Algebra through practice
Book 4: Linear algebra
Algebra through practice A collection of problems in algebra with solutions
Book 4
Linear algebra T. S. BLYTH o E. F. ROBERTSON University of St Andrews
-
r. The right f ,h, University of Cambridge to print and sell all manner of books as granted by
Henry Vlll in 1534. The University has printed and published continuously since 1584.
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Library of Congress Catalogue Card Number: 83-24013 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 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
18
Preface
The aim of this series of problem-solvers is to provide a selection of worked examples in algebra designed to supplement undergraduate algebra courses. We have attempted, mainly with the average student in mind, to produce a varied selection of exercises while incorporating a few of a more challenging nature. Although complete solutions are included, it is intended that these should be consulted by readers only after they have attempted the questions. In this way, it is hoped that the student will gain confidence in his or her approach to the art of problem-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 n this should not discourage them from attempting problem n+ 1. A great many of the ideas involved in these problems have been used in examination papers of one sort or another. Some test papers (without solutions) are included at the end of each book ; these contain questions based on the topics covered.
TSB, EFR St 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 of sophistication. Here is a list of some major texts that are widely used and to which the reader may refer for background material. The subject matter of these texts covers all six of the present volumes, and in some cases a great deal more. For the convenience of the reader there is given overleaf an indication of which parts of which of these texts are most relevant to the appropriate sections of this volume. [1] I. T. Adamson, Introduction to Field Theory, Cambridge
University 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 University Press, 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 4 1: 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 some elementary results are proved for modules. In [5] the author uses `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 sum A ® B of subspaces A, B of a vector space V. Recall that V = A ® B if and only if every x E V can be expressed uniquely in the form a + b where a E A and b E B; equivalently, if V = A+B and AnB = {0}. For every 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 that
spans a subspace B such that V = A ®B. If f : V -* V is a linear transformation then a subspace W of V is said to be f-invariant if f maps W into itself. If W is f-invariant then there is an ordered basis of V with respect to which the matrix of V is of 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 the subspace Imf is f-invariant (and the subspace Kerf is trivially so). A vector space V is the direct sum of subspaces W1, .. . , Wk if and only if there 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 M1
M2
Mk Of particular importance is the situation where each Ms is of the form A
1
0
0 0
A
1
0
A
... ... ...
0 0
0 0
0 0
... ...
0 0 0
0 0 0
A
l
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 f is the monic polynomial of least degree of which f is a zero. When the minimum polynomial of f factorises into a product of linear polynomials
then there is a basis of V with respect to which the matrix of f is a Jordan matrix. This matrix is unique (up to the sequence of the diagonal blocks), the diagonal entries A above are the eigenvalues of f, and the number of M1 associated with a given A is the geometric multiplicity of A. The corresponding basis is called a Jordan basis. We mention here that, for space considerations in the solutions, we shall 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
1: Direct sums and Jordan forms (ii) If A is a finite set of linearly independent vectors then the dimension of the subspace spanned by A is equal to the number of vectors in A.
(iii) The subspace {(x, x, x) 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 vectors in A. (vi) If A is a subset of IRn and the subspace spanned by A is IRn itself then A contains exactly n vectors. I
(vii) If A and B are subspaces of IRn then we can find a basis of IRn which contains a basis of A and a basis of B. (viii) An n-dimensional vector space contains only finitely many subspaces.
(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 represented 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 such
that P-1 AP = B. (xviii) There is a bijection between the set of linear transformations from IRn 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 represented by 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
Linear algebra
Book 4 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 t1 for 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 such
that sot = idv. Prove that to s = W. Prove also that a subspace of V is t-invariant if and only if it is s-invariant. Are these results true when V is infinite-dimensional? 1.5
Let V,, be the vector space of polynomials of degree less than n over the field IR. If D E £(V,,, V,,) is the differentiation map, find Im D and Ker 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 field Z2? 4
1: Direct sums and Jordan forms 1.6
Let V be a finite-dimensional vector space and let t E C (V, V) - Establish the 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 and deduce that (Vk > 1)
Imp = Im tP+k and Ker tP = Ker
tP+k.
Show also that V = 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 --+ V
be a non-zero linear transformation such that f o f = 0. Show that if Im f is of dimension r then 2r < n. Suppose now that W is a subspace of V.such that V = Ker f ® W. Show that W is of dimension r and that if {wl,... , wr} is a basis of W then { f (wl ), ... , f (wr)} is a linearly independent subset of Ker f. Deduce that n - 2r elements XI, ... , xn-2r can 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 be B = {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
Linear algebra
Book 4
(2) {b1 + b4, b2 + b3} is a basis of V1 and {b1 - b4, b2 - b3} is a basis of 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 is
i 2 0 0
0
i
1
2
2
i
1
0
0
2-
2
0
-2
2 0
i
2
1
0
1
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 matrix of the form
1.9
Q
0
0
ry
S
0
0
0
0
E
S
0
0
n
tg
Let V be a vector space of dimension n over a field F. Suppose first that F is not of characteristic 2 (i.e. that 1F + 1F 54 OF). If f : V --+ V is 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 only if A is similar to a matrix of the form
Suppose now that F is of characteristic 2 and that f o f = idv. If g = idv +f show that
xEKerg b x= f(x), 6
1: Direct sums and Jordan forms and that g o g = 0. Deduce that an n x n matrix A over F is such that A2 = 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 of
Img 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 such that t # idv and t # 0. Is it possible to have Imt n Kert # {0}? Is it possible to have Im t = Ker t? Is it possible to have Im t C Ker t? Is it possible to have Kert C Imt? Which of these are possible if t is a projection?
1.11 Is it possible to have projections e, f E £(V,V) with Kere = Ken f and Im e # Im f ? Is it possible to have Im e = Im f and Ker e # Ker f? Is it possible 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. Let el,e2 E .C(V,V) be projections. Prove that el + e2 is a projection if and only if el o e2 = e2 o el = 0. If el + e2 is a projection, find Im(el + e2) and Ker(el + e2) in terms of the images and kernels of el, e21.13 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 a projection e E C(IR3, IR3) such that Ime = V and Kere = U. Find also a projection f E C(IR3, IR3) such that Im f = U and Ker f = V. 7
Linear algebra
Book 4
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
Ek 1 ai = 1. Prove that
e =ale, + A2e2 + ...+ Akek is a projection with Im e = Im ei. Is it necessarily true that if fl, ... , fk E £(V, V) are projections and 1 ai = 1 then Ek 1 ai fi is also a projection? 1.15 A net over the interval [0, 11 of IR is a finite sequence (ai)o
rk that
0=ao
A step function on [0, 1[ is a mapping f : [0,1[--+ IR for which there exists a net (ai)o
Show that the set E of step functions on [0, 1[ is a vector space over IR and that a basis of E is the set {ek k E [0, 1 [} of functions ek : [0, 1 [-> IR I
given by
if0<x
ek(x)= J0 1
A piecewise linear function on [0, 1[ is a mapping f : [0,1[-> IR for which there exists a net (ai)o
Let F be the set of piecewise linear functions on [0, 1[ and let G be the subset of F consisting of the piecewise linear functions g that are continuous with g(0) = 0. Show that F, G are vector spaces over IR and
that F=E®G. Show that a basis of G is the set {gk
I
k E [0, 1[} of functions given
by
gk(x)
if0<x
0
x-k
Finally, show that the assignment
fi-+I(f)= jf(t)dt describes an isomorphism from E to G. 8
1: Direct sums and Jordan forms 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. Is it 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 consisting of eigenvectors of t. Find such a basis, and the matrix of t with respect to this basis.
1.18 Suppose that t E C(V,V) has zero as an eigenvalue. Prove that t is not invertible. Is it true that t is invertible if and only if all the eigenvalues of t are non-zero? If t is invertible, how are the eigenvalues of t related to 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 t are 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, a - b). a
Find the minimum polynomial of t.
1.21 Let F be a field and let F,+1 [X] be the vector space of polynomials of degree 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 that t2 = 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
Linear algebra
Book 4
For those matrices A that are diagonalisable, find an invertible matrix P such that P-1 AP is diagonal.
(a)
l
L-'
3J, (b)
(d)
2
1
0
2
0
0
3
-1
1
-1
5
-1
1
-1
3
-1
-1
-1 -2
7
2
2
10
1
0
0
2
1.
-1
0
3
1, (e) 1
(c)
,
7
2 ,
1
1.24 Consider the sequence described by 3
1
an
7
5,..., bn,. 11
21
where an+1 = an + 2bn and bn+1 = an + bn.
