 1; 0; 1/

D .�1; 0;�1; 0/

Now

T 2 .u/ D �u

D 0 � T .u/� u;

and therefore�1 D fu; T .u/g D f.1; 0; 1; 0/ ; .0; 1; 0; 1/g

is a basis for U:

(b) By the Invariant Subspace Test it suf�ces to show that both

T .0; 0; 1; 0/ and T .0; 0; 0; 1/

lie in W: According to the formula for T;

T .0; 0; 1; 0/ D .0; 0; 1; 1/

D .0; 0; 1; 0/C .0; 0; 0; 1/

which lies in W; and

T .0; 0; 0; 1/ D .0; 0;�1; 1/

D � .0; 0; 1; 0/C .0; 0; 0; 1/ ;

which also lies in W: Thus, W is T �invariant. Finally,

�2 D f.0; 0; 1; 0/ ; .0; 0; 0; 1/g

is a basis for W:

(c) From (a),

T .1; 0; 1; 0/ D .0; 1; 0; 1/

D 0 � .1; 0; 1; 0/C 1 � .0; 1; 0; 1/C 0 � .0; 0; 1; 0/C 0 � .0; 0; 0; 1/ ;

98

and

T .0; 1; 0; 1/ D T 2 .u/ D �u

D �1 � .1; 0; 1; 0/C 0 � .0; 1; 0; 1/C 0 � .0; 0; 1; 0/C 0 � .0; 0; 0; 1/

From (b),

T .0; 0; 1; 0/

D 0 � .1; 0; 1; 0/C 0 � .0; 1; 0; 1/C 1 � .0; 0; 1; 0/C 1 � .0; 0; 0; 1/ ;

and

T .0; 0; 0; 1/

D 0 � .1; 0; 1; 0/C 0 � .0; 1; 0; 1/� 1 � .0; 0; 1; 0/C 1 .0; 0; 0; 1/

Hence,

A D [T ]� D

2666640 �1 0 01 0 0 00 0 1 �10 0 1 1

377775 ;which is in block�diagonal form.

(d)

f .t/ D det .A � t I /

D det

266664�t �1 0 01 �t 0 00 0 1� t �10 0 1 1� t

377775D det

"�t �11 �t

#� det

"1� t �1

1 1� t

#(see SG: Section 4.3)

D�t2 C 1

� �t2 � 2t C 2

�Part 2 SummaryIn this part we focused on a special class of operators, namely, those which can be diagonalised. The investigationresulted in the important concepts of eigenvalue and eigenvector, which led to the development of a test for diag-onalisability, and a procedure for diagonalisation in cases where this is possible. A generalization of the conceptof eigenspace led to the concept of invariant subspace, which enabled us to localise the investigation of operatorsto parts of a vector space and to derive properties which can be transferred to the whole space. Matrix limits werestudied using results on diagonalisable matrices to calculate powers of A; and this was then used to study Markovchains. In the process, Gerschgorin's Disk Theorem for the location of the eigenvalues of a complex matrix waspresented.

99 MAT3701/1

.

PART 3

INNER PRODUCT SPACES

Page

Study unit 10 (Friedberg: Section 6.1): Inner Products and Norms 101

Study unit 11 (Friedberg: Section 6.2): The Gram�Schmidt OrthogonalisationProcess and Orthogonal Complements

107

Study unit 12 (Friedberg: Section 6.3): The Adjoint of a Linear Operator 117

Study unit 13 (Friedberg: Section 6.4): Normal and Self�Adjoint Operators 126

Study unit 14 (Friedberg: Section 6.5): Unitary and Orthogonal Operatorsand Their Matrices

136

Study unit 15 (Friedberg: Section 6.6): Orthogonal Projections and theSpectral Theorem

149

Study unit 16 (F2 and F3: Section 6.7 / F4: Section 6.8): Bilinear and Quadratic Forms 161

Study unit 17 (F2 and F3: Section 6.9 / F4: Section 6.10): Conditioning and the Rayleigh Quotient 165

100

Part 3 OverviewIn this part we focus on the introduction of a concept of measure for a vector space by means of an inner productwhich resulted in notions of length and angle, among others. To take advantage of these geometric notions, specialoperators are introduced which preserve not only the usual vector-space structure, but also the geometric structureof the new enriched space. Much of this part is devoted to the study of these special operators. We remind you that,in this course, the �eld F is assumed to be either the real �eld R; or the complex �eld C:


� solve a range of problems in a general vector space related to inner products, orthogonalisation, adjoint op-erators, least�squares approximation, minimal solution, normal operators, self�adjoint operators, unitary op-erators, rigid motion, orthogonal projections, the Spectral Theorem, bilinear and quadratic forms, Sylvester'slaw of inertia, conditioning, and the Rayleigh quotient

� reproduce basic theoretic arguments, including proofs, and solve basic theoretic problems related to the abovetopics over the real or complex numbers.

101 MAT3701/1

.

STUDY UNIT 10.

INNER PRODUCTS AND NORMS


10.1 OverviewIn this study unit we introduce the de�nition of an inner product on a real or complex vector space and derive thebasic properties of an inner�product space. You are familiar with inner products over the real numbers. However,the generalisation to complex numbers is not as obvious as one might have expected. The �rst sign of this appearsin condition (c) of the de�nition of inner product, where complex conjugation unexpectedly appears. Beware!

10.2 What to StudyStudy Friedberg: Section 6.1. The �rst twenty exercises are at a suitable level for you to try. The solutions to someof them appear in what follows.

10.3 Worked Examples

1. Friedberg: Example 1

For x D .a1; : : : ; an/ and y D .b1; : : : ; bn/ in Fn; de�ne

hx; yi DnXiD1aibi

Show that h�; �i satis�es the conditions for an inner product.

SOLUTION

We must verify conditions (a) � (d) in the de�nition of inner product. Condition (a) has already been veri�edby Friedberg.

Condition (b): For c 2 F

hcx; yi D h.ca1; : : : ; can/ ; .b1; : : : ; bn/i

D .ca1/ b1 C : : :C .can/ bnD c

�a1b1 C : : :C anbn

�D chx; yi

102

Condition (c): Using the properties of complex conjugation,

hx; yi D a1b1 C : : :C anbnD a1b1 C : : :C anbnD a1b1 C : : :C anbnD a1b1 C : : :C anbnD b1a1 C : : :C bnanD hy; xi

Condition (d): Suppose x 6D 0:

hx; xi D a1a1 C : : :C ananD ja1j2 C : : :C janj2 > 0;

since at least one component of x; say ai ; is nonzero, so that jai j2 > 0: Thus, all four conditions are satis�ed,and therefore h�; �i is an inner product on Fn (called the standard inner product).

(Remark: If F D R; then h�; �i takes the form

hx; yi DnXiD1aibi :/

2. If x and y are viewed as column vectors in Fn; then the standard inner product on Fn (previous example) canbe expressed as hx; yi D y�x :

SOLUTION

Let

x D

2664a1:::

an

3775 and y D

2664b1:::

bn

3775By the de�nition of the standard inner product,

hx; yi D a1b1 C a2b2 C : : :C anbn

D�b1 : : : bn

�2664a1:::

an

3775

D

2664b1:::

bn

3775t 2664

a1:::

an

3775D y�x

103 MAT3701/1

3. Consider the vector space V D C2 over the �eld F D C of complex numbers, and let h; i : C2 � C2 ! C bede�ned by

h.x1; x2/ ; .y1; y2/i D 2x1 Ny1 C x1 Ny2 C x2 Ny1 C x2 Ny2;

where .x1; x2/ ; .y1; y2/ 2 C2:

(a) Show that h�; �i is an inner product on V :

(b) Show that x D .0; 1/ is a unit vector in V :

(c) Find a unit vector y in V which is orthogonal to x D .0; 1/ :

SOLUTION

(a) We verify conditions (a) � (d) in the de�nition of inner product.Condition (a)Put z D .z1; z2/ ; then xC z D .x1 C z1; x2 C z2/ ; hence:

hxC z; yi D h.x1 C z1; x2 C z2/ ; .y1; y2/i

D .x1 C z1/ .2 Ny1 C Ny2/C .x2 C z2/ . Ny1 C Ny2/

D x1 .2 Ny1 C Ny2/C z1 .2 Ny1 C Ny2/C x2 . Ny1 C Ny2/C z2 . Ny1 C Ny2/

D fx1 .2 Ny1 C Ny2/C x2 . Ny1 C Ny2/g C fz1 .2 Ny1 C Ny2/C z2 . Ny1 C Ny2/g

D hx; yi C hz; yi; as required.

Condition (b)

hcx; yi D h.cx1; cx2/ ; .y1; y2/i

D cx1 .2 Ny1 C Ny2/C cx2 . Ny1 C Ny2/

D c fx1 .2 Ny1 C Ny2/C x2 . Ny1 C Ny2/g

D chx; yi; as required.

Condition (c)

hy; xi D y1 .2 Nx1 C Nx2/C y2 . Nx1 C Nx2/

D Ny1 .2x1 C x2/C Ny2 .x1 C x2/

D 2x1 Ny1 C Ny1x2 C Ny2x1 C Ny2x2D x1 .2 Ny1 C Ny2/C x2 . Ny1 C Ny2/

D hx; yi; as required.

Condition (d)Given x D .x1; x2/ ; we have

hx; xi D h.x1; x2/ ; .x1; x2/i

D 2x1 Nx1 C x1 Nx2 C x2 Nx1 C x2 Nx2D x1 Nx1 C .x1 C x2/ . Nx1 C Nx2/

D x1 Nx1 C .x1 C x2/�x1 C x2

�D jx1j2 C jx1 C x2j2 ... (i)

104

Hence, hx; xi � 0 for all x in C2: Also, if hx; xi D 0; then jx1j2 C jx1 C x2j2 D 0; hence jx1j Djx1 C x2j D 0: Thus, x1 D x2 D 0; so that x D 0: Thus, if x 6D 0; then hx; xi > 0:

(b) It is required to verify that h.0; 1/ ; .0; 1/i D 1: Now

h.0; 1/ ; .0; 1/i D 02 C .0C 1/2 ; according to (i)

D 1; as required.

(c) We seek y D .y1; y2/ such that hy; yi D 1 and hx; yi D 0: These conditions yield the following:

h.y1; y2/ ; .y1; y2/i D jy1j2 C jy1 C y2j2 D 1 (from (i)) ...(ii)

h.0; 1/ ; .y1; y2/i D 0 � .2 Ny1 C Ny2/C 1 � . Ny1 C Ny2/ D 0 ... (iii)

From (iii), we get y1 C y2 D 0; so that, substituting in (ii), jy1j2 D 1: Thus, jy1j D 1: When y1 D 1;then y2 D �1; since y1 C y2 D 0I hence .y1; y2/ D .1;�1/ is a solution. (Note that there are actuallyin�nitely many solutions, namely, .y1; y2/ D .cos � C i sin �;� .cos � C i sin �// D

�ei� ;�ei�

�; 0 �

� < 2� .)


In C2; show that hx; yi D x Ay� is an inner product, where

A D

1 i�i 2

!

Compute hx; yi for x D .1� i; 2C 3i/ and y D .2C i; 3� 2i/ :

SOLUTION

Note that here the vectors in C2 are viewed as row vectors. We verify conditions (a) � (d) of the de�nition ofinner product. Let x; y; z 2 C2 and c 2 C:

Condition (a)

hx C y; zi D .x C y/ Az� D x Az� C yAz� D hx; zi C hy; zi:

Condition (b)

hcx; yi D .cx/ Ay� D c .x Ay�/ D chx; yi:

Condition (c)

To proceed as in the previous two conditions using matrix arithmetic, we need a special result on the conjuga-tion of matrix products which we do not wish to derive here. Instead, we rather take the following approach.Suppose x D .x1; x2/ and y D .y1; y2/ ; then

hx; yi D x Ay� D .x1; x2/

1 i�i 2

! y1y2

!;

sohx; yi D x1y1 C i x1y2 � i x2y1 C 2x2y2 ... (i)

105 MAT3701/1

Hence,

hx; yi D x1y1 C i x1y2 � i x2y1 C 2x2y2D x1y1 C i x1y2 � i x2y1 C 2x2y2D x1y1 � i x1y2 C i x2y1 C 2x2y2D y1x1 C iy1x2 � iy2x1 C 2y2x2D hy; xi; according to (i)

Condition (d)

From (i),

hx; xi D x1x1 C i x1x2 � i x2x1 C 2x2x2D .x1 � i x2/ .x1 C i x2/C x2x2D .x1 � i x2/ .x1 � i x2/C x2x2D jx1 � i x2j2 C jx2j2 � 0

Also,hx; xi D 0, x1 � i x2 D x2 D 0, x1 D x2 D 0;

so hx; xi > 0 if x 6D 0:

It follows that h�; �i is an inner product. Finally, to compute hx; yi for the values of x and y given in thestatement, it follows from (i) that

hx; yi D .1� i/ .2C i/C i .1� i/ .3� 2i/� i .2C 3i/ .2C i/C 2 .2C 3i/ .3� 2i/

D .1� i/ .2� i/C i .1� i/ .3C 2i/� i .2C 3i/ .2� i/C 2 .2C 3i/ .3C 2i/

D 1� 3i C i .5� i/� i .7C 4i/C 2 .13i/

D 1� 3i C 1C 5i C 4� 7i C 26i

D 6C 21i


Let V be an inner product space, and suppose that x and y are orthogonal vectors in V : Prove that kx C yk2 Dkxk2 C kyk2 : Deduce the Pythagorean theorem in R2:

SOLUTION

kx C yk2 D hx C y; x C yi

D hx; xi C hy; yi C hx; yi C hy; xi

D hx; xi C hy; yi; since x and y are orthogonal

D kxk2 C kyk2 ... (i)

106

To deduce the Pythagorean theorem, let OAB denote the right�angled triangle

Put�!OA D x and

�!AB D y; then

�!OB D x C y: Since x and y are orthogonal, �!OB 2 D �!OA 2 C �!AB 2 by equation (i);

that isjOBj2 D jOAj2 C jABj2 ;

which is the statement of the Pythagorean theorem.

