algebra through practice volume 4_blyth t.s., robertson e.f

Algebra through practice

Book 4: Linear algebra

Algebra through practiceA collection of problems in algebra with solutions

Book 4Linear algebra

T. S. BLYTH o E. F. ROBERTSONUniversity of St Andrews

- r.The right f ,h,

University of Cambridgeto print and sell

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Henry Vlll in 1534.The University has printedand published continuously

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ISBN 978-0-521-27289-6 paperback

Contents

Preface vi

Background reference material vii

1: Direct sums and Jordan forms 1

2: Duality and normal transformations 18

Solutions to Chapter 1 31

Solutions to Chapter 2 67

Test paper 1 96

Test paper 2 98

Test paper 3 100

Test paper 4 102

v

Preface

The aim of this series of problem-solvers is to provide a selection ofworked examples in algebra designed to supplement undergraduatealgebra courses. We have attempted, mainly with the average studentin mind, to produce a varied selection of exercises while incorporatinga few of a more challenging nature. Although complete solutions areincluded, it is intended that these should be consulted by readers onlyafter they have attempted the questions. In this way, it is hoped thatthe student will gain confidence in his or her approach to the art ofproblem-solving which, after all, is what mathematics is all about.

The problems, although arranged in chapters, have not been`graded' within each chapter so that, if readers cannot do problem nthis should not discourage them from attempting problem n+ 1. Agreat many of the ideas involved in these problems have been used inexamination papers of one sort or another. Some test papers (withoutsolutions) are included at the end of each book ; these contain questionsbased on the topics covered.

TSB, EFRSt Andrews

Background reference material

Courses on abstract algebra can be very different in style and content.Likewise, textbooks recommended for these courses can vary enorm-ously, not only in notation and exposition but also in their level ofsophistication. Here is a list of some major texts that are widely usedand to which the reader may refer for background material. Thesubject matter of these texts covers all six of the present volumes, andin some cases a great deal more. For the convenience of the reader thereis given overleaf an indication of which parts of which of these textsare most relevant to the appropriate sections of this volume.

[1] I. T. Adamson, Introduction to Field Theory, CambridgeUniversity Press, 1982.

[2] F. Ayres, Jr, Modern Algebra, Schaum's Outline Series,McGraw-Hill, 1965.

[3] D. Burton, A first course in rings and ideals, Addison-Wesley,1970.

[4] P. M. Cohn, Algebra Vol. I, Wiley, 1982.[5] D. T. Finkbeiner II, Introduction to Matrices and Linear

Transformations, Freeman, 1978.[6] R. Godement, Algebra, Kershaw, 1983.[7] J. A. Green, Sets and Groups, Routledge and Kegan Paul,

1965.[8] I. N. Herstein, Topics in Algebra, Wiley, 1977.[9] K. Hoffman and R. Kunze, Linear Algebra, Prentice Hall,

1971.[10] S. Lang, Introduction to Linear Algebra, Addison-Wesley, 1970.[11] S. Lipschutz, Linear Algebra, Schaum's Outline Series,

McGraw-Hill, 1974.

vii

[12] I. D. Macdonald, The Theory of Groups, Oxford UniversityPress, 1968.

[13] S. MacLane and G. Birkhoff, Algebra, Macmillan, 1968.[14] N. H. McCoy, Introduction to Modern Algebra, Allyn and

Bacon, 1975.[15] J. J. Rotman, The Theory of Groups: An Introduction, Allyn

and Bacon, 1973.[16] I. Stewart, Galois Theory, Chapman and Hall, 1975.[17] I. Stewart and D. Tall, The Foundations of Mathematics,

Oxford University Press, 1977.

References useful for Book 41: Direct sums and Jordan forms [4, Sections 11.1-11.4],[5, Chapter 7], [8, Sections 6.1-6.6], [9, Chapters 6, 7],[11, Chapter 10].2: Duality and normal transformations [4, Chapter 8,Section 11.4], [5, Chapter 9], [8, Sections 4.3, 6.8, 6.10],[9, Chapters 8, 9], [11, Chapters 11, 12].In [4] and [6] some ring theory is assumed, and someelementary results are proved for modules. In [5] the authoruses `characteristic value' where we use `eigenvalue'.

viii

1: Direct sums and Jordan forms

In this chapter we take as a central theme the notion of the direct sumA ® B of subspaces A, B of a vector space V. Recall that V = A ® Bif and only if every x E V can be expressed uniquely in the form a + bwhere a E A and b E B; equivalently, if V = A+B and AnB = {0}. Forevery subspace A of V there is a subspace B of V such that V = A® B.In the case where V is of finite dimension, this is easily seen; take a basis

-{V1, .. , vk} of A, extend it to a basis {vi,. .. , v,,} of V, then note thatspans a subspace B such that V = A ®B.

If f : V -* V is a linear transformation then a subspace W of V issaid to be f-invariant if f maps W into itself. If W is f-invariant thenthere is an ordered basis of V with respect to which the matrix of V isof the form

0 X

where M is of size dim W x dim W.If f : V --+ V is such that f o f = f then f is called a projection.

For such a linear transformation we have V = Im f ® Ker f where thesubspace Imf is f-invariant (and the subspace Kerf is trivially so). Avector space V is the direct sum of subspaces W1, .. . , Wk if and only ifthere are non-zero projections p1, ... , Pk : V - V such that

k

Epi = idv and pi o pj= 0 for i # j.i=1

In this case Wi = Im pi for each i, and relative to given ordered bases of

Book 4 Linear algebra

W1, ... , Wk the matrix of f is of the diagonal block form

M1M2

Mk

Of particular importance is the situation where each Ms is of the form

A 1 0 ... 0 0

0 A 1 ... 0 0

0 0 A ... 0 0

0 0 0 ... A l0 0 0 ... 0 A

in which case the diagonal block matrix is called a Jordan matrix.The Cayley-Hamilton theorem says that a linear transformation f is

a zero of its characteristic polynomial. The minimum polynomial of fis the monic polynomial of least degree of which f is a zero. When theminimum polynomial of f factorises into a product of linear polynomialsthen there is a basis of V with respect to which the matrix of f is aJordan matrix. This matrix is unique (up to the sequence of the diagonalblocks), the diagonal entries A above are the eigenvalues of f, and thenumber of M1 associated with a given A is the geometric multiplicity ofA. The corresponding basis is called a Jordan basis.

We mention here that, for space considerations in the solutions, weshall often write an eigenvector

x1

X2

xn

as [x1, x2, ... , xn].

1.1 Which of the following statements are true? For those that are false,give a counter-example.(i) If {al, a2, a3} is a basis for IR3 and b is a non-zero vector in IR then

{b + a1, a2, a3} is also a basis for IR3.

2


(ii) If A is a finite set of linearly independent vectors then the dimensionof the subspace spanned by A is equal to the number of vectors inA.

(iii) The subspace {(x, x, x) I x E IR} of IR3 has dimension 3.(iv) If A is a linearly dependent set of vectors in IRn then there are more

than n vectors in A.(v) If A is a linearly dependent subset of IRn then the dimension of the

subspace spanned by A is strictly less than the number of vectorsin A.

(vi) If A is a subset of IRn and the subspace spanned by A is IRn itselfthen A contains exactly n vectors.

(vii) If A and B are subspaces of IRn then we can find a basis of IRnwhich contains a basis of A and a basis of B.

(viii) An n-dimensional vector space contains only finitely many sub-spaces.

(ix) If A is an n x n matrix over Q2 with A3 = I then A is non-singular.(x) If A is an n x n matrix over Q with A3 = I then A is non-singular.

(xi) An isomorphism between two vector spaces can always be repre-sented by a square singular matrix.

(xii) Any two n-dimensional vector spaces are isomorphic.(xiii) If A is an n x n matrix such that A2 = I then A = I.(xiv) If A, B and C are non-zero matrices such that AC = BC then

A=B.(xv) The identity map on IRn is represented by the identity matrix with

respect to any basis of IRn.(xvi) Given any two bases of IRn there is an isomorphism from IRn to

itself that maps one basis onto the other.(xvii) If A and B represent linear transformations f, g : IRn -* IRn with

respect to the same basis then there is a non-singular matrix P suchthat P-1 AP = B.

(xviii) There is a bijection between the set of linear transformations fromIRn to itself and the set of n x n matrices over IR.

(xix) The map t : IR2 --4 IR2 given by t(x, y) = (y, x+y) can be representedby the matrix

with respect to some basis of IR2.(xx) There is a non-singular matrix P such that P-1 AP is diagonal for

any non-singular matrix A.

3


1.2 Let t1i t2, t3, t4 E ,C(IR3, IR3) be given by

t1(a,b,c) = (a+b,b+c,c+a);t2 (a, b, c) = (a - b, b - c, 0);

t3 (a, b, c) = (-b, a, c);

t4 (a, b, c) = (a, b, b).

Find Ker t1 and Im t; for i = 1, 2, 3, 4. Is it true that IR3 = Ker t1®Im t1for any of i = 1, 2, 3, 4?

Is Im t2 t3-invariant? Is Ker t2 t3-invariant?Find t3 o t4 and t4 o t3. Compute the images and kernels of these

composites.

1.3 Let V be a vector space of dimension 3 over a field F and let t E £ (V, V)be represented by the matrix

3 -1 1

-1 5 -1

1 -1 3

with respect to some basis of V. Find dim Ker t and dim Im t when

(i) F = IR;(ii) F = 12;(iii) F = 73.

Is V = Ker t ® Im t in any of cases (i), (ii) or (iii)?

1.4 Let V be a finite-dimensional vector space and let s, t E .C (V, V) be suchthat sot = idv. Prove that to s = W. Prove also that a subspace of Vis t-invariant if and only if it is s-invariant. Are these results true whenV is infinite-dimensional?

1.5 Let V,, be the vector space of polynomials of degree less than n overthe field IR. If D E £(V,,, V,,) is the differentiation map, find Im D andKer D. Prove that Im D _- V,,_ 1 and that Ker D = IR. Is it true that

V,y=ImDa) KerD?

Do the same results hold if the ground field IR is replaced by the fieldZ2?

4


1.6 Let V be a finite-dimensional vector space and let t E C (V, V)- Establishthe chains

V D Im t D Im t2 D ... D Im to D Im to+1 Q

{0} C Ker t C Ker t2 C ... C Ker to C Ker to+1 C ....

Show that there is a positive integer p such that ImtP = ImtP+1 anddeduce that

(Vk > 1) Imp = Im tP+k and Ker tP = Ker tP+k.

Show also thatV = Im tP ® Ker tP

and that the subspaces Imt' and KertP are t-invariant.

1.7 Let V be a vector space of dimension n over a field F and let f : V --+ Vbe a non-zero linear transformation such that f o f = 0. Show that ifIm f is of dimension r then 2r < n. Suppose now that W is a subspaceof V.such that V = Ker f ® W. Show that W is of dimension r andthat if {wl,... , wr} is a basis of W then { f (wl ), ... , f (wr)} is a linearlyindependent subset of Ker f. Deduce that n - 2r elements XI, ... , xn-2rcan be chosen in Ker f such that

('wl)...,wr,f(wl),..., f(wr),xl,...,xn-2r}

is a basis of V.Hence show that a non-zero n x n matrix A over F is such that A2 = 0

if and only if A is similar to a matrix of the form

1.8 Let V be a vector space of dimension 4 over IR. Let a basis of V beB = {bl, b2, b3i b4}. Writing each x E V as x = E l x;b;, let

VI = {x E V I x3 = x2 and x4 = XI),

V2 = {x E V I x3 = -x2 and x4 = -x1 }.

Show that(1) V1 and V2 are subspaces of V;

5


(2) {b1 + b4, b2 + b3} is a basis of V1 and {b1 - b4, b2 - b3} is a basisof V2i

(3) V = V1 ® V2;

(4) with respect to the basis B. and the basis

C = {bl + b4, b2 + b3, b2 - b3, bl - b4}

the matrix of idv isi20

0i

2

0 0i 1

2 2i 1

2 -20 0

12

0

01

2-

A 4 x 4 matrix M over IR is said to be centro-symmetric if

mij = m5-i,5-j

for all i, j. If M is centro-symmetric, show that M is similar to a matrixof the form

Q 0 0ry S 0 0

0 0 E S

0 0 n tg

1.9 Let V be a vector space of dimension n over a field F. Suppose firstthat F is not of characteristic 2 (i.e. that 1F + 1F 54 OF). If f : V --+ Vis a linear transformation such that f o f = idv prove that

V = Im(idv +f) ® Im(idv -f).

Deduce that an n x n matrix A over F is such that A2 = In if and onlyif A is similar to a matrix of the form

Suppose now that F is of characteristic 2 and that f o f = idv. Ifg = idv +f show that

xEKerg b x= f(x),6


and that g o g = 0. Deduce that an n x n matrix A over F is such thatA2 = In if and only if A is similar to a matrix of the form

In-2p

[Hint. Observe that Img C Kerg. Let {g(cl),...,g(cp)} be a basis ofImg and extend this to a basis {b1, ... , bn-2p, g(cl ), ... , g(cp) } of Ker g.Show that

{b1,. .. , bn-2p, 9(Cl ), C1, ... , 9(Cp), Cp}

is a basis of V.]

1.10 Let V be a finite-dimensional vector space and let t E C(V, V) be suchthat t # idv and t # 0. Is it possible to have Imt n Kert # {0}? Isit possible to have Im t = Ker t? Is it possible to have Im t C Ker t? Isit possible to have Kert C Imt? Which of these are possible if t is aprojection?

1.11 Is it possible to have projections e, f E £(V,V) with Kere = Ken f andIm e # Im f ? Is it possible to have Im e = Im f and Ker e # Ker f? Is itpossible to have projections e, f with e o f = 0 but f o e 0?

1.12 Let V be a vector space over a field of characteristic not equal to 2. Letel,e2 E .C(V,V) be projections. Prove that el + e2 is a projection if andonly if el o e2 = e2 o el = 0.

If el + e2 is a projection, find Im(el + e2) and Ker(el + e2) in termsof the images and kernels of el, e2-

1.13 Let V be the subspace of IR3 given by

V = {(a, a, 0) I a E IR}.

Find a subspace U of IR3 such that IR3 = V ® U. Is U unique? Find aprojection e E C(IR3, IR3) such that Ime = V and Kere = U. Find alsoa projection f E C(IR3, IR3) such that Im f = U and Ker f = V.

7


1.14 If V is a finite-dimensional vector space over a field F and e, f E .C(V, V)are projections prove that Im e = Im f if and only if eof = f and foe = e.

Suppose that e1,...,ek E .C(V,V) are projections with

Let A 1 , A2, ... , Ak E F be such that Ek1 ai = 1. Prove that

e =ale, + A2e2 + ...+ Akekis a projection with Im e = Im ei.

Is it necessarily true that if fl, ... , fk E £(V, V) are projections andrk1 ai = 1 then Ek1 ai fi is also a projection?

1.15 A net over the interval [0, 11 of IR is a finite sequence (ai)o<i<n.+1 suchthat

0=ao <a1 < <an <an+1 = 1.A step function on [0, 1[ is a mapping f : [0,1[--+ IR for which there existsa net (ai)o<i<n+1 over [0, 1] and a finite sequence (bi)o<,<n of elementsof IR such that

(Vx E jai, ai+1 [) f (x) = bi.Show that the set E of step functions on [0, 1[ is a vector space over IRand that a basis of E is the set {ek I k E [0, 1 [} of functions ek : [0, 1 [-> IR

by

ek(x)= J0 if0<x<k;1 ifk<x<1.

A piecewise linear function on [0, 1[ is a mapping f : [0,1[-> IR forwhich there exists a net (ai)o<i<n+1 and sequences (b1)o<i<n, (c,)o<i<nof elements of IR such that

given by

(Vx E [ai, ai+1 [) f (x) = bix + ci.

Let F be the set of piecewise linear functions on [0, 1[ and let G bethe subset of F consisting of the piecewise linear functions g that arecontinuous with g(0) = 0. Show that F, G are vector spaces over IR andthat F=E®G.

Show that a basis of G is the set {gk I k E [0, 1[} of functions given

0 if0<x<k;gk(x) x - k ifk<x<1.

Finally, show that the assignment

fi-+I(f)= jf(t)dtdescribes an isomorphism from E to G.

8


1.16 Let V be a vector space over a field F and let t E ,C(V,V). Let ). and)2 be distinct eigenvalues of t with associated eigenvectors v1 and v2. Isit possible for al + .12 to be an eigenvalue of t? What about .11 )2?

1.17 Let t E C((E2, d2) be given by

t(a, b) = (a + 2b, b - a).

Find the eigenvalues of t and show that there is a basis of C2 consistingof eigenvectors of t. Find such a basis, and the matrix of t with respectto this basis.

1.18 Suppose that t E C(V,V) has zero as an eigenvalue. Prove that t is notinvertible. Is it true that t is invertible if and only if all the eigenvaluesof t are non-zero? If t is invertible, how are the eigenvalues of t relatedto those of t-1?

1.19 Let V be a vector space of finite dimension over Q and let t E C(VV)be such that t' = 0 for some m > 0. Prove that all the eigenvalues of tare zero. Deduce that if t # 0 then t is not diagonalisable.

1.20 Consider t E C(IR2, IR2) given by

t(a, b) _ (a + 4b,aa - b).

Find the minimum polynomial of t.

1.21 Let F be a field and let F,+1 [X] be the vector space of polynomials ofdegree less than or equal to n over F. Define t : F,,+1 [X] -' F,,+1 [X]by t(f(X)) = f(X + 1). Show that t is linear.

Find the matrix of t relative to the basis {1,X,. .. , X'j of F,,+1 [X].Find also the eigenvalues of t. If g(X) = (X -1)"+1 show that g(t) = 0.Hence find the minimum polynomial of t.

1.22 If V is a finite-dimensional vector space and t E C(V, V) is such thatt2 = idv prove that the sum of the eigenvalues of t is an integer.

1.23 For each of the following real matrices, determine

(i) the eigenvalues;(ii) the geometric multiplicity of each eigenvalue;(iii) whether it is diagonalisable.

9


For those matrices A that are diagonalisable, find an invertible matrixP such that P-1 AP is diagonal.

l 3 -1 1 7 -1 2

(a)L-'

3J, (b) -1 5 -1 , (c) -1 7 2 ,

1 -1 3 -2 2 10

2 1 -1 1 0 1

(d) 0 2 1, (e) 0 2 1.0 0 1 -1 0 3

1.24 Consider the sequence described by

1 3 7 an11 21

5,..., bn,.

where an+1 = an + 2bn and bn+1 = an + bn.