Find a matrix A such that A bn
[bn+1]
By diagonalising A, obtain explicit formulae for an and bn and hence show that an lim = 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. Let a = f (t) and b = g(t). Prove that every eigenvector of a o b that is associated with the eigenvalue 0 is the sum of an eigenvector of a associated with the eigenvalue 0 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 eigenvalues and 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 an eigenvalue of t, the subspace Ca; has dimension 1. Hence show that the matrix of s with respect to the basis of eigenvectors of t is diagonal. 10
1: Direct sums and Jordan forms 1.27 Let u be a non-zero vector in an n-dimensional vector space V over a field 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. Show also that U is t-invariant. Show that there is a non-zero monic polynomial f (X) E F[X] of degree r such that [f (t)](u) = 0. If to : U --* U is the linear transformation induced 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) respectively. 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, every column 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 + y Show that {M(a,Q, y) a,Q, y E Q} is a subspace of Matax3(C) and 1
that 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 the matrix of f relative to the basis {el + e2 + e3, e2, e3}. Hence find the eigenvalues of M(a, Q, y). 11
Linear algebra
Book 4
1.30 Let V be a vector space of dimension n over a field F. A linear transformation f : V -> V (respectively, an n x n matrix A over F) is said to be nilpotent of index p if there is an integer p > 1 such that fP-1 # 0 and 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 that fp-1(x) # 0 then
{x,f(x),...,f"-1(x)} is a linearly independent subset of V. Hence show that f is nilpotent of index n if and only if there is an ordered basis of V with respect to which the matrix of f is 0
0
In-1
0'
Deduce that an n x n matrix A over F is nilpotent of index n if and only if A is similar to this matrix. 1.31 Let V be a finite-dimensional vector space over IR and let f : V -+ V be
a linear transformation such that f o f = -idv. Extend the external law IR x V -* V to an external law (E x V -p V by defining, for all x E V
and 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
t=1
r
r
1:(at - iµt)vt =
lttf(vt)
Atvt + 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 independent subset of the IR-vector space V. Deduce that the dimension of V as 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 = -I2n if and only if A is similar to the matrix
1.32 Let A be a real skew-symmetric matrix with eigenvalue A. Prove that the real part of A is zero, and that A is also an eigenvalue. 12
1: Direct sums and Jordan forms 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 without repeated 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 that Av = -au.
Au = av,
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) have eigenvalues -2,1,2. Use the Cayley-Hamilton theorem to express ten as 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
..
P12
Qln
M
0
Observe that in general Qi2...... in are non-zero, so that W is not finvariant. Let it be the projection of V onto W and let g = it of. Show that W is g-invariant and that if g' is the linear transformation induced on W by g then Mat (g', (bf)) = M. Show also that all the zeros of the characteristic polynomial of g' lie in F. Deduce that f is triangularisable in the sense that there is a basis B of V relative to which the matrix of f is upper triangular with diagonal entries the eigenvalues of f. 13
Linear algebra
Book 4
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 is not diagonalisable. Find a basis of IR3 with respect to which the matrix of t is upper triangular. 1.36 Let V = Q3[X] be the vector space of polynomials of degree less than or equal to 2 over the field Q. If t E L (V, V) is given by
t(1)=-5-8X-5X2 t(X) = 1 + X + X2 t(X2) = 4 + 7X + 4X2,
show that t is nilpotent. Find a basis of V with respect to which the matrix of t is upper triangular. 1.37 For each of the following matrices A find a Jordan normal form and an invertible matrix P such that P-1 AP is in Jordan normal form.
-64 -41 ]'
239
(a)
[
(c)
5 1
3
0
7
0
9
-2 -4 -5
-1 -1
1
(b)
,
(d)
I
-0 3
0
1
0 0
3
0
0
3
J'
1.38 Find a Jordan normal form J of the matrix
A=
2
1
1
1
0
0
2
0
0
0
0
0
2
1
0
0
0
0
1
1
-1 -1 -1
0
0
Find also a Jordan basis and hence an invertible matrix P such that
P-'AP = J.
1.39 For each of the following matrices A find a Jordan normal form J, a Jordan basis, and an invertible matrix P such that P-1AP = J. 22
(a)
20 30
-2 -12 0
(b)
-12
-3 -16 14
-13 -22
8
1
13
0
3
8
-5
0
-1
-22
13
5
5
2
1: Direct sums and Jordan forms 1.40 Find a Jordan normal form and a Jordan basis for the matrix 5
0 1
0 1
-1 -3
2
2
0
0
0
0
1
1
-2
3
1
1
1
-1 0 -1 -1
-5
1.41 Find the minimum polynomial of the matrix
-1 -3
1
0
2
1
0
1
1
-3 -1 -3 4 -8 -2 -7 5
4
1
0
-4
1
A = -2 -1
1
From only the information given by the minimum polynomial, how many
essentially different Jordan normal forms are possible? How many linearly independent eigenvectors are there? Does the number of linearly independent eigenvectors determine the Jordan normal form J? If not, does the information given by the minimum polynomial together with the number of linearly independent eigenvectors determine J? 1.42 Find a Jordan normal form of the differentiation map D on the vector space IR4[X) of polynomials of degree less than or equal to 3 with real coefficients. 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 Jordan normal forms? Which of these are similar?
1.44 Which of the following are true? If A, B E Mat,,,,(() then AB and BA have 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 let A be an eigenvalue of t. Let J be a matrix that represents t relative to some Jordan basis of V. Show that there are dim Ker(t - A ides) 15
Linear algebra
Book 4
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 matrix 0
1
0
_ -2
3
0
-1 -1
-2
1
2
-1
2
-1
0
3
`1
Find also a Jordan basis and an invertible matrix P such that P-1 AP =
J. Hence solve the system of differential equations 0
1
3
0 0
X3
_ -2 -2
1
2
-1 -1 -1
x4
2
-1
0
3
x1 X1
x1
x2 x3 x4
1.47 Solve each of the following systems of differential equations : = 5x1 + 4x2
dt
=
dx2
dt_
4x1 - x2 - x3 x1 + 2x2 - x3
dx3
dt =
dxj
j-j-=x1+3x2-2x3
dIi_5x'+6x2+6x3 = 0
(c)
dx2
dt dX3
1.48
+ x1 - 4x2 - 2x3 = 0
(d)
- 3x1 + 6x2 + 4x3 = 0
dx2
dt
dt
- 3 dx2 + 2 dx3
= y1
-5dx2+4d3
= Y2
-9dx +7d =y3 16
= 7x2 - 4x3
dt = 9x2 - 5x3
Solve the system of differential equations dxl
xl - x2 + 2x2
1: Direct sums and Jordan forms 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 the method of the Jordan normal form, solve the equation given the initial conditions x(0) = 0,
x'(0) = 0,
17
x"(0) = 16.
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 dimension and B = {v1,...,vn} is a basis of V then the basis that is dual to B is B4 = { va, ... , vn } where each 0 : V -+ F is given by 1
vd(vj)
0
ifi=j; 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)t where 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 matrix from Bd to (ed)n is given by (M-l)t. We can therefore usefully denote the dual basis by
B = {[all, ... , aln], [all, ... , a2n], ... , [anl, ... , an-[}
2: Duality and normal transformations ain] denotes the ith row of M-1, so that
where
[ail,...,ain](xl) ...,xn) =ailxl + ...+atnxn, The bidual of an element x is x^ : Vd -* F where x^(yd) = yd(x). It is common practice to write yd(x) as (x, yd) and say that yd annihilates x 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 linear mapping ft : Wd --+ Vd described by yd ,--* yd o f. When V is of finite dimension we can identify V and its bidual (Vd)d, in which case we have that (f t )t = f. Moreover, if f : V -+ W is represented relative to fixed ordered bases by the matrix A then f' : Wd -+ V d is represented relative to the corresponding dual bases by the transpose At of A.
If V is a finite-dimensional inner product space then the mapping 19v : x ,--* xd describes a conjugate isomorphism from V to Vd, by which
we mean that (x + y)d = xd + yd and
The adjoint f* : W -+ V of f : V
(Ax)d = .1xd.
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 matrix of f relative to a given ordered basis is A then that of f* is V. We say that A is normal if it commutes with T. A matrix is normal if and only if it is unitarily similar to a diagonal matrix, i.e. if there is a matrix U with U-1 = U such that U-1 AU is diagonal. A particularly important type of normal transformation occurs when the vector space in question is a real inner product space, and topics dealt with in this section reach as far as the orthogonal reduction of real symmetric matrices and its application to finding the rank and signature of quadratic forms. 19
Linear algebra
Book 4
2.1
Determine which of the following mappings are linear functionals on the vector space IR3 [X] of all real polynomials of degree less than or equal to 2 :
(a) f '-' f';
(b) f '-+
f f;
(c) f '-+ f (2);
(e) f '-'
(d) f '--' f'(2);
2.2
1
/
1
f2.