6. (cf Friedberg: Example 5)

Let V D Mn�n .F/ be the inner product space with

hA; Bi D tr�B�A

�for A; B 2 V

(see Friedberg: Section 6.1, Example 5).

Show that hA; Bi can also be expressed as

hA; Bi DnX

i; jD1ai j Nbi j where A D

�ai j�and B D

�bi j�

This inner product on Mn�n .F/ is known as the Frobenius inner product. Note the similarity with the standardinner product on Fn:

SOLUTION

hA; Bi D tr�B�A

�D

nXjD1

nXiD1

�B��j i Ai j

DnXjD1

nXiD1

Nbi jai j

DnX

i; jD1ai j Nbi j

(Note that if F D R; then

hA; Bi DnX

i; jD1ai jbi j .)

107 MAT3701/1

.

STUDY UNIT 11.

THE GRAM�SCHMIDT ORTHOGONA�LISATION PROCESS AND ORTHOGONALCOMPLEMENTS


11.1 OverviewThe vectors in a basis can be considered as the building blocks of a vector space. In the case of inner-productspaces, orthonormal bases turn out to be more useful, as they also take into account the geometric nature of suchspaces. The Gram�Schmidt orthogonalisation process describes how an ordinary basis for a �nite�dimensionalinner-product space can be transformed into one which is orthonormal.

11.2 What to StudyStudy Friedberg: Section 6.2. In the discussion which follows, the main points are highlighted, and then illustratedwith examples.

11.3

The Coordinates of a Vector Relative to an Orthogonal (Orthonormal) BasisSuppose � D fv1; v2; : : : ; vng is an orthogonal basis for an inner product space V :A vector v 2 V can be expressed as

v Dhv; v1i

kv1k2 v1 C

hv; v2i

kv2k2 v2 C : : :C

hv; vni

kvnk2 vn

If � is an orthonormal basis, then the linear combination simpli�es tov D hv; v1iv1 C hv; v2iv2 C : : :C hv; vnivn

In this case the coordinates hv; vi i of v are also referred to as the Fourier coef�cientsof v relative to �:

(See Friedberg: Theorems 6.3 and 6.5)The following examples illustrate the simpli�cation which results with this formula.

108

Examples

1. Consider the basis� D f.1; 1; 0/ ; .1;�1; 1/ ; .�1; 1; 2/g

for C3; and suppose you must write v D .1; 2; 3/ as a linear combination of �I that is, you must �nd a; b andc such that

.1; 2; 3/ D a .1; 1; 0/C b .1;�1; 1/C c .�1; 1; 2/

To get a; b and c; you must solve a system of three equations in the three unknowns a; b and c; namely

a C b � c D 1a � b C c D 2

b C 2c D 3

9>=>;Gaussian elimination yields the following solution:

a D32I b D

23I c D

76

This looks easy enough � but that is because we have chosen a simple basis and have left out a few steps.

To use the above formula, we notice that � is an orthogonal basis. The formula then yields the coef�cientsdirectly, thus:

.1; 2; 3/ � .1; 1; 0/ D 3 and k.1; 1; 0/k2 D 2

.1; 2; 3/ � .1;�1; 1/ D 2 and k.1;�1; 1/k2 D 3;

.1; 2; 3/ � .�1; 1; 2/ D 7 and k.�1; 1; 2/k2 D 6

Hence,a D

32I b D

23I c D

76

Now, that is easy!

(If � is an orthonormal basis, the calculations are easier still.)

2. Consider the inner�product space V D C2 over the �eld of complex numbers with inner product de�ned by

h.x1; x2/ ; .y1; y2/i D 2x1 Ny1 C x1 Ny2 C x2 Ny1 C x2 Ny2

You are given that � D f.0; 1/ ; .1;�1/g is an orthonormal basis of V with respect to h; i:

(a) Show thatf.1; 0/ ; .0; 1/g is not orthonormal with respect to the given inner product.

(b) Express v D .a; b/ as a linear combination of � by solving for x and y in terms of a and b in the system

.a; b/ D x .0; 1/C y .1;�1/

(c) Now use the above formula to express v D .a; b/ as a linear combination of �:

109 MAT3701/1

SOLUTION

(a) h.1; 0/ ; .1; 0/i D 2 � 1 � N1C 1 � N0C 0 � N1C 0 � N0 D 2

So .1; 0/ is not normal, and therefore the given set is not orthonormal.

(b) Since

.a; b/ D .0 � x C y; x � y/ ;

equating the corresponding components yields

0 � x C y D ax � y D b

)

The solution to this is

y D a and x D y C b D a C b

Therefore,

.a; b/ D .a C b/ .0; 1/C a .1;�1/

(c) According to the above formula

.a; b/ D h.a; b/ ; .0; 1/i .0; 1/C h.a; b/ ; .1;�1/i .1;�1/

D .a C b/ .0; 1/C .2a � a C b � b/ .1;�1/

D .a C b/ .0; 1/C a .1;�1/ ;

where the inner�products are calculated by the de�nition in the statement.

11.4 Friedberg: Theorem 6.5, Corollary shows how the above formula can be used to compute the matrixrepresentation of an operator.

The Matrix of an Operator Relative to an Orthonormal BasisConsider the operator T : V ! V; where V is an inner�product space withorthonormal basis � D fv1; v2; : : : ; vng : Let

A D [T ]� ; then Ai j D hT�v j�; vi i for i � i; j � n

(Note that the indices i and j are swapped in the inner product.)

ExampleCompute the matrix representation of

T : C2 ! C2 de�ned by T .a; b/ D .a C 2b; 2a C b/

in the basis

� D

�v1 D

1p2.1; 1/ ; v2 D

1p2.1;�1/

�

110

SOLUTION

hT .v1/ ; v1i D hT�1p2.1; 1/

�;1p2.1; 1/i D h

1p2.3; 3/ ;

1p2.1; 1/i D 3;

hT .v2/ ; v1i D hT�1p2.1;�1/

�;1p2.1; 1/i D h

1p2.�1; 1/ ;

1p2.1; 1/i D 0;

hT .v1/ ; v2i D hT�1p2.1; 1/

�;1p2.1;�1/i D h

1p2.3; 3/ ;

1p2.1;�1/i D 0;

hT .v2/ ; v2i D hT�1p2.1;�1/

�;1p2.1;�1/i D h

1p2.�1; 1/ ;

1p2.1;�1/i D �1

Thus,

[T ]� D

"3 00 �1

#

11.5 The Gram�Schmidt Orthogonalisation Process

This process describes how a basis in a �nite�dimensional inner�product space can be converted to an orthonormalbasis (see Friedberg: Theorem 6.4).

The Gram�Schmidt Orthogonalisation ProcessA basis � D fw1; w2; : : : ; wng for an inner�product space V can be convertedto an orthogonal basis � 0 D fv1; v2; : : : ; vng by letting

v1 D w1;

v2 D w2 �hw2; v1i

kv1k2 v1;

v3 D w3 �hw3; v1i

kv1k2 v1 �

hw3; v2i

kv2k2 v2

:::

vn D wn �n�1PkD1

hwn;vki

kvkk2 vk

To obtain an orthonormal basis from this, divide each vector in � 0 by its norm.

Example

LetU be the subspace ofC4 spanned by the linearly independent vectorsw1 D .1C i; 1; 0; 0/ ; w2 D .2; 1� i; 1; 0/and w3 D .2; 1� i; 1; 1/ : Use the Gram�Schmidt Orthogonalisation Process to �nd an orthogonal basis for U:

SOLUTION

We shall follow the procedure outlined above, and illustrated in Friedberg: Example 4.

111 MAT3701/1

Let:

v1 D w1 D .1C i; 1; 0; 0/ ;

v2 D w2 � hw2; v1i �v1

kv1k2 ;

v3 D w3 �

�hw3; v1i

v1

kv1k2 C hw3; v2i

v2

kv2k2

�To determine v2; we need to know hw2; v1i and hv1; v1i D kv1k2 :

kv1k2 D h.1C i; 1; 0; 0/ ; .1C i; 1; 0; 0/i

D .1C i/ .1C i/C 1 � 1C 0 � 0C 0 � 0

D 2C 1

D 3; so that kv1k Dp3

hw2; v1i D h.2; .1� i/ ; 1; 0/ ; .1C i; 1; 0; 0/i

D 2.1C i/C .1� i/ � 1

D 2� 2i C 1� i D 3� 3i D 3 .1� i/

Hence,

v2 D .2; 1� i; 1; 0/� 3 .1� i/ �13.1C i; 1; 0; 0/

D .2; 1� i; 1; 0/� .2; 1� i; 0; 0/

D .0; 0; 1; 0/

Nowv3 D w3 �

�hw3; v1i

v1

kv1k2 C hw3; v2i

v2

kv2k2

�;

so to determine v3; we need to know hw3; v1i; hw3; v2i and kv2k2 :

hw3; v1i D 2.1C i/C .1� i/ � 1

D .2� 2i/C .1� i/

D 3� 3i D 3 .1� i/

hw3; v2i D 2 � 0C .1� i/ � 0C 1 � 1C 1 � 0

D 1

kv2k2 D hv2; v2i

D 0 � 0C 0 � 0C 1 � 1C 0 � 0

D 1

112

Hence,

v3 D .2; 1� i; 1; 1/��3 .1� i/ �

.1C i; 1; 0; 0/3

C 1 �.0; 0; 1; 0/

2

�D .2; 1� i; 1; 1/� .2; 1� i; 1; 0/

D .0; 0; 0; 1/

Conclusion: An orthogonal basis for U is

f.1C i; 1; 0; 0/ ; .0; 0; 1; 0/ ; .0; 0; 0; 1/g

Hence, an orthonormal basis for U is�1p3.1C i; 1; 0; 0/ ; .0; 0; 1; 0/ ; .0; 0; 0; 1/

�

11.6 We will return to orthogonal projections later on where the following formula will play an important role.

The Orthogonal Projection of a Vector on a SubspaceIf W is a �nite�dimensional subspace of an inner�product space V; thenV D W �W?I hence any y 2 V can be uniquely expressed as

y D u C z where u 2 W and z 2 W?