Find a matrix A such that

A[bn+1]

bn

By diagonalising A, obtain explicit formulae for an and bn and henceshow that

liman

= V2.n-oo bn

1.25 Let t be a singular transformation on a real vector space V. Let f (X)and g(X) be real polynomials whose highest common factor is 1. Leta = f (t) and b = g(t).

Prove that every eigenvector of a o b that is associated with the eigen-value 0 is the sum of an eigenvector of a associated with the eigenvalue0 and an eigenvector of b associated with the eigenvalue 0.

1.26 Suppose that s, t E C (V, V) each have n = dim V distinct eigenvaluesand that s o t = t o s. Prove that, for every A in the ground field F,

CA = {v E V I t(v) = Av}

is a subspace of V. Show that C,\ is s-invariant and that, when A is aneigenvalue of t, the subspace Ca; has dimension 1.

Hence show that the matrix of s with respect to the basis of eigenvec-tors of t is diagonal.

10


1.27 Let u be a non-zero vector in an n-dimensional vector space V over afield F, let t E Z (V, V), and let U be the subspace spanned by

{u,t(u),t2(u),...,tr-1(u)}.

Show that there is a greatest integer r such that the set

{u,t(u),t2(u),...,tr-1(u)}is linearly independent and deduce that this set is a basis for U. Showalso that U is t-invariant.

Show that there is a non-zero monic polynomial f (X) E F[X] of de-gree r such that [f (t)](u) = 0. If to : U --* U is the linear transformationinduced by t, show that its minimum polynomial is f (X).

In the case where u = (1,1,0) E IR3 and t is given by

t(x, y, z) = (x + y, x - y, z),

find the minimum polynomial of iu.

1.28 Let r, s, t be non-zero linear transformations on a finite-dimensional vec-tor spaceVsuchthatrotor=0. Let p=rosandq=ro(s+t)and suppose that the minimum polynomials of p, q are p(X), q(X) re-spectively. Prove that (with composites written as products)(1) q' = p'-lq and p' = q'-lp for all n > 1;(2) p(X) and q(X) are divisible by X;(3) q satisfies Xp(X) = 0, and p satisfies Xq(X) = 0.

Deduce that one of the following holds

(i) p(X) = q(X);(ii) p(X) = Xq(X);(iii) q(X) = Xp(X).

1.29 A 3 x 3 complex matrix M is said to be magic if every row sum, everycolumn sum, and both diagonal sums are equal to some 16 E C.

If M is magic, prove that 6 = 3m22. Deduce that, given a,8, y E d,there is a unique magic matrix M(a,)6,, -j) such that

m22 = of m1 = a +Q, m31 = a + yShow that {M(a,Q, y) 1 a,Q, y E Q} is a subspace of Matax3(C) andthat

B = {M(1, 0, 0), M(0,1, 0), M(0, 0, 1)}is a basis of this subspace.

If f : C3 -+ C3 represents M(a, /3, y) relative to the canonical basis{e1, e2, e3 b show that e1 + e2 + e3 is an eigenvector of f. Determine thematrix of f relative to the basis {el + e2 + e3, e2, e3}. Hence find theeigenvalues of M(a, Q, y).

11


1.30 Let V be a vector space of dimension n over a field F. A linear trans-formation f : V -> V (respectively, an n x n matrix A over F) is said tobe nilpotent of index p if there is an integer p > 1 such that fP-1 # 0and fP = 0 (respectively, AP-' 54 0 and AP = 0).

Show that if f is nilpotent of index p and x E V\{0} is such thatfp-1(x) # 0 then

{x,f(x),...,f"-1(x)}

is a linearly independent subset of V. Hence show that f is nilpotentof index n if and only if there is an ordered basis of V with respect towhich the matrix of f is

0

In-10

0'

Deduce that an n x n matrix A over F is nilpotent of index n if andonly if A is similar to this matrix.

1.31 Let V be a finite-dimensional vector space over IR and let f : V -+ V bea linear transformation such that f o f = -idv. Extend the externallaw IR x V -* V to an external law (E x V -p V by defining, for all x E Vand alla+if3 E(,

(a +i8)x = ax - P f(x).

Show that in this way V becomes a vector space over C. Use the identity

r r r

1:(at - iµt)vt = Atvt + lttf(vt)t=1 t=1 t=1

to show that if {v1i... , vr} is a linearly independent subset of the (-vector space V then {v1, ... , vr, f (v1), ... , f (vr)} is a linearly indepen-dent subset of the IR-vector space V. Deduce that the dimension of Vas a vector space over Q is finite, n say, and that dimR V = 2n.

Hence show that a 2n x 2n matrix A over IR is such that A2 = -I2nif and only if A is similar to the matrix

1.32 Let A be a real skew-symmetric matrix with eigenvalue A. Prove thatthe real part of A is zero, and that A is also an eigenvalue.

12


If (A - AI)2Z = 0 and Y = (A - AI)Z show, by evaluating V Y,that Y = 0. Hence prove that A satisfies a polynomial equation withoutrepeated roots, and deduce that A is similar to a diagonal matrix.

If x is an eigenvector corresponding to the eigenvalue A = is and if

u=x+x,v=i(x-x) show thatAu = av, Av = -au.

Hence show that A is similar to a diagonal block matrix

fo

Al

A2

Ak

where each Ai is real and of the form

cl 0 J.

1.33 Let V be a vector space of dimension 3 over IR and let t E Z (V, V) haveeigenvalues -2,1,2. Use the Cayley-Hamilton theorem to express tenas a real quadratic polynomial in t.

1.34 Let V be a vector space of dimension n over a field F and let f E £ (V, V)be such that all the zeros of the characteristic polynomial of f lie in F.

Let Al be an eigenvalue of f and let bl be an associated eigenvector.Let W be such that V = Fbl ®W and let (b;)2<i<n be an ordered basisof W. Show that the matrix of f relative to the basis {bl, b2i ... , bn} isof the form

[Ai P12 .. Qln

0 M

Observe that in general Qi2...... in are non-zero, so that W is not f-invariant. Let it be the projection of V onto W and let g = it of. Showthat W is g-invariant and that if g' is the linear transformation inducedon W by g then Mat (g', (bf)) = M. Show also that all the zeros of thecharacteristic polynomial of g' lie in F.

Deduce that f is triangularisable in the sense that there is a basis Bof V relative to which the matrix of f is upper triangular with diagonalentries the eigenvalues of f.

13


1.35 Suppose that t E £(IR3, IR3) is given by

t(a, b, c) = (2a + b - c, -2a - b + 3c, c).

Find the eigenvalues and the minimum polynomial of t. Show that t isnot diagonalisable. Find a basis of IR3 with respect to which the matrixof t is upper triangular.

1.36 Let V = Q3[X] be the vector space of polynomials of degree less thanor equal to 2 over the field Q. If t E L (V, V) is given by

t(1)=-5-8X-5X2t(X) = 1 + X + X2

t(X2) = 4 + 7X + 4X2,

show that t is nilpotent. Find a basis of V with respect to which thematrix of t is upper triangular.

1.37 For each of the following matrices A find a Jordan normal form and aninvertible matrix P such that P-1 AP is in Jordan normal form.

(a) [239 -64

-1 -1

5 -41 ]' (b) I 0 -1 J'

1 3 -2 3 0 1

(c) 0 7 -4 , (d) 0 3 0

0 9 -5 0 0 3

1.38 Find a Jordan normal form J of the matrix

A=

Find

2 1 1 1 0

0 2 0 0 0

0 0 2 1 0

0 0 0 1 1

0 -1 -1 -1 0

also a Jordan basis and hence an invertible matrix P such thatP-'AP = J.

1.39 For each of the following matrices A find a Jordan normal form J, aJordan basis, and an invertible matrix P such that P-1AP = J.

-13 8 1 222 -2 -12

(a) 20 0 -12 (b) -22 13 0 3

8 -5 0 -130 -3 -16

-22 13 5 5

14


1.40 Find a Jordan normal form and a Jordan basis for the matrix

5 -1 -3 2 -50 2 0 0 0

1 0 1 1 -20 -1 0 3 1

1 -1 -1 1 1

1.41 Find the minimum polynomial of the matrix

1 0 -1 1 0

-4 1 -3 2 1

A = -2 -1 0 1 1

-3 -1 -3 4 1

-8 -2 -7 5 4

From only the information given by the minimum polynomial, how manyessentially different Jordan normal forms are possible? How many lin-early independent eigenvectors are there? Does the number of linearlyindependent eigenvectors determine the Jordan normal form J? If not,does the information given by the minimum polynomial together withthe number of linearly independent eigenvectors determine J?

1.42 Find a Jordan normal form of the differentiation map D on the vectorspace IR4[X) of polynomials of degree less than or equal to 3 with realcoefficients. Find also a Jordan basis for D on IR4[X].

1.43 If a 3 x 3 real matrix has eigenvalues 3,3,3 what are the possible Jordannormal forms? Which of these are similar?

1.44 Which of the following are true? If A, B E Mat,,,,(() then AB and BAhave the same Jordan normal form

(i) if A and B are both invertible;(ii) if one of A, B is invertible;(iii) if and only if A and B are invertible;(iv) if and only if one of A, B is invertible.

1.45 Let V be a vector space of dimension n over C. Let t E C(V, V) and letA be an eigenvalue of t. Let J be a matrix that represents t relative tosome Jordan basis of V. Show that there are

dim Ker(t - A ides)

15


blocks in J with diagonal entries A.More generally, if ni is the number of i x i blocks with diagonal entries

A and di = dim Ker(t - A ides)`, show that

di = nl + 2n2 + ... + (i - 1)nti_1 + i(ni + n;+1 + ...).Deduce that ni = 2d1 - d;_1 - di+1

1.46 Find a Jordan normal form J of the matrix0 1 0 -1

_ -2 3 0 -1

`1 -2 1 2 -12 -1 0 3

Find also a Jordan basis and an invertible matrix P such that P-1 AP =J.

Hence solve the system of differential equations

x1 0 1 0 -1 x1

X1 _ -2 3 0 -1 x2

X3 -2 1 2 -1 x3

x4 2 -1 0 3 x4

1.47 Solve each of the following systems of differential equations :

= 5x1 + 4x2dt = 4x1 - x2 - x3

dx2 dt_

dx3

x1 + 2x2 - x3

dt =xl - x2 + 2x2

dxj

dx2 dx2

dIi_5x'+6x2+6x3 = 0 j-j-=x1+3x2-2x3

(c)dt

+ x1 - 4x2 - 2x3 = 0 (d)dt

= 7x2 - 4x3

dX3

dt- 3x1 + 6x2 + 4x3 = 0 dt = 9x2 - 5x3

1.48 Solve the system of differential equations

dxl - 3 dx2 + 2dx3 = y1

-5dx2+4d3 = Y2

-9dx +7d =y3

16


1.49 Solve the system of differential equations

dxl_dt = x1 + x2

dx2

dt = 2x1 + 3x2

given that x1(0) = 0 and x2(0) = 1.

1.50 Show how the differential equation

x"'-2x"-4x'+8x=0

can be written as a first-order matrix system X' = AX. By using themethod of the Jordan normal form, solve the equation given the initialconditions

x(0) = 0, x'(0) = 0, x"(0) = 16.

17

2: Duality and normal transformations

The dual of a vector space V over a field F is the vector space Vd =,C(V, F) of linear functionals f : V -> F. If V is of finite dimensionand B = {v1,...,vn} is a basis of V then the basis that is dual to B isB4 = { va, ... , vn } where each 0 : V -+ F is given by

vd(vj)1 ifi=j;0 ifi#j.

For every x E V we have

x = of (x)v1 +7J2(x)v2 + ... + vn(x)vn

and for every f E Vd /we have

f = f(vl)vf + f(v2)v2 + ... + f(vn)vn.

If (Vi)n, (w:)n are ordered bases of V and (va)n, (wa)n the correspond-ing dual bases then the transition matrix from (va)n to (wa)n is (p-1)twhere P is the transition matrix from (vi)n to (wt)n. In particular,consider V = IRS. Note that if

B = {(all,...,aln),(a21,...) a2n),...,(anl) ...7an.)}

is a basis of IRS then the transition matrix from B to the canonical basis(ei)n of IRS is M = [mj,]nxn where mij = aji. The transition matrixfrom Bd to (ed)n is given by (M-l)t. We can therefore usefully denotethe dual basis by

B = {[all, ... , aln], [all, ... , a2n], ... , [anl, ... , an-[}


where ain] denotes the ith row of M-1, so that

[ail,...,ain](xl) ...,xn) =ailxl + ...+atnxn,

The bidual of an element x is x^ : Vd -* F where x^(yd) = yd(x). Itis common practice to write yd(x) as (x, yd) and say that yd annihilatesx if (x, yd) = 0. For every subspace W of V the set

W1={ydEVd I (Vx E W) (x, yd) = 0)

is a subspace of W and

dim W + dim W J- = dim V.

The transpose of a linear transformation f : V -+ W is the linearmapping ft : Wd --+ Vd described by yd ,--* yd o f. When V is of finitedimension we can identify V and its bidual (Vd)d, in which case we havethat (f t )t = f. Moreover, if f : V -+ W is represented relative to fixedordered bases by the matrix A then f' : Wd -+ V d is represented relativeto the corresponding dual bases by the transpose At of A.

If V is a finite-dimensional inner product space then the mapping19v : x ,--* xd describes a conjugate isomorphism from V to Vd, by whichwe mean that

(x + y)d = xd + yd and (Ax)d = .1xd.

The adjoint f* : W -+ V of f : V W is defined by

f* _ V1 ° ft o t9w

and is the unique linear transformation such that

(Vz,y E V) (f(x)Iy) = (xlf*(y))

We say that f is normal if it commutes with its adjoint. If the matrixof f relative to a given ordered basis is A then that of f* is V. We saythat A is normal if it commutes with T. A matrix is normal if and onlyif it is unitarily similar to a diagonal matrix, i.e. if there is a matrix Uwith U-1 = U such that U-1 AU is diagonal. A particularly importanttype of normal transformation occurs when the vector space in questionis a real inner product space, and topics dealt with in this section reachas far as the orthogonal reduction of real symmetric matrices and itsapplication to finding the rank and signature of quadratic forms.

19


2.1 Determine which of the following mappings are linear functionals on thevector space IR3 [X] of all real polynomials of degree less than or equalto 2 :

1(a) f '-' f'; (b) f '-+ f f; (c) f '-+ f (2);

1(d) f '--' f'(2); (e) f '-' / f2.

2.2 Let C[0,1] be the vector space of continuous functions f : [0, 11 -+ IR. Iffo is a fixed element of C[O,1], prove that (p : C[0,1] -+ IR given by

o(t) f (t) dt1'P(f) = fo f

is a linear functional.

2.3 Determine the basis of (IR3)d that is dual to the basis

{(1'O' -1), (-1, 1, 0), (01 111))

of 1R3.

2.4 Let A = {x1i x2} be a basis of a vector space V of dimension 2 and letAd = {V1i a2) be the corresponding dual basis of Vd. Find, in terms ofepl, (p2 the basis of V d that is dual to the basis A' = {xl +2x2, 3x1 +4x2}of V.

2.5 Which of the following bases of (IR2)d is dual to the basis {(-1, 2), (0, 1)}of IR2?

(a) {[-1, 2], [0, 1]}; (b) {[-1, 0], [2, 1]};(c) {[-1, 0], [-2,1]}; (d) {[l, 0], [2, -1]}.

2.6 (i) Find a basis that is dual to the basis

{(4,5,-2,11), (3, 4, -2,6),(2,3,-1,4), (1,1,-1,3)1

of IR4.(ii) Find a basis of IR4 whose dual basis is

([2,-1,1'0]) [-1, 0, -2)0]j[-2,2,1,0j, [-8,3,-3, 1]).

2.7 Show that if V is a finite-dimensional vector space over a field F and ifA, B are subspaces of V such that V = A ® B then V d =A' a B1.

Is it true that if V = A ® B then V d = Ad ® Bd?

20


2.8 Let IR3[X] be the vector space of polynomials over IR of degree lessthan or equal to 2. Let t1, t2, t3 be three distinct real numbers and fori = 1, 2, 3 define mappings fi : IR3 [X] -+ IR by

MAX)) = p(ti)

Show that Bd = {fl, f2, f3} is a basis for the dual space (IR3[X])d anddetermine a basis B = {pl(X),p2(X),p3(X)} of IR3[X] of which Bd isthe dual.

2.9 Let a = (1, 2) and Q = (5, 6) be elements of IR2 and let cp = [3,4] be anelement of (IR2)d. Determine

(a) a^ ((p); (b) 01 (`p);(c) (2a+3,0)1 (,p); (d) (2a+3,(3)^([a,b]).

2.10 Prove that if S is a subspace of a finite-dimensional vector space V then

dim S + dim Sl =dim V.

If t E C(U,V) and td E ,C(Vd, Ud) is the dual of t, prove that

Kertd = (Imt)1.

Deduce that if v E V then one of the following holds(i) there exists u E U such that t(u) = v;(ii) there exists (p E Vd such that td((p) = 0 and <p(v) = 1.

Translate these results into a theorem on solving systems of linearequations.

Show that (i) is not satisfied by the system

3x+ y=2x+2y=1

-x+3y= 1.

Find the linear functional 'p whose existence is guaranteed by (ii).

2.11 If s, t : U -+ V are linear transformations, show that

(sot)d=td0Sd.

Prove that the dual of an injective linear transformation is surjective,and that the dual of a surjective linear transformation is injective.

21


2.12 Let t E C(IR3, IR3) be given by the prescription

t(a, b, c) = (2a+Ill a+b+c,-c).

If X = {(1, 0, 0), (1,1, 0), (1,1,1)} and yd = {[1, 0, 0], [1,1, 0], [1,1,1]},find the matrix of td with respect to the bases yd and Xd.

2.13 Let {al, 02) 013} and {al, a2i a3} be bases of IR3 that differ only in thethird basis element. Suppose that {<pi, ,p2, ,p3 } and {, 'p' , V3 } are thecorresponding dual bases. Prove that V3 is a scalar multiple of V3-

2.14 Let CIO, 11 denote the space of continuous functions on the interval [0, 11.Given g E C[0,1], define Lg : C[0,11 -+ IR by

Lg(f) = f f (t) g(t) dt.1

Show that Lg is a linear functional.Let x be a fixed element of [0, 1] and define F,, : CIO, 11 --+ IR by

FF(f) = f (x). Show that F., is a linear functional. Show also that thereis no g E CIO, 11 such that F. = Lg.