Let C[0,1] be the vector space of continuous functions f : [0, 11 -+ IR. If fo is a fixed element of C[O,1], prove that (p : C[0,1] -+ IR given by
'P(f) = fo fo(t) f (t) dt 1
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 let Ad = {V1i a2) be the corresponding dual basis of Vd. Find, in terms of epl, (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?
2.6
(a) {[-1, 2], [0, 1]};
(b) {[-1, 0], [2, 1]};
(c) {[-1, 0], [-2,1]};
(d) {[l, 0], [2, -1]}.
(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 if A, 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: Duality and normal transformations 2.8
Let IR3[X] be the vector space of polynomials over IR of degree less than or equal to 2. Let t1, t2, t3 be three distinct real numbers and for i = 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 and determine a basis B = {pl(X),p2(X),p3(X)} of IR3[X] of which Bd is the dual.
2.9
Let a = (1, 2) and Q = (5, 6) be elements of IR2 and let cp = [3,4] be an element 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
Translate these results into a theorem on solving systems of linear equations. Show that (i) is not satisfied by the system
3x+ y=2 x+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
Linear algebra
Book 4
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 the third basis element. Suppose that {
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 there is no g E CIO, 11 such that F. = Lg.
2.15 By a canonical isomorphism V -* Vd we mean an isomorphism such 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 except when n = 2 and F has two elements.
If S is such an isomorphism show that, for y # 0, the subspace Kerr(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 # 0 let {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 transformation 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.
Take
x = xl, y = t.] If, on the other hand, t 0 Ker S(t) for all t
0 let
22
2: Duality and normal transformations {z1,...,xn-1} be a basis of Kers(t) so that {z1i...,xn_1,t} is a basis of 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 if f : 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 these observations that we must have n = 2. Suppose now that F has more than two elements and let A E F be such that 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 - V
is the (unique) linear transformation such that f (t) = t, f (z) = Az show that f is an isomorphism that does not satisfy (*). [Hint. Take x = z, y = t.] If, on the other hand, t 0 Ker S(t) for all t 0 0 let {z} be a 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 that f 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. Only four of these are linear transformations from F2 to F; and each of these is determined by its action on the natural basis of F2. Compute (F2)d and 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 subspace of V of dimension k - 1 (a hyperplane). Show that there exists a unit vector n in V such that
U= {x E V I (n]x) = 0}. Given v E V, define
v' = 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 mapping t : V --4V defined by
t(v)=v' is linear and orthogonal. What can you say about its eigenvalues and eigenvectors? 23
Linear algebra
Book 4
Ifs
IR3 -> IR3 and t : IR4 --, IR4 are respectively reflections in the
plane 3x - y + z = 0 and in the hyperplane 2x - y + 2z - t = 0, show that the matrices of s and t are respectively -7 6 11
6
-6
-6 9
2
2
9
1
2
-4
2
2
4
2
-1
-4
2
1
2
2
-1
2
4
1 ,
5
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 --+ V be linear. Show that if A is the matrix of f relative to an orthonormal basis B of V then the matrix of the adjoint f* of f relative to B is the transpose 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 the mapping
(AIB) = tr(B*A),
(A, B)
where B* denotes the transpose of the complex conjugate of B, is an inner 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 every f E Vd there is a unique ,0 E V such that (bx E V)
f (x) = (xI,8). 24
2: Duality and normal transformations [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 inner product spaces of infinite dimension. Let V be the vector space of polynomials over Q. Show that the mapping (P, q) H (PIq) =
f
1
p (t) q(t) dt
0
is an inner product on V. Let z be a fixed element of C and let f E Vd be the `evaluation at z' map given by
f(P)=P(z)
(VPEV)
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 - z and show that, for every p E V, 0=
f
1
r(t) p(t) q(t) dt.
Now let p be given by p(t) = r(t)q(t) and deduce the contradiction
q=0.]
For the rest of this question let V continue to be the vector space of polynomials over C with the above inner product. If p E V is given by
p(t) =
aktk define p E V by f(t) _
aktk, and let fp : V -* V be
given 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 and q(1) = 1. Use the previous part of the question (with z = 1) to obtain the required contradiction.] 25
Linear algebra
Book 4
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 the integral 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 2
fn(x) = xn -
n+2*
Show that fn is an eigenfunction of K with associated eigenvalue 0. Use the Gram-Schmidt orthonormalisation process to find two orthogonal eigenfunctions of K with associated eigenvalue 0. Prove that K has only one non-zero eigenvalue. Find this eigenvalue and an associated eigenfunction.
2.22 Let t be a skew-adjoint transformation on a unitary space V. Prove that id ±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 matrix of 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-symmetric matrix of the same order. Suppose that A and S commute and that det(A - S) # 0. Prove that
(A+ S)(A - S)-1 is orthogonal. 26
2: Duality and normal transformations 2.25 A complex matrix A is such that A A = -A. Show that the eigenvalues of A are either 0 or -1. 2.26 Let A and B be orthogonal n x n matrices with det A that A + B is singular.
det B. Prove
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 that g(A) = 0, prove that g(-A) = 0. Deduce that the minimum polynomial of A contains only terms of even degree. Deduce that if A is skew-symmetric and f (X), g(X) are polynomials whose terms are respectively odd and even then f (A), g(A) are respectively skew-symmetric and symmetric.
2.29 For every complex n x n matrix A let n
N(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 then g(A) is normal. 2.32 If A and B are real symmetric matrices prove that A + iB is normal if and 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 that every quadratic form xtAx is identically zero. Prove that the non-zero eigenvalues of A are of the form l i where µ E IR. If x = y + iz where y, z E IRn is an eigenvector associated with
the eigenvalue iµ, show that Ay = -juz and Az = µy. Show also that yty = ztz and that ytz = 0. If Au = 0 show also that uty = utz = 0. 27
Book 4
Linear algebra
Find the eigenvalues of the matrix 0
2
-2
2
1
0
A= -2 0 -1 and an orthogonal matrix P such that 0
0
PAP = 0
0
0
-3
0 3 . 0
2.34 Consider the quadratic form q(x) = xtAx on IR". Prove that q(x) > 0 for 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 = 0 if 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 represented by the matrix
A=
1
1
-1
1
1
0
-1
.
0 -1
Is q positive definite? Is q positive semi-definite? Find a basis of IR3 with 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 h
such 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 the rank 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: Duality and normal transformations 2.38 By completing squares, find the rank and signature of the following quadratic 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 symmetric matrix A for which the form is expressible as xt Ax. Diagonalise each of the forms and in each case find a real non-singular matrix P for which the 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 all zero such that 2
Q
a
2.43 If the real square matrix A is such that det A # 0 show that the quadratic form xtAtAx is positive definite.
2.44 Let f : IRn x IRn --+ IR be a symmetric bilinear form and let Q f be the associated quadratic form. Suppose that Qf is positive definite and let g : IRn X IRn --> IR be a symmetric bilinear form with associated quadratic
form Qg. Prove that there is a basis of IRn with respect to which Q f and 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). Call f degenerate if there exists x E IRn with ff = 0. Determine the scalars A E IR such that g - A f is degenerate. Show that such scalars are the 29
Linear algebra
Book 4
roots of the equation det(B - AA) = 0 where A, B represent f, g relative to some basis of IR'. By considering the quadratic forms 2xy + 2yz and x2 - y2 + 2xz show that the result in the first paragraph fails if neither f nor g is positive definite.
2.45 Evaluate (xz+y'+xa+xN+xz+yz)dx
1. 01.0 f 00
8
30
dy dz.
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) x E IR} is a subspace of IR2 for every A E IR. (ix) True. I
(x) True.
(xi) False. An isomorphism is always represented by a non-singular matrix. (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, 1
[0
0
1
0][1 0]- [0
1
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
Linear algebra
Book 4
that f is represented by the zero matrix and g is represented by the identity matrix; and there is no invertible matrix P such that P-142P = 0. (xviii) True. (xix) False. The transformation t is non-singular (an isomorphism), but r1
21
1
2J
is singular. r
1
(xx) False. The matrix I 0 1.2
1J
is not diagonalisable.