The vector u is called the orthogonal projection of y on W; and is alsoreferred to as the vector in W "closest" to y:If fv1; v2; : : : ; vkg is an orthonormal basis for W; then

u D hy; v1iv1 C hy; v2iv2 C : : :C hy; vkivk

This is a slight reformulation of Friedberg: Theorem 6.6, since we have the concept of direct sum at our disposal.From the proof of the theorem it already follows that V D W C W?I but W \ W? D f0g also holds, sincev 2 W \W? implies v 2 W and v 2 W?; hence hv; vi D 0 so that v D 0: Thus, V D W �W?:

Example (F2 and F3: Section 6.2, Exercises 17(b) and 18(b) / F4: Section 6.2, Exercises 19(b) and 20(b))Let V D R3; y D .2; 1; 3/ and W D f.x; y; z/ : x C 3y � 2z D 0g :

(a) Find the orthogonal projection of y on W:

(b) Find the distance from y to W:

SOLUTION

(a) In order to apply the above formula, we need an orthonormal basis for W: Since the de�ning equation for Wcan be expressed as x D �3y C 2z; we have

W D f.�3y C 2z; y; z/ j y; z 2 Rg

D f.�3; 1; 0/ y C .2; 0; 1/ z j y; z 2 Rg

D span f.�3; 1; 0/ ; .2; 0; 1/g

113 MAT3701/1

The two vectors in the span are linearly independent (show), and therefore form a basis for W: Apply Gram�Schmidt to obtain an orthogonal basis for W :

w1 D .�3; 1; 0/

w2 D .2; 0; 1/�h.2; 0; 1/ ; .�3:1; 0/ih.�3; 1; 0/ ; .�3; 1; 0/i

.�3; 1; 0/

D .2; 0; 1/C610.�3; 1; 0/

D15.1; 3; 5/

So, an orthonormal basis of W is the following:

v1 Dw1

kw1kD

1p10.�3; 1; 0/

v2 Dw2

kw2kD

1p35.1; 3; 5/

According to the above formula, the orthogonal projection of y on W is given by

hy; v1iv1 C hy; v2iv2

D h.2; 1; 3/ ;1p10.�3; 1; 0/i

1p10.�3; 1; 0/C h.2; 1; 3/ ;

1p35.1; 3; 5/i

1p35.1; 3; 5/

D �510.�3; 1; 0/C

2035.1; 3; 5/

D114.29; 17; 40/

(b) The distance is given by

.2; 1; 3/� 114.29; 17; 40/

D

114 .�1;�3; 2/

D114k.�1;�3; 2/k

D1p14

114

11.7Extending an Orthonormal Subset to an Orthonormal BasisSuppose S D fv1; v2; : : : ; vkg is an orthonormal set in an n�dimensional innerproduct space V :To extend S to an orthonormal basis for V; �rst extend it to an arbitrary basisS0 D fv1; v2; : : : ; vk; wkC1; : : : ; wng for V; and then apply Gram�Schmidt. Thiswill yield an orthonormal basis fv1; v2; : : : ; vk; vkC1; : : : ; vng for V in which the�rst k vectors remain unchanged.Also, if we put W D span.S/ ; then S1 D fvkC1; : : : ; vng is an orthonormalbasis of W?:

(Friedberg: Theorem 6.7)

Examples

1. F2 and F3: Section 6.2, Exercise 8 / F4: Section 6.2, Exercise 9

Let W D span.f.i; 0; 1/g/ in C3: Find orthonormal bases for W and W?:

SOLUTION

Following the above procedure, �rst extend f.i; 0; 1/g to an arbitrary basis for C3I for example,

fw1 D .i; 0; 1/ ; w2 D .0; 1; 0/ ; w3 D .0; 0; 1/g

Then apply Gram�Schmidt

v1 D w1 D .i; 0; 1/ ;

v2 D w2 �hw2; v1i

kv1k2 v1

D .0; 1; 0/� 0 � v1D .0; 1; 0/ ;

v3 D w3 �hw3; v1i

kv1k2 v1 �

hw3; v2i

kv2k2 v2

D .0; 0; 1/�12.i; 0; 1/� 0 � v2

D12.�i; 0; 1/

Hence, an orthonormal basis for W isn1p2.i; 0; 1/

oand for W? is�

v2

kv1k;v3

kv3k

�D

�.0; 1; 0/ ;

1p2.�i; 0; 1/

�

115 MAT3701/1

2. Let U be the subspace of C4 spanned by the linearly independent vectors w1 D .1C i; 1; 0; 0/ ; w2 D.2; 1� i; 1; 0/ and w3 D .2; 1� i; 1; 1/ :

(a) Use the Gram�Schmidt Orthogonalisation Process to �nd an orthonormal basis for U:

(b) Find the vector in U closest to y D .1; 2C i; 1; 1/ :

(c) Extend your basis for U to an orthogonal basis for C4:

SOLUTION

(a) According to SG: Example 11.5, an orthonormal basis for U is given by�u1 D

1p3.1C i; 1; 0; 0/ ; u2 D .0; 0; 1; 0/ ; u3 D .0; 0; 0; 1/

�: ... (i)

(b) The closest vector is given by the orthogonal projection u of y on U I that is

u D h.1; 2C i; 1; 1/ ;�1C ip3;1p3; 0; 0

�i

�1C ip3;1p3; 0; 0

�

Ch.1; 2C i; 1; 1/ ; .0; 0; 1; 0/i .0; 0; 1; 0/C h.1; 2C i; 1; 1/ ; .0; 0; 0; 1/i .0; 0; 0; 1/

Hence,

u D

�1� ip3C2C ip3

��1C ip3;1p3; 0; 0

�C .0; 0; 1; 0/C .0; 0; 0; 1/

D .1C i; 1; 1; 1/

(c) According to SG: Section 11.6, C4 D U �U? and

z D y � u

D .1; 2C i; 1; 1/� .1C i; 1; 1; 1/

D .�i; 1C i; 0; 0/

lies in U?:

From (i), dim .U / D 3I hence fzg is basis for U?; since dim .U / C dim�U?

�D 4: An orthogonal basis for

C4 is therefore fz; u1; u2; u3g :

3. Let S be a nonempty subset of an inner product space V : Then S? is a subspace of V :

SOLUTION

We verify the three conditions in the Subspace Test. 0 2 S?; since h0; si D h0 � 0; si D 0h0; si D 0 for alls 2 S: Suppose x; y 2 S? and a 2 F: Then

hx; si D hy; si D 0 for all s 2 S

116

Hence,hx C y; si D hx; si C hy; si D 0C 0 D 0

andhax; si D ahx; si D a � 0 D 0

for all s 2 S; so that x C y 2 S? and ax 2 S?:

We conclude that S? is a subspace of V :

4. (Not in F2 and F3 / F4: Section 6.2, Exercise 7)

Let W be a subspace of an inner�product space V :

(a) Then W \W? D f0g :

(b) If W D span.S/ ; where S D fw1; v2; : : : ; wng ; then W? D S?:

SOLUTION

(a) Let x 2 W \W?: Then x 2 W and x 2 W?; so that hx; xi D 0: By Friedberg: Theorem 6.1 this impliesthat x D 0I thus W \W? D f0g :

(b) Let x 2 W?; then hx; wi D 0 for all w 2 W I in particular, hx; wi i D 0 for i D 1; 2; : : : ; n: ThusW? � S?:

Conversely, if x 2 S?; then hx; wi i D 0 for i D 1; 2; : : : ; n: Let w denote an arbitrary vector in W:Since W D span.S/ ; there exist scalars a1; a2; : : : ; an; such that w D a1w1 C a2w2 C : : :C anwn: So

hx; wi D hx; a1w1 C a2w2 C : : :C anwni

D a1hx; w1i C a2hx; w2i C : : :C anhx; wni

D 0; since hx; wi i D 0 for i D 1; : : : ; n

It follows that x 2 W?; and therefore S? � W?:

117 MAT3701/1

.

STUDY UNIT 12.

THE ADJOINT OF A LINEAR OPERATOR


12.1 OverviewIn Study Units 12�15 we investigate the special operators associated with inner�product spaces. These are operatorswhich, apart from preserving the normal vector�space structure, also preserve certain aspects of the geometricstructure inherent in inner�product spaces.In this study unit we focus on adjoints, which allows for an operator to be shifted from one side of the inner productto the other.

12.2 What to StudyStudy Friedberg: Section 6.3. In the discussion which follows the main points are highlighted, and then illustratedby examples.

12.3 Worked Example (Linear Functionals)Let V D C3: For each of the following linear transformations g : V ! C , �nd a vector y 2 C3 such thatg .x/ D hx; yi for all x 2 V :

(a) g .z1; z2; z3/ D i z1 C 2z2 C .1� i/ z3

(b) g .z1; z2; z3/ D .z1 � z2 C z3/C .z1 C 2z2 C 3z3/ i

SOLUTIONLet � =fv1; v2; v3g denote the standard ordered basis of C3; which is orthonormal with respect to the standard innerproduct.

(a) Using the construction in Friedberg: Theorem 6.8,

y D3XiD1g .vi /vi

D g .1; 0; 0/ .1; 0; 0/C g .0; 1; 0/ .0; 1; 0/C g .0; 0; 1/ .0; 0; 1/

D i .1; 0; 0/C 2 .0; 1; 0/C .1� i/ .0; 0; 1/

D .�i; 2; 1C i/

118

Alternative direct method

g .z1; z2; z3/ D i z1 C 2z2 C .1� i/ z3D z1.�i/C z22C z3

�1C i

�D h.z1; z2; z3/ ; .�i; 2; 1C i/i;

so y D .�i; 2; 1C i/ :

(b) Using the construction in Friedberg: Theorem 6.8,

y D3XiD1g .vi /vi

D g .1; 0; 0/ .1; 0; 0/C g .0; 1; 0/ .0; 1; 0/C g .0; 0; 1/ .0; 0; 1/

D 1C i .1; 0; 0/C .�1C 2i/ .0; 1; 0/C 1C 3i .0; 0; 1/

D .1� i;�1� 2i; 1� 3i/

Alternative direct method

g .z1; z2; z3/ D .z1 � z2 C z3/C .z1 C 2z2 C 3z3/ i

D z1 .1C i/C z2 .�1C 2i/C z3 .1C 3i/

D z1.1� i/C z2.�1� 2i/C z3.1� 3i/

D h.z1; z2; z3/ ; .1� i;�1� 2i; 1� 3i/i;

so y D .1� i;�1� 2i; 1� 3i/ :

12.4 Worked Example (Adjoints)Friedberg: Section 6.3, Exercise 3(b)Evaluate T � at x D .3� i; 1C 2i/ for

T : C2 ! C2 de�ned by T .z1; z2/ D .2z1 C i z2; .1� i/ z1/

SOLUTIONFor all z D .z1; z2/ in C2;

hz; T � .x/i D hT .z/ ; xi

D h.2z1 C i z2; .1� i/ z1/ ; .3� i; 1C 2i/i

D .2z1 C i z2/ .3� i/C .1� i/ z11C 2i

D .2z1 C i z2/ .3C i/C .1� i/ z1 .1� 2i/

D .2z1 C i z2/ .3C i/C z1 .�1� 3i/

D z1 .5� i/C z2 .�1C 3i/

D z1.5C i/C z2.�1� 3i/

D hz; .5C i;�1� 3i/iI

119 MAT3701/1

hence,T � .x/ D T � .3� i; 1C 2i/ D .5C i;�1� 3i/

12.5

Computing the AdjointSuppose T : V ! V is linear, and � is an orthonormal basis for V(a �nite�dimensional inner product space).Then

[T �]� D [T ]��(Friedberg: Theorem 6.10)

In words: "The matrix of T star is star of the matrix of T :"Following this method, the matrix of T relative to � is calculated �rst, and then the conjugate transpose is formed.This yields the matrix of T � relative to �; from which the formula for T � can be determined, if required.

ExampleLet V D C2 be the inner product space over C with respect to the inner product de�ned by

h.x1; x2/ ; .y1; y2/i D 2x1 Ny1 � x1 Ny2 � x2 Ny1 C x2 Ny2

Let T : C2 ! C2 be the linear operator de�ned by

T .a; b/ D .b;�a C 2b/

Find the formula for T �:Hint: � D f.0; 1/ ; .1; 1/g is orthonormal with respect to h; i:

SOLUTIONThe following linear expression for .a; b/ 2 C2 in terms of � will be used repeatedly:

.a; b/ D .b � a/ .0; 1/C a .1; 1/ : ... (i)

We proceed by calculating [T ]� :

T .0; 1/ D .1; 2/ ; formula for T

D 1 � .0; 1/C 1 � .1; 1/ ; from (i)

T .1; 1/ D .1; 1/

D 0 � .0; 1/C 1 � .1; 1/

Thus,

[T ]� D

"1 01 1

#According to Friedberg: Theorem 6.10,

�T ��D

"1 01 1

#�D

"1 10 1

#

120

Hence, according to the de�nition of [T �]� ,

T � .0; 1/ D 1 � .0; 1/C 0 � .1; 1/ D .0; 1/

andT � .1; 1/ D 1 � .0; 1/C 1 � .1; 1/ D .1; 2/

Thus, from (i)

T � .a; b/ D .b � a/ T � .0; 1/C aT � .1; 1/

D .b � a/ .0; 1/C a .1; 2/

D .a; a C b/

12.6Least�Squares Approximation

Let A 2 Mm�n .F/ and y 2 Fm :x0 2 Fn minimises kAx � yk as x ranges over Fn

iff x D x0 is a solution of A�Ax D A�y:(Friedberg: Theorem 6.12)

This result is an application of the theory of closest vector developed in Friedberg: Theorem 6.6 and its corollary.When rank.A/ D n (ie, A is of full column rank) then A�A is invertible and x0 is given by x0 .A�A/�1 A�y:However, to actually �nd x0; it is still better, even in this case, to solve for it from A�Ax D A�y by Gaussianelimination, rather than to calculate .A�A/�1 ; which amounts to more work.The following is an application of the above theorem:

The Least�Squares Polynomial FitThe least�squares �t of a polynomial

antn C an�1tn�1 C : : :C a0in Pn .F/ .n � 1/ to a set of data

f.t1; y1/ ; .t2; y2/ ; : : : ; .tm; ym/gis obtained by letting

A D

266664tn1 tn�11 : : : 1tn2 tn�12 : : : 1:::

::::::

tnm tn�1m : : : 1

377775 ; y D266664y1y2:::

ym

377775 and x D

266664anan�1:::

a0

377775 ;and solving for the optimal coef�cients x0 in

A�Ax D A�y

ExampleF2 and F3: Section 6.3, Exercise 18 / F4: Section 6.3, Exercise 20(a)For the set of data

f.�3; 9/ ; .�2; 6/ ; .0; 2/ ; .1; 1/g ;

�nd the line and the parabola that provide the least�squares �ts. Compute the error E in both cases.