2.15 By a canonical isomorphism V -* Vd we mean an isomorphismsuch that, for all x, y E V and all isomorphisms f : V -r V, we have

(*) (X, SW) = (f(x),S[f(y)]),

where the notation (x, S(y)) means [S(y)](x).In this exercise we indicate a proof of the fact that if V is of dimension

n > 1 over F then there is no canonical isomorphism V -> Vd exceptwhen n = 2 and F has two elements.

If S is such an isomorphism show that, for y # 0, the subspaceKerr(y) = {S(y)}1 is of dimension n- 1.

Suppose first that n > 3. If there exists t E Ker S(t) for some t # 0let {t, xl, ... , xn_2 } be a basis of Ker S(t) and extend this to a basis{t, xl, ... , xn_2i z} of V. Let f : V -* V be the (unique) linear trans-formation such that

f (t) = t, f (xi) = z, f (z) = xi i and f (xi) = xi for i ,A 1.

Show that f is an isomorphism that does not satisfy (*). [Hint. Takex = xl, y = t.] If, on the other hand, t 0 Ker S(t) for all t 0 let

22


{z1,...,xn-1} be a basis of Kers(t) so that {z1i...,xn_1,t} is a basisof V. Show that

{xi + x2,x2,x3,...,xn_1,t}

is also a basis of V. Show also that x2 E Kers(x1). Now show that iff : V -* V is the (unique) linear transformation such that

f(xl) = x2, f(x2) = xl + x2, f(t) = t, f(xi) = xi (i 54 1,2)

then f is an isomorphism that does not satisfy (*). Conclude from theseobservations that we must have n = 2.

Suppose now that F has more than two elements and let A E F be suchthat A # 0,1. If there exists t 0 such that t E Ker S(t) observe that {t}is a basis of Ker S(t) and extend this to a basis It, z} of V. If f : V - Vis the (unique) linear transformation such that f (t) = t, f (z) = Azshow that f is an isomorphism that does not satisfy (*). [Hint. Takex = z, y = t.] If, on the other hand, t 0 Ker S(t) for all t 0 0 let {z} bea basis for Kers(t) so that {z,t} is a basis for V. If f : V -. V is the(unique) linear transformation such that f (z) = Az, f (t) = t show thatf is an isomorphism that does not satisfy (*). [Hint. Take x = y = z.]Conclude from these observations that F must have two elements.

Now examine the vector space F2 where F = {0, 1).[Hint. (F' )d is the set of linear transformations f : F x F -+ F. Since

F2 has four elements there are 24 = 16 laws of composition on F. Onlyfour of these are linear transformations from F2 to F; and each of theseis determined by its action on the natural basis of F2. Compute (F2)dand determine a canonical isomorphism from F2 onto (F2)a ]

2.16 Let V be an inner product space of dimension k and let U be a subspaceof V of dimension k - 1 (a hyperplane). Show that there exists a unitvector n in V such that

U= {x E V I (n]x) = 0}.

Given v E V, definev' = v - 2(nlv)n.

Show that v - v' is orthogonal to U and that i (v + v') E U, so that v'is the reflection of v in the hyperplane U. Show also that the mappingt : V --4V defined by

t(v)=v'is linear and orthogonal. What can you say about its eigenvalues andeigenvectors?

23


Ifs IR3 -> IR3 and t : IR4 --, IR4 are respectively reflections in theplane 3x - y + z = 0 and in the hyperplane 2x - y + 2z - t = 0, showthat the matrices of s and t are respectively

-7 6 -6 1

6 9 2 ,11 -6 2 9

5

1 2 -4 2

2 4 2 -1-4 2 1 2

2 -1 2 4

2.17 Find the equations of the principal axes of the hyperbola

-x2 + 6xy - y2 = 1.

Find also the equations of the principal axes of the ellipsoid

7x2 + 6y2 + 5z2 + 4xy - 4yz = 1.

2.18 Let V be a finite-dimensional inner product space and let f : V --+ Vbe linear. Show that if A is the matrix of f relative to an orthonormalbasis B of V then the matrix of the adjoint f* of f relative to B is thetranspose of the complex conjugate of A.

2.19 For every A E Mat,i*x,,(() define the trace of A by tr(A) _ i ate.Show that if V is the vector space of n x n matrices over d then themapping

(A, B) (AIB) = tr(B*A),

where B* denotes the transpose of the complex conjugate of B, is aninner product on V.

Consider, for every M E V, the mapping fm : V --+ V defined by

fm (A) = MA.

Show that, relative to the above inner product,

(fM)* = .fM*.

2.20 Let V be a finite-dimensional inner product space. Show that for everyf E Vd there is a unique ,0 E V such that

(bx E V) f (x) = (xI,8).

24


[Hint. Let {al,...,an} be an orthonormal basis of V and consider

i=1

Show as follows that this result does not necessarily hold for innerproduct spaces of infinite dimension. Let V be the vector space of poly-nomials over Q. Show that the mapping

(P, q) H (PIq) = f p (t) q(t) dt10

is an inner product on V. Let z be a fixed element of C and let f E Vdbe the `evaluation at z' map given by

(VPEV) f(P)=P(z)

Show that there is no q E V such that (Vp E V) f (p) = (pjq).[Hint. Suppose that such a q exists. Let r E V be given by r(t) = t - zand show that, for every p E V,

10 = f r(t) p(t) q(t) dt.

Now let p be given by p(t) = r(t)q(t) and deduce the contradictionq=0.]

For the rest of this question let V continue to be the vector space ofpolynomials over C with the above inner product. If p E V is given byp(t) = aktk define p E V by f(t) _ aktk, and let fp : V -* V begiven by

(dq E V) fp(q) = Pq

where, as usual, (pq)(t) = p(t)q(t). Show that (fp)* exists and is fp.Now let D : V -, V be the differentiation map. Show that D does

not admit an adjoint.[Hint. Suppose that D* exists and show that, for all p, q E V,

(p I D(q) + D*(q)) = P(1)9(1) - P(0)4(0)

Suppose now that q is a fixed element of V such that q(0) = 0 andq(1) = 1. Use the previous part of the question (with z = 1) to obtainthe required contradiction.]

25


2.21 Let C[O,1] be the inner product space of real continuous functions on[0, 11 with the integral inner product. Let K : C[O,1] -+ C[0,1] be theintegral operator defined by

K(f) = f xyf(y) dy.1

Prove that K is self-adjoint.For every positive integer n let f, be given by

fn(x) = xn -2

n+2*

Show that fn is an eigenfunction of K with associated eigenvalue 0. Usethe Gram-Schmidt orthonormalisation process to find two orthogonaleigenfunctions of K with associated eigenvalue 0.

Prove that K has only one non-zero eigenvalue. Find this eigenvalueand an associated eigenfunction.

2.22 Let t be a skew-adjoint transformation on a unitary space V. Prove thatid ±t is a bijection and that the transformation

s = (id -t)(id +t)- 1

is unitary. Show also that s cannot have -1 as an eigenvalue.

2.23 If S is a real symmetric matrix and T is a real skew-symmetric matrixof the same order, show that

det(I - T - iS) 0 0.

Show also that the matrix

U=(I+T+iS)(I-T-iS)-1

is unitary.

2.24 Let A be a real symmetric matrix and let S be a real skew-symmetricmatrix of the same order. Suppose that A and S commute and thatdet(A - S) # 0. Prove that

(A+ S)(A - S)-1

is orthogonal.

26


2.25 A complex matrix A is such that A A = -A. Show that the eigenvaluesof A are either 0 or -1.

2.26 Let A and B be orthogonal n x n matrices with det A det B. Provethat A + B is singular.

2.27 Let A be an orthogonal n x n matrix. Prove that(1) if det A = 1 and n is odd, or if det A = -1 and n is even, then 1 is

an eigenvalue of A;(2) if det A = -1 then -1 is an eigenvalue of A.

2.28 If A is a skew-symmetric matrix and g(X) is a polynomial such thatg(A) = 0, prove that g(-A) = 0. Deduce that the minimum polynomialof A contains only terms of even degree.

Deduce that if A is skew-symmetric and f (X), g(X) are polynomialswhose terms are respectively odd and even then f (A), g(A) are respec-tively skew-symmetric and symmetric.

2.29 For every complex n x n matrix A let

nN(A) = tr(A A) = E[A A]ii.

i=1

Prove that, for every unitary n x n matrix U,

N(UA) = N(AU) = N(A) and N(A - U) = N(In - U-1A).

2.30 If the matrix A is normal and non-singular prove that so is A-1.Prove that p(A) for some polynomial p(X) if and only if A is

normal.

2.31 Prove that if A is a normal matrix and g(X) is any polynomial theng(A) is normal.

2.32 If A and B are real symmetric matrices prove that A + iB is normal ifand only if A, B commute.

2.33 Let A be a real skew-symmetric n x n matrix. Show that det(-A) _(-1)1 det A and deduce thst if n is odd then det A = 0. Show also thatevery quadratic form xtAx is identically zero.

Prove that the non-zero eigenvalues of A are of the form l i whereµ E IR. If x = y + iz where y, z E IRn is an eigenvector associated withthe eigenvalue iµ, show that Ay = -juz and Az = µy. Show also thatyty = ztz and that ytz = 0. If Au = 0 show also that uty = utz = 0.

27


Find the eigenvalues of the matrix

0 2 -2A= -2 0 -1

2 1 0

and an orthogonal matrix P such that

0 0 0

PAP = 0 0 3 .

0 -3 0

2.34 Consider the quadratic form q(x) = xtAx on IR". Prove that q(x) > 0for all x E IR" if and only if the rank of q equals the signature of q.Prove also that q(x) > 0 for all x E IR" with q(x) = 0 only when x = 0if and only if the rank and signature of q are each n.

2.35 With respect to the standard basis for IR3, a quadratic form q is repre-sented by the matrix

1 1 -1A= 1 1 0 .

-1 0 -1Is q positive definite? Is q positive semi-definite? Find a basis of IR3with respect to which the matrix representing q is in normal form.

2.36 Let f be the bilinear form on IR2 x IR2 given by

f((xi,x2),(yi,y2)) = xiyi +xiy2+2x2y1 +x2Y2

Find a symmetric bilinear form g and a skew-symmetric bilinear form hsuch that f = g + h.

Let q be the quadratic form given by q(x) = f (x, x) where x E IR2.Find the matrix of q with respect to the standard basis. Find also therank and signature of q. Is q positive definite? Is q positive semi-definite?

2.37 Write the quadratic form

4x2 +4y 2 + 4z2 - 2yz + 2xz - 2xy

in matrix notation and show that there is an orthogonal transformation(x, y, z) H (u, v, w) which transforms the quadratic form to

3u2 + 3v2 + 6w2.

Deduce that the original form is positive definite.

28


2.38 By completing squares, find the rank and signature of the followingquadratic forms :

(1) 2y2 - z2 + xy + xz;(2) 2xy - xz - yz;(3) yz+xz+xy+xt+yt+zt.

2.39 For each of the following quadratic forms write down the symmetricmatrix A for which the form is expressible as xt Ax. Diagonalise each ofthe forms and in each case find a real non-singular matrix P for whichthe matrix Pt AP is diagonal with entries in {1, -1,0}.(1) x2 + 2ya + 9z2 - 2xy + 4xz - 6yz;(2) 4xy + 2yz;(3) x2 + 4y2 + z2 - 4t2 + 2xy - 2xt + 6yz - 8yt - 14zt.

2.40 Find the rank and signature of the quadratic form

Q(x1,...,x,) = >(xr - x8)2.

r<8

2.41 Show that the rank and signature of the quadratic form

n

E (Ars + r + s)xrx,r'8=1

are independent of A.

2.42 Let A be the matrix associated with the quadratic form Q(x1, ... , xn)and let A be an eigenvalue of A. Show that there exist a1, ... , an not allzero such that

Q 2 a

2.43 If the real square matrix A is such that det A # 0 show that the quadraticform xtAtAx is positive definite.

2.44 Let f : IRn x IRn --+ IR be a symmetric bilinear form and let Q f be theassociated quadratic form. Suppose that Qf is positive definite and letg : IRn X IRn --> IR be a symmetric bilinear form with associated quadraticform Qg. Prove that there is a basis of IRn with respect to which Q fand Qg are each represented by sums of squares.

For every x E IRn let f, E (IRn)d be given by f,, (y) = f (x, y). Callf degenerate if there exists x E IRn with ff = 0. Determine the scalarsA E IR such that g - A f is degenerate. Show that such scalars are the

29


roots of the equation det(B - AA) = 0 where A, B represent f, g relativeto some basis of IR'.

By considering the quadratic forms 2xy + 2yz and x2 - y2 + 2xz showthat the result in the first paragraph fails if neither f nor g is positivedefinite.

2.45 Evaluate

1. 01.0 f 008

(xz+y'+xa+xN+xz+yz)dxdy dz.

30

Solutions to Chapter 1

1.1

(i) False. For example, take b = -al.(ii) True.

(iii) False. {(1,1,1)} is a basis, so the dimension is 1.(iv) False. For example, take A = {0} or A = {v, 2v}.(v) True.(vi) False. For example, take A = IR'.

(vii) True.(viii) False. {(x, \x) I x E IR} is a subspace of IR2 for every A E IR.

(ix) True.(x) True.

(xi) False. An isomorphism is always represented by a non-singularmatrix.

(xii) False. Consider, for example, IR2 and d2. The statement is true,however, if the vector spaces have the same ground field.

(xiii) False. 0 0 is a counter-example.

(xiv) False. For example,

0 1 1

[0

1

0][1 0]- [0 4][0 0]

(xv) True.(xvi) True.(xvii) False. Take, for example, f, g : IR" -* IR' given by f (x, y) (0, 0)

and g(x, y) = (x, y). Relative to the standard basis of IR' we see


that f is represented by the zero matrix and g is represented bythe identity matrix; and there is no invertible matrix P such thatP-142P = 0.

(xviii) True.(xix) False. The transformation t is non-singular (an isomorphism), but

r 1 21is singular.

1 2J

r 1

(xx) False. The matrix I 01 J

is not diagonalisable.

1.2 We have that

(a,b,c) E Kert1 (a + b, b + c, c + a) = (0,0,0)

4=#- a=b=c=O

and so Kert1 = {0}. It follows from the dimension theorem that Imt1 =IR3.

As for t2, we have

(a,b,c)EKert2 b a-b=0,b-c=0b a=b=c

and so Kert2 = {(a, a, a) I a E IR}. It is clear from the definition of t2that Imt2 = {(a, b, 0) I a, b E IR}.

Likewise, it is readily seen thatKer t3 = {0}, Imt3 = IR3,Kert4={(0,0,a) I aEIR}, Imt4={(a,b,b) I a,bEIR}.

If Ker t; fl Im ti = {0} then by the dimension theorem we have

dim(Kerti + Imti) = dim IR3

and so Ker ti + Im ti = IR3. Now for i = 1, 2, 3, 4 we have from the abovethat Ker ti n Im ti = {0}. Thus we see that IR3 = Ker ti ® Im ti holds inall cases.

Im t2 is t3-invariant. For, if v E Im t2 then v = (a, b, 0) and so

t3(v) = t3(a,b,0) = (-b,a,0) E Imt2.

However, Kert2 is not t3-invariant. For (1, 1, 1) E Kert2 but t3(1,1,1) _(-1,1,1) Kert2.

32


For the last part, we have that (t3 o t4) (a, b, c) _ (-b, a, b) and that(t4 o t3) (a, b, c) = (-b, a, a). Consequently,

Ker(t3 o t4) _ {(0, 0, a) I a E IR};

Im(t3 0 t4) = {(-b, a, b) I a, b E IR};

Ker(t4 o t3) = {(0, 0, a) I a E IR};

Im(t4 o t3) = {(-b, a, a) I a, b (=- IR}.

1.3 Reducing the matrix to row-echelon form we obtain

3 -1 1 1 -1 3 1 -1 3

-1 5 -1 -> -1 5 -1 -> 0 4 2

1 -1 3 3 -1 1 0 2 -8

1 -1 3 1 -1 3- 0 2 -8 -> 0 2 -8 .

0 4 2 0 0 18

Note that we have been careful not to divide by any number that isdivisible by either 2 or 3 (since these will be zero in 74 and 713 respec-tively).

(i) When F = IR the rank of the row echelon matrix is 3, in which casedim Im t = 3 and hence dim Ker t = 0.

(ii) When F = 712 we have that 2,18, -8 are zero so that the rank is 1,in which case dim Im t = 1 and dim Ker t = 2.

(iii) When F = 713 we have that 18 is zero so that the rank is 2, inwhich case dim Im t = 2 and dim Ker t = 1.

1.4

V = Ker t ® Im t holds in cases (i) and (ii), but not in case (iii); for incase (iii) we have that (1, 1, 1) belongs to both Kert and Imt.

If s o t = idv then s is surjective, hence bijective (since V is of finitedimension). Then t = s-1 and so t o s = idv.

Suppose that W is t-invariant, so that t(W) C_ W. Since t is anisomorphism we must have dimt(W) = dim W and so t(W) = W. HenceW = s[t(W)] = s(W) and W is s-invariant.

The result is false for infinite-dimensional spaces. For example, con-sider the real vector space IR[X] of polynomials over IR. Let s be thedifferentiation map and t the integration map. We have s o t = id buttoe 54 id.

33


1.5 KerD = {a I a E F} and ImD = {p(X) I degp(X) < n - 2}. Clearly,Im D is isomorphic to Vn_ 1 and Ker D is isomorphic to F. Now Ker D nImD # {0} since if a E F with a # 0 then the constant polynomial abelongs to both.

The same results do not hold when the ground field is Z2. For exam-ple, in this case we see that the polynomial X2 belongs to the kernel ofD.

1.6 Let s, t E £(V, V). Then if w E Im(s o t) we have w = a[t(u)] for someu E V which shows that w E Im s. Thus Im(s o t) C Im s. The firstchain now follows by taking s = t'.

Similarly, if u E Kertn then s[tn(u)] = s(0) = 0 gives u E Ker(s o tn)and so Ker t' C_ Ker(s o tn). The second chain now follows by takings=t.

Now we cannot have an infinite number of strict inclusions in the firstchain since X C Y implies that dim X < dim Y, and the dimension of Vis finite. Hence the chain is finite. It follows that there exists a positiveinteger p such that ImtP = ImtP+k for all positive integers k. Sincedim Im tP + dim Ker tP = dim V the corresponding results for the kernelchain are easily deduced.