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=0
b a=b=c
and so Kert2 = {(a, a, a) a E IR}. It is clear from the definition of t2 that Imt2 = {(a, b, 0) a, b E IR}. Likewise, it is readily seen that Ker t3 = {0}, Imt3 = IR3, Kert4={(0,0,a) I aEIR}, Imt4={(a,b,b) a,bEIR}. If Ker t; fl Im ti = {0} then by the dimension theorem we have I
I
I
dim(Kerti + Imti) = dim IR3 and so Ker ti + Im ti = IR3. Now for i = 1, 2, 3, 4 we have from the above that Ker ti n Im ti = {0}. Thus we see that IR3 = Ker ti ® Im ti holds in all 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
Solutions to Chapter 1 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) a, b E IR}; I
Ker(t4 o t3) = {(0, 0, a)
I
Im(t4 o t3) = {(-b, a, a)
1.3
a E IR}; I
a, b (=- IR}.
Reducing the matrix to row-echelon form we obtain 3
-1
-1
5
1
-1
1
-1
-1 -> -1
5
1
3
3
-1 4
2
0
2
-8
3
-1
-0
-1
0
4
1
1
-1 -> 0
2
1
3
1
-8 -> 0 2
0
3
-1
3
2
-8
0
18
.
Note that we have been careful not to divide by any number that is divisible by either 2 or 3 (since these will be zero in 74 and 713 respectively). (i) When F = IR the rank of the row echelon matrix is 3, in which case dim 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, in which case dim Im t = 2 and dim Ker t = 1. V = Ker t ® Im t holds in cases (i) and (ii), but not in case (iii); for in case (iii) we have that (1, 1, 1) belongs to both Kert and Imt. 1.4
If s o t = idv then s is surjective, hence bijective (since V is of finite dimension). 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 an isomorphism we must have dimt(W) = dim W and so t(W) = W. Hence W = s[t(W)] = s(W) and W is s-invariant. The result is false for infinite-dimensional spaces. For example, consider the real vector space IR[X] of polynomials over IR. Let s be the differentiation map and t the integration map. We have s o t = id but toe 54 id. 33
Linear algebra
Book 4 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 n
ImD # {0} since if a E F with a # 0 then the constant polynomial a belongs to both. The same results do not hold when the ground field is Z2. For example, in this case we see that the polynomial X2 belongs to the kernel of D.
1.6
Let s, t E £(V, V). Then if w E Im(s o t) we have w = a[t(u)] for some
u E V which shows that w E Im s. Thus Im(s o t) C Im s. The first chain 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 taking
s=t.
Now we cannot have an infinite number of strict inclusions in the first chain since X C Y implies that dim X < dim Y, and the dimension of V is finite. Hence the chain is finite. It follows that there exists a positive
integer p such that ImtP = ImtP+k for all positive integers k. Since dim Im tP + dim Ker tP = dim V the corresponding results for the kernel chain 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 then tP(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, if
x E KertP then tP(x) = 0 gives tP+1(x) = 0 so tP[t(x)] = 0 whence t(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
Solutions to Chapter 1 If {wl,...,wr} is a basis of W then for i = 1,...,r we have f(w;) E Im f C Ker f. Moreover, { f (wl), ... , f (wr)} is linearly independent since
r
Atf ('wi) = 0
f
/ r
Aiwt) = 0
t)LiwiEKerf 1=1
=i' r
EAtwtEKerfnW={0} :=1
r
> A1w;=0 i=1
(i = 1,...,r) Ai = 0.
Every linearly independent subset of a vector space can be enlarged to form a basis so, since dim Ker f = n - r, we can enlarge the independent subset (f (wl ), ... , f (wr)} of Ker f to form a basis of Ker f. Thus we may 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 = 0 and if f : V -+ V is represented by A relative to some fixed ordered basis then f o f = 0 and, from the above, there is a basis of V with respect to which the matrix of f is of the above form. Thus A is similar to this matrix. Conversely, if M denotes the above matrix then clearly M2 = 0. So if A is similar to M there is an invertible matrix P such
that A = P-'MP whence A2 = P-1M2P = P-lOP = 0. 35
Linear algebra
Book 4
1.8
(1) Sums and scalar multiples of elements of V1, V2 are clearly elements of V1, V2 respectively.
(2) If z E V1 then x = xl(bl + b4) + x2(b2 + b3) shows that V1 is generated by {b1 + b4, b2 + b3}. Also, if xl (bl + b4) + x2(b2 + b3) = 0 then, 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 basis of V2-
(3) It is clear from the definitions of V1 and V2 that we have V1 n V2 = {0}. Consequently, the sum V1 + V2 is direct. Since V1,V2 are of dimension 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 =
2
(b1 + b4) + 0(b2 + b3) + 0(b2 - b3) -
2
A
0
0
0 12 2
2
0 1
2 0
-2
0
0
2
1
0
2(b1 - b4)-
It is readily seen that 11
0
0
1
0
1
1
0
0
1
-1
0
1
0
0
-1
A-1 =2A=
Suppose now that M is centro-symmetric; i.e. of the form c
d
h
g
g f
c
b
a
a
b
e
f
h d
36
e
Solutions to Chapter 1 Let f represent M relative to the basis B. Then the matrix of f relative to the basis C is given by AMA-1, which is readily seen to be of the form
K
_
a
/3
0
ry
6
0
0 0
0 0
E
0 0 S
$
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 + 1F we 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, since f 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 since a1 = 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 some c E V so f (ap+,) = f (c) - f If (c)] = f (c) - c = -ap+1, and similarly for ap+2, ... , an. Hence the matrix off relative to the basis {a,,... , an } is
11611P
Ip 0
- I0n- pJ
and A is then similar to this matrix. Conversely, if A is similar to a matrix of the form AP then there is an invertible matrix Q such that Q-1 AQ = Ap. Then A2
=
(QAPQ-1)2 = QO2PQ-1 = QjnQ-1 37
= In.
Linear algebra
Book 4
Suppose now that F is of characteristic 2, so that x + x = 0 and hence
x= -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 and hence g o g = 0.
Suppose now that A2 = In and let f represent A relative to some fixed ordered basis. Let g = idv +f and note from the above that Im g C Kerg. Let {g(c1),... , g(cp)} be a basis of Im g and extend this to a basis {b1, ... , bn-2p) 9(C1) , ... , g(cp)}
of Ker g (which is of dimension n - dim Im g = n - p). Consider now the set
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. It is 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 that vjg(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 compute the matrix of f relative to the basis B we observe that since bi E Ker g we have, 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.ct f (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
Solutions to Chapter 1 Consequently A is similar to this matrix. Conversely, if A is similar to a matrix of the form VP then there is an invertible matrix Q such that Q-1 AQ = V p and so
= (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 A2
{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. 1.12
Also, t1 o t2 = 0 but t2 o ti 96 0. (Note that t2 o t1 is not a projection.) Clearly, ei + e2 is a projection if and only if (denoting composites by
juxtaposition) e1e2 + e2e1 = 0. Thus if e1e2 = 0 and e2e1 = 0 then the 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 obtain ele2+e1e2e1 = 0, and multiplying each side on the right by e1 we obtain ele2el + e2e1 = 0. It follows that e1e2 = e2e1. But e1e2 + e2e1 = 0 also gives 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)
a, b E IR}. Then IR3 = V ® U since it is clear that V n U = {0} and that I
(a,b,c) _ (a, a, 0) + (0, b - a, c).