121 MAT3701/1

SOLUTIONLeast�squares line: Denote the line by y D ct C d: In this case,

A D

266664�3 1�2 10 11 1

377775 ; x D"cd

#and y D

2666649621

377775Hence, the system .A�A/ x D A�y becomes"

14 �4�4 4

#"cd

#D

"�3818

#We leave it to you to solve this system by Gaussian elimination. The solution is

x0 D

"cd

#D

"�252

#;

hence the line is y D �2t C 52 .

The estimation error E in this case is E D ky � Ax0k2 ; where

y � Ax0 D

2666649621

377775�266664

1721325212

377775 D 12

2666641

�1�11

377775 ; so that E D 1

Least�squares parabola: Denote the parabola by y D ct2 C dt C e: In this case,

A D

2666649 �3 14 �2 10 0 11 1 1

377775 ; x D264 cde

375 and y D

2666649621

377775Hence, the system .A�A/ x D A�y becomes264 98 �34 14

�34 14 �414 �4 4

375264 cde

375 D264 106�3818

375Solving this system by Gaussian elimination, it turns out that

x0 D

26413

� 432

375 ; so that y D13t2 �

43t C 2

In this case, y � Ax0 D 0; which means that we have an exact �t. Consequently, E D 0:

12.7Minimal Solutions to Systems of Linear Equations

Suppose Ax D b is a consistent system, where A 2 Mm�n .F/ and b 2 Fm :The minimal solution to Ax D b is given by s D A�u; where u is a solutionto AA�u D b:(Friedberg: Theorem 6.13)

122

Notice that the coef�cient matrix for the minimal solution is AA�; while the coef�cient matrix for a polynomial �tis A�A:

Examples

(1) Find the minimal solution to

x � y � z D 0

y � z D 2

x � y � z D 0

with respect to the standard inner product in C3:

SOLUTION

Follow the method outlined in Friedberg: Section 6.3, Example 3.

A D

264 1 �1 �10 1 �11 �1 �1

375 and b D

264 020

375Hence,

AA� D

264 1 �1 �10 1 �11 �1 �1

375264 1 0 1�1 1 �1�1 �1 �1

375 D264 3 0 30 2 03 0 3

375 ;so that

AA�x D b

iff

264 3 0 30 2 03 0 3

375264 x1x2x3

375 D264 020

375

iff3x1 C 3x3 D 0

2x2 D 23x1 C 3x3 D 0

9>=>;A solution of this system is, for example, the following:

x1 D 0; x2 D 1; x3 D 0

Therefore, the minimal solution is

s D A�x D

264 1 0 1�1 1 �1�1 �1 �1

375264 010

375 D264 0

1�1

375

123 MAT3701/1

(2) Consider the linear system

x1 � x2 � x3 D �1x2 � x3 D 0

x1 � 2x3 D �1

9>=>; ; ... (i)

which can be expressed as Ax D b; where

A D

264 1 �1 �10 1 �11 0 �2

375 ; x D

264 x1x2x3

375 and b D

264 �10

�1

375(a) Express the solution to (i) in parametric form, that is, �nd vectors v;w 2 R3 such that a general solution

x can be expressed as

x D v C wt with t 2 R ... (ii)

Test your answer by checking whether v and v C w are solutions to (i).

(b) Using the standard inner product in R3; express kxk2 ; with x as in (ii), as a quadratic polynomial in tIand then �nd the value for t which makes kxk2 a minimum. Substitute this value for t in (ii) to obtainthe minimal solution to (i).

(c) Verify your answer in (b) by using the method described in Friedberg: Section 6.3, Example 3 tocompute the minimal solution to (i) in a different way, based on Friedberg: Theorem 6.13.

SOLUTION

(a) Use Gaussian elimination to solve (i).

In matrix form:264 1 �1 �1 : �10 1 �1 : 01 0 �2 : �1

375!264 1 �1 �1 : �10 1 �1 : 00 1 �1 : 0

375R3 � R1

!

264 1 �1 �1 : �10 1 �1 : 00 0 0 : 0

375R3 � R2

Letting x3 D t in the last matrix, it follows from the second equation that x2 D t; and from the �rst thatx1 D 2t � 1: Thus, the general solution to (i) is, in parametric form,

x D .x1; x2; x3/ D .2t � 1; t; t/ ; t 2 R

D .�1; 0; 0/C .2; 1; 1/ t; t 2 R ... (iii)

D v C wt; t 2 R;

where

v D .�1; 0; 0/ and w D .2; 1; 1/ ... (iv)

Direct substitution in (i) shows that both v D .�1; 0; 0/ and v C w D .1; 1; 1/ are solutions to (i).

124

(b) From (ii),

kxk2 D hv C wt; v C wti

D hv; vi C hv;wti C hwt; vi C hwt; wti

D hv; vi C hv;wit C hw; vit C hw;wit2; t belongs to R

D kvk2 C 2hv;wit C kwk2 t2; since hv;wi D hw; vi

With v;w as in (iv), this becomeskxk2 D 1� 4t C 6t2

This equation represents a parabola in t; whose minimal value is obtained at

t D�b2a

D412D13

(using matric maths or �rst�year calculus).Substituting this in (iii) yields the minimal solution, namely

x D .�1; 0; 0/C .2; 1; 1/ �13

D13.�1; 1; 1/

(c) Following the method in Friedberg: Section 6.3, Example 3,

AA� D

264 1 �1 �10 1 �11 0 �2

375264 1 0 1�1 1 0�1 �1 �2

375

D

264 3 0 30 2 23 2 5

375 Iso we consider the system

3x C 3z D �1

2y C 2z D 0

3x C 2y C 5z D �1

A solution to this system is, for example,

u D

264 � 43

�11

375Hence, the minimal solution is given by

s D A�u D

264 1 0 1�1 1 0�1 �1 �2

375264 � 4

3�11

375 D264 � 1

31313

375 ;which agrees with the solution in (b).

125 MAT3701/1


Let A be an n � n matrix. Prove that det .A�/ D det .A/:

(This exercise is needed for future use.)

SOLUTION

For an n � m matrix A; de�ne A by�A�i j D Ai j : That is, A is obtained from A by conjugating each entry.

For example, if

A D

"1C i i 1

2 2C i 3i

#; then A D

"1� i �i 1

2 2� i �3i

#

We �rst show for an n � n matrix A that det�A�D det .A/ by induction on the order n of A: If n D 1; then

A D [A11] I sodet .A/ D A11 D det

�A�; as required

Assume n > 1; and the result holds for n � 1: Using the cofactor expansion of det .A/ along the �rst row,

det .A/ DnXjD1.�1/1C j A1 j � det

�eA1 j�D

nXjD1.�1/1C j A1 j � det

�gA1 j�; property of conjugation

DnXjD1.�1/1C j A1 j � det

�gA1 j� ; by induction

DnXjD1.�1/1C j A1 j � det

�gA1 j�D det

�A�; de�nition of determinant

This completes the induction step, and concludes the proof that

det .A/ D det�A�

Note that in the above proof we used the fact thatgA1 j D gA1 j : This follows readily, by keeping in mind that,in general, eAi j is obtained from A by deleting the i th row and j th column. So, eAi j is obtained by conjugatingeach entry in the resulting matrix eAi j: On the other hand, A is obtained from A by conjugating each entry.Then eAi j is obtained from the resulting matrix A by deleting the i th row and j th column. This results in thesame matrix as with the �rst process.

Finally, observe that, by de�nition, A� D�A�tI so

det�A��D det

��A�t�

D det�A�

D det .A/; by the result above

126

.

STUDY UNIT 13.

NORMALANDSELF�ADJOINTOPERATORS


13.1 OverviewRead the introduction to Friedberg: Section 6.4. Diagonalisation of an operator by means of an orthonormal basisis another situation where the underlying scalar �eld has a signi�cant impact on the resulting theory. If F D C; thenthe operator must be normal; if F D R; then it must be self�adjoint.

13.2 What to StudyStudy Friedberg: Section 6.4. You may try the exercises preceding F2 and F3: Section 6.4, Exercise 12 / F4: Section6.4, Exercise 13.

13.3 Worked Example (related to the proof of Schur's Theorem)Let

A D

264 1 1 0�1 1 10 1 1

375(a) Calculate the characteristic polynomial of A and show that it splits over R:

(b) Find an eigenvector v of A�; and let W D span.fvg/ :

(c) Show that W? is A�invariant.

127 MAT3701/1

SOLUTION

(a)

f .t/ D det .A � t I /

D det

0B@ 1� t 1 0�1 1� t 10 1 1� t

1CAD .1� t/ det

1� t 1

1 1� t

!C det

1 01 1� t

!(expand along the �rst column)

D .1� t/�.1� t/2 � 1

�C .1� t/

D .1� t/�.1� t/2 � 1C 1

�D .1� t/3

(b) According to (a), � D 1 is the only eigenvalue of A: Hence, from the proof of the �rst lemma of Friedberg:Section 6.4, it follows that � D 1 is the only eigenvalue of A�: To �nd a corresponding eigenvector, we solvethe homogeneous linear system with coef�cient matrix

A� � I D

0B@ 1 �1 01 1 10 1 1

1CA�0B@ 1 0 00 1 00 0 1

1CA

D

0B@ 0 �1 01 0 10 1 0

1CAIt follows that

v D .1; 0;�1/

is a required eigenvector.

(c) First calculate the subspace W? :

x D .x1; x2; x3/ 2 W?

, hx; vi D 0

, h.x1; x2; x3/ ; .1; 0;�1/i D 0

, x1 � x3 D 0

, x1 D x3

So,

W? D f.x1; x2; x1/ j x1; x2 2 Rg

D fx1 .1; 0; 1/C x2 .0; 1; 0/ j x1; x2 2 Rg

D span f.1; 0; 1/ ; .0; 1; 0/g

128

To show that W? is A�invariant, it suf�ces to show that both

A

264 101

375 and A

264 010

375lie in W?:

Now,

A

264 101

375 D264 1 1 0�1 1 10 1 1

375264 101

375 D264 101

375 2 W?

and

A

264 010

375 D

264 1 1 0�1 1 10 1 1

375264 010

375

D

264 111

375 D264 101

375C264 010

375 2 W?

Thus, W? is A�invariant.

13.4 Example (Properties of a Normal Operator)Let T : C3 ! C3 denote the linear operator on C3 (over the complex �eld) with matrix

A D

264 1 i 0i 1 00 0 1C i

375with respect to the standard basis.

(a) Show that T is normal.

(b) Calculate T .1; 1; 1/ and T � .1; 1; 1/, and verify that kT .1; 1; 1/k D kT � .1; 1; 1/k :

(c) Show that v D .1; 1; 1/ is an eigenvector of T corresponding to � D 1C i: Verify that T � .v/ D �v:

(d) Find an eigenvector u of T corresponding to 1� i: Verify that hu; vi D 0, with v as in (c).

(e) Find an orthonormal basis for C3 consisting of eigenvectors of T :

129 MAT3701/1

SOLUTION

(a)

AA� D

264 1 i 0i 1 00 0 1C i

375264 1 �i 0�i 1 00 0 1� i

375 ... (i)

D

264 2 0 00 2 00 0 2

375D 2I3

Similarly,A�A D 2I3;

hence,AA� D A�A

This equation can be written as[T ]� [T ]�� D [T ]

�� [T ]� ;

where � denotes the standard basis of C3: Hence,

[T ]��T ��D�T ��[T ]� ;

since � is orthonormal with respect to the standard inner product. Hence,�T T �

��D�T �T

��;

and thereforeT T � D T �T;

that is, T is normal.

(b)

[T .1; 1; 1/]� D [T ]� .1; 1; 1/�

D A

264 111

375

D

264 1C i1C i1C i

375 ;

130

so thatT .1; 1; 1/ D .1C i/ .1; 1; 1/ ... (ii)

Similarly, �T � .1; 1; 1/

��D

�T ��[.1; 1; 1/]�

D A�

264 111

375

D

264 1� i1� i1� i

375 ;from the second matrix in (i). Hence,

T � .1; 1; 1/ D .1� i/ .1; 1; 1/ ... (iii)

From (ii)

kT .1; 1; 1/k2 D h.1C i/ .1; 1; 1/ ; .1C i/ .1; 1; 1/i

D .1C i/ .1C i/h.1; 1; 1/ ; .1; 1; 1/i

D 3 .1C i/ .1� i/

D 6;

and similarly, from (iii), T � .1; 1; 1/ 2 D 6Hence,

kT .1; 1; 1/k D T � .1; 1; 1/

(c) The �rst part follows from (ii), and the second part�T � .v/ D �v

�follows from (iii).