To show that V = Im tP ®Ker tP it suffices, by the dimension argument,to prove that ImtP n KertP = {0}. Now if x E ImtP n KertP thentP(x) = 0 and there exists v E V such that x = tP(v). Consequently

0 = tP(x) = t2p(v)

and so v E Kert2p = KertP whence x = tP(v) = 0.For the last part, observe that if x E Im tP then x = tP(v) gives

t(x) = tP+1(v) E Im tP+1 C_ Im tP and so Im tP is t-invariant. Also, ifx E KertP then tP(x) = 0 gives tP+1(x) = 0 so tP[t(x)] = 0 whencet(x) E KertP and so KertP is t-invariant.

1.7 If f of = 0 then (Vx E V) f (X) E Ken f and so Imf C Ken.We know that

n= dim V = dim Im f + dim Ker f = r + dim Ker f

and, by the above, dim Ker f > dim Im f = r. Hence 2r < n.If W is a subspace such that V = Ken f ® W then we have that

dim V = dim Ker f + dim W and so

dim W = dim V - dim Ker f = dim Im f = r.

34


If {wl,...,wr} is a basis of W then for i = 1,...,r we have f(w;) EIm f C Ker f. Moreover, { f (wl), ... , f (wr)} is linearly independentsince

r / r

Atf ('wi) = 0 f Aiwt) = 0

t)LiwiEKerf1=1

r=i'

EAtwtEKerfnW={0}:=1r

> A1w;=0i=1

(i = 1,...,r) Ai = 0.

Every linearly independent subset of a vector space can be enlarged toform a basis so, since dim Ker f = n - r, we can enlarge the independentsubset (f (wl ), ... , f (wr)} of Ker f to form a basis of Ker f. Thus wemay choose n - 2r elements x1, ... , xn-2r of Ker f such that

{f(wl),...,f(wr),xi,...,xn_2r}

is a basis for Ker f . Since V = W ® Ker f it follows that

{wl,...,Wr, f(wl),..., f(wr),xl,.... xn_2r}

is a basis for V.Using the fact that f o f = 0 and each x; E Ker f it is readily seen

that the matrix of f relative to this basis is of the form

Suppose now that A is a non-zero n x n matrix over F. If A2 = 0and if f : V -+ V is represented by A relative to some fixed orderedbasis then f o f = 0 and, from the above, there is a basis of V withrespect to which the matrix of f is of the above form. Thus A is similarto this matrix. Conversely, if M denotes the above matrix then clearlyM2 = 0. So if A is similar to M there is an invertible matrix P suchthat A = P-'MP whence A2 = P-1M2P = P-lOP = 0.

35


1.8 (1) Sums and scalar multiples of elements of V1, V2 are clearly elementsof V1, V2 respectively.

(2) If z E V1 then x = xl(bl + b4) + x2(b2 + b3) shows that V1 isgenerated by {b1 + b4, b2 + b3}. Also, if xl (bl + b4) + x2(b2 + b3) = 0then, since {b1, b2, b3, b4} is a basis of V, we have x1 = x2 = 0. Thus{b1 + b4i b2 + b3} is a basis of V1. Similarly, {b1 - b4i b2 - b3} is a basisof V2-

(3) It is clear from the definitions of V1 and V2 that we have V1 nV2 = {0}. Consequently, the sum V1 + V2 is direct. Since V1,V2 are ofdimension 2 and V is of dimension 4, it follows that V = V1 ® V2.

(4) To find the matrix of idv relative to the bases B = {b1, b2, b3, b4}and C = {b1 + b4, b2 + b3, b2 - b3, b1 - b4} we observe that

bl = 2(b1 + b4) + 0(b2 + b3) + 0(b2 - b3) + 2cb1 - b4)

b2 = 0(b1 + b4) + 2(b2 + b3) + 1#2 - b3) + 0(bl - b4)

b3 = 0(bi + b4) + 2(b2 + b3) - 2(b2 - b3) + 0(b1 - b4)

b4 = (b1 + b4) + 0(b2 + b3) + 0(b2 - b3) - 2(b1 - b4)-2

2 0 0 20 1 1 0

A 0 2 10

2 -22 0 0 2

It is readily seen that

11 0 0 1

A-1 =2A= 0 1 1 0

0 1 -1 0

1 0 0 -1

Suppose now that M is centro-symmetric; i.e. of the form

a b c d

e f g hh g f e

d c b a

36


Let f represent M relative to the basis B. Then the matrix of f relativeto the basis C is given by AMA-1, which is readily seen to be of theform

a /3 0 0

_ ry 6 0 0K

0 0 E S

0 0 $

Thus if M is centro-symmetric it is similar to a matrix of the form K.

1.9 If F is not of characteristic 2 then 1F + 1F # OF. Writing 2 = 1F + 1Fwe have that 2 E F. Given x c V we then observe that

x = 2x+ 2f(x) + 2x - 2f(x)= 2x+f(ax)+2x-f(2x)=(idv+f)(ax)+(idv-f)(ax)

so V = Im(idv +f) + Im(idv -f). Also, if x E Im(idv +f) n Im(idv -f)then x = y + f (y) = z - f (z) for some y, z E V and hence, sincef o f = idv by hypothesis,

f(x) = f(y) + f[f(n)] = f(y) + y = x;f (x) = f (z) - f If (z)] = f (z) - z = -x,

whence x = 0. Thus V = Im(idv +f) ® Im(idv -f).If A2 = In, let f represent A relative to some fixed ordered basis. Then

f of = idv. Let {a,,.. . , ap} be a basis of Im(idv +f) and {ap+1, ... , an}be a basis of Im(idv -f). Then {a,, ... , an} is a basis of V. Now sincea1 = b+ f (b) for some b E V we have f (al) = f (b) + f If (b)] = f (b) +b =a1, and similarly for a2, ... , ap. Likewise, ap+1 = c - f (c) for somec E V so f (ap+,) = f (c) - f If (c)] = f (c) - c = -ap+1, and similarly forap+2, ... , an. Hence the matrix off relative to the basis {a,,... , an } is

11611P

IpI0 pJ0 - n-

and A is then similar to this matrix. Conversely, if A is similar to amatrix of the form AP then there is an invertible matrix Q such thatQ-1 AQ = Ap. Then

A2= (QAPQ-1)2 = QO2PQ-1 = QjnQ-1 = In.

37


Suppose now that F is of characteristic 2, so that x + x = 0 and hencex= -x for every x E F. Let f of =idv and let g = idv +f . Then

(*) g(x)=0 x+f(z)=0 b x=-f(x)= f(x).Moreover, for every x E V we have g[g(x)] = g[x + f (x)] = z + f (z) +f[x + f(x)] = x + f(x) + f(x) + f[f(x)] = x + f(z) + f(x) +x= 0 andhence g o g = 0.

Suppose now that A2 = In and let f represent A relative to somefixed ordered basis. Let g = idv +f and note from the above thatIm g C Kerg. Let {g(c1),... , g(cp)} be a basis of Im g and extend thisto a basis

{b1, ... , bn-2p) 9(C1) , ... , g(cp)}

of Ker g (which is of dimension n - dim Im g = n - p). Consider now theset

B = {bl,... , bn-2p, 9(C1), CI) ... , 9(Cp), Cp}.

This set has n elements; for c; = bj gives the contradiction g(c;) _g(b2) = 0, and c; = g(c2) gives the contradiction g(ci) = g[g(cj)] = 0. Itis also linearly independent; for if we had

EAibi+Ep3g(c1)+Evjcj = 0

then, applying g and using the fact that bi, g(cj) E Ker g, we deduce thatvjg(cj) = 0 whence each vj = 0, and then from E Aib1+E ujg(cj) =

0 we obtain a; = 0 = u for all i, j. Thus B is a basis of V. To computethe matrix of f relative to the basis B we observe that since bi E Ker g wehave, by (*), that f (b;) = b= for every i. Also, f [g(ci)] = g[g(cj)]+g(c;) _g(cf) so that we have

f [9(ci)] = l.g(ci) + O.ctf (c;) = l.g(ct) + l.c1

It follows from these observations that the matrix of f relative to B is

f In-2p 1

1 1

0 1

vp =

38


Consequently A is similar to this matrix. Conversely, if A is similar toa matrix of the form VP then there is an invertible matrix Q such thatQ-1 AQ = Vp and so

A2 = (QVPQ-1)2 = QV Q-1 = QInQ-1 = In.

1.10 If t E £(IR2, IR2) is given by t(a,b) = (b,0) then clearly Imt = Kert{0}.

If t E C(IR3,IR3) is given by t(a,b,c) = (c,0,0) then Imt C Kert.If t E £(IR3, IR3) is given by t(a,b, c) = (b,c,0) then Kert C Imt.If t is a projection then Im t f1 Ker t = {0} and none of the above are

possible.

1.11 Consider the elements of £(IR3, IR3) given by

ti (a, b, c) _ (a, a, 0);

t2 (a, b, c) _ (0, b, 0);

t3 (a, b, c) _ (0, b, c);

t4(a,b,c)=(0,b-a,c);tb(a,b,c) _ (a,0,0).

Each of these transformations is a projection. We have

Ker tb = Ker t 1 but Im tb Im t1 f

Im t3 = Im t4 but Ker t3 ¢ Ker t4.

Also, t1 o t2 = 0 but t2 o ti 96 0. (Note that t2 o t1 is not a projection.)

1.12 Clearly, ei + e2 is a projection if and only if (denoting composites byjuxtaposition) e1e2 + e2e1 = 0. Thus if e1e2 = 0 and e2e1 = 0 thenthe property holds. Conversely, suppose that e1 + e2 is a projection.Then multiplying each side of e1 e2+e2e1 = 0 on the left by e1 we obtainele2+e1e2e1 = 0, and multiplying each side on the right by e1 we obtainele2el + e2e1 = 0. It follows that e1e2 = e2e1. But e1e2 + e2e1 = 0 alsogives e1e2 = -e2e1. Hence we have that each composite is zero.

When ei + e2 is a projection, we have that

Ker(ei + e2) = Kerel fl Kere2i

Im(ei +e2) = Imel ®Ime2.

1.13 Take U = {(0, a, b) I a, b E IR}. Then IR3 = V ® U since it is clear thatV n U = {0} and that

(a,b,c) _ (a, a, 0) + (0, b - a, c).

For the last part refer to question 1.11; take e = t1 and f = t4-

39


1.14 Suppose that Im e = Im f . Then for every v E V we have f (v) E Im f =Ime so, since e acts as the identity map on its image, elf (v)] = AV)and hence e o f = f. Likewise, f o e= e. Conversely, if e o f= f andfoe = e, let z E Im e. Then e(x) = x gives e(x) = f [e(x)] = f (x) E Im fand so Im e C Im f . The reverse inclusion is obtained similarly.

Since Im el = = Im ek we have ei o ej = ej for all i, j. Now

e2 (?iAjej)2=

_ A e + ... + akek + A1A2e1e2 + A2.1e2e1(k)+..+

(k)sG-Jl

+ + AkAk-lekek-1

= e,

and so e is also a projection. To show that Ime = Imel it suffices toprove that e o el = el and e1 o e = e. Now

(A1e1 + + .kek)el = Ale, + + Akel = el

gives the first of these, and the second is similar.For the last part, consider e, f E .C(IR2, IR2) given by

e(a, b) = (a, 0), f (a, b) = (0, b).

Then e and f are projections but clearlya

e +2if is not.

1.15 Since sums and scalar multiples of step functions are step functions it isclear that E is a subspace of the real vector space of all mappings fromIR to IR. Given 6 E E, the step function t9i whose graph is

ai ai+1

i.e. the function that agrees with t9 on [ai, ai+1 [ and is zero elsewhere,

40


is given by the prescription

Oi(x) = O(ai)[ear(x) - ear+,(x)]-It follows that {ek I k E [0, 1[} generates E since then

n+1

tf = E 9gi.i=0

Since the functions ek clearly form an independent set, they thereforeform a basis of E.

It is likewise clear that F is a vector space and that G is a subspaceof F. Consider now an element of F, as depicted in the diagram

From geometric considerations it is clear that every element of F can bewritten uniquely as the sum of a function e E E and a function g E G.[It helps to think of the above strips as pieces of wood that can slide upand down.] Thus it is clear that F = E ® G.

To show that {gk I k E [0, 1[} is a basis for G, observe first that thegraph of gar - gar+1 is of the form

ai+1- ai

41


Given p E F, consider the function pi whose graph is

Zai ai+1

i.e. the function that agrees with µ on [ai, ai+1 [ and is zero elsewhere.Let the gradient in the interval [ai, ai+i [ be bi, so that di = p(ai) +bi(ai+l - ai). Then it can be seen that

Ai = bi(gai - ga,+i) + li(ai)ea; - diea;+i.

Consequently µ = Enof µi gives the expression for it as a sum of g E Gand e E E. It now follows that {gk I k E [0, 1[} must be a basis for G.

Finally, observe that

I(ek) = foo ek(t) dt = gk

so that I carries a basis to a basis and therefore extends to an isomor-phism from E to G.

1.16 Let t1it2 E C(IR3,IR3) be given by

tj (a, b, c) = (a, 2b, 3c), t2(a, b, c) _ (2a, 3b, 6c).

Then t1 has eigenvalues 1, 2, 3 and t2 has eigenvalues 2, 3, 6. So bothquestions can be answered in the affirmative.

1.17 The eigenvalues of t are 1 + if and 1 - if. Associated eigenvectorsof any matrix representing t are [1, if /2] and [1, -if /2]. Since theeigenvalues are distinct, the eigenvectors of t form a basis of d2. Thematrix of t with respect to this basis is

1+i/ 0

0 1-ice

42


1.18 Since t has 0 as an eigenvalue we have t(v) = 0 for some non-zero v E Vand hence t is not injective, so not invertible. Thus if t is invertiblethen all its eigenvalues are non-zero. For the converse, suppose that t isnot invertible and hence not injective. Then there is a non-zero vectorV E Ker t, and t(v) = 0 shows that 0 is an eigenvalue of t.

If now t is invertible and t(v) = av with A # 0 then v = t-1 [t(v)] _t-1(Av) = Jet-1(v) gives t-1(v) = A' 1v and so A-1 is an eigenvalue oft-1 with the same associated eigenvector. (Remark. Note that we haveassumed that V is finite-dimensional (where?)-in fact the result is falsefor infinite-dimensional spaces.)

1.19 Suppose that A is a non-zero eigenvalue of t. Then t(v) = Av for somenon-zero v E V and

0 = tm(v) = tm-,[t(v)] = t'-' (AV) = ... = Amv,

and we have the contradiction A' = 0. Hence all the eigenvalues of tare zero.

If t is diagonalisable then the matrix A of t is similar to the diagonalmatrix

Al

An

where A 1, . . . , an are the eigenvalues of t. But we have just seen thatall the eigenvalues are zero. Thus, for some invertible matrix P we haveP-1AP = 0 which gives A = 0 and hence the contradiction t = 0.

1.20 The matrix of t with respect to the canonical basis {(1, 0), (0, 1)} is

The characteristic equation is (A- f)(A+f) = 0 and, since t-f id 00, t + f id 0 0, the minimum polynomial is (X - V3_) (X + f).

1.21 That t is linear follows from

t(f (X) + g(X)) = t((f + g)(X))=(f+g)(X+1)= f (X + 1) + g(X + 1) = t(f (X)) + t(g(X)),

43


t(Af(X)) = t((Af)(X)) = Af(X + 1) = At(f(X)).The matrix of t relative to { 1, X, ... , XI I is

1 1 1 1 ... 1

0 1 2 3 ... n0 0 1 3 ... an(n - 1)

0 0 0 0 ... 1

The eigenvalues of t are all 1. The characteristic polynomial is

g(X) = (X - 1)n+1

Hence, by the Cayley-Hamilton theorem, g(t) = 0. The minimum poly-nomial of t is then m(X) = (X - 1)' for some r with 1 < r < n + 1.A simple check using the above matrix shows that (t - idv)" 54 0 for1 < r < n. Consequently we have that m(X) = (X - 1)n+1

1.22 Let A1, ... , An be the eigenvalues of t. Then A', . . . , An are the eigenval-ues of t2 = idv and so

az =...=Azl n =1.

Consequently, at = ±1 for each i and hence the sum of the eigenvaluesof t is an integer.

1.23 (a) We have

3-A -1-1 3-A =Az-6A+8=(A-4)(A-2)

so the eigenvalues are 2 and 4, each of geometric multiplicity 1. For theeigenvectors associated with the eigenvalue 2, solve

[-1 1][ y)- 1001

to obtain the eigenspace {[x, x] I x E iR}. For the eigenvectors associ-ated with the eigenvalue 4, solve

-1 y1=[01

44


to obtain the eigenspace {[x, -z] I x E IR}. Since A has distinct eigen-values it is diagonalisable. A suitable matrix P is

(b) The eigenvalues are 2, 3, 6 and are all of geometric multiplicity 1.Associated eigenvectors are [1,0, -1], [1,1,1], [1, -2,1]. A is diagonalis-able; take, for example,

1 1 1

P 0 1 -2 .

-1 1 1

(c) The eigenvalues of A are 6, 6, and 12. For the eigenvalue 6 weconsider

1 -1 -2 x 0

-1 1 2 y = 0 .

-2 2 4 z 0

We obtain -x + y + 2z = 0. We can therefore find two linearly indepen-dent eigenvectors associated with the eigenvalue 6, for example [1, 1,01and [2, 0, 1]. Hence this eigenvalue has geometric multiplicity 2. Theeigenvalue 12 has geometric multiplicity 1 and an associated eigenvectoris [-1, 1, 2]. Hence A is diagonalisable and a suitable matrix P is

(d) The eigenvalues are 2, 2, 1. For the eigenvalue 2 we consider

0 1 -1 x 0

0 0 1 y= 00 0 -1 z 01 l1

from which we see that the corresponding eigenspace is spanned by[1, 0, 0]. Hence the eigenvalue 2 has geometric multiplicity 1. The eigen-value 1 also has geometric multiplicity 1, the corresponding eigenspacebeing spanned by [-2, 1, -1]. In this case A is not diagonalisable.

45


(e) The eigenvalues are 2, 2, 2. From

-1 0 1 x 0

00 0 1 y=-1 0 1 z 0

we see that the corresponding eigenspace is spanned by [0, 1, 0] and hasdimension 1. Thus the geometric multiplicity of the eigenvalue 2 is 1,and A is not diagonalisable.

1.24 The matrix in question is

and we have

[:1=An-1

The characteristic polynomial of A is X2 - 2X -1 and its eigenvalues area1 = 1 + f and )2 = 1 - v'-2. Corresponding eigenvectors are [f, 1]and [-vi, 1]. The matrix

1 1P =is such that P-1 AP = diag{A1,A2}.