For the last part refer to question 1.11; take e = t1 and f = t439
Book 4 1.14
Linear algebra
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 and foe = e, let z E Im e. Then e(x) = x gives e(x) = f [e(x)] = f (x) E Im f and 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
(k)+ (k)
_ A e + ... + akek + A1A2e1e2 + A2.1e2e1
+
+ AkAk-lekek-1
..+
l
sG-J
= e,
and so e is also a projection. To show that Ime = Imel it suffices to prove 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 clearly e + if is not. a 2 1.15 Since sums and scalar multiples of step functions are step functions it is clear that E is a subspace of the real vector space of all mappings from IR to IR. Given 6 E E, the step function t9i whose graph is
ai+1
ai
i.e. the function that agrees with t9 on [ai, ai+1 [ and is zero elsewhere, 40
Solutions to Chapter 1 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 therefore form a basis of E. It is likewise clear that F is a vector space and that G is a subspace of F. Consider now an element of F, as depicted in the diagram
From geometric considerations it is clear that every element of F can be written 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 up and down.] Thus it is clear that F = E ® G. To show that {gk k E [0, 1[} is a basis for G, observe first that the graph of gar - gar+1 is of the form I
ai+1- ai
41
Book 4
Linear algebra
Given p E F, consider the function pi whose graph is
Z ai+1
ai
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 G and e E E. It now follows that {gk k E [0, 1[} must be a basis for G. Finally, observe that I
I(ek) = foo ek(t) dt = gk
so that I carries a basis to a basis and therefore extends to an isomorphism 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 both questions can be answered in the affirmative. 1.17 The eigenvalues of t are 1 + if and 1 - if. Associated eigenvectors
of any matrix representing t are [1, if /2] and [1, -if /2]. Since the eigenvalues are distinct, the eigenvectors of t form a basis of d2. The matrix of t with respect to this basis is
1+i/ 0 42
0
1-ice
Solutions to Chapter 1 1.18
Since t has 0 as an eigenvalue we have t(v) = 0 for some non-zero v E V
and hence t is not injective, so not invertible. Thus if t is invertible then all its eigenvalues are non-zero. For the converse, suppose that t is not invertible and hence not injective. Then there is a non-zero vector V 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 of t-1 with the same associated eigenvector. (Remark. Note that we have assumed that V is finite-dimensional (where?)-in fact the result is false for infinite-dimensional spaces.) 1.19
Suppose that A is a non-zero eigenvalue of t. Then t(v) = Av for some non-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 t are zero. If t is diagonalisable then the matrix A of t is similar to the diagonal matrix Al
An
where A 1, . . . , an are the eigenvalues of t. But we have just seen that all the eigenvalues are zero. Thus, for some invertible matrix P we have
P-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 0 0, 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
Linear algebra
Book 4
t(Af(X)) = t((Af)(X)) = Af(X + 1) = At(f(X)). The matrix of t relative to { 1, X, ... , XI I is
...
1
3
... ...
an(n - 1)
0
...
1
1
1
1
1
0 0
1
2
3
0
1
0
0
0
n
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 for 1 < 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
azl =...=Azn =1. Consequently, at = ±1 for each i and hence the sum of the eigenvalues of 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 the eigenvectors associated with the eigenvalue 2, solve
[-1
1][ y)- 1001
to obtain the eigenspace {[x, x] x E iR}. For the eigenvectors associated with the eigenvalue 4, solve I
-1 44
y1=[01
Solutions to Chapter 1 to obtain the eigenspace {[x, -z] x E IR}. Since A has distinct eigenvalues it is diagonalisable. A suitable matrix P is I
(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 diagonalisable; take, for example,
P
1
1
1
0
1
-2
-1
1
1
.
(c) The eigenvalues of A are 6, 6, and 12. For the eigenvalue 6 we consider 1
-1
-2
-1
1
2
y=0
-2
2
4
z
x
0 .
0
We obtain -x + y + 2z = 0. We can therefore find two linearly independent eigenvectors associated with the eigenvalue 6, for example [1, 1,01 and [2, 0, 1]. Hence this eigenvalue has geometric multiplicity 2. The eigenvalue 12 has geometric multiplicity 1 and an associated eigenvector is [-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 0 0
1
0 0
-1 1
-1
x
0
y= 0
1 l1 z
0
from which we see that the corresponding eigenspace is spanned by [1, 0, 0]. Hence the eigenvalue 2 has geometric multiplicity 1. The eigenvalue 1 also has geometric multiplicity 1, the corresponding eigenspace being spanned by [-2, 1, -1]. In this case A is not diagonalisable. 45
Linear algebra
Book 4
(e) The eigenvalues are 2, 2, 2. From
-1
0 0 0
0
-1
x
1 1
y=
1
z
0 0 0
we see that the corresponding eigenspace is spanned by [0, 1, 0] and has dimension 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 An-1
[:1= The characteristic polynomial of A is X2 - 2X -1 and its eigenvalues are
a1 = 1 + f and )2 = 1 - v'-2. Corresponding eigenvectors are [f, 1] and [-vi, 1]. The matrix 1
P=
1
is such that P-1 AP = diag{A1,A2}. In the new coordinate system, I i ] becomes P-1 { i ] = [1'1]where P2
L
P1=
f+1 2f
P2=
,
f-1 2/
We then have
an =P1Ai-1vi-P2.A2-1vi, from which we see that an
__
bn
P,An-,\/2 I PlAi-1
-P2-
a2 +P2\a-1
f-
(P2/P1)(A2/a1)n-1vi
1 +
_
bn =p1Ai-1
(p2/P1)(A2/A1)n-1
1 - (P2/P1)(A2/A1)n-1
(1
+ (P2/P1)(A2/)t1)n-1
46
+
P2,\2-1
Solutions to Chapter 1 Now since 0 < 1A2/1\1I < 1 we deduce that
hm
n-.oo
1.25
an b ,,,
=f.
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). Then ac + 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.) Let u = 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 independent vectors, 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 independent and {u, t(u), ... , tr(u)} linearly dependent. Then
tr(u) = aou + alt(u) +
+ ar-ltr-1(u),
so that, applying t,
tr+1(u) =aot(u) a+ alt2(u) + . + ar-itr(u), l 47
Linear algebra
Book 4
Continuing in this way we see that {u, t(u), ... , tr-1(u)} spans U and ajt3(u), so is a basis. By the above argument we have t[t`(u)] _ so U is t-invariant.
Let f(X)=-ao-a1X- -a,._1X''-1+X'. Then f (X) has degree r and [f(t)](u) = 0. Since U is t-invariant the restriction to of t to U induces 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 on U. 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. Hence f (X) is the (monic) polynomial of least degree such that f (tu) = 0, so f (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 have that 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 example it is necessary, in order to apply the second principle of induction, to include a proof for n = 2
q2 = 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 established in a similar way. (2) If r is non-singular then r-' exists and consequently from rtr = 0 we obtain the contradiction t = 0. Hence r is singular, so both p and q are singular and hence have 0 as an eigenvalue. Consequently we see that 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 which shows 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
Solutions to Chapter 1 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 middle column, 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-7
a
a+Q+ry
M(a,Q)ry)= a-Q- 1
a-,0
a+Q -ry
CL +ry
It is readily seen that sums and scalar multiples of magic matrices are also magic. Hence the magic matrices constitute a subspace of Mat3X3(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 if a = /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 3a
1
1 = 3a = 3a
1
3a
1
1
M(a,Q, 7)
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
Linear algebra
Book 4
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 - ry L=
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 similar 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 be such 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 that Ao fP-1(x) = 0 whence we deduce that Ao = 0. Deleting the first term in (*) 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 is linearly independent. It follows from the above that if f is nilpotent of index n = dim V then
B. = {x, f(x),..., fn-1(x)} 50
Solutions to Chapter 1 is a basis of V. The matrix of f relative to Bn is readily seen to be 0
I* = In-1
0 0
.
Consequently, if A is an n x n matrix over F that is nilpotent of index n then A is similar to I. Conversely, if A is similar to I* then there is an 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 concerning 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 the d-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 equation in 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. Consequently, (V1) ...,Vr,f(VI),.... f(Vr)}
is linearly independent in the IR-vector space V. Since V is of finite dimension over IR it must then be so over C. The given identity shows that every complex linear combination of {v1, ... , vn} can be written as a real linear combination of
V1i...,Vnif(Vl),..., f(Vn). 51
Book 4
Linear algebra
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 matrix of f relative to this basis is 0
In
In
0
1
].
Clearly, it follows from the above that if A is a 2n x 2n matrix over IR
such that A2 = -I2n then A is similar to r. Conversely, if A is similar
to r then there is an invertible matrix P such that P-' AP = r and hence A2 = (PrP-1)2 = Pr2P-1 = P(-I2n)P-1 = -12n1.82
Let x be an eigenvector corresponding to A. Then from Ax = Ax we have that xtAt = Axt and hence e T = A. Since A = A and At = -A we deduce that -Y A = ax . Thus -tAx = -ax x. But we also have
x Ax = iAx = Aix. It follows that A = -A, so the real part of A is zero. 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 hence -2'(-A + XI) = -Z (A - AI). Consequently,
Y=
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 of the form a + ib
= a 2 + b2
[a - ib ... x - iy]
+ +x2+y2
x+iy and a sum of squares is zero if and only if each summand is zero. Hence we see that Y = 0. The minimum polynomial of A cannot have repeated roots. For, if this
were of the form m(X) = (X - a)2p(X) then from (A - aI)2p(A) = 0 we 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
Solutions to Chapter 1 Thus the minimum polynomial has simple roots and so A is similar to a 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]- I a 0][ v]' The last part follows by choosing is 1 i ... , iak to be the non-zero eigenvalues 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 see that t2n
= t2 + 4(11 + 4 +
- + 4n-2) (t2 - id)
14(t2-id)
=t2+4(1_4n_1)
= t2 + 3 (4n-1 - 1)(t2 - id). 1.34
We have that
f(b1) = AIbl;
(i>2) f
Qiib1 + E mjibj' j>2 53
Linear algebra
Book 4
Thus the matrix of f relative to the basis {b1, b'2, ... , b,n} is of the form
A=
Pin
012
1 Al
M
0
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 is g-invariant.