(d) Calculate E1�i .A/ : The corresponding homogeneous system to solve is, in matrix form,264 1� .1� i/ i 0i 1� .1� i/ 00 0 .1C i/� .1� i/

375 D264 i i 0i i 00 0 2i

375 ;hence,

E1�i .A/ D

8><>:264 t�t0

375 : t 2 C9>=>;

D span

8><>:264 1�10

3759>=>;

131 MAT3701/1

Since A D [T ]� ; with � the standard basis,

E1�i .T / D span f.1;�1; 0/g

Hence, u D .1;�1; 0/ is a suitable candidate, and clearly,

hu; vi D h.1;�1; 0/ ; .1; 1; 1/i

D 0

(e) The characteristic polynomial of T is given by

det .A � t I / D

��1� t i 0

i 1� t 00 0 .1C i/� t

��D ..1C i/� t/

�� 1� t ii 1� t

��D ..1C i/� t/

�.1� t/2 � i2

�D ..1C i/� t/ .1� t � i/ .1� t C i/

D ..1C i/� t/2 ..1� i/� t/ ;

hence the eigenvalues of T are� D 1C i and � D 1� i

From (d), E1�i .T / D spanf.1;�1; 0/g :We follow the same procedure for � D 1C i; by �rst calculating theeigenspace E1Ci .A/ of A: The corresponding system to solve is, in matrix form,264 �i i 0

i �i 00 0 0

375 ;hence,

E1Ci .A/ D

8><>:264 tts

375 : t; s 2 C9>=>;

D span

8><>:264 110

375 ;264 001

3759>=>;

Hence,E1Ci .T / D span f.1; 1; 0/ ; .0; 0; 1/g

132

Hence, an orthonormal basis for E1Ci .T / is�1p2.1; 1; 0/ ; .0; 0; 1/

�

Thus, �1p2.1; 1; 0/ ; .0; 0; 1/ ;

1p2.1;�1; 0/

�is an orthonormal basis for C3 consisting of eigenvectors of T :

13.5Diagonalising an Operator by means of an Orthonormal Basis

Let T be an operator on V; a �nite�dimensional inner�product space over the�eld F D R or F D C: If T is self�adjoint if F D R or normal if F D C; thenthere exists an orthonormal basis � for V consisting of eigenvectors of T suchthat [T ]� is a diagonal matrix.The basis � is obtained by computing an orthonormal basis � i (using Gram�Schmidt, if necessary) for each of the eigenspaces E�i .T / corresponding to thedistinct eigenvalues �1; �2; : : : ; �k of T; and taking the ordered union

� D �1 [ �2 [ : : : [ �k(Friedberg: Theorems 6.16 and 6.17)

Example(Not in F2 and F3 / F4: Section 6.4, Exercise 2(e))Consider the real inner�product space V D M2�2 .R/ with

hA; Bi D tr�B�A

�for A; B 2 V;

and de�neT : V ! V by T .A/ D At

(a) Show that T is self�adjoint

(b) Find an orthonormal basis � for V consisting of eigenvectors of T :

(c) Write down [T ]� :

SOLUTIONThis inner�product on V D M2�2 .R/ is de�ned in Friedberg: Section 6.1, Example 5. Let

A D

"a1 a2a3 a4

#and B D

"b1 b2b3 b4

#;

then it follows from SG: Example 10.3.6 that

hA; Bi D a1b1 C a2b2 C a3b3 C a4b4;

133 MAT3701/1

so the inner product acts like the standard inner product on R4; by multiplying the corresponding entries of A andB and adding them up. For example, if

A D

"1 23 4

#and B D

"4 23 1

#;

thenhA; Bi D 1 � 4C 2 � 2C 3 � 3C 4 � 1 D 21

It is now easy to see that the standard basis for V D M2�2 .R/ ; namely,

D

("1 00 0

#;

"0 10 0

#;

"0 01 0

#;

"0 00 1

#)is orthonormal with respect to h�; �i:

(a) Since is an orthonormal basis for V; T is self�adjoint iff [T ] is self�adjoint. Let us therefore compute[T ] :

T

"1 00 0

#!D

"1 00 0

#D 1 �

"1 00 0

#C 0 �

"0 10 0

#C 0 �

"0 01 0

#C 0 �

"0 00 1

#;

T

"0 10 0

#!D

"0 01 0

#D 0 �

"1 00 0

#C 0 �

"0 10 0

#C 1 �

"0 01 0

#C 0 �

"0 00 1

#;

T

"0 01 0

#!D

"0 10 0

#D 0 �

"1 00 0

#C 1 �

"0 10 0

#C 0 �

"0 01 0

#C 0 �

"0 00 1

#;

T

"0 00 1

#!D

"0 00 1

#D 0 �

"1 00 0

#C 0 �

"0 10 0

#C 0 �

"0 01 0

#C 1 �

"0 00 1

#;

so

[T ] D

2666641 0 0 00 0 1 00 1 0 00 0 0 1

377775 ... (i)

Since [T ] D [T ]� D [T ]t ; the operator T is self�adjoint.

(b) Begin by calculating the eigenvalues of T; which are the same as the eigenvalues of [T ] :

det�[T ] � t I4

�D

��1� t 0 0 0

0 �t 1 00 1 �t 00 0 0 1� t

��from (i)

D .1� t/2�t2 � 1

�(check)

D .t � 1/3 .t C 1/

Thus, the eigenvalues of T are � D 1 (with multiplicity 3/ and � D �1 (with multiplicity 1/:

134

Next, we compute the eigenspaces of T : For E1 .T / ; we �rst compute E1�[T ]

�: The corresponding homo-

geneous system to solve is, according to (i),

[T ] � I D

2666640 0 0 00 �1 1 00 1 �1 00 0 0 0

377775 ;

so thatE1�[T ]

�D span .f.1; 0; 0; 0/ ; .0; 1; 1; 0/ ; .0; 0; 0; 1/g/ (check)

A spanning set for E1 .T / is obtained from this by �nding the 2�2 matrices whose coordinate vectors relativeto are 266664

1000

377775 ;2666640110

377775 and

2666640001

377775 ; respectively,

(the spanning vectors of E1�[T ]

�/:

Thus,

E1 .T / D span

("1 00 0

#;

"0 11 0

#;

"0 00 1

#)

Since the spanning set is orthogonal with respect to the inner product on V; it is linearly independent, andtherefore a basis for E1 .T / : Thus, an orthonormal basis for E1 .T / is obtained by dividing each vector by itsnorm: ("

1 00 0

#;1p2

"0 11 0

#;

"0 00 1

#)... (ii)

Repeat the whole process for E�1 .T / : To compute E�1�[T ]

�; the corresponding system to solve is, ac-

cording to (i),

[T ] C I D

2666642 0 0 00 1 1 00 1 1 00 0 0 2

377775 ;so that

E�1�[T ]

�D span .f.0; 1;�1; 0/g/ (check)

Thus,

E�1 .T / D span

("0 1

�1 0

#)I

hence an orthonormal basis for E�1 .T / is(1p2

"0 1

�1 0

#)... (iii)

135 MAT3701/1

The desired basis � is obtained by forming the ordered union of (ii) and (iii):

� D

("1 00 0

#;1p2

"0 11 0

#;

"0 00 1

#;1p2

"0 1

�1 0

#)... (iv)

(c) The diagonal matrix [T ]� is constructed from the eigenvalues of T corresponding to the basis vectors in �; inthe same order:

[T ]� D

2666641 0 0 00 1 0 00 0 1 00 0 0 �1

377775

136

.

STUDY UNIT 14.

UNITARY AND ORTHOGONALOPERATORS AND THEIR MATRICES


14.1 OverviewRead the introduction to Friedberg: Section 6.5. In this study unit we consider the operators which preserve thestructure of an inner�product space. An operator T on an inner�product space V which preserves the inner�product(i.e.,

hv;wi D hT .v; / T .w/i for all v;w in V /

also preserves the norm, sincekvk2 D hv; vi D hT .v/ ; T .v/i D kT .v/k2

This necessarily makes T one�to�one (since nonzero vectors have nonzero norms), and therefore also onto (andconsequently an isomorphism) in the case of �nite�dimensional inner�product spaces.Thus, the operators on a �nite�dimensional inner�product space which preserve the inner product are exactly thosewhich preserve the structure of the inner�product space. They are called unitary operators if F D C; and orthogonaloperators if F D R:

14.2 What to StudyStudy Friedberg: Section 6.5. Note that the subsections on rigid motions, orthogonal operators on R2; and conicsections should all be studied for this module. See SG: Section 14.5 below, which repeats the slightly differentpresentation of rigid motions and orthogonal operators on R2 as it appears in the fourth edition of Friedberg.

14.3 Example 3

(1) (Friedberg: Section 6.5, Exercise 4)

Let V be a complex inner�product space, and de�ne T : V ! V by T .v/ D cv; with c 2 C �xed. Find all csuch that

(a) T is normal

(b) T is self�adjoint

137 MAT3701/1

(c) T is unitary

SOLUTION

For all v;w 2 V;hT .v/ ; wi D hcv;wi D hv; cwi D hv; T � .w/i;

so T � .w/ D cw for all w 2 V :

(a) For all v 2 V; �T �T

�.v/ D T � .cv/ D cT � .v/ D ccv D jcj2 v; ... (i)

and �T T �

�.v/ D T .cv/ D cT .v/ D ccv D jcj2 v ... (ii)

So, T is normal for all values of c:

(b) T D T �() c D c() c is real.

(c) From (i) and (ii), T is unitary() jcj2 D 1() jcj D 1:

(2) Let A D

"a bb 0

#; where a; b 2 C and b 6D 0: Find all a; b such that

(a) A is self�adjoint

(b) A is unitary

(c) A is normal

SOLUTION

A D

"a bb 0

#; A� D

"a bb 0

#... (i)

AA� D

"aa C bb ab

ab bb

#D

"jaj2 C jbj2 ab

ab jbj2

#... (ii)

A�A D

"aa C bb ab

ab bb

#D

"jaj2 C jbj2 ab

ab jbj2

#... (iii)

(a) According to (i), A D A� (i.e., A is self�adjoint)() a D a and b is arbitrary() a is real and b isarbitrary.

(b) AA� D I in (ii)() ab D 0 and jaj2 C jbj2 D jbj2 D 1() a D 0 and jbj D 1:

(c) From (ii) and (iii), AA� D A�A (i.e., A is normal)() a D a (since b 6D 0, given) and b is arbitrary() a is real and b is arbitrary.

(3) Statement preceding F2 and F3: Section 6.5, Example 3 / F4: Section 6.5, Example 4

Suppose T is a linear operator on a �nite�dimensional inner�product space V : Show that T is unitary [or-thogonal] if and only if [T ]� is unitary [orthogonal] for some orthonormal basis � for V :

138

SOLUTION

T is unitary iff T T � D T �T D Iiff [T T �]� D [T �T ]� D [I ]�iff [T ]� [T �]� D [T �]� [T ]� D I (Friedberg: Theorem 2.11)iff [T ]� [T ]�� D [T ]

�� [T ]� D I .� is orthonormal)

iff [T ]� is unitary

The same argument applies exactly if T is orthogonal, keeping in mind that in the second last step, [T ]��becomes [T ]t� since [T ]� is real.


For each of the following pairs of matrices, explain whether or not they are unitarily equivalent:

(a)

1 00 1

!and

0 11 0

!

(b)

0 11 0

!and

0 1

212 0

!

(c)

0B@ 0 1 0�1 0 00 0 1

1CA and

0B@ 2 0 00 �1 00 0 0

1CA

(d)

0B@ 0 1 0�1 0 00 0 1

1CA and

0B@ 1 0 00 i 00 0 �i

1CA

(e)

0B@ 1 1 00 2 20 0 3

1CA and

0B@ 1 0 00 2 00 0 3

1CASOLUTION

Since unitary equivalence implies similarity, we may apply the Test for Nonsimilarity in SG: Section 6.5.2.

(a) No, since their determinants differ.

(b) No, since their determinants differ.

(c) No, since their determinants differ.

(d) Yes. The �rst matrix is normal with eigenvalues 1; i and �i (check); hence by Friedberg: Theorem6.19, it is unitarily equivalent to the second matrix, which is diagonal and contains the eigenvalues ofthe �rst.

(e) No, the �rst matrix is not normal (check); hence by Friedberg: Theorem 6.19, it cannot be unitarilyequivalent to the second matrix, which is diagonal.

139 MAT3701/1

14.4Unitary (resp., Orthogonal) Diagonalisation of Matrices

If A is an n � n complex normal (resp., real symmetric) matrix, then there exists acomplex unitary (resp., real orthogonal) matrix P such that P�AP D D (resp., P t AP D D/;where D is a complex (resp., real) diagonal matrix.The matrices P and D are obtained by computing an orthonormal basis � i(using Gram�Schmidt, if necessary) for each of the eigenspaces E�i .A/ corres�ponding to the distinct eigenvalues �1; �2; : : : ; �k of A; and taking the orderedunion � D �1 [ �2 [ : : : [ �k : The columns of P consist of the vectors in �; andD is constructed from the eigenvalues of A corresponding to the columns of P;in the same order.(Friedberg: Theorems 6.19 and 6.20)

Examples

(1) Let

A D

264 0 1 0�1 0 00 0 i

375(a) Show that A is normal.