[1'1]whereIn the new coordinate system, I i ] becomes P-1 { i ] =L

P2

f+1 f-1P1= 2f , P2= 2/

We then have

an =P1Ai-1vi-P2.A2-1vi, bn =p1Ai-1

from which we see that

+ P2,\2-1

an __ P,An-,\/2 a2I -P2-

bn PlAi-1 +P2\a-1

f -(P2/P1)(A2/a1)n-1vi

1 + (p2/P1)(A2/A1)n-1

1 - (P2/P1)(A2/A1)n-1_ (1

+ (P2/P1)(A2/)t1)n-1

46


Now since 0 < 1A2/1\1I < 1 we deduce that

hmanb

=f.n-.oo ,,,

1.25 Since f (X) and g(X) are coprime there are polynomials p(X) and q(X)such that f (X)p(X) + g(X)q(X) = 1. Let c = p(t) and d = q(t). Thenac + bd = idv.

Suppose now that v is an eigenvector of ab associated with the eigen-value 0. (Note that ab has 0 as an eigenvalue since t is singular.) Letu = a(cv) and w = b(dv). Then since a, b, c commute we have

bu = bacv = cabv = 0,

and since a, b, d commute we have

aw = abdv = dabv = 0.

Also, u + w = (ac + bd)v = v since ac + bd = idv.

1.26 Ifu,vE CA then t(u)=Au and t(v) = Av and so

t(au + bv) = at(u) + bt(v) = aAu + bAv = A(au + bv)

and hence Ca is a subspace of V.Let v E CA. Then, since s and t commute, we have

t(s(v)] = s[t(v)] = s(Av) = As(v)

from which it follows that Ca is s-invariant.If A is an eigenvalue then Ca; # {0}. If dim Ca, > 1 then since t has

n distinct eigenvalues this would give more than n linearly independentvectors, which is impossible. Hence CA, is spanned by a single vector,v; say. Since Ca, is s-invariant we have s(vi) = uivi for some µ; E F.Hence the matrix of s with respect to the basis {v1i...,v,,} is diagonal.

1.27 There is an integer r _< n with {u, t(u), ... , tr-1(u)} linearly independentand {u, t(u), ... , tr(u)} linearly dependent. Then

tr(u) = aou + alt(u) + + ar-ltr-1(u),

so that, applying t,

tr+1(u) = a+ alt2(u) + . + ar-itr(u),l aot(u)

47


Continuing in this way we see that {u, t(u), ... , tr-1(u)} spans U andso is a basis. By the above argument we have t[t`(u)] _ ajt3(u),so U is t-invariant.

Let f(X)=-ao-a1X- -a,._1X''-1+X'. Then f (X) has degreer and [f(t)](u) = 0. Since U is t-invariant the restriction to of t to Uinduces a linear transformation from U to itself. Now [f (tu)] (u) = 0 so[f(tu)](v) = 0 for all v E U. Hence f(tu) is the zero transformation onU. If now g(X) is a polynomial of degree less than r with g(tu) = 0 then[g(tu)](u) = 0 implies that {u, t(u), ... , t''-1(u)} is dependent. Hencef (X) is the (monic) polynomial of least degree such that f (tu) = 0, sof (X) is the minimum polynomial of tv.

When u = (1, 1, 0) E IR3 and t(x, y, z) = (x + y, x - y, z) we havethat U = (1, 1, 0), (2, 0, 0)}. Also, t2 (u) = (2, 2, 0) = 2u and so f (X) _X2-2.

1.28 (1) Proceed by induction. The result clearly holds for n = 1. In this ex-ample it is necessary, in order to apply the second principle of induction,to include a proof for n = 2

2q = r(s + t)r(s + t)= (rs + rt)2= rsrs + rtrs + rsrt + rtrt

= rsr(s + t)

= pq.

Suppose now that it is true for all r < n where n > 2. Then

qn+1 = qn q = pn-1q2 = pn-1pq = pnq,

which shows that it holds for n + 1. The second equality is establishedin a similar way.

(2) If r is non-singular then r-' exists and consequently from rtr = 0we obtain the contradiction t = 0. Hence r is singular, so both p andq are singular and hence have 0 as an eigenvalue. Consequently we seethat p(X) and q(X) are divisible by X.

(3) Let q(X) = a1X+a2X2+ +a,.Xr. Then a1q+a2g2+ +a,.qr' _0 and so(a1Q + + a,.q'')p = 0, i.e. a1p2 + + arp''}1 = 0 whichshows that p satisfies Xq(X) = 0. Similarly, q satisfies Xp(X) = 0.

By (3), p(X) divides Xq(X), and q(X) divides Xp(X), so we have

Xq(X) = p1(X)p(X), Xp(X) = g1(X)q(X)

48


for monic polynomials pi(X),gl(X). Now X2q(X) = Xpl(X)p(X) _pl(X)gl(X)q(X) so, since q(X) # 0, we have pi(X)gl(X) = V. Con-sequently, either (i) pi(X) = ql(X) = X, or (ii) ql(X) = X2, or(iii) P, (X) = X2.

1.29 Clearly, adding together the elements in the middle row, the middlecolumn, and both diagonals, we obtain

E m{j + 3m22 = 4$,i,a

so that 3$ + 3m22 = 4$ and hence t = 3m22.If m22 = a, ml l = a + Q and m31 = a + 'y then

m21 =3a-m11 -m31 =3a-a-Q-a-7=a-Q-'y,m23 =3a-m21 -m22 =3a-a+p+ry-a=a+p+j,

and so on, and we obtain

a+Q a-P+7 a-7M(a,Q)ry)= a-Q- 1 a a+Q+ry

CL +ry a+Q -ry a-,0

It is readily seen that sums and scalar multiples of magic matricesare also magic. Hence the magic matrices constitute a subspace ofMat3X3(Q). Also,

M(a, Q, 'y) = aM(1, 0, 0) + QM(0,1, 0) + iM(0, 0, 1)

so that B generates this subspace. Since M(a, Q, ry) = 0 if and only ifa = /3 = ry = 0, it follows that B is a basis for this subspace.

That el + e2 + e3 is an eigenvector of f follows from the fact that

1 3a 1

M(a,Q, 7) 1 = 3a = 3a 1

1 3a 1

To compute the matrix of f relative to the basis {el + e2 + e3, e2, e3}we observe that, by the above,

f(el + e2 + e3) = 3a(el + e2 + e3);

49


and that

f (e2) = (a - /3 +'y)e1 + ae2 + (a + /3 - 'y)e3

=(a-/.3+7)(ei+e2+e3-e2-e3)+ae2+(a+p-ry)e3= (a - Q + ry)(el + e2 + e3) + (/3 -'y)e2 + (2Q - 2ry)e3i

f(e3) = (a - ry)el + (a + Q + ry)e2 + (a - /3)e3

=(a-'r)(e1+e2+e3-e2-e3)+(Q+27)e2+ (7-Q)e3=(a-ry)(e1+e2+e3)+(/3+2ry)e2+(7-9)e3

The matrix of f relative to {e1 + e2 + e3, e2, e3) is then

3a a-+ry a - ryL = 0 /3-i /3+2ry

0 2/3-try ry-/3

Since L and M(a, /3, -y) represent the same linear mapping they are sim-ilar and therefore have the same eigenvalues. It is readily seen that

det (L - AI3) = (3a - A)(A2 - 3/32 + 372),

so the eigenvalues are 3a and f 3(/32 - i2).

1.80 If f is nilpotent of index p then fP = 0 and fP-1 0 0. Let x E V besuch that fP-1(x) 54 0 and consider the set

Bp = {x, f(x),...,fp-1(x)}

Suppose that

(*) .lox + A11(x) + ... + AP-1 p-1(x) = 0.

On applying fP-1 to (*) and using the fact that fP = 0, we see thatAo fP-1(x) = 0 whence we deduce that Ao = 0. Deleting the first termin (*) and applying fp-2 to the remainder, we obtain similarly Al = 0.Repeating this argument, we see that each Al = 0 and hence that Bp islinearly independent.

It follows from the above that if f is nilpotent of index n = dim Vthen

B. = {x, f(x),..., fn-1(x)}

50


is a basis of V. The matrix of f relative to Bn is readily seen to be

0 0I* = In-1 0 .

Consequently, if A is an n x n matrix over F that is nilpotent of indexn then A is similar to I. Conversely, if A is similar to I* then there isan invertible matrix P such that P-i AP = I*, so that A = PI*P-1.Computing the powers of A we see that(i) An = 0;

(ii) [An-1]n1 = 1, so An-1 # 0.Hence A is nilpotent of index n.

1.31 To see that V is a Q-vector space it suffices to check the axioms con-cerning the external law. For example,

(a + if)[(ry + iS)x] = (a + i8)[ryx - 51(x)]

= a['Yx- 6f W1 -Qf[7x-Sf(x)]= a'Yx - aS f (x) - ,(dryf (x) - QSx

= (ary -,QS)x - (aS +Q7)f(x)= [(a+i/)(ry+iS)]x.

Suppose now that {V1,...,yr} is a linearly independent subset of thed-vector space V and that in the IR-vector space V we have

Eajva+E# f(Vj)=0.Using the given identity, we can rewrite this as the following equationin the Q-vector space V :

E(aj - if1)v, = 0.

It follows that aj - i3j = 0 for every j, so that a3 = 0 = P j. Conse-quently,

(V1) ...,Vr,f(VI),.... f(Vr)}

is linearly independent in the IR-vector space V. Since V is of finitedimension over IR it must then be so over C. The given identity showsthat every complex linear combination of {v1, ... , vn} can be written asa real linear combination of

V1i...,Vnif(Vl),..., f(Vn).

51


If dime V = n it then follows that dimR V = 2n.By considering a basis of V (over IR) of the form

{vi,...,vn,f(vl),.... f(vn)}

we deduce immediately from the fact that f o f = - idv that the matrixof f relative to this basis is

0 In1

In0

].

Clearly, it follows from the above that if A is a 2n x 2n matrix over IRsuch that A2 = -I2n then A is similar to r. Conversely, if A is similarto r then there is an invertible matrix P such that P-' AP = r andhence A2 = (PrP-1)2 = Pr2P-1 = P(-I2n)P-1 = -12n-

1.82 Let x be an eigenvector corresponding to A. Then from Ax = Ax wehave that xtAt = Axt and hence e T = A. Since A = A and At = -Awe deduce that -Y A = ax . Thus -tAx = -ax x. But we also havex Ax = iAx = Aix. It follows that A = -A, so the real part of A iszero. We also deduce from Ax = Ax that Ax = ax, i.e. that AT = ax,so A is also an eigenvalue.

Y = (A - AI)Z gives Yt = Zt(At - AI) = -Zt(A+ AI) and henceY = -2'(-A + XI) = -Z (A - AI). Consequently,

V Y = -Z (A - AI).(A - AI)Z = 0

since it is given that (A - AI)2Z = 0. Now the elements of V Y are ofthe form

a + ib

[a - ib ... x - iy] = a2 + b2 + +x2+y2x+iy

and a sum of squares is zero if and only if each summand is zero. Hencewe see that Y = 0.

The minimum polynomial of A cannot have repeated roots. For, if thiswere of the form m(X) = (X - a)2p(X) then from (A - aI)2p(A) = 0we would have, by the above applied to each column of p(A) in turn,(A - aI)p(A) = 0 and m(X) would not be the minimum polynomial.

52


Thus the minimum polynomial has simple roots and so A is similar toa diagonal matrix.

Suppose now that Ax = iax. Then Ax = -iax and

Au = A(x+x) =iax-iax=ia(x-x) =av,Av=Ai(x-x)=-ax-ax=-a(x+x) = -au.

These equalities can be written in the form

[Av]- Ia 0][ v]'

The last part follows by choosing is 1 i ... , iak to be the non-zero eigen-values of A.

1.33 Since t satisfies its characteristic equation we have

(t - id)(t + 2id)(t - lid) = 0,

which gives t3 = t2 + 4t - 4 id. It is now readily seen that

t4 = t2 + 4(t2 - id);t6 = t2 +4(1+4) (t2 - id);is = t2 +4(1+4+4 2) (t2 - id).

This suggests that in general

t2p=t2+4(1+4+42+ ..+4p-2)(t2-id).

It is easy to see by induction that this is indeed the case. Thus we seethat

1.34 We have that

t2n = t2 + 4(11 + 4 + - + 4n-2) (t2 - id)=t2+4(1_4n_1)14(t2-id)

= t2 + 3 (4n-1 - 1)(t2 - id).

f(b1) = AIbl;

(i>2) f Qiib1 + E mjibj'j>2

53


Thus the matrix of f relative to the basis {b1, b'2, ... , b,n} is of the form

1 Al 012 Pin

A=0 M

If wEW,say w=w1b1+E1>2w;b; then

g(w) ir[f(w)] = ir(wiAlbl +>wif(b:))i>2

wif(bi)EW,i>2

since b1 E Ker it and it acts as the identity on Im it = W. Thus W isg-invariant.

It is clear that Mat (g', M. Also, the characteristic equationof f is given by det (A - XIn) = 0, i.e. by

(A1 -X)det(M-XIn)=0.

So the eigenvalues of g' are precisely those of f with the algebraic mul-tiplicity of Al reduced by 1. Since all the eigenvalues of f belong to Fby hypothesis, so then do all those of g'.

The last part follows from the above by a simple inductive argument;if the result holds for (n - 1) x (n - 1) matrices then it holds for M andhence for A.

1.35 The eigenvalues of t are 0, 1, 1. The minimum polynomial is eitherX(X - 1) or X(X - 1)2. But t2 - t 54 0 so the minimum polynomial isX(X - 1)2. We have that

V = Ker t ® Ker(t - idv)2.

We must find a basis {w1, w2, w3} with

t(wl) = 0, (t - idv)(w2) = 0, (t - idv)(w3) = Aw2.

A suitable basis is {(-1, 2, 0), (1, -1, 0), (1,1,1)}, with respect to whichthe matrix of t is

0 0 0

0 1 1 .

0 0 1

54


1.36 We have that

t(1) = -5 - 8X - 5X2,t2(1) = -5(-5 - 8X - 5X2) - 8(1 + X + X2) - 5(4 + 7X + 4X2),

t3(1) = 0.

Similarly we have that t3(X) = 0 and 13(X2) = 0. Consequently t3 = 0and so t is nilpotent.

Take vl = 1 + X + X2. Then we have t(vl) = 0. Now take v2 =5 + 8X + 5X2. Then we have

t(v2) = 3(1 + X + X2) E span (vi).

Finally, take v3 = 1 and observe that

t(1) = -5 - 8X - 5X2 E span {vl, v2}.

It is now clear that {1+X+X2, 5+8X+5X2,1} is a basis with respectto which the matrix of t is upper triangular.

1.37 (a) The characteristic polynomial is X2 + 2X + 1 so the eigenvalues are-1 (twice). The corresponding eigenvector satisfies

40

5 -40][y]-[00]

[2

so -1 has geometric multiplicity 1 with [8, 51 as an associated eigenvec-tor. Hence the Jordan normal form is

A Jordan basis can be found by solving

(A+ I2)vl = 0, (A+ I2)v2 = vi.

Take vi = [8, 5]. Then a possible solution for v2 is [5, 3], giving

55


(b) The characteristic polynomial is (X+1)2. The eigenvalues are -1(twice) with geometric multiplicity 1, and a corresponding eigenvectoris [1, 01. The Jordan normal form is

A Jordan basis satisfies

(A+I2)v1=0, (A+I2)v2=v1

Take v1 = [1, 0] and v2 = [0, -1]; then

P=

[1O

(Any Jordan basis is of the form {[c, 0], [d, -c]} with P =L

c dl0

-c

(c) The characteristic polynomial is (X - 1)3, so the only eigenvalue is1. It has geometric multiplicity 2 with {[I, 0, 01, [0, 2,3]) as a basis forthe eigenspace. The Jordan normal form is then

1 1 0

0 1 0 .

0 0 1


(A - I3)vl = 0, (A - I3)v2 = v1, (A - I3)v3 = 0.

Now (A- I3)2 = 0 so choose v2 to be any vector not in ([1, 0, 01, [0, 2, 3]),for example v2 = [0, 1, 0]. Then v1 = (A-I3)v2 = [3, 6, 9]. For v3 chooseany vector in ([1, 0, 0], [0, 2,31) that is independent of [3,6,9], for exampleV3 = [1, 0, 0]. This gives

3 0 1

P= 6 1 0 .

9 0 0

56


(d) The Jordan normal form is

3 1 0

0 3 0 .

0 0 3


(A-313)vl = 0, (A-313)v2=v1i (A-313)v3=0.

Choose v2 = [0, 0, 11. Then vl = [1, 0, 0] and a suitable choice for v3 is[0, 1, 0]. Thus

1 0 0P= 0 0 1

0 1 0

1.88 The characteristic polynomial of A is (X- 1)'(X-2)1. For the eigenvalue2 we solve

0 1 1 1 0 x 0

0 0 0 0 0 y 0

0 0 0 1 0 z 0

0 0 0 -1 1 t 0

0 -1 -1 -1 - 2 w 0

to obtain w = t = 0, y + z = 0. Thus the general eigenvector associatedwith the eigenvalue 2 is [x, y, -y, 0, 01 with x, y not both zero. TheJordan block associated with the eigenvalue 2 is

[ 0

2

2]

For the eigenvalue 1 we solve

1 1 1 1 0 x 0

0 1 0 0 011

0

0 0 1 1 0 z 0

0 0 0 0 1 t 0

-0 -1 - 1 -1 - 1 w 0

57


to obtain w = y = x = 0, z + t = 0. Thus the general eigenvectorassociated with the eigenvalue 1 is [0, 0, z, -z, 0] with z # 0. The Jordanblock associated with the eigenvalue 1 is

1 1 0

0 1 1 .

0 0 1

The Jordan normal form of A is therefore

2 0 0 0 0

0 2 0 0 0

0 0 1 1 0

0 0 0 1 1

0 0 0 0 1

Take [0, 0, 1, -1, 0] as an eigenvector associated with the eigenvalue 1.Then we solve

1 1 1 1 0 x 0

0 1 0 0 0 y 0

0 0 1 1 0 z = 1

0 0 0 0 1 t -10 -1 -1 -1 -1 w 0

to obtain y = 0, w = -1, z + t = 1, x = -1, so we take [-1, 0, 0, 1, -11.Next we solve

1 1 1 1 0 x -10 1 0 0 0 y 0

0 0 1 1 0 z = 0

0 0 0 0 1 t 1

0 -1 -1 -1 -1 w -1

to obtain y = 0, t + z = 0, w = 1,x = -1, so we consider [-1,0,0,0,11.A Jordan basis is therefore

{[1, 0, 0, 0, o], [0, 1, -1,0,011 10,0, 1,-1, o], [-1,0,0, 11 -1], [-1101010, ill

and a suitable matrix is1 0 0 -1 -1

0 1 0 0 0

P= 0 -1 1 0 0

0 0 -1 1 0

0 0 0 -1 1

58


1.39 (a) The Jordan form and a suitable (non-unique) matrix P are

J=2 1 00 2 0 , P=

2 -5 5

2 -3 8 .