It is clear that Mat (g', M. Also, the characteristic equation of 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 multiplicity of Al reduced by 1. Since all the eigenvalues of f belong to F by 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 and hence for A. 1.35 The eigenvalues of t are 0, 1, 1. The minimum polynomial is either X(X - 1) or X(X - 1)2. But t2 - t 54 0 so the minimum polynomial is
X(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 which the matrix of t is 0
0
0
0
1
1
0
0
1
54
.
Solutions to Chapter 1 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 = 0 and 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 respect to 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
[2 5
[00]
-40][y]-
so -1 has geometric multiplicity 1 with [8, 51 as an associated eigenvector. 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
Linear algebra
Book 4
(b) The characteristic polynomial is (X+1)2. The eigenvalues are -1 (twice) with geometric multiplicity 1, and a corresponding eigenvector is [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
= c
(Any Jordan basis is of the form {[c, 0], [d, -c]} with P =
-cdl
0 L
(c) The characteristic polynomial is (X - 1)3, so the only eigenvalue is 1. It has geometric multiplicity 2 with {[I, 0, 01, [0, 2,3]) as a basis for the eigenspace. The Jordan normal form is then 1
1
0
0
1
0
0
0
1
.
A Jordan basis satisfies
(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 choose any vector in ([1, 0, 0], [0, 2,31) that is independent of [3,6,9], for example V3 = [1, 0, 0]. This gives 3
P= 6 9
56
0
1
1
0
0
0
.
Solutions to Chapter 1 (d) The Jordan normal form is 3
1
0
0
3
0
0
0
3
.
A Jordan basis satisfies
(A-313)v3=0.
(A-313)v2=v1i
(A-313)vl = 0,
Choose v2 = [0, 0, 11. Then vl = [1, 0, 0] and a suitable choice for v3 is [0, 1, 0]. Thus
1.88
P= 0
1
0 0
0
0
1
0
1
The characteristic polynomial of A is (X- 1)'(X-2)1. For the eigenvalue 2 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
0
0
-1
-1
t w
-1 - 2
0
to obtain w = t = 0, y + z = 0. Thus the general eigenvector associated with the eigenvalue 2 is [x, y, -y, 0, 01 with x, y not both zero. The Jordan block associated with the eigenvalue 2 is 2
2]
[0
For the eigenvalue 1 we solve 1
1
1
1
0
x
0
0
1
0
0
0
11
0
0
0
1
1
0
z
0
0
0
0
0
1
t
0
-0
-1
-1
-1 - 1
w
0
57
Linear algebra
Book 4
to obtain w = y = x = 0, z + t = 0. Thus the general eigenvector associated with the eigenvalue 1 is [0, 0, z, -z, 0] with z # 0. The Jordan block 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
0 0
1
1
0 0 0
0 0
1
1
0
1
Take [0, 0, 1, -1, 0] as an eigenvector associated with the eigenvalue 1. Then we solve x y
0
1
1
1
1
0 0
1
0
0
0 0
1
1
0 0 0
0
0
1
t
-1
-1 -1 -1 -1
w
0
0 0
0
z=
1
to obtain y = 0, w = -1, z + t = 1, x = -1, so we take [-1, 0, 0, 1, -11. Next we solve 0 0 0
x
-1
y
0 0
1
1
1
1
0 0
1
0
0
0
1
1
0
0
0
1
t
1
-1 -1 -1
-1
w
-1
0 0
z=
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 is
P=
1
0
0
-1
-1
0
1
0
0
0
0
-1
1
0
0
0
0
-1
1
0
0
0
0
-1
1
58
Solutions to Chapter 1 1.39
(a) The Jordan form and a suitable (non-unique) matrix P are
J=
2
1
0
2
0 0
0
0
2
2
-5 -3
3
8
2
P=
,
5 8
.
-7
(b) The Jordan form and a suitable P are
1.40
2
0
0
0
4
3
2
07
0
1
1
0
5
4
3
0
0
0
1
1
-2 -2
-1
0
0
0
0
1
6
4
P=
11
1j
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
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. 1.41
The minimum polynomial is (X - 2)3. There are two possibilities for the Jordan normal form, namely 2
1
0
0
0
2
1
0
0
01
0
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 two linearly independent eigenvectors; e.g., [0, -1,1,1, 0] and [0, 1, 0, 0, 11. The number of linearly independent eigenvectors does not determine the 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
Linear algebra
Book 4
have two linearly independent eigenvectors. Both pieces of information are required in order to determine the Jordan form. For the given matrix this is J2. 1.42
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 represented by 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
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}. 1.43
The possible Jordan forms are 3
0
0
0
3
0
0
0
3
,
3
1
0
3
1
0
0
3
1
0
3
0
0
0
3
0
0
3
,
3
0
0
3
1.
0
0
3
0
The last two are similar. 1.44
(i) and (ii) are true : use the fact that AB and BA = A-' (AB)A are similar. (iii) and (iv) are false; for example, 0
[0
1
0][0
0]
and
1
clearly have the same Jordan normal form. 60
0
[0 0][0 01
Solutions to Chapter 1 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 one eigenvalue of t. Without loss of generality we can assume that t has a single eigenvalue A. Consider an i x i Jordan block. Corresponding to this 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. For every 2 x 2 block there correspond two basis elements in Ker(t - A idv)j if j > 2 and 1 basis element if j < 2. In general, to each i x i block there correspond i basis elements in Ker(t - A idv)' if j > i and j basis elements if j < i. It follows that
dj=n1+2n2+-..+(j-1)nj-1+9(nj+nj+l+...) 1.46
and a simple calculation shows that 2di - dt_1 - di+1 = ni. The characteristic polynomial of A is (X - 2)4, and the minimum polynomial 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 Jordan form is 2
1
0
0
0
2
0
0
0
0
2
0
0
0
0
2
61
Linear algebra
Book 4
Now Ker(A - 214) = {[x, y, z, t]
2x - y + t = 0}, and we must choose
I
v2 such that (A - 214)2v2 = 0 but v2 0 Ker(A - 214). So we take v2 = [1,0,0,0], and then v1 = (A - 214)v2 = [-2,-2, -2,21. We now wish 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
-2 P= -2 -2 2
1
0
0
0
1
0
0
0
1
0
1
0
To solve the system X' = AX we first solve the system Y' = JY, namely
yi = 2yi + Y2 y2' = 2y2
ys=2y3 y4 = 2y4
The solution to this is clearly Zt y4 = c4e 2t
y3=c3e
2t
y2 = c2e
Yi = c2te2t +
cle2t.
Since now
-2 X=PY= -2 -2 2
L
we deduce that
0
0
c2te2t + cie2t
0
1
0
c2e2t
0
0
1
c3e2t
0
1
0
c4e2t
1
xi = -2c2te2t - 2c1e2t + c2e2t x2 = -2c2te2t - 2c1e2t +
C3e2t
x3 = -2c2te2t - 2cie2t + C4e2t x4 = 2c2te2t + 2cie2t + c3e2t.
62
Solutions to Chapter 1 1.47 (a) The system is X' = AX where
[-1
`4 =
0].
The characteristic polynomial is (X - 1)(X - 4). The eigenvalues are therefore 1 and 4, and associated eigenvectors are El = [1, -1] and E4 = [4, -1]. The solution is aEjet + bE4e4t, i.e.
xl = aet
+ 4be4t
x2 = -aet - befit
(b) The system is X' = AX where 4
A= 1
-1 -1 2 -1 -1
1
.
2
The characteristic polynomial is (X - 3)2(X - 2). The eigenvalues are therefore 3 and 2. An eigenvector associated with 2 is [1,1,1] so take 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 that xi = ae2t + (b + c)e3t
x2 = ae2t + be 3t
x3 = ae2t +
ce3t.
(c) The system is X' = AX where
-6 -6
5
A= -1
4
2
.
-6 -4
3
The characteristic polynomial is (X - 1)(X - 2)2. The eigenvalues are therefore 1 and 2. An eigenvector associated with 1 is El = [3, -1,31, 63
Linear algebra
Book 4
and independent eigenvectors associated with 2 are E2 = [2, 1, 0] and E2 = [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
0 0
0
1
1
1
J= 0 0
and an invertible matrix P such that P-1 AP = J is 0
1
6
1
0
9
0
0
3
P=
.