(b) Find a unitary matrix U and a diagonal matrix D such that U �AU D D:

SOLUTION

(a)

AA� D

264 0 1 0�1 0 00 0 i

375264 0 �1 01 0 00 0 �i

375 D264 1 0 00 1 00 0 1

375 D I3Similarly, A�A D I3I hence AA� D A�A, and thus A is normal.

(b) The characteristic polynomial f .t/ of A is given by

f .t/ D

��t 1 0�1 �t 00 0 i � t

��D .i � t/

�t2 C 1

�D � .t � i/2 .t C i/

Thus, the eigenvalues are �i:

For a basis for Ei ; the corresponding system to solve is, in matrix form,264 �i 1 0�1 �i 00 0 0

375!264 i �1 00 0 00 0 0

375 �R1R2 C i R1

140

Thus,Ei D span f.1; i; 0/ ; .0; 0; 1/g ;

and hence an orthonormal basis for Ei is given by�1p2.1; i; 0/ ; .0; 0; 1/

�... (i)

For E�i ; the corresponding system is264 i 1 0�1 i 00 0 2i

375!264 i 1 00 0 00 0 1

375 R2 � i R112i R3

Thus,E�i D span f.1;�i; 0/g ;

and hence an orthonormal basis for E�i is given by�1p2.1;�i; 0/

�... (ii)

The union of (i) and (ii) yields an orthonormal basis for C3; namely�1p2.1; i; 0/ ; .0; 0; 1/ ;

1p2.1;�i; 0/

�Thus,

U D1p2

264 1 0 1i 0 �i0p2 0

375 and D D

264 i 0 00 i 00 0 �i

375

(2) Let

A D

264 �1 2 22 �1 22 2 �1

375Given the eigenvalues of A are �3 (with multiplicity 2/ and 3; with corresponding eigenspaces

E�3 D span f.1; 0;�1/ ; .1;�1; 0/g

andE3 D span f.1; 1; 1/g ;

�nd an orthogonal matrix P and a diagonal matrix D such that P t AP D D:

141 MAT3701/1

SOLUTION

Orthonormalise the basis for E�3 using Gram�Schmidt:

v1 D .1; 0;�1/

v2 D .1;�1; 0/�h.1;�1; 0/ ; .1; 0;�1/ih.1; 0;�1/ ; .1; 0;�1/i

.1; 0;�1/

D .1;�1; 0/�12.1; 0;�1/

D

�12;�1;

12

�Thus,

�1 D

�v1

kv1k;v2

kv2k

�D

�1p2.1; 0;�1/ ;

1p6.1;�2; 1/

�is an orthonormal basis for E�3: Since

�2 D

�1p3.1; 1; 1/

�is an orthonormal basis for E3; an orthonormal basis for R3 consisting of eigenvectors of A is

� D �1 [ �2 D

�1p2.1; 0;�1/ ;

1p6.1;�2; 1/ ;

1p3.1; 1; 1/

�The columns of P consist of the vectors in �

P D

26641p2

1p6

1p3

0 � 2p6

1p3

� 1p2

1p6

1p3

3775 ;and D is the diagonal matrix formed by the eigenvalues of A corresponding to the columns in P; in the sameorder

D D

264 �3 0 00 �3 00 0 3

37514.5 Rigid MotionsFriedberg presents this material in a slightly different way in the fourth edition, which we repeat here for complete-ness' sake. We restrict ourselves to V D R2: Recall that if v0 is a vector in V; then the function

g : V ! V de�ned by g .x/ D x C v0 for all x 2 V

is called the translation by v0:We then have the following result:

F4: Theorem 6.22 If f : V ! V is a rigid motion, then there exists a unique orthogonal operator T on V and aunique translation g on V such that f D g � T :This is basically F2 and F3: Section 6.5, Lemma 2 (the translation g is by f .0//:

ExampleLet f : R2 ! R2 be the function de�ned by

f .a; b/ D .1C b; 2C a/ ; where .a; b/ 2 R2

142

(a) Show that f is a rigid motion on R2:

(b) Express f as f D g � T; where g is a translation and T an orthogonal operator on R2:

SOLUTION

(a) To verify that f is a rigid motion, it is required to show that

k f .x/� f .y/k D kx� yk ; for all x; y in R2

Write x D .a; b/ ; y D .A; B/ ; then

k f .x/� f .y/k D k.1C b; 2C a/� .1C B; 2C A/k

D k.b � B; a � A/k

Dq.b � B/2 C .a � A/2

D k.a � A; b � B/k

D k.a; b/� .A; B/k

D kx� yk ;

so that f is a rigid motion.

(b)

f .a; b/ D .1; 2/C .b; a/

D .1; 2/C T .a; b/

D g � T .a; b/ ;

whereg .a; b/ D .1; 2/C .a; b/ ;

which is the translation by .1; 2/ ; andT .a; b/ D .b; a/ ;

which is an orthogonal operator.

(Note that f .0/ D .1; 2/ and T .a; b/ D f .a; b/� f .0/ ; which is in accordance with the theory.)

14.5.1 Orthogonal Operators on R2

Basically, the difference in presentation between the fourth and the earlier editions is in the classi�cation of theorthogonal operators on R2: According to F2 and F3: Section 6.5, Lemma 4, an orthogonal operator T on R2 iseither

(a) a rotation by an angle � .0 � � < 2�/ ;

in which case the matrix of T relative to the standard ordered basis � for R2 is given by

[T ]� D

"cos � � sin �sin � cos �

#; and det

�[T ]�

�D 1I or

143 MAT3701/1

(b) a re�ection about the x�axis followed by a rotation by � .0 � � < 2�/ ;

in which case

[T ]� D

"cos � sin �sin � � cos �

#; and det

�[T ]�

�D �1

In the fourth edition of Friedberg, the following is noted:

The Re�ection About a Line in R2 through the Origin(F4: Section 6.5, Exercise 5)Let T be the re�ection about the line L through the origin which makes an angle � with the positive x�axis. Then

[T ]� D

"cos 2� sin 2�sin 2� � cos 2�

#;

where � is the standard ordered basis for R2:

ProofThe vector v1 D .cos�; sin�/ lies on L ; hence

T .v1/ D v1I

and the vector v2 D .� sin�; cos�/ is perpendicular to L ; hence

T .v2/ D �v2

Thus,

[T ] D

"1 00 �1

#;

where D fv1 D .cos�; sin�/ ; v2 D .� sin�; cos�/g :Let

Q D [I ]� D

"cos� � sin�sin� cos�

#;

then

[T ]� D Q [T ] Q�1

D

"cos� � sin�sin� cos�

#"1 00 �1

#"cos� sin�

� sin� cos�

#

D

"cos� sin�sin� � cos�

#"cos� sin�

� sin� cos�

#

D

"cos2 � � sin2 � 2 sin� cos�2 sin� cos� sin2 � � cos2 �

#

D

"cos 2� sin 2�sin 2� � cos 2�

#

144

RemarkOn the basis of this result, Friedberg, in the fourth edition, describes case (b) above as a re�ection about the linethrough the origin making an angle �=2 with the positive x�axis.

A Classi�cation of the Orthogonal Operators on R2

Suppose T is an orthogonal operator on R2 and A D [T ]� ; where � denotesthe standard ordered basis for R2:(a) If det .A/ D 1; then T is a rotation through an angle � .0 � � < 2�/ ; and

[T ]� D

"cos � � sin �sin � cos �

#(b) If det .A/ D �1; then T is a re�ection about the line through the origin making

an angle �=2 .0 � � < 2�/ with the positive x�axis, and

[T ]� D


#(See F2 and F3: Section 6.5, Lemma 4 / F4: Theorem 6.23)

ExampleClassify the orthogonal operator T on R2 in the previous example according to the above classi�cation, where T isde�ned by

T .a; b/ D .b; a/ for all .a; b/ in R2

SOLUTIONLet � denote the standard ordered basis for R2; then

[T ]� D

"0 11 0

#Since det

�[T ]�

�D �1; T is a re�ection about a line L through the origin. To determine the angle which L makes

with the positive x�axis, express [T ]� in the form

[T ]� D

"0 11 0

#D


#with 0 � � < 2�

It follows that � D �2 I hence the angle � which L makes with the positive x�axis is

� D�

2D�

4With the above classi�cation of orthogonal operators on R2 it is possible to provide the following classi�cation ofrigid motions on R2:

14.5.2A Classi�cation of the Rigid Motions on R2

A rigid motion f on R2 can be expressed as f D g � T; where g is thetranslation on R2 by f .0/ and T is an orthogonal operator on R2; whichcan be either a rotation, or a re�ection about a line through the origin.The complete description of T can be obtained from the above Classi��cation of Orthogonal Operators.(See F2 and F3: Section 6.5, Theorem 6.22 and Lemmas 2 and 4 /F4: Theorems 6.22 and 6.23)

145 MAT3701/1

ExampleExpress the rigid motion f : R2 ! R2 de�ned by

f .a; b/ D .1� b; 1� a/ for all .a; b/ 2 R2

in the form f D g � T; where g is a translation and T an orthogonal operator on R2: If T is a rotation, �nd the angle� through which it rotates; and if T is a re�ection about a line L through the origin, �nd the equation of L and theangle it makes with the positive x�axis.

SOLUTION

f .a; b/ D .1; 1/C .�b;�a/

D .g � T / .a; b/ ;

whereg .a; b/ D .1; 1/C .a; b/ ;

hence the translation is by .1; 1/ D f .0/ ; and

T .a; b/ D .�b;�a/

Let � be the standard ordered basis for R2; then

[T ]� D

"0 �1

�1 0

#I

hence T is orthogonal, since[T ]�� [T ]� D [T ]� [T ]

�� D I2

Also,det

�[T ]�

�D �1;

so T is a re�ection about a line L through the origin.Since

[T ]� D

"0 �1

�1 0

#D

"cos 3�2 sin 3�2sin 3�2 � cos 3�2

#;

L makes an angle � D 3�4 with the positive x�axis. Hence, the equation of L is

y D .tan�/ x D �x

Alternatively, L can be determined as follows. Note that L is determined by the eigenspace E1 .T / of T corre-sponding to the eigenvalue 1; since the vectors lying on L (i.e., having L as their carrier) are exactly the vectors inE1 .T / : To compute E1 .T /, the corresponding homogeneous system to solve is, in matrix form,

[T ]� � I D

"�1 �1�1 �1

#I

hence,E1 [T ] D span f.1;�1/g

146

Thus, the gradient of L is �1; hence its equation is y D �x; and the angle � makes with the positive x�axis isgiven by tan� D �1; that is, � D 3�

4 :

14.6 Conic SectionsThe orthogonal diagonalisation of a real symmetric matrix can be used to eliminate the xy�term in a quadratic formby a rotation of the x� and y�axes.

Eliminating the Cross Term by RotationIn the quadratic form ax2 C bxy C cy2; the xy�term (cross term) can beeliminated by rotating the x� and y�axes as follows:If

A D

"a b=2

b=2 c

#and X D

"xy

#;

then the quadratic form is given by X t AX:Compute a 2� 2 real orthogonal matrix P such that P t AP D D with D diagonal.Make det .P/ D 1 by interchanging its columns, if necessary. Then P represents therotation of R2 through an angle � (calculated as in SG: 14.5.1) and X D PX 0 yieldsthe desired change of variables.The columns of P D [v1 v2] are eigenvectors of A; and the diagonal of D consists of the

corresponding eigenvalues �1 and �2 of A; respectively. If X 0 D

x 0

y0

!; then the

x 0�axis is in the direction of v1; and the y0�axis is in the direction of v2: The x 0� andy0�axes are obtained by rotating the x� and y�axes through �; and the change ofvariables X D PX 0 transforms the quadratic form to �1

�x 0�2C �2

�y0�2:

ExampleEliminate the xy�term (cross term) in

3x2 � 2xy C 3y2 D 1 ... (i)

by a rotation of the axes, that is, �nd a rotation matrix P 2 M2�2 .R/ such that (i), expressed in terms of x 0; y0 givenby "

xy

#D P

"x 0

y0

#;

contains no cross term.Express x and y in terms of x 0 and y0; and specify the counterclockwise angle of rotation.

SOLUTIONThe quadratic form (i) can be expressed as

X t AX D 1;

where

A D

"3 �1

�1 3

#and X D

"xy

#

147 MAT3701/1

The characteristic polynomial of A is given by

det .A � t I2/ D

�� 3� t �1�1 3� t

��D .3� t/2 � 1

D .3� t � 1/ .3� t C 1/

D .t � 2/ .t � 4/ ;

yielding the eigenvalues t D 2 and t D 4:To �nd E2; we solve the homogeneous system with coef�cient matrix"

1 �1�1 1

#!

"1 �10 0

#R2 C R1

Thus,E2 D span f.1; 1/g ;

with orthonormal basis �1p2.1; 1/

�To �nd E4; we solve "

�1 �1�1 �1

#!