0 0 2 3 8 -7

(b) The Jordan form and a suitable P are

2 0 0 0 4 3 2 07

0 1 1 0P=

5 4 3 0

0 0 1 1 -2 -2 -1 0

0 0 0 1 11 6 4 1j

1.40 The Jordan normal form is

2 0 0 0 0

0 2 1 0 0

0 0 2 0 0

0 0 0 3 1

0 0 0 0 3

1.41

A Jordan basis is

{[2, 1, 0, 0,1], [1,0,1,0,0], [0, 1, 0,1, 0], [-1,0,0,1,0], [2, 0, 0, 0, 1]j.

The minimum polynomial is (X - 2)3. There are two possibilities forthe Jordan normal form, namely

2 1 0 0 0 2 1 0 0 010 2 1 0 0 0 2 0 0 0

J1= 0 0 2 0 0, J2 = 0 0 2 1 0

0 0 0 2 0 0 0 0 2 1

0 0 0 0 2 0 0 0 0 2

Each of these has (X - 2)3 as minimum polynomial. There are twolinearly independent eigenvectors; e.g., [0, -1,1,1, 0] and [0, 1, 0, 0, 11.The number of linearly independent eigenvectors does not determinethe Jordan form. For example, the matrix J2 above and the matrix

2 0 0 0 0

0 2 1 0 0

0 0 2 1 0

0 0 0 2 1

0 0 0 0 2

59


1.42

have two linearly independent eigenvectors. Both pieces of informationare required in order to determine the Jordan form. For the given matrixthis is J2.

A basis for IR4[X] is {1,X,X2,X3} and D(1) = 0,D(X) = 1,D(X2) _2X, D(X3) = 3X2. Hence, relative to the above basis, D is representedby the matrix

0 1 0 0

0 0 2 0

0 0 0 3

0 0 0 0

The characteristic polynomial of this matrix is X4, the only (quadruple)eigenvalue is 0, and the eigenspace of 0 is of dimension 1 with basis {1}.So the Jordan normal form is

f0 1 0 0

0 0 1 0

0 0 0 1

0 0 0 0

1.43

A Jordan basis is {fl, f2, f3, f4} where

Df, = 0, Df2 = fl, Df3 = f2, Df4 = f3.

Choose f1 = 1; then f2 = X, f3 = 2X2, f4 = 6X3 so a Jordan basis is{6,6X,3X2,X3}.

The possible Jordan forms are

3 0 0 3 1 0 3 1 0 3 0 0

0 3 0 , 0 3 1 0 3 0 , 0 3 1.0 0 3 0 0 3 0 0 3 0 0 3

The last two are similar.

1.44 (i) and (ii) are true : use the fact that AB and BA = A-' (AB)A aresimilar.

(iii) and (iv) are false; for example,

1 100][0 0]

and[0 0][0 01

0[0

clearly have the same Jordan normal form.

60


1.45 V decomposes into a direct sum of t-invariant subspaces, say V =Vi ® . ® V,., and each summand is associated with one and only oneeigenvalue of t. Without loss of generality we can assume that t has asingle eigenvalue A. Consider an i x i Jordan block. Corresponding tothis block there are i basis elements of V, say v1i ... , vt, with

(t - A idv)vl = 0;

(t-Aidv)2v2 = (t-Aidv)v1 =0;

(t-Aidv)ivi=(t-Aidv)i-1vt_i = ...=0.Thus there is one eigenvector associated with each block, and so there

are

dim Ker(t - A idv )

blocks.Consider Ker(t - A idv)j. For every 1 x 1 block there corresponds

a single basis element which is an eigenvector in Ker(t - A idv)j. Forevery 2 x 2 block there correspond two basis elements in Ker(t - A idv)jif j > 2 and 1 basis element if j < 2. In general, to each i x i blockthere correspond i basis elements in Ker(t - A idv)' if j > i and j basiselements if j < i.

It follows that

1.46

dj=n1+2n2+-..+(j-1)nj-1+9(nj+nj+l+...)

and a simple calculation shows that 2di - dt_1 - di+1 = ni.

The characteristic polynomial of A is (X - 2)4, and the minimum poly-nomial is (X - 2)2. A has a single eigenvalue and is not diagonalisable.The possible Jordan normal forms are

2 1 0 01 2 1 0 0

0 2 0 0 0 2 0 0

0 0 2 0 0 0 2 1

0 0 0 2 0 0 0 2

Now dimIm(A - 214) = 1 so dim Ker(A - 214) = 3 and so the Jordanform is

2 1 0 0

0 2 0 0

0 0 2 0

0 0 0 2

61


Now Ker(A - 214) = {[x, y, z, t] I 2x - y + t = 0}, and we must choosev2 such that (A - 214)2v2 = 0 but v2 0 Ker(A - 214). So we takev2 = [1,0,0,0], and then v1 = (A - 214)v2 = [-2,-2, -2,21. We nowwish to choose v3 and v4 such that {vi, v3i v4} is a basis for Ker(A-214).So we take v3 = [0, 1, 0, 1] and v4 = [0, 0, 1, 0]. Then we have

P=-2 1 0 0

-2 0 1 0

-2 0 0 1

2 0 1 0

To solve the system X' = AX we first solve the system Y' = JY,namely

yi = 2yi + Y2y2' = 2y2

ys=2y3y4 = 2y4

The solution to this is clearly

Zty4 = c4e

Since now

y3=c3e2t

y2 = c2e2t

Yi = c2te2t +cle2t.

-2 1 0 0 c2te2t + cie2t

X=PY= -2 0 1 0 c2e2t

L

-2 0 0 1 c3e2t

2 0 1 0 c4e2t

we deduce thatxi = -2c2te2t - 2c1e2t + c2e2t

x2 = -2c2te2t - 2c1e2t +C3e2t

x3 = -2c2te2t - 2cie2t + C4e2t

x4 = 2c2te2t + 2cie2t + c3e2t.

62


1.47 (a) The system is X' = AX where

`4 = [-1 0].

The characteristic polynomial is (X - 1)(X - 4). The eigenvalues aretherefore 1 and 4, and associated eigenvectors are El = [1, -1] andE4 = [4, -1]. The solution is aEjet + bE4e4t, i.e.

xl = aet + 4be4t

x2 = -aet - befit

(b) The system is X' = AX where

4 -1 -1A= 1 2 -1 .

1 -1 2

The characteristic polynomial is (X - 3)2(X - 2). The eigenvaluesare therefore 3 and 2. An eigenvector associated with 2 is [1,1,1] sotake E2 = [1,1,1]. The eigenvalue 3 has geometric multiplicity 2 and[1, 1, 0], [1, 0, 1] are linearly independent vectors in the eigenspace of 3.The general solution vector is therefore

all, 1,1]e2t + b[1,1, 0]e3t + c[l, 0,1]e3t

so thatxi = ae2t + (b + c)e3t

x2 = ae2t + be 3t

x3 = ae2t + ce3t.

(c) The system is X' = AX where

5 -6 -6A= -1 4 2 .

3 -6 -4

The characteristic polynomial is (X - 1)(X - 2)2. The eigenvalues aretherefore 1 and 2. An eigenvector associated with 1 is El = [3, -1,31,

63


and independent eigenvectors associated with 2 are E2 = [2, 1, 0] andE2 = [2, 0, 1]. The solution space is then spanned by

{[3et, -et, 3et], [2e2t, e2t, 0], [2e2t, 0, e2t]}.

(d) The system is X' = AX where

1 3 -2

A= 0 7 -4 .

0 9 -5

Now A has Jordan normal form

1 1 0

J= 0 1 0

0 0 1

and an invertible matrix P such that P-1 AP = J is

P=3 0 1

6 1 0 .

9 0 0

First we solve Y' = JY to obtain y' = yi + y2, y2 = Y2, y3 = yshence

1.48

ys = cett

ya = be

yl = btet + aet.

Thus X' = AX has the general solution

X = PY = a [3, 6, 9]e t + b([3,6, 9] tet + [0, 1, 0] et) + c[1, 0, 0] et .

The system is AY' = Y where

Now A is invertible with

1 -3 2

A= 0 -5 4 .

0 -9 7

1 3 -2A-' = 0 7 -4

0 9 -5

and the system Y' = A-'Y is that of question 1.47(d).

and

64


1.49 The system is X' = AX where

The eigenvalues are 2 + v and 2 - f, with associated eigenvectors

[-1, -1 - f] and [-1, -1 + f]. The general solution is

a[-1, -1 - v"3]e(2+-,13-)t + b[-1, -1 + V]ei2-flt.

Since x1(0) = 0 and x2(0) = 1 we have a+b = 0 and a- f a-b+sb =

1, giving a = -2

and b = 2 , so the solution is

3te2t f t2Nf3-

+ [-1, -1 + -V3]e- ).([1,1 + V31e

1.50 Let x = xl,xi = x2,4' = x2 = X3, X'1" = x3 = 2x3 + 4x2 - 8x1. Then

the system can be written in the form X' = AX where

0 1 0

A= 0 0 1 .

-8 4 2

The characteristic polynomial is (X - 2) (X2 - 4) so the eigenvalues are2 and -2. The Jordan normal form is

2 1 0

J= 0 2 0 .

0 0 -2

A Jordan basis {v1i v2, v31 satisfies

(A - 213)vl = 0

(A - 213)v2 = VI

(A+213)v3 = 0.

Take v1 = [1, 2, 4] and v3 = [1, -2, 4]. Then v2 = [0, 1, 4]. Hence aninvertible matrix P such that P-1AP = J is

1 0 1

P= 2 1 -2 .

4 4 4

65


Now solve the system Y' = JY to get

yi = 2yl + y2, y2= 2y2, y3 = -2y3

so that Y2 = c2e2t, y3 = c3e-2t and hence yl = 2yi + c2e2t which givesyl = cle2t + c2te2t.

Now observe that

1 0 1

X=PY= 2 1 -24 4 4

cle2t + c2te2t

c2 e2t

c3e-2t

Hence x = xl = cl e2t + c2te2t + c3e-2t. Now apply the initial conditionsto obtain

x = (4t - 1)e2t +e-2t.

66


2.1 (a) f H f' does not define a linear functional since f' 0 IR in general.(b),(c),(d) These are linear functionals.(e) 99 : f F-+ 1J f 2 is not a linear mapping; for example, we have

0 = t9[ f + (-f)] whereas in general

1,O(,f)+0(-f)=2 f f2#0.0

2.2 That cp is linear follows from the fact that

<p(a f +'8g) = f fo (t) [a.f (t) + Q9(t)] dt1

0

= a fo fo(t) f (t) dt +,Q f 1 .fo (t) 9(t) dt1

0

= c (f) + &M.

2.3 The transition matrix from the given basis to the standard basis is

1 -1 0P= 0 1 1

-1 0 1

The inverse of this is readily seen to be

2 2 2

1 1 1

2 2 -2P-1 = _ 1 1 1

2 2 21 1 1


Hence the dual basis is

{[2, 2, 21, [-2, 2, 21, [2, 2]}.

2.4 The transition matrix from the basis A' to the basis A is

P-I3 4].

Its inverse isp-1 = [-2 -2 ].

Consequently, (A')d = {-2<pl + 21(P2, rpl - 2 cp2}.

2.5 The transition matrix from the given basis to the standard basis is

P=

Its inverse is

P-1- 2

1]so the dual basis is (b), namely { [-1, 0], [2, 1] }.

2.6 (i) {[2, -1, 1,0j,[7,-3,1,-1], [-10,5,-2, 1], [-8,3,-3, 1]};(ii) {(4,5,-2,11),(3,4,-2,6), (2, 3, -1,4), (0, 0,0, 1)}.

2.7 Since V = A (D B we have

Al + B-L = (A n B)1 = {0}1 = Vd

Al n B1 = (A + B)1 =V1 = {0}.

Consequently, Vd = Al ® B1.The answer to the question is `no' : Ad is the set of linear functionals

f : A -> F so if A $ V we have that Ad is not a subset of Vd. What istrue is : if V = A ® B then V d = A' ® B' where A', B' are subspacesof Vd with A' Ad and B' Bd. To see this, let f E Ad and definef : V -+ F by f (v) = f (a) where v = a + b. Then V : Ad --> Vdgiven by o(f) = f is an injective linear transformation and V(Ad) is a

68


subspace of Vd that is isomorphic to Ad. Define similarly µ : Bd -+ Vdby µ(g) = g where g(v) = g(b). Then we have

2.8

V d = co(Ad) ® µ(Bd).

{ fl, f2i f3} is linearly independent. For, if Al f1 + A2f2 + )t3f3 = 0 thenwe have

0 = (A1f{1 + A2f2 + A3f3)(1) = Al + A2 + A3;

0 = (A1!1 +A2f2 +A3f3)(X) _ Alt1 +A2t2+A3t3;0 = (A1f1 + A2 f2 + 3f3)(X2) _ ltl + A2t2 + )3t3.

Since the coefficient matrix is the Vandermonde matrix and since the t;are given to be distinct, the only solution is Al = A2 = A3 = 0. Hence{ fl, f2i f3} is linearly independent and so forms a basis for (IR3[X])d

If {p1i p2i p3} is a basis of V of which {fl, f2i f3} is the dual then wemust have fg(p3) = Sgt; i.e.

pi(ts) = Std.

It is now easily seen that

pl(X) - (tl - t2)(tl - t3)' p2(X) - (t2 - tl)(t2- t3)'

p3(X) - ( t2)

2.9 (a) a^ ('p) _ cp(a) = [ 3

(b) Q" ('P) = p(Q) = [ 3

4][2]= 11;

4][6]= 39;

(c) (2a+3Q)^(,p) = p(2a+3Q) _ [3 4](2I 2]+3[6] I = 139;

(d)(2a+3Q)^([a b])=[a b](2{ ]+3I6]l=17a+22b.6\\\9 L J


2.10 Let V be of dimension n and S of dimension k. Take a basis {v1,. .. , vk}of S and extend it to a basis

{V1,...,vk,vk+1,.... vn}

of V. Let {(p1,...,Vn) be the basis of Vd dual to {v1i...,v.}. GivencpEVd we have

p= al p, + ... + a.Pn -

Since V(vi) = a; we see that <p(v) = 0 for every v E S if and only ifa1 = = ak = 0. Thus

<p E S1 (p = ak+1(Pk+1 + + anPn

and so {cpk+1,... , V , , is a basis for S1. Hence dim S + dim S1 = n.If 0 E Kertd then td(O) = 0 and so [td(ji)](u) = 0 for all u E U. But

[td('O)](u) = ['b(t)](u) =O[t(u)]

Thus 0 E (Imt)1. Conversely, let V E (Imt)1-. Then for all u E U wehave

[td(o)](u) = si[t(u)] = 0

and so td(1b) = 0 whence 0 E Kertd. Thus Kert' = (Imt)1.(i) means that v c Imt while (ii) means v f (Kertd)1. In terms of

linear equations, this says that either (i) the system AX = B has asolution, or (ii) there is a row vector C with CA = 0 and CB = 1.

It is readily seen that the given system of equations has no solution,so (ii) holds. The linear functional satisfying (ii) is [1, -2, 1].

2.11 If O E Wd then for all u E U we have

(S o t)d(<,)(u) ='P[(S o t)(u)] = ps[t(u)] = SdOP)[t(u)] = [td(SdOP))](u)

from which the result follows. The final statements are immediate fromthe fact that Imt = (Kertd)1 (see question 2.10).

2.12 To find Y we find the dual of {[1, 0, 0], [1, 1, 0], [1,1,1]}. The transitionmatrix and its inverse are

1 1 1 1 -1 0P= 0 1 1, P'1= 0 1 -1 .

0 0 1 0 0 1

70


Thus Y = {(1, -1, 0), (0, 1, -1), (0, 0,1)}.The matrix of t with respect to the standard basis is

2 1 0

1 1 1

0 0 -1

and the transition matrices relative to X, Y are respectively

1 0 0 1 1 1

-1 1 0 , 0 1 1.0 -1 1 0 0 1

By the change of basis theorem, the matrix of t relative to X, Y is then

1 0 0 2 1 0 1 1 1 2 3 3

1 1 0 1 1 1 0 1 1 = 3 5 6.1 1 1 0 0 -1 0 0 1 3 5 5

The required matrix is then the transpose of this one.

2.13 The annihilator (al, a2 ) 1 is of dimension 1 and contains both V andcp'3. Thus cp3 is a scalar multiple of (p3.

2.14 It is immediate from properties of integrals that

Lg(.\lfl +)'2f2) = A,Lg(fl) +A2Lg(f2)

and so Lg is linear.To show that Fx is a linear functional, we must check that

Fx(Alfl +A2f2) = )1F.(fl)+A2FF(f2),

i.e. that (A1f1 + A2f2)(x) = A1fl(x) + )t2f2(x), which is clear.Suppose that x is fixed and that Fy = Lg for some g. Then

I

W E C[0,11) f f(t) g(t) dt = f(x)0

Now, by continuity, the left hand side depends on the values of f atpoints other than x whereas the right hand side does not. Thus Fr 54 Lgfor any g.

71


2.15 The method of solution is outlined in the question.

2.16 Let {v1,. .. , vk_1 } be a basis of U and extend this to a basis { v 1 , . . . , vk}of V. Apply the Gram-Schmidt process to obtain an orthonormal basis{u1,...,uk} of V, where {ui,...,uk_1} is an orthonormal basis of U.Let n = Uk. Then if x = _i=1 a;uj E U we have

k-1(njx) = (ukJAlul + ... +Ak-luk-1) = E Ai(uklui) = 0-

i=1

Conversely, if x = l aiui E V and if (nix) = 0 then

k0 = (uki Al ul + ... + Akuk) Ai (uk lui) = Ak.

i=1

Consequently we see that x E U and so

U= {x E V I (nix) = 0}.

Now v - v' = 2(nly )n, a scalar multiple of n, and so is orthogonal toU. Then 1(v + v') = v - (nl v) n E U since

(nlv- (nlv)n) = (niv) - (niv) (nIn)=(niv)-(niv)=0.