First we solve Y' = JY to obtain y' = yi + y2, y2 = Y2, y3 = ys hence ys = cet ya = be
t
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 . 1.48
The system is AY' = Y where
-3 A= 0 -5 1
-9
0
2
4
.
7
Now A is invertible with
A-' = 0
3 7
-2 -4
0
9
-5
1
and the system Y' = A-'Y is that of question 1.47(d). 64
and
Solutions to Chapter 1 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 = and b = 2 , so the solution is 2
e2t
([1,1 + V31e
f3t + [-1, -1 + -V3]e- t ).
2Nf3-
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 0
A=
-8
1
0
0
1
4
2
.
The characteristic polynomial is (X - 2) (X2 - 4) so the eigenvalues are 2 and -2. The Jordan normal form is 2
1
J= 0
2
0
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 an invertible matrix P such that P-1AP = J is 1
P= 2 4 65
0
1
1
-2
4
4
.
Linear algebra
Book 4
Now solve the system Y' = JY to get yi = 2yl + y2, y2= 2y2, y3 = -2y3
so that Y2 = c2e2t, y3 = yl = cle2t + c2te2t. Now observe that
c3e-2t and hence yl
1
0
1
X=PY= 2
1
-2
4
4
4
= 2yi + c2e2t which gives
cle2t + c2te2t
c2 e2t
c3e-2t
Hence x = xl = cl e2t + c2te2t + c3e-2t. Now apply the initial conditions to obtain x = (4t - 1)e2t
66
+e-2t.
Solutions to Chapter 2
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-+ J 1f 2 is not a linear mapping; for example, we have 0 = t9[ f + (-f)] whereas in general
f f2#0.
,O(,f)+0(-f)=2 2.2
1
0
That cp is linear follows from the fact that
= a fo 1fo(t) f (t) dt +,Q f 1 .fo (t) 9(t) dt 0
= c
2.3
(f) + &M.
The transition matrix from the given basis to the standard basis is
P=
1
-1
0
1
1
-1
0
1
0
The inverse of this is readily seen to be 1
P-1 = _ 12
1
2 1
1
-2 1
2
2
2
1
1
1
2
2
2
Linear algebra
Book 4 Hence the dual basis is {[2, 2, 2.4
21, [-2, 2,
21, [2,
2]}.
The transition matrix from the basis A' to the basis A is
P-I3 Its inverse is
4].
p-1 = [-2
-2 ].
Consequently, (A')d = {-2
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 is true is : if V = A ® B then V d = A' ® B' where A', B' are subspaces of Vd with A' Ad and B' Bd. To see this, let f E Ad and define f : V -+ F by f (v) = f (a) where v = a + b. Then V : Ad --> Vd given by o(f) = f is an injective linear transformation and V(Ad) is a 68
Solutions to Chapter 2 subspace of Vd that is isomorphic to Ad. Define similarly µ : Bd -+ Vd by µ(g) = g where g(v) = g(b). Then we have V d = co(Ad) ® µ(Bd). 2.8
{ fl, f2i f3} is linearly independent. For, if Al f1 + A2f2 + )t3f3 = 0 then we 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 we must have fg(p3) = Sgt; i.e. pi(ts) = Std.
It is now easily seen that
pl(X) - (tl
p2(X) - (t2
- t2)(tl - t3)' p3(X) - (
2.9
(a) a^ ('p) _ cp(a) = [ 3
4][2]= 11;
(b) Q" ('P) = p(Q) = [ 3
4][6]= 39;
(c) (2a+3Q)^(,p) = p(2a+3Q) _ [3
(d)(2a+3Q)^([a
-
tl)(t2- t3)'
t2)
4](2I 2]+3[6]
I = 139;
b])=[a b](2{ ]+3I6]l=17a+22b. 6\\\9
L
J
Linear algebra
Book 4
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.}. Given cpEVd we have
p= al p, + ... + a.Pn Since V(vi) = a; we see that
+ 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 we have
[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 a solution, 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 from the 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 transition matrix and its inverse are 1
P= 0 0
1
1
1
1,
0
1
1
P'1= 0 0
70
-1
0
1
-1
0
1
.
Solutions to Chapter 2 Thus Y = {(1, -1, 0), (0, 1, -1), (0, 0,1)}. The matrix of t with respect to the standard basis is 0
2
1
1
1
1
0
0
-1
and the transition matrices relative to X, Y are respectively 1
0
-1
1
0 0
0
-1
1
,
1
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 and cp'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 0f(t) g(t) dt = f(x)
Now, by continuity, the left hand side depends on the values of f at points other than x whereas the right hand side does not. Thus Fr 54 Lg for any g. 71
Linear algebra
Book 4
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) = 0i=1
Conversely, if x =
l aiui E V and if (nix) = 0 then k
0 = (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 to U. 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 multiplicity
k - 1. Also, t(n) = -n and so -1 is also an eigenvalue with associated eigenvector n. Since the sum of the geometric multiplicities is k it follows
that t is diagonalisable. 72
Solutions to Chapter 2 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 gives 1
mats =11-
-7 6 -6 6
9
2
_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
matt = 5 I
2.17
1
2
-4
2
2
4
2
-1
-4
2
1
2
2
-1
2
4
The hyperbola is represented by the equation
yJ[ 3 -11[y]= 1.
Lx
The eigenvalues of A = [
3
-11 are 2 and -4 with associated nor-
malised eigenvectors
1/V 1/vf2
[1/v, 1/-,42
Thus PAP = diag{2, -4} where
P1/f -1// 1// 1/f 73
Linear algebra
Book 4
Now change coordinates by defining
xl
Pt 1
- [ X, I Y1
X yI
= Ptx.
Then we have x = Pxj and the original equation becomes 1 = xt Ax = (Pxl )t A(Pxl )
=xiPtAPxj
_ [xi
yi
0
]I
yi,
O4lIx
= 2z2 - 4y1.
Thus the principal axes are given by xl = 0 and yl = 0, i.e. y = -x and y = x. The ellipsoid is represented by the equation [x
7
2
0
0
6 -2
5
zJ
y
The eigenvalues of A =
7
2
0
2
6 -2
-2
0
][;]. z
a re 3, 6, 9 with associated nor-
5
malised eigenvectors
-1/3 2/3
2/3
-1/3
,
2/3 2/3
,
-1/3
2/3
2/3
Thus Pt AP = diag{3, 6, 9) where
P=3
-1
2
2
-1
2 2
2
2
-1
.
Now change coordinates by defining
xi =
xl
x
Yi =
y = Ptx.
zl
pt 74
z
Solutions to Chapter 2 Then we have x = Pxj and the original equation becomes (with a calculation similar to the above) X2
2 = 1. + 6y12 + 9z1
Since
x1 = y1 = z1 =
(-x + 2y + 2z), s (2x - y + 2z), s (2x + 2y - z), s
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 V we have x = >'=1(xIvk ) vk so in particular
= (i = i,...,n)
f(vi)
n
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 the matrix of f* then likewise we have that mjk = (f*(vj)Ivk ). The result now 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 that tr(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
Linear algebra
Book 4 2.20
Let Q be as stated and let fp E V* be given by fp(a) = (aJQ ). Then since {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
P(z) = f (P) = fo1 P (t) q(t) dt for every p. Writing rp for p we then have
r(t) p(t) q(t) dt = r(z)p(z) = 0. f In particular, this holds when p = rq. So 1
1
Ir(t)I2 Iq(t)I2 dt = 0
fo
whence we have rq = 0. Since r 54 0 we must therefore have q = 0, a contradiction. For the next part of the question note that
4
(fp(q)Ir) = (pglr)
= f0 =
f
1
p(t) q(t) r(t) dt
1
q(t) p(t) r(t) dt
0
_ (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
Solutions to Chapter 2 2.21
That K is self-adjoint follows from the fact that xy is symmetric in x and y. We have
f -f
K(fn) =
1
xy fn(y) dy
0
2
n+2)
xy (yn
0
1
= f xyn+1 dy xyn+2 Y-1
n+2
dy
- n 2 2 Jo 2
1
xydy xy2I!-1
2
n+2 2
x-0
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
a=- (x2- 2, x-
(x-2 ,x-2)3 3
3
Since
(x2-a,x-3)- f (x2
)(x
)dx-9,
0
(x-3,x-3f =(x - 3)2) dx= 1
it follows that a = -1 and hence that e2(x) = x2 - x +
g.
Thus e1, e2 are orthogonal eigenfunctions associated with the eigenvalue 0.