"1 10 0

#�R1R2 � R1

Thus,E4 D span f.1;�1/g ;

with orthonormal basis �1p2.1;�1/

�Thus,

P t AP D D;

where

P D1p2

"1 1

�1 1

#; that is, P t D

1p2

"1 �11 1

#(note that det .P/ D 1/; and

D D

"4 00 2

#If we let "

xy

#D P

"x 0

y0

#D

1p2

"1 1

�1 1

#"x 0

y0

#;

that is,

x D1p2

�x 0 C y0

�y D

1p2

��x 0 C y0

�;

148

then (i) becomes4�x 0�2C 2

�y0�2D 1

Since

P D

"1p2

1p2

� 1p2

1p2

#D

"cos

��4

�� sin

��4

�sin��4

�cos

��4

� # ;it means the x� and y�axes are rotated through �45o:

149 MAT3701/1

.

STUDY UNIT 15.

ORTHOGONAL PROJECTIONS AND THESPECTRAL THEOREM


15.1 Overview

In Study Unit 13 (Friedberg: Section 6.4), normal (if F D C/ or self�adjoint (if F D R/ operators are characterisedas exactly those operators for which orthonormal bases consisting of eigenvectors exist. In this study unit, anothercharacterisation of such operators is presented, namely, as operators which can be expressed as a linear combinationof orthogonal projections. The coef�cients in the linear combination are the distinct eigenvalues of the operator �thus, the name spectral decomposition, as the set of distinct eigenvalues of an operator, is referred to as its spectrum.

15.2 What to Study

Study Friedberg: Section 6.6. This study unit assumes a knowledge of direct sums (SG: Sections 2.7, 2.8 and 7.5)and projections (SG: Sections 3.5 and 11.6).

15.3 Orthogonal Projections

We restrict our investigation of orthogonal projections to �nite�dimensional inner�product spaces V : Suppose Wis a subspace of V : By SG: Section 11.6, V D W � W?: An orthogonal projection T : V ! V is a projectionon a subspace W of V along its orthogonal complement W?: By SG: Example 3.5.1.2, W D R .T / and W? D

N .T / : Hence, an orthogonal projection can also be described as a projection T : V ! V such that R .T /? DN .T / (equivalently R .T / D N .T /? ; for �nite�dimensional inner�product spaces V /: If fv1; v2; : : : ; vkg is anorthonormal basis of W; then it follows from Friedberg: Theorem 6.6 that

T .y/ D hy; v1iv1 C hy; v2iv2 C : : :C hy; vkivk for y 2 V

In what follows, this result is re�established independently:

150

First Characterisation of an Orthogonal ProjectionLet V be a �nite�dimensional inner�product space, and let W be a subspace of Vwith orthonormal basis fv1; : : : ; vkg : De�ne T : V ! V by

T .y/ DkPiD1hy; vi ivi

Then(a) V D W �W?

(b) T is a linear transformation(c) T .w/ D w for all w 2 W(d) T .u/ D 0 for all u 2 W?

(e) T is the orthogonal projection on W I that is, for y 2 V with y D w C u; wherew 2 W and u 2 W?; we have T .y/ D w

Proof

(a) W \ W? D f0g ; since x 2 W \ W? implies x 2 W and x 2 W?I hence hx; xi D 0; so that x D 0 byFriedberg: Theorem 6.1. Thus, the result follows, since by Friedberg: Theorem 6.6, V D W CW?:

(b) For x; y 2 V and r 2 F;

T .x C y/ DkXiD1hx C y; vi ivi ; de�nition of T

DkXiD1

.hx; vi i C hy; vi i/ vi ; property of h; i

DkXiD.hx; vi ivi C hy; vi ivi /

DkXiD1hx; vi ivi C

kXiD1hy; vi ivi

D T .x/C T .y/ ;

and

T .r x/ DkXiD1hr x; vi ivi ; de�nition of T

DkXiD1rhx; vi ivi ; property of h; i

D rkXiD1hx; vi ivi

D rT .x/

151 MAT3701/1

(c) Since fv1; : : : ; vkg is an orthonormal basis of W and w 2 W;

w D hw; v1iv1 C hw; v2iv2 C : : :C hw; vkivk by Friedberg: Theorem 6.5

DkXiD1hw; vi ivi ; summation notation

D T .w/ ; according to the de�nition of T

(d)

T .u/ DkXiD1hu; vi ivi ; de�nition of T

DkXiD10 � vi ; since u 2 W?; hence hu; vi i D 0 for each 1 � i � k

D 0

(e)

T .y/ D T .w C u/

D T .w/C T .u/

D w C 0 according to (c) and (d) above

D w

ExampleFind a formula for P .a; b; c/ where P : C3 ! C3 denotes the orthogonal projection on

W D span f.1; 1; 0/ ; .0; 1; 1/g

SOLUTIONFirst we use Gram�Schmidt to calculate an orthonormal basis for W:

v01 D .1; 1; 0/

v02 D .0; 1; 1/�h.0; 1; 1/ ; .1; 1; 0/ih.1; 1; 0/ ; .1; 1; 0/i

.1; 1; 0/

D .0; 1; 1/�12.1; 1; 0/

D

��12;12; 1�

152

Hence, an orthonormal basis for W is given by fv1; v2g ; where

v1 Dv01 v01

D1p2.1; 1; 0/

and

v2 Dv02 v02

D1p6.�1; 1; 2/

According to the above characterisation, P .a; b; c/ is given by

P .a; b; c/ D h.a; b; c/ ; v1iv1 C h.a; b; c/ ; v2iv2

D

�.a; b; c/ ;

1p2.1; 1; 0/

�1p2.1; 1; 0/C

�.a; b; c/ ;

1p6.�1; 1; 2/

�1p6.�1; 1; 2/

D12.a C b/ .1; 1; 0/C

16.�a C b C 2c/ .�1; 1; 2/

D16.4a C 2b � 2c; 2a C 4b C 2c;�2a C 2b C 4c/

D13.2a C b � c; a C 2b C c;�a C b C 2c/

15.4

Second Characterisation of Orthogonal ProjectionsLet V be a �nite�dimensional inner�product space, and let T be a linearoperator on V : Then T is an orthogonal projection if and only if T 2 D T D T �:(F2 and F3: Theorem 6.23 / F4: Theorem 6.24)

ExampleLet V D C3 denote the complex inner�product space with respect to the standard inner product. Let

W D span f.1; 1; 1/ ; .1; 1; 0/g ;

and let T : C3 ! C3 denote the orthogonal projection on W:

(a) Use the Gram�Schmidt Process to calculate an orthonormal basis of W:

(b) Use Friedberg: Theorem 6.6 to �nd the matrix of T with respect to the standard basis.

(c) Verify that T satis�es the condition in F2 and F3: Theorem 6.23 / F4: Theorem 6.24 for an orthogonalprojection, namely

T 2 D T D T �

153 MAT3701/1

SOLUTION

(a)

w1 D .1; 1; 1/

w2 D .1; 1; 0/�h.1; 1; 0/ ; .1; 1; 1/ih.1; 1; 1/ ; .1; 1; 1/i

.1; 1; 1/

D .1; 1; 0/�23.1; 1; 1/

D

�13;13;�23

�

D13.1; 1;�2/

Normalising both these vectors yields an orthonormal basis of W; namely

� D

�v1 D

1p3.1; 1; 1/ ; v2 D

1p6.1; 1;�2/

�

(b) Use Friedberg: Theorem 6.6 to express

T .1; 0; 0/ ; T .0; 1; 0/ and T .0; 0; 1/

as linear combinations of �; that is,

T .1; 0; 0/ D h.1; 0; 0/ ; v1iv1 C h.1; 0; 0/ ; v2iv2

D13.1; 1; 1/C

16.1; 1;�2/

D

�12;12; 0�

T .0; 1; 0/ D h.0; 1; 0/ ; v1iv1 C h.0; 1; 0/ ; v2iv2

D13.1; 1; 1/C

16.1; 1;�2/

D

�12;12; 0�

154

T .0; 0; 1/ D h.0; 0; 1/ ; v1iv1 C h.0; 0; 1/ ; v2iv2

D13.1; 1; 1/�

26.1; 1;�2/

D .0; 0; 1/

Thus,

[T ] D

0B@12

12 0

12

12 0

0 0 1

1CA ;where

D f.1; 0; 0/ ; .0; 1; 0/ ; .0; 0; 1/g

(c) Since is orthonormal with respect to the standard inner product, it suf�ces to verify the result for [T ] :

�T 2� D [T ]2 D

0B@12

12 0

12

12 0

0 0 1

1CA0B@

12

12 0

12

12 0

0 0 1

1CA

D

0B@12

12 0

12

12 0

0 0 1

1CAD [T ]

Hence, T 2 D T : Finally, �T �� D [T ]� D [T ] (check),

so that T D T �:

15.5 Diagonalising a ProjectionLet V be a �nite�dimensional vector space (not necessarily an inner�product space), and let T be a projection onV : By SG: Example 3.5.1.2,

V D R .T /� N .T / ; and R .T / D fx 2 V : T .x/ D xg

Hence, R .T / D E1 .T / ; the eigenspace of T corresponding to the eigenvalue 1; and

N .T / D E0 .T / D fx 2 V : T .x/ D 0 � x D 0g ;

the eigenspace of T corresponding to the eigenvalue of 0: If �1 is a basis for E1 .T / and �0 a basis for E0 .T / ; then� D �1 [ �0 is a basis for V consisting of eigenvectors of T by SG: Example 2.8.3.1. Also,

[T ]� D

"Ik 00 0

#; ... (i)

where k D dim .E1 .T // :

155 MAT3701/1

ConclusionA projection T on a �nite�dimensional vector space V is diagonalisable. It has the eigenvalues 0 and 1 unlessT D OV (in which case it has only the eigenvalue 0) or T D IV (in which case it has only the eigenvalue 1). Thus,there exists a basis � for V such that [T ]� has the form (i).

Diagonalising a ProjectionTo diagonalise a projection T on a �nite�dimensional vectorspace V; compute an ordered basis �1 for E1 .T / and an orderedbasis �0 for E0 .T / ; and form the ordered union � D �1 [ �0: Then� is a basis for V consisting of eigenvectors of T such that

[T ]� D

"Ik 00 0

#; where k D dim .E1 .T //

ExampleDiagonalise the projection T on C2 de�ned by

T .a; b/ D .�a � b; 2a C 2b/ for all .a; b/ 2 C2

SOLUTIONBy SG: Example 3.5.3(b), �1 D f.�1; 2/g is a basis for E1 .T / ; and �0 D f.1;�1/g is a basis for E0 .T / : Hence,

� D �1 [ �0 D f.�1; 2/ ; .1;�1/g

is a basis for C2 such that

[T ]� D

"1 00 0

#As a check, note that by the formula for T;

T .�1; 2/ D .�1; 2/ and T .1;�1/ D .0; 0/ ;

as expected.

15.6 Diagonalising an Orthogonal ProjectionIf T is an orthogonal projection on a �nite�dimensional inner�product space V; then T is also self�adjoint, inaddition to being a projection. Thus, if the ordered bases �1 and �0 for E1 .T / and E0 .T /, respectively, are chosento be orthonormal, then � D �1 [ �0 is an orthonormal basis for V consisting of eigenvectors of T by SG: 13.5.

Diagonalising an Orthogonal Projection by means ofan Orthonormal Basis

To diagonalise an orthogonal projection T on a �nite�dimensional inner�product space V; compute (using Gram�Schmidt, if necessary) orthonormalordered bases �1 and �0 for E1 .T / and E0 .T / ; respectively, and form theordered union � D �1 [ �0: Then � is an orthonormal basis for V consistingof eigenvectors of T such that

[T ]� D

"Ik 00 0

#; where k D dim .E1 .T //

156

ExampleConsider the operator L A : C3 ! C3 where

A D12

264 1 0 10 2 01 0 1

375(a) Show that L A is an orthogonal projection.

(b) Find a unitary matrix Q such that

Q�AQ D

"Ik 00 0

#where k D dim .E1 .L A//

SOLUTION

(a) Let denote the standard ordered basis for C3: Then [L A] D A: Since is orthonormal, it suf�ces to showthat A2 D A D A�: That A D A�; follows from the fact A is a real symmetric matrix. Also,

A2 D14

264 1 0 10 2 01 0 1

375264 1 0 10 2 01 0 1

375

D14

264 2 0 20 4 02 0 2

375 D A(b) For E1 .L A/ ; the corresponding homogeneous system to solve is, in matrix form,

A � I D

264 � 12 0 1

20 0 012 0 � 1

2

375 Ihence,

E1 .L A/ D span f.1; 0; 1/ ; .0; 1; 0/g ;

with orthonormal basis�1 D

�1p2.1; 0; 1/ ; .0; 1; 0/

�For E0 .L A/ ; the corresponding homogeneous system to solve is, in matrix form,

A D12

264 1 0 10 2 01 0 1

375 Ihence,

E0 .L A/ D span f.1; 0;�1/g ;

157 MAT3701/1

with orthonormal basis�0 D

�1p2.1; 0;�1/

�Thus,

� D �1 [ �0 D

�1p2.1; 0; 1/ ; .0; 1; 0/ ;

1p2.1; 0;�1/

�is an orthonormal basis of eigenvectors of L A such that

[L A]� D

264 1 0 00 1 00 0 0

375Finally,

[L A]� D�IR2�� [L A]

�IR2� �

D Q�1AQ

D Q�AQ;

since

Q D1p2

264 1 0 10p2 0

1 0 �1

375is unitary.