We have

t(v+w) = v + w -2(nly+w)n=v+w-2((nly)+(nlw))n= (v - 2(nlv)n) + (w - 2(nlv)n)= t(v) + t(w);

t(Av) = Av - 2(nlav )n

=Av-2A(nlv)n= At(v),

and so t is linear.Note that t(u) = u for every u E U and so 1 is an eigenvalue and

Ker(t - id) = U is of dimension k - 1. Thus 1 has geometric multiplicityk - 1. Also, t(n) = -n and so -1 is also an eigenvalue with associatedeigenvector n. Since the sum of the geometric multiplicities is k it followsthat t is diagonalisable.

72


In the last part we have that n = it (3, -1,1) so if v = (a,b, c) then(n,v) = iI(3a-b+c). Thus

s(a, b, c) = (a, b, c) - ii (3a - b + c)(3, -1,1)

= 1j-(-7a + 6b - 6c, 6a + 9b + 2c, -6a + 2b + 9c),

which gives1 -7 6 -6

mats = - 6 9 2 .11 _6 2 9

As for t, we have n = 711-0 (2, -1,2, -1) and so if v = (a, b, c, d) then

t(a, b, c, d) _ (a, b, c, d) -5

(2a - b + 2c - d)(2,-1,2, -1)= 6(a+2b-4c+2d,2a+4b+2c-d,

-4a+2b+c+2d,2a-b+2c+4d),

which gives

1 1 2 -4 21

2 4 2 - 1matt = 5 -4 2 1 2

I 2 - 1 2 4

2.17 The hyperbola is represented by the equation

Lx yJ[ 3 -11[y]= 1.

The eigenvalues of A = [ 3 -11 are 2 and -4 with associated nor-

malised eigenvectors

[1/v, 1/V1/-,42

1/vf2

Thus PAP = diag{2, -4} where

P1/f -1//1// 1/f73


Now change coordinates by defining

xl - [ X, IPt 1

X I= Ptx.

Y1 y

Then we have x = Pxj and the original equation becomes

1 = xt Ax = (Pxl )t A(Pxl )

=xiPtAPxj

_ [xi yi ]I 0O4lIx

yi,= 2z2 - 4y1.

Thus the principal axes are given by xl = 0 and yl = 0, i.e. y = -xand y = x.

The ellipsoid is represented by the equation

[x y z J

7 2

6

0 -2

0

5

][;].z

7 2 0

The eigenvalues of A = 2 6 -2 a re 3, 6, 9 with associated nor-

malised eigenvectors

0 -2 5

-1/3 2/3 2/3

2/3 , -1/3 , 2/3

2/3 2/3 -1/3

Thus Pt AP = diag{3, 6, 9) where

-1 2 2

P=3 2 -1 2 .

2 2 -1

Now change coordinates by defining

xl x

xi = Yi =

pt

y = Ptx.

zl z

74


Then we have x = Pxj and the original equation becomes (with acalculation similar to the above)

X2 2 2+ 6y1 + 9z1 = 1.

Sincex1 =

s(-x + 2y + 2z),

y1 =s

(2x - y + 2z),z1 =

s(2x + 2y - z),

the xl-axis is given by y1 = z1 = 0 and has direction numbers (-1, 2,2);the yl-axis is given by x1 = z1 = 0 and has direction numbers (2,-1,2);the z1-axis is given by x1 = y1 = 0 and has direction numbers (2, 2, -1).

2.18 Let {v1,.. . , vn} be an orthonormal basis of V. Then for every x E Vwe have x = >'=1(xIvk ) vk so in particular

(i = i,...,n)

n

f(vi)=

E(f(vi)Ivk)vk.k=1

If A is the matrix of f we thus see that a,;k = (f(v?)Ivk ). If M is thematrix of f* then likewise we have that mjk = (f*(vj)Ivk ). The resultnow follows from the observation that

m3k = (f*(vj)Ivk) = (vklf*(vj))

= (f(vk)Ivj)= akj.

2.19 The first part is a routine check of the axioms. Using the fact thattr(AB) = tr(BA) we have

(fM(A)IB) = tr[B*(MA)] = tr[MAB*]

= tr[B*MA]

= tr[(M*B)*A]

= (AI fM-(B) )

from which it follows that (fM)* = fm*.

75


2.20 Let Q be as stated and let fp E V* be given by fp(a) = (aJQ ). Thensince {al, ... , an} is an orthonormal basis we have

n

fp(ak) _ (akI E f (ai)ai) _ .f (ak).i=1

Since fp and f coincide on this basis, it follows that f = fp.For the next part suppose that such a q exists. Then we have

(t) q(t) dt1P(z) = f (P) = fo P

for every p. Writing rp for p we then have

f 1 r(t) p(t) q(t) dt = r(z)p(z) = 0.

In particular, this holds when p = rq. So

fo1 Ir(t)I2 Iq(t)I2 dt = 0

whence we have rq = 0. Since r 54 0 we must therefore have q = 0, acontradiction.

For the next part of the question note that

(fp(q)Ir) = (pglr)4= f 1 p(t) q(t) r(t) dt

0

=0f 1 q(t) p(t) r(t) dt

_ (glpr)_ (ql fp(r) )

so (fP)* = f.Integration by parts gives

(D(P)I q) = P(1)q(1) - P(0)q(0) - (PID(q) ).If D* exists then we have

(PI D*(q)) = P(l)q(1) - P(0)q(0) - (PI D(q) )so that

(PI D(q) + D*(q)) = P(1)q(1) - P(0)q(0)Now fix q such that q(0) = 0, q(1) = 1. Then we have

(PID(q) + D*(q)) = p(1),which is impossible (take z = 1 in the previous part of the question).Thus D* does not exist.

76


2.21 That K is self-adjoint follows from the fact that xy is symmetric in xand y.

We have

K(fn) = f xy fn(y) dy1

0

- f xy (yn2

n+2) dy0

1 1

= f xyn+1 dy - 2

n 2 2 Jo xydyxyn+2 Y-1

2xy2I!-1

n+2 x-0 n+2 2 y=o

= 0,

so K(fn) = Of,, as required.Apply the Gram-Schmidt process to {f1, f2). Let e1(x) = f, (x) _

x - 3 and define e2(x) = f2(x) + af1(x) _ (x2 - 2) + a(x - 3 where

(x2- 2, x-3a=-

(x-2 ,x-2)3 3

Since

(x2-a,x-3)- f (x2 )(x )dx-9,0

(x-3,x-3f 3)2)=(x - dx=1

it follows that a = -1 and hence that

e2(x) = x2 - x + g.

Thus e1, e2 are orthogonal eigenfunctions associated with the eigenvalue0.

If K(f) = Af then

1

Xf(x) = x fo yf(y) dy0

If A # 0 then f must then be of the form x '-+ ax for some constant a.Substituting ax for f (x), it clearly follows that A = 3. An associatedeigenfunction is given by f (x) = x.

77


2.22 Since t is skew-adjoint its eigenvalues are purely imaginary so ±1 arenot eigenvalues. Hence id ±t cannot be singular. Now

s* = (id +t*)-1(id -t*) = (id -t)-' (id +t).

But (id+t)(id-t) = id -t' = (id-t)(id+t) so

(id-t)-1(id+t) _ (id+t)(id-t)-1.

Hence s* = (id+t)(id-t)-1 and so

ss* = (id-t)(id+t)-1(id+t)(id-t)-1 = id.

To see that s cannot have -1 as an eigenvalue, consider

id +s = id +(id -t) (id +t)-1.

We have that

(id +s) (id +t) = (id +t) + (id -t) = 2 id,

and so (id +s)-1 = a (id +t) whence id +s does not have 0 as an eigen-value and hence -1 is not an eigenvalue of s.

2.23 St = S and Tt = -T, so

(T + iS)t = (T - iS)t = Tt - iSt = -(T + iS).

Thus T + IS is skew-adjoint. But the eigenvalues of a skew-adjointmatrix are purely imaginary, so 1 is not an eigenvalue of T + IS, so

det(T + iS - I) # 0.

As for the second part, we have

U = (I + T + iS)(I - T - iS)-1_ [I + (T + iS)] [I - (T + iS)]-i.

The fact that U is unitary now follows from the previous question.

78


2.24 It is given that At = A, St = -S, AS = SA,det(A - S) # 0. LetB = (A + S)(A - S)-1. Then we have

BtB = [(A + S)(A - S)-1]t(A+ S)(A - S)-1= [(A - S)-'It (A+S)t(A+S)(A- S)-1= (At - St)-' (At + St)(A + S)(A - S)-1= (A+S)-1(A- S)(A + S)(A - S)-1_ (A + S)-1(A + S)(A - S)(A - S)-1,

the last equality following from the fact that since A, S commute so doA + S and A - S. Hence Bt B = I and B is orthogonal.

2.25 Since

(1)

we have, taking transposes,

AA+A=0

AtA+At=0

and hence, taking complex conjugates,

(2) AA+A =0.

It follows from (1) and (2) that A = A, so that A is self-adjoint. LetA1, ... , An be the distinct non-zero eigenvalues of A. Then Al, ... , Anare necessarily real.

The relation (1) can now be written in the form

A2=-A

from which it follows that the distinct non-zero eigenvalues of -A,namely are precisely the distinct non-zero eigenvalues ofA2, namely A I. ... , An. It follows that Al, ... , An are all negative. Letai = -Ai for each i, and suppose that.

a1 < a2 < ... < an.

Then this chain must coincide with the chain

ai < a2 < < an.

Consequently, a; = a for every i and, since by hypothesis ai # 0, weobtain A; = -ai = -1.

79


2.26 A and B are given to be orthogonal with det A + det B = 0. Now wehave that

(1) det(A + B) = det[(ABt + I)B]

from which we see that det(A+B) = 0 if and only if -1 is an eigenvalueof ABt. But ABt is orthogonal since

(ABt)t(ABt) BAtABt = I.Also, det(ABt) = det Adet B-1 = -1 since it is given that detA =- det B. Thus C = ABt is orthogonal with det C = -1. It follows that-1 is an eigenvalue of C; for

Ct(I+C) = Ct +I= (C+I)tand so, taking determinants,

det(I+ C)[det C - 1] = 0whence det(I+ C) = 0. It now follows from (1) that det(A + B) = 0.

2.27 (1) Since At(A- I) = I - At = -(At - I) we have thatdet Adet(A - I) = (-1)' det(A - I)

and sodet(A - I)[det A - (-1)'] = 0.

If det A = 1 and n is odd then it follows that det(A - I) = 0 and hence1 is an eigenvalue of A. If det A = -1 and n is even then likewisedet(A - I) = 0 and again 1 is an eigenvalue of A.

(2) Since At(I + A) = At + I = (I + A)t we have that

det A det(I + A) = det(I + A)

and so if det A = -1 then det(I+ A) = 0 whence -1 is an eigenvalue ofA.

2.28 The first part follows from the observation that

g(A) = 0 g(At) = 0 g(-A) = 0.Suppose now that g(X) is the minimum polynomial of A, say

g(X) = ao + a1X + ... + ar_1Xr-1 + Xr.Since g(-A) = 0 we have that

ao - a1X + ....+ (-1)rXris also the minimum polynomial of A. Thus a1 = a3 = = 0.

Since (A")t = (-1)' A for a skew-symmetric matrix A we see that A'is skew-symmetric if n is odd, and is symmetric if n is even. Hence f(A)is skew-symmetric, and g(A) is symmetric.

80


2.29 We haveN(UA) = tr(UA UA)

= tr(A U UA)

= tr(A A)

= N(A).

Similarly, using the fact that tr(XY) = tr(YX),

N(AU) = tr[(AU)tAU]

= tr[AU(AU)t]

= tr(AUU A )

= tr(AA )

= N(A).

Finally, by the above,

N(In - U-1A) = N[U(In - U-'A)]=N(U-A)= N(A - U).

2.80 Since TA= AT we have that A-1(A)-1 = (T)-'A-1. But

A-'A =I==>. A-'A =IAA-1 =A-1.It follows that

A-1A l= A-1(A)-1 = /A)-1A-1 = X---41A-1

and so A-1 is normal.l l

If A = aoI + a, A + + an An then clearly AA = A A. Supposeconversely that A is normal. Then there is a unitary matrix P and adiagonal matrix D such that

A=P-1DP=P DP, A =P DP=P-1DP.Let A1, ..., A,. be the distinct elements of D. Consider the equations

1 = ao + al A, + a2A1 + ... + a,._1 Jai-1

A2 =ao+a1A2+a2.\22 +

ao + a,,\, + a2A2 r + ... + a,.-1r-1.

81


Since A1,...,A,. are distinct the (Vandermonde) coefficient matrix hasnon-zero determinant and so the system has a unique solution. We thenhave

D=aoI+a1D+a2D2+ +a,._1D''-1

and consequently

A =P-1DP=P-1(aoI+a1D+ +a,._1D''-1)P=aoI+ajA+ +a,._1A''-1.

2.31 Suppose that A is normal and let B = g(A). There is a unitary matrixP and a diagonal matrix D such that

A=P DP=P-1DP.

Consequently we have

B = g(A) = P g(D)P = P-lg(D)P,

and so

and similarlyBB = P g(D)g(D)P.

Since g(D) and g(D) are diagonal matrices, it follows that B B = BBand so B is normal.

2.32 We have that

(A + Bi)*(A + Bi) = (A* - B*i)(A + Bi)

= (A - Bi)(A+ Bi)= A2 - (BA - AB)i + B2,

and similarly (A + Bi)(A+ Bi)* = A2 - (AB - BA)i + B2. It followsthat A + Bi is normal if and only if AB = BA.

2.33 To get -A multiply each row of A by -1. Then clearly det(-A) _(-1)" det A. If n is odd then

det A = det(At) = det(-A) _ (-1)" det A = - det A

and so det A = 0.

V B = P g(D)PPg(D)P = Pg(D)g(D)P

82


Since xtAx is a 1 x 1 matrix we have

xtAx = (xtAx)t = xtAtx = -xtAx

and so xt Ax = 0.Let Ax = Ax and let stars denote transposes of complex conjugates.

Then we have x*Ax = Ax*x. Taking the star of each side and usingA* = At = -A, we obtain

ax*x = (x*Ax)* = x*A*x = -x*Ax = -Ax*x.

Since x*x # 0 it follows that A = -A. Thus A = iu where p E IR \ {0}.If x = y + iz then from Ax = ipx we obtain A(y + iz) = ip(y + iz)

and so, equating real and imaginary parts, Ay = -pz, Az = py. Now

pyty = ytAz = (ytAz)t = ztAty = -ztAy = pztz

and so yty = ztz. Also, pytz = -ytAy = 0 (by the first part of thequestion). If, therefore, At = 0 then

put y = utAz = -(Au)tz = 0

and similarlyputz = -utAy = (Au)ty = 0.

For the last part, we have

det(A - AI) = det-A 2 -2

-2 -A -1 = -A(A2 + 9),2 1 -A

so the eigenvalues are 0 and ±3i.A normalised eigenvector corresponding to 0 is

1 -1U=3 2 .

2

To find y, z as above, choose y perpendicular to u, say

83


Then we have

which gives

-4-3z=Ay= 1 -1

1

fRelative to the basis {u, y, z} the representing matrix is now

0 0 0

0 0 3 .

0 -3 0

The required orthogonal matrix P is then

-1/3 0 4/3fP = 2/3 -1/f 1/3f

2/3 1/f 1/3f2.84 Let Q be the matrix that represents the change to a new basis with

respect to which q is in normal form. Then xtAx becomes ytBy wherex = Qy and B = QtAQ. Now

q(x)=9(y)=yz 1+...+ypz -yp+21-...-ypa+m

where p - m is the signature of q and p + m is the rank of q. Notice thatthe rank of q is equal to the rank of the matrix B which is in turn equalto the rank of the matrix A (since Q is non-singular), and

y = (yi ) ... , yp, yp+i , ... , yp+m , yp+m+ 1 , ... , Yn)t .

Now if q has the same rank and signature then clearly m = 0. HenceytBy > 0 for all y E IRn since it is a sum of squares. ConsequentlyxtAx > 0 for all x E IRn.

Conversely, if xtAx > 0 for all x c IRn then ytBy > 0 for all y E IRn.Choose y = (0, ... , 0, yj, 0, ... , 0). Now the coefficient of yE must be 0 or1, but not -1. Therefore there are no terms of the form -y?, so m = 0and q has the same rank and signature.

84


If the rank and signature are both equal to n then m = 0 and p = n.Hence

ytBy=yi+...+yn,

But a sum of squares is zero if and only if each term is zero, so xtAx > 0and is equal to 0 only when x = 0.

Conversely, if xtAx _> 0 for x E IR' then ytBy > 0 for y E IR" som = 0, for otherwise we can choose

y = (0) ... ,0,yp+1,0,...,0)

with yp+1 = 1 to obtain ytBy < 0. Also, xtAx = 0 only for x = 0 givesytBy = 0 only for y = 0. If p < n then, since we have m = 0, choosey = (0,..., 0, 1) to get yt By = 0 with y 0 0. Hence p = n as required.

2.35 The quadratic form q can be reduced to normal form either by complet-ing squares or by row and column operations. We solve the problem bycompleting squares. We have

q(x) = xi + 2x1x2 + x2 - 2x1x3 - x3

= (x1 + x2)2 + xi - (xl + x3)2

and so the normal form of q is

1 0 0

0 1 0 .

0 0 -1

Since the rank of q is 3 and its signature is 1, q is neither positive definitenor positive semi-definite.

Coordinates (x1, x2, x3) with respect to the standard basis become(z1 +x2i x1, x1 +x3) in the new basis. Therefore the new basis elementscan be taken as the columns of the inverse of

i.e. {(O, 1, 0), (1, -1, -1), (0, 0,1)}.

85


2.86 Take

9((xi, x2), (y1, y2)) = 2 [f ((xi, z2), (y1, y2)) + f ((yi, y2), (z1, z2))]

= x1y1 + 2(x1y2 + x2y1) + X2!2

and

h((z1, x2), (y1, y2)) = a [f ((xi, x2), (yi, y2)) - f ((yl, y2), (x1, x2))I

=-ZX1Y2+ZX2Yl.

We have &1, z2) = f ((xii x2), (x1i x2)) = zi + 3x1 x2 + z2 and so thematrix of q relative to the standard basis is

1

32

321

Completing squares gives (x1 + 2x2)2 - 4x2. The signature is then 0and the rank is 2. The form is neither positive definite nor positivesemi-definite.

2.87 In matrix notation, the quadratic form is

4 -1 1 xxtAx=[x y z, -1 4 -1 y .