If K(f) = Af then 1
Xf(x) = x fo yf(y) dy 0
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 associated eigenfunction is given by f (x) = x. 77
Linear algebra
Book 4 2.22
Since t is skew-adjoint its eigenvalues are purely imaginary so ±1 are not 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, +s)-1 and so (id = a (id +t) whence id +s does not have 0 as an eigenvalue 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-adjoint matrix 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
Solutions to Chapter 2 2.24
It is given that At = A, St = -S, AS = SA,det(A - S) # 0. Let B = (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 do A + S and A - S. Hence Bt B = I and B is orthogonal. 2.25
Since
AA+A=0
(1)
we have, taking transposes,
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. Let A1, ... , An be the distinct non-zero eigenvalues of A. Then Al, ... , An are 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 of A2, namely A I. ... , An. It follows that Al, ... , An are all negative. Let
ai = -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, we
obtain A; = -ai = -1. 79
Linear algebra
Book 4 2.26
A and B are given to be orthogonal with det A + det B = 0. Now we have 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 eigenvalue of 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)t and so, taking determinants,
det(I+ C)[det C - 1] = 0 whence det(I+ C) = 0. It now follows from (1) that det(A + B) = 0.
(1) Since At(A- I) = I - At = -(At - I) we have that det Adet(A - I) = (-1)' det(A - I)
2.27
and so
det(A - I)[det A - (-1)'] = 0. If det A = 1 and n is odd then it follows that det(A - I) = 0 and hence 1 is an eigenvalue of A. If det A = -1 and n is even then likewise det(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 of A.
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)rXr is 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
Solutions to Chapter 2 2.29
We have
N(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 l l
= X---41A-1
and so A-1 is normal.
If A = aoI + a, A + + an An then clearly AA = A A. Suppose conversely that A is normal. Then there is a unitary matrix P and a diagonal matrix D such that
A =P DP=P-1DP.
A=P-1DP=P DP,
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 +
r-1.
ao + a,,\, + a2A2 r + ... + a,.-1
81
Linear algebra
Book 4
Since A1,...,A,. are distinct the (Vandermonde) coefficient matrix has non-zero determinant and so the system has a unique solution. We then have
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 matrix P 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
V B = P g(D)PPg(D)P = Pg(D)g(D)P and similarly
BB = P g(D)g(D)P. Since g(D) and g(D) are diagonal matrices, it follows that B B = BB and 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 follows that 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. 82
Solutions to Chapter 2 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 using A* = 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 the question). If, therefore, At = 0 then
put y = utAz = -(Au)tz = 0 and similarly
putz = -utAy = (Au)ty = 0. For the last part, we have -A
det(A - AI) = det
2
-2 -A 2
-2
-1 = -A(A2 + 9), -A
1
so the eigenvalues are 0 and ±3i. A normalised eigenvector corresponding to 0 is 1
U=3
-1 2
.
2
To find y, z as above, choose y perpendicular to u, say
83
Linear algebra
Book 4 Then we have
-4
-3z=Ay= 1 -1 1
which gives
f Relative 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
0 4/3f
-1/3
2/3 -1/f 1/3f 2/3 1/f 1/3f
P= 2.84
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 where x = 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 that the rank of q is equal to the rank of the matrix B which is in turn equal to 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. Hence ytBy > 0 for all y E IRn since it is a sum of squares. Consequently xtAx > 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 or
1, but not -1. Therefore there are no terms of the form -y?, so m = 0 and q has the same rank and signature. 84
Solutions to Chapter 2 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 > 0 and is equal to 0 only when x = 0.
Conversely, if xtAx _> 0 for x E IR' then ytBy > 0 for y E IR" so m = 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 gives ytBy = 0 only for y = 0. If p < n then, since we have m = 0, choose y = (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 completing squares or by row and column operations. We solve the problem by completing 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
1
0 0
0
0
-1
.
Since the rank of q is 3 and its signature is 1, q is neither positive definite nor 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 elements can be taken as the columns of the inverse of
i.e. {(O, 1, 0), (1, -1, -1), (0, 0,1)}. 85
Linear algebra
Book 4 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 the matrix of q relative to the standard basis is 3
1
2
3 2
1
Completing squares gives (x1 + 2x2)2 The signature is then 0 and the rank is 2. The form is neither positive definite nor positive 4x2.
semi-definite.
2.87 In matrix notation, the quadratic form is
-1
4
xtAx=[x
z, -1
y
4 -1
1
1
x
-1
y
4
z
.
It is readily seen that the eigenvalues of A are 3 (of algebraic multiplicity
2) and 6. An orthogonal matrix P such that PAP is diagonal is
1/f 1/f _1/-,/3 1/f
P = 2/f
0
1/v
1/\ -1/V2_
Changing coordinates by setting u
11
V =Pt y w
z X
transforms the original quadratic form to
3u2 + 3v2 +6w 2
which is positve definite. 86
Solutions to Chapter 2 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 - Z)2 - 2 Z2
-
W.
2
Thus the rank is 3 and the signature is -1. (3) In yz + xz + xy + xt + yt + zt put
x=X+Y, y=X-Y, z=Z, t=T. Then we obtain
(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
A= -1
2
-3
2
2
-3 . 9
Now
x2 + 2y2 + 9z2 - 2xy + 4xz - 6yz
=(x-y+2z)2+y2+5z2-2yz = (Cx- y+2z)2 +(y-z)2 +4z2 = [2 + Q2 + S2 87
Linear algebra
Book 4
where E=x-y+2z,, =y-z,s=2z. Then
z=1 y=rI+2S
X+q - ZS 1
1
P= 0
21
1
z
0
0
2
and PtAP = diag{1,1,1}. (2) Here the matrix is 0
2
A= 2
0
0 1
0
1
0
.
Now
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 let 1
1
1
2
2
-2
0
0
P= 2 -2 88
0 1
Solutions to Chapter 2 then we have
and Pt AP = diag{1, -1, 0}. (3) Here we have
A-
1
1
1
4
0 3
3
1
0
-1 -4 -7
-4 -7 -4
-1
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.
x=-7s-2r
Then
Y=
1n-+3T
z=
S-2r
t = T
and so 1
P=
-1/f 1/f -2
0
1/\ -1/'
o
o
o
0
1/
gives
x
y
z
=P
71
Ti
t
and Pt AP = diag{1, 1,-1,0}. 89
3
2
-2
0
1
Book 4
2.40
Linear algebra
Here we have 2
(xr - xa)
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
n-1
-1
-1 -1
n-1
-1
-1
A=
-1 -1
-1 -1 -1
... ... n - 1 ...
-1
-1
... n - 1
1
Now, by adding n 1 1 times the first column to columns 2,... , n and adding n 1 times the first row to rows 2, ... , n, then multiplying rows 2, ... , n and columns 2, ... , n by the matrix
n- 1
0
0
n-2
0
-1
0
-1
nn l , we see that A is congruent to
... ... n - 2 ... 0
0
-1
-1 -1
-1
...
n-2
Repeating this process we can show that A is congruent to the matrix
n-1
L
0
0
0
n-2
0
0
0
n-3
0
0
-1
0
0
-1
... ... -1 ... n - 3 ... 0 0
90
-1
...
0 0
-1 -1
n-3
Solutions to Chapter 2 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 have n
E (Ars+r+s)xrx, r,e= 1 n
_
n
E (rxr)(sx,) + >2 (rxr)x. + E xr(sxe) r,8=1 r,8=1 r,e=1
l(n 12 rxr) +2 (n > xr) I > rxr
n
_ .1
n
I
r-1
r=1
r,s=1 _ A(x1 + 2x2 +
+ nxn)2
+ 2(x1 +
+ xn)(x1 + 2x2 +
+ nxn).
Now let =x1+2x2+...+nxn
y1
y2=xl+x2+
+xn
y3 = x3
yn = xn
Then the form is Ayi + 2y1112 which can be written as
A(yl + *y2)2 - 1 y2
if A
0;
if A = 0. I(yl + 112)2 - 4(y, - y2)2 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 such that Ax = Ax where x = [a1 ... an]t. Then xtAx = Axtx and so, if Q = xtAx then we have Q(a1,... , an) = A(ai + ... + an),
2.44
Let Q(x,x) = xtAtAx = (Ax)tAx. Since detA $ 0 we may apply the non-singular linear transformation described by y = Ax so that Q(x, x) - Q(y, y) where
Q(yfY) =VY=yl + "'+ynThus Q is positive definite. 91
Linear algebra
Book 4 2.44
Let {u1, ... , un} be an orthonormal basis of the real inner product