15.7 The Spectral Decomposition for a Complex Normal (resp., Real Symmetric) MatrixSuppose A 2 Mn�n .F/ is normal if F D C; or symmetric if F D R: Let �1; �2; : : : ; �k be the distinct eigenvaluesof A with multiplicities m1;m2; : : : ;mk; respectively. By SG: 14.4, there exists a complex unitary (resp., realorthogonal) matrix P such that

P�AP D

266664�1 Im1 0 : : : 0

0 �2 Im2 : : : 0:::

::::::

0 0 : : : �k Imk

377775

D �1

266664Im1 0 : : : 00 0 : : : 0:::

::::::

0 0 : : : 0

377775C �22666640 0 : : : 00 Im2 : : : 0:::

::::::

0 0 : : : 0

377775C : : :C �k2666640 0 : : : 00 0 : : : 0:::

::::::

0 0 : : : Imk

377775So,

A D �1P

266664Im1 0 : : : 00 0 : : : 0:::

::::::

0 0 : : : 0

377775 P� C �2P2666640 0 : : : 00 Im2 : : : 0:::

::::::

0 0 : : : 0

377775 P� C : : :C �kP2666640 0 : : : 00 0 : : : 0:::

::::::

0 0 : : : Imk

377775 P�

158

Thus,A D �1P1 C �2P2 C : : :C �kPk;

where

P1 D P

266664Im1 0 : : : 00 0 : : : 0:::

::::::

0 0 : : : 0

377775 P�; P2 D P

2666640 0 : : : 00 Im2 : : : 0:::

::::::

0 0 : : : 0

377775 P�; etcSince P is complex unitary (resp., real orthogonal), it follows that

Pi Pj D �i j Pi .1 � i; j � k/ and I D P1 C P2 C : : :C Pk

The Spectral Decomposition in Matrix FormSuppose A 2 Mn�n .F/ is normal if F D C; or symmetric if F D R: If�1; �2; : : : ; �k; denote the distinct eigenvalues of A; then A can beexpressed as

A D �1P1 C �2P2 C : : :C �kPk with Pi 2 Mn�n .F/ ;such that

Pi Pj D �i j Pi .1 � i; j � k/ and I D P1 C P2 C : : :C Pk

ExampleFind the spectral decomposition of

A D

264 1 i 0i 1 00 0 2

375SOLUTIONIt can be shown (check) that the eigenvalues of A are � D 2; 1� i and 1C i; with corresponding eigenspaces

E2 D span f.0; 0; 1/g

E1�i D span�1p2.1;�1; 0/

�

E1Ci D span�1p2.1; 1; 0/

�Thus,

P�AP D

264 2 0 00 1� i 00 0 1C i

375 ;where

P D1p2

264 0 1 10 �1 1

p2 0 0

375

159 MAT3701/1

is unitary. Following the above procedure, we have

A D 2P

264 1 0 00 0 00 0 0

375 P� C .1� i/ P264 0 0 00 1 00 0 0

375 P� C .1C i/ P264 0 0 00 0 00 0 1

375 P� ... (i)

Now

P1 D P

264 1 0 00 0 00 0 0

375 P� D 1p2

264 0 0 00 0 0

p2 0 0

375 1p2

264 0 0p2

1 �1 01 1 0

375 D264 0 0 00 0 00 0 1

375 ;

P2 D P

264 0 0 00 1 00 0 0

375 P� D 1p2

264 0 1 00 �1 00 0 0

375 1p2

264 0 0p2

1 �1 01 1 0

375 D 12

264 1 �1 0�1 1 00 0 0

375 ;

P3 D P

264 0 0 00 0 00 0 1

375 P� D 1p2

264 0 0 10 0 10 0 0

375 1p2

264 0 0p2

1 �1 01 1 0

375 D 12

264 1 1 01 1 00 0 0

375Substitution in (i) yields the desired spectral decomposition, namely,

A D 2

264 0 0 00 0 00 0 1

375C .1� i/264

12 � 1

2 0� 12

12 0

0 0 0

375C .1C i/264

12

12 0

12

12 0

0 0 0

375

15.8 The Spectral Decomposition for Matrices with Two EigenvaluesIn this case, we can use a short method as illustrated by the following example:

ExampleFind the spectral decomposition of

A D

264 0 i 0i 0 00 0 i

375 ;given that A is normal and its eigenvalues are i and �i:

SOLUTIONAccording to the spectral theorem, we need to �nd matrices P1 and P2 such that

A D i P1 � i P2I D P1 C P2

)... (i)

Multiplying the last equation by i and adding it to the �rst yields

A C i I D 2i P1;

160

so that

P1 D12i.A C i I /

D12i

264 i i 0i i 00 0 2i

375

D12

264 1 1 01 1 00 0 2

375Substituting this in I D P1 C P2 yields

P2 D I � P1

D

26412 � 1

2 0� 12

12 0

0 0 0

375

D12

264 1 �1 0�1 1 00 0 0

375RemarkIf the complex normal or real symmetric matrix A has only one eigenvalue �; then A D �I , which is its spectraldecomposition already.

161 MAT3701/1

.

STUDY UNIT 16.

BILINEAR AND QUADRATIC FORMS

(F2 & F3: Section 6.7 / F4: Section 6.8)

16.1 Overview

Like inner products, a bilinear form on a vector space V is also a special type of mapping from pairs of vectors inV to elements in the scalar �eld F: In fact, over the real �eld, an inner product is also a bilinear form, but not viceversa. We investigate the basic properties of bilinear forms, and apply them to quadratic surfaces and multivariablecalculus.

16.2 What to Study

Study F2 & F3: Section 6.7 / F4: Section 6.8. Note, the restriction to �elds that are not of characteristic two (whichis required in some places in the material) does not concern us, since we assume throughout that F denotes eitherR or C; and both are of characteristic zero. It is also for this reason that F2 & F3: Section 6.7, Example 4 / F4:Section 6.8, Example 5 may be ignored.

16.3 Bilinear Forms

The algorithm following F2 & F3: Section 6.7, Example 5 / F4: Section 6.8, Example 6 may be used to diagonalise,with respect to congruence, a symmetric matrix over a �eld F;which is not of characteristic two .

16.3.1

Diagonalising a Symmetrix Matrix with respect to CongruenceSuppose A 2 Mn�n .F/ is a symmetrtic matrix over a �eld F; which is not ofcharacteristic two. Then A is congruent to a diagonal matrix D; i.e., there exists anonsingular matrix Q such that

Qt AQ D D:Q and D may be obtained from the following reduction

[A j I ]!�D j Qt

�by performing elementary row and corresponding column operations.

162

ExampleLet H : R2 � R2 ! R be de�ned by

H ..a1; a2/ ; .b1; b2// D a1b2 C a2b1

and let

A D

"0 11 0

#

(a) Show that H is a symmetric bilinear form.

(b) Show that

� .H/ D

"0 11 0

#;

where � denotes the standard basis for R2:

(c) Use the above algorithm to �nd an invertible matrix Q and a diagonal matrix D such that Qt AQ D D:

(d) Find a basis of R2 such that .H/ is diagonal.

SOLUTION

(a) SinceH .x; y/ D x t Ay for all x; y 2 R2; ::: .i/

it follows from Friedberg: Example 2 that H is a bilinear form which is symmetric by F2 & F3: Theorem6.28 / F4: Theorem 6.34.

(b) If follows from (i) that 2 .H/ D A:

(c) 24 0 1::: 1 0

1 0::: 0 1

35 !

24 2 1::: 1 1

1 0::: 0 1

35 [R1 ! R1 C R2] and[C1 ! C1 C C2]

!

24 2 0::: 1 1

0 � 12

::: � 12

12

35 �R2 ! R2 � 1

2 R1�and�

C2 ! C2 � 12C1

�Thus,

Q D

"1 1

� 12

12

#tD

"1 � 1

21 1

2

#and D D

"2 00 � 1

2

#

(d) consists of the columns of Q; thus

D

�.1; 1/ ;

��12;12

��

163 MAT3701/1

16.4 Quadratic Forms

ExampleContinuing with the previous example,

(e) Write down the quadratic form K .x/ D H .x; x/ associated with H:

(f) Find an orthonormal basis � for R2 such that � .H/ is a diagonal matrix. Calculate � .H/ :

(g) Find the eigenvalues and corresponding eigenvectors of A:

(h) Show that the cross term in K .x/ can be eliminated by a counterclockwise rotation through � degrees.

Find � and express the old variables in terms of the new.

SOLUTION

(e) K

t1t2

!D H

t1t2

!;

t1t2

!!D 2t1t2

(f) Note that in (d) is already orthogonal, thus we simply normalise to get � :

� D

�1p2.1; 1/ ;

1p2.�1; 1/

�:

� .H/ D�[I ]��

�t � .H/ [I ]�� (F2 & F3: Theorem 6.27/ F4: Theorem 6.33)

D

"1p2

� 1p2

1p2

1p2

#t "0 11 0

#"1p2

� 1p2

1p2

1p2

#

D12

"1 1

�1 1

#"1 11 �1

#

D

"1 00 �1

#

As an exercise, do the above calculation of � .H/ again, but use instead of �:

(g) Since P D [I ]�� D

"1p2

� 1p2

1p2

1p2

#is orthogonal, it follows from (f) that

AP D P

"1 00 �1

#

Thus, the eigenvalues of A are 1;with corresponding eigenvector

"1p21p2

#; and�1;with corresponding eigen-

vector

"� 1p

21p2

#:

164

(h)

K

"t1t2

#D

ht1 t2

iA

"t1t2

#

Dht1 t2

iP

"1 00 �1

#P t"t1t2

#; from (g)

Dhs1 s2

i " 1 00 �1

#"s1s2

#D s21 � s

22 ;

where "t1t2

#D P

"s1s2

#

i.e.

t1 D1p2.s1 � s2/

t2 D1p2.s1 C s2/

Since

P D

"1p2

� 1p2

1p2

1p2

#D

"cos �4 � sin �4sin �4 cos �4

#

it follows that � D 45�;which represents a rotation of the axes through an angle of 45� in the counterclockwisedirection.

16.5 Sylvester's Law of Inertia

ExampleContinuing with the previous example,

(i) Write down the index and signature of A:

(j) Find an invertible matrix R such that Rt AR D Jpr ;where p and r denote the index and rank of A; respectively.

SOLUTION

(i) It follows from (f) that the index of A is 1; and the signature is 0:

(j) According to (f), we may choose R D [I ] � D P:

165 MAT3701/1

.

STUDY UNIT 17.

CONDITIONING AND THE RAYLEIGHQUOTIENT

(F2 & F3: Section 6.9 / F4: Section 6.10)

17.1 Overview

Read the introduction to this section in Friedberg.

17.2 What to Study

Study F2 & F3: Section 6.9 / F4: Section 6.10, and try as many of the exercises as possible. The last exercise inthe 4th edition may be left out as it is based on material not covered in this course.

17.3 Examples

1. Friedberg: Exercise 5)

Suppose that x is the actual solution of Ax D b and that a computer arrives at an approximate solutionex : Ifcond.A/ D 100; kbk D 1; and kb � Aexk D 0:1; obtain upper and lower bounds for kx �exk = kxk :SOLUTION

Letex D x C �x and Aex D b C �b; then kx �exk D k�xk and kb � Aexk D k�bk D 0:1:From F2: Theorem 6.37(a) / F3: Theorem 6.38(a) / F4: Theorem 6.44(a)

0:001 D0:1100

�kx �exkkxk

� 100 .0:1/ D 10

2. (Friedberg: Exercise 7)

Let B be a (real) symmetric matrix. Prove that minx 6D0

R .x/ equals the smallest eigenvalue of B:

166

SOLUTION

Let � D fv1; : : : ; vng be an orthonormal basis for Rn consisting of eigenvectors of B; say Bvi D �ivi ; where�i is real since B is symmetric, and suppose �1 � �2 � : : : �n: For any x D �1v1 C : : : �nvn in Rn;

R .x/ DhBx; xikxk2

DhB .�1v1 C : : :C �nvn/ ; �1v1 C : : :C �nvni

kxk2

Dh�1B .v1/C : : :C �nB .vn/ ; �1v1 C : : :C �nvni

kxk2

Dh�1�1v1 C : : :C �n�nvn; �1v1 C : : :C �nvni

kxk2

D�1�

21 C : : :C �n�

2n

kxk2

��n��21 C : : :C �

2n�

kxk2D �n

But R .x/ actually assumes the value �n; since for x D vn;

R .x/ DhB .vn/ ; vnikvnk

2

D h�nvn; vni

D �n

Thus,minx 6D0

R .x/ D �n

Part 3 Summary

Geometry is introduced in vector spaces by means of the inner product. To utilise the geometric properties of theseenhanced spaces it turns out that special bases and special types of operators are needed, which led to orthonormalbases and a variety of operators designed to preserve various aspects of inner-product spaces.

mathematics linear algebra - gimmenotesstudy unit friedberg title page 1 overview 1 part 1 familiar...

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