1 -1 4 z

It is readily seen that the eigenvalues of A are 3 (of algebraic multiplicity2) and 6. An orthogonal matrix P such that PAP is diagonal is

1/f 1/f 1/fP = 2/f 0

_1/-,/3

1/\ -1/V2_ 1/vChanging coordinates by setting

V =Pt yu

11Xw z

transforms the original quadratic form to

3u2 + 3v2 +6w 2

which is positve definite.

86


2.38 (1) We have

2y2-z2+xy+xz=2(y+4x)2-$x2+xz-z2= 2(y + 4x)2 - '(x - 4z)2 + z2.

Thus the rank is 3 and the signature is 1.(2) In 2xy-xz-yz put x=X+Y,y=X-Y,z=Ztoobtain

2(X2 - Y2) - (X+ Y)Z - (X_ Y)Z= 2X2 - 2Y2 - 2XZ= 2(X -

2

Z)2 - 2 Z2 - W.

Thus the rank is 3 and the signature is -1.(3) In yz + xz + xy + xt + yt + zt put

Then we obtain

x=X+Y, y=X-Y, z=Z, t=T.

(X2 - Y2) + (X - Y)Z + (X + Y)Z + (X + Y)T + (X - Y)T + ZT=X2-Y2+2XZ+2XT+ZT=(X+Z+T)2-Y2-Z2-T2-ZT=(X+Z+T)2-(T+2Z)2-4Z2-Y2.

Thus the rank is 4 and the signature -2.

2.39 (1) The matrix in question is

1 -1 2

A= -1 2 -3 .2 -3 9

Nowx2 + 2y2 + 9z2 - 2xy + 4xz - 6yz

=(x-y+2z)2+y2+5z2-2yz= (Cx- y+2z)2 +(y-z)2 +4z2

= [2 + Q2 + S2

87


where E=x-y+2z,, =y-z,s=2z. Thenz=1y=rI+2SX+q - ZS

1 1 12

P= 0 1 z0 0 2

and PtAP = diag{1,1,1}.(2) Here the matrix is

Now

0 2 0

A= 2 0 1 .

0 1 0

4xy+2yz=(x+y)2-(x-y)2+2yz=X2-Y2+(X-Y)z [X=x+y,Y=x-y]=(X+Zz)2-Y2-Yz-4z2

=(X+Zz)2_(Y+Zz)2

= r2 - n2,

where e=x+y+Zz,tl=x-y+Zzand S=z, say. Then

x= W+4-S)y = 2( - n)z=S

so if we let1 1 1

2 2 -2P= 2 -2 0

0 0 1

88


then we have

and Pt AP = diag{1, -1, 0}.(3) Here we have

1 1 0 -1

A- 1 4 3 -40 3 1 -7

-1 -4 -7 -4

The quadratic form is

x2 + 4y2 + z2 - 4t2 + 2xy - 2xt + 6yz - 8yt - 14zt

=(x+y-t)2+3y2+z2-5t2+6yz-6yt-14zt= (x+y-t)2+3(y+z-t)2 -2z2 -8t2 -8zt= (x + y - t)2 + 3(y + z - t)2 - 2(z + 2t)2

=52+172-52,

where x+y-t,,?=/(y+z-t),5=-(z+2t) and T=tsay.Then x=-7s-2r

Y= 1n-+3Tz= S-2rt = T

and so1 -1/f 1/f -2

P= 0 1/\ -1/' 3

o o 1/ 2 -2o 0 0 1

gives

x

y =P 71

z

t Ti

and Pt AP = diag{1, 1,-1,0}.

89


2.40 Here we have

(xr - xa)2

r<8

= (x1 - xn)2 + (x2 - xn)2 + ... + (xn_1 - x,4)2

+ (x1 - xn_1)2 + (x2 - xn_1)2 + ... + (xn_2 - xn_1)2

+ (XI - X2)'

= (n - 1)(x2 + ... +.. xn)-2(xlx2+ ...+xlxn+x2x3+ ...+x2xn+ ...+.xn_lxn)

= xt Ax

where

A=

n - 1 -1 -1 ... -1-1 n - 1 -1 ... -1-1 -1 n - 1 ... -1

-1 -1 -1 ... n - 1

Now, by adding n1 1times the first column to columns 2,... , n and

adding n11 times the first row to rows 2, ... , n, then multiplying rows

2, ... , n and columns 2, ... , n by nn l , we see that A is congruent tothe matrix

n- 1 0 0 ... 0

0 n - 2 -1 ... -10 -1 n - 2 ... -1

0 -1 -1 ... n-2Repeating this process we can show that A is congruent to the matrix

n-1 0 0 0 ... 0

0 n-2 0 0 ... 0

0 0 n-3 -1 ... -10 0 -1 n - 3 ... -1

L 0 0 -1 -1 ... n-3

90


Continuing in this way, we see that A is congruent to the diagonal matrix

diag{n-1,n-2,n-3,...,2,1,0}.Consequently the rank is n - 1 and the signature is n - 1.

2.41 We haven

E (Ars+r+s)xrx,r,e= 1

n n n

_ E (rxr)(sx,) + >2 (rxr)x. + E xr(sxe)r,e=1 r,8=1 r,8=1

Now let

n 12 (n l ( n_ .1 I rxr) +2 > xr) I > rxr

r,s=1 r=1 r-1_ A(x1 + 2x2 + + nxn)2

+ 2(x1 + + xn)(x1 + 2x2 + + nxn).

y1 =x1+2x2+...+nxn

y2=xl+x2+ +xny3 = x3

yn = xn

Then the form is Ayi + 2y1112 which can be written as

A(yl + *y2)2 - 1 y2I(yl + 112)2 - 4(y, - y2)2

if A 0;

if A = 0.

Hence in either case the rank is 2 and the signature is 0.

2.42 Since A is an eigenvalue of A there exist a1, ... , an not all zero suchthat Ax = Ax where x = [a1 ... an]t. Then xtAx = Axtx and so, ifQ = xtAx then we have

Q(a1,... , an) = A(ai + ... + an),

2.44 Let Q(x,x) = xtAtAx = (Ax)tAx. Since detA $ 0 we may applythe non-singular linear transformation described by y = Ax so thatQ(x, x) - Q(y, y) where

Q(yfY) =VY=yl + "'+yn-Thus Q is positive definite.

91


2.44 Let {u1, ... , un} be an orthonormal basis of the real inner product spaceIRn under the inner product given by (x1y) = f (x, y). Let x = E 1 xiu1and y = n

i_1 yiuiThen

/n n n

f (x) y) _ (xIy) = (E xiui I > yiui) = E xiyii=1 i=1 i=1

and, for some real symmetric matrix B = [bij[nxn,

n n

g(x, y) = > > bi? xi yj = xt Byi=1 j=1

where

x=

Lxn yn

Now we know that there is an orthogonal matrix P such that

PtBP =

i.e. that there is an ordered basis {v1, ... , vn} that is orthonormal (rel-ative to the inner product determined by f) and consists of eigenvectorsof B. Let x = EE ;vi and y = 1 pivi. Then, relative to thisorthonormal basis, we have

n

Consequently,

xl Yi

y=

f (x, y) _ e£fli;ti=1n

g(x,y) _ eirlii=1

n

Qf (x) = f (x, x) = E d;i=1

n

Qg(x) = g(x, x)i=1

Observe now that

g - Af degenerate b (3x)(Vy) (g - Af)(x, y) = 0

4=> (9x)(Vy) g(x, y) - )f (x, y) = 0.

92


Since, from the above expressions for f (x, y) and g(x, y) computed rel-ative to the orthonormal basis {v1, ... , vn),

n

g(x, y) - Af (x, y) _ E(A1- A)esrl+4=1

it follows that g - A f is degenerate if and only if A = A, for some i.Suppose now that A, B are the matrices of f, g respectively with re-

spect to some ordered basis of IRn. If

x -

xl y l

Y=xn yn

are the coordinate vectors of x, y relative to this basis then we have

g(x, y) - Af(x, y) = xt(B - AA)y.

Thus we see that g - A f is degenerate if and only if A is a root of theequation det(B - AA) = 0.

For the last part, observe that the matrices of 2xy + 2yz and x2 -y2+2xz relative to the canonical basis of IR3 are respectively

0

101,0 1

A= 1 B= 0 -1 0.0 1 0 1 0 0

Since1 -A 1

det(B - AA) = det -A -1 -A1 -A 0

=A2+1

the equation det(B-AA) = 0 has no solutions. But, as observed above, ifa simultaneous reduction to sums of squares were possible, such solutionswould be the coefficients in one of these sums of squares. Since neitherof the given forms is positive definite, the conclusion follows.

2.45 The exponent is -xeAx where

1 1 1

2 2

A = 2 1 21 1 1

93


The quadratic form xtAx is positive definite since

x2+y2+z2+zy+xz+yz=(x+2y+az)2+4y2+4z2+ayz=(x+2y+az)2+4(y+3z)2+ 3z2

which is greater than 0 for all x # 0. So the integral converges toir3/2/ det A, i.e. to 2P.

94

Test paper 1

Time allowed : 3 hours(Allocate 20 marks for each question)

Let V be a finite-dimensional vector space. Prove that if f E C (V, V)then

(a) dim V = dim Im f + dim Ker f ;(b) the properties

(i) f is surjective,(ii) f is injective,

are equivalent;(c) V = Im f ® Ker f if and only if Im f = IM f2;(d) Im f = Ker f if and only if the following properties are satisfied

(i) f2=0,(ii) dim V = n is even,

(iii) dim Im f = an.

2 Suppose that t E Q5) is represented with respect to the basis

W, 010, 010), (1, 1101 0, 0), (1,1,1, 0, 0), (1,1,1,1, 0), (1,1,1,1,1) }

by the matrix1 8 6 4 1

0 1 0 0 0

0 1 2 1 0

0 -1 -1 0 1

0 -5 -4 -3 -2

Find a basis of Q5 with respect to which the matrix of t is in Jordannormal form.

S Let rP1, ... , rPn E (IRn)d. Prove that the solution set C of the linearinequalities

rP1(x)>0, rai(x)>0, ... ,rpn(x)>0satisfies

(a) a,QEC=a+QEC;(b) aEC,tEIR,t>O==> taEC.

Show that if VI, ... , rPn form a basis of (IR')d then

C = {t1 al + - + tnan I t1 E IR, t{ > 0}

where {a1, ... , an} is the basis of IRn dual to the basis {(pl,... , rPn }.Hence write down the solution of the system of inequalities

(P1 (X) > 0, rP2(x) >_ 0, S03 (X) > 0, 94 (X) > 0

where rP1 = [4,5,-2, 111, rP2 = [3, 4, -2,6], ro3 = 12,3,-1,4] and rP4 =[0, 0, 0, 1].

4 Let A be a real orthogonal matrix. If (A - AI)2x = 0 and y = (A - AI)xshow, by considering yty, that y = 0. Hence prove that an orthogonalmatrix satisfies an equation without repeated roots.

Prove that a real orthogonal matrix with all its eigenvalues real isnecessarily symmetric.

5 Prove that if a real quadratic form in n variables is reduced by a realnon-singular linear transformation to a form

having p positive, q negative, and n - p - q zero coefficients then p andq do not depend on the choice of transformation.

For the form

a 1 x 1x2 + a2 x2 x3 + ... + an-1 xn _ 1 xn

in which each a; # 0, show that p = q; and for the form

a1x1x2 + a2x2x3 + + an_1xn_1xn + anxnxl

in which each a; 0, show that

I0 if n is even;

Ip - ql =1 if n is odd.

97

Test paper 2


Let V be a finite-dimensional vector space and let e E £(V, V) be aprojection. Prove that

Kere = Im(idv -e).

If t E £(V, V) show that Im e is t-invariant if and only if e o toe = toe;and that Kere is t-invariant if and only if e o t o e = e o t. Deduce thatIm e and Ker e are t-invariant if and only if e and t commute.

2 If U = [UT8] E Matnxn(Q) is given by

_ J1 ifs=r+ura = l 0 otherwise,

and J = [jre] E Matnxn(C) is given by

ifr+s=11jr° - 0 otherwise,

n+1;

show that Ut = JUJ. Deduce that if A E Matnxn(() then there is aninvertible matrix P such that P-1 AP = At.

Find such a matrix P when A is the matrix

0 4 4

2 2 1 .

-3 -6 -5

3 Let V be a vector space of dimension n over a field F. Suppose that Wis a subspace of V with dim W = m. Show that(a) dim W-'- = n - m;(b) (Wl)1 = W.

If f, g E Vd are such that there is no A E F \ {0} with f = Ag, showthat Ker f n Ker g is of dimension n - 2.

4 Let V be a finite-dimensional complex inner product space and let f :V -+ V be a normal transformation. Prove that

f2(x) = 0 f(x) = 0

and deduce that the minimum polynomial of f has no repeated roots.If e : V - V is a projection, show that the following statements are

equivalent

(a) e is normal;(b) e is self-adjoint;(c) e is the orthogonal projection of V onto Im e.

Show finally that a linear transformation h : V V is normal if andonly if there are complex scalars Al, ... , Ak and self-adjoint projectionsel,...,ek on V such that(1) f = A1e1 + + Akek;(2) ides = e1 + + ek;(3) (i34 j) eioej=0.

b (a) Show that the quadratic form xtAx is positive definite if andonly if there exists a real non-singular matrix P such that A = PP.Show also that if Eij_1 bijxixj > 0 for all non-zero vectors x then

bijpixipjxj > 0 for all x. Hence show that if xtAx and xtBx areboth positive definite then so is

nE at,bijxtx,'i,j=1

(b) For what values of k is the quadratic form

n

xr+k>xixjr=1 i<j

positive definite?

99

Test paper 3


If U, W are subspaces of a finite-dimensional vector space V prove that

dimU+dimW = dim(U + W) + dim(U n W).

Suppose now that V = U ® W. If S is any subspace of V prove that

2dimS-dimV <dim[(UnS)®(WnS)] <dimS.

In the case where V = ® 1 U1 find similar upper and lower boundsfor

2 For the matrix

n

dim ®(U1 n S).i=1

0 1 0

A= -1 1 1

-1 0 2

find a non-singular matrix P such that P-'AP is in Jordan normalform.

If A is positive, obtain real values of x;j such that C2 = D where

X101 X12 x13 A I U

0 x22 x23 , D = 0 A 1

0 0 x33 0 0 a

Hence, or otherwise, find a real matrix X such that X2 = A.

3 Let V be a vector space of dimension n over a field F and let W, X besubspaces of V. Prove that

(W+X)-L=W1nXl and (WnX)1=W1+Xl.

Given gi, ... , gn E V d, prove that the following conditions concerningf E Vd are equivalent :

(1) n 1 Ker gE C Ker f ;(2) f is a linear combination of gl,... , ga.

4 Show that the matrix

1 1 1 1

2 2 2 21 1 1 1

A=1

2 2 2 2

i 11

2 -2 2 -21 1 1 1

2 2 2 2

is orthogonal. Find its eigenvalues and show that the matrix A2 - I4has characteristic equation

X2(X2 - 3X + 3) = 0.

Find a unitary matrix U such that U-1AU is diagonal.

5 Let Q(k, r) be the quadratic form

k(zi + x2 + ... + xr) - (z1 + x2 + + xr)2

Show that(k - 1)Q(k, r) = kQ(k - 1, r - 1) + yr

where Yr is a homogeneous linear function of z1,.. . , xr.Hence find the rank and signature of

n(z1 + z2 + ... + xn) - (x1 + x2 + ... + xn)2.

101

Test paper 4


Let F be the vector space of infinitely differentiable complex functionsand let Pn be the subspace of complex polynomial functions of degreeless than n. For every A E d define Pn,a = {eazp I p E Pn}. Show thatPn,.\ is a subspace of F and that

zkB = {eaz

k!I k = 0,...,n - 1}

is a basis of Pn,A. If D denotes the differentiation map prove that(1) Dn(e az f) = e'az(D - A id)nf;(2) Pn,.\ = Ker(D - A id)n;(3) Pn,A is D-invariant;(4) B is a cyclic basis for D - A id.

If Dn,A denotes the restriction of D to Pn,a find the characteristicpolynomial of Dn,A. If µ iA A show, by considering the characteristicpolynomial of Dn,A + (µ - A) id, that (Dn,A - µ id)n is invertible.

2 Given A = I ab use the Cayley-Hamilton theorem and euclidean

division to show that every positive power of A can be written in theform

An=th12+Q2A.

If the eigenvalues of A are Al, A2 show that

- A2A1 - A1A2I2 +

\n-

\nnl A if Al # A2,

A AZ - Al A2 - Al(1-n)AnI2+nAi-1A if Al=A2.

Hence solve the system of difference equations

xn+1 = xn + 2yn

yn+1 = 2xn + yn

where x1=0andy1=1.3 Suppose that f E C(C , Qn) and that every eigenvalue of f is 0. Show

that f is nilpotent and explain how to find dimKer f from the Jordannormal form of f.

Let f, g E C((6, Q8) be nilpotent with the same minimum polynomialand dim Ker f = dim Ker g. Show that f, g have the same Jordan normalform. By means of an example show that this fails in general for f, g EC(d', (7).

Deduce that if s, t E C(Cn, Qn) have the same characteristic polyno-mial

(X - a1)k'(X - a2)k' ... (X - ar)krand the same minimum polynomial, and if

dim Ker(s - at id) = dim Ker(t - ai id)

for 1 < i < r, then s and t have the same Jordan normal form providedki < 6 for 1 < i < r.

4 Let V be a vector space of dimension n over a field F.(i) If s E C (V, V) show that s o s = 0 if and only if Im s C Ker s, in

which case dim Im s <ain.

(ii) Let p E C(VV) be such that p' = 0 and pn-1 # 0. Showthat there is a basis B = {z1,...,xn} of V such that p(xj) = xj+1 forj = 1,...,n - 1 and p(xn) = 0.

Show that if t = > 1 Aipi-1 where each Ai E F then t commuteswith p. Conversely, suppose that t E C (V, V) commutes with p and isrepresented relative to the basis B by the matrix [a=j]nxn Prove byinduction that

n-j+l(9 = 1, ... , n) t(xj) _ ail xi+j-l

i=1

and deduce that t is a linear combination of id, p, ... , pn-1

103

b Show that each of the quadratic forms

6x?+5x2+7x3-4fx2x3i7yi + 6y2 + 5y3 + 4yi y2 + 4y2Y3

can be reduced by an orthogonal transformation to the same form

2 2aIzl+a2z2+a3z3.

Obtain an orthogonal transformation which will convert the first of theabove forms into the second.

104

algebra through practice volume 4_blyth t.s., robertson e.f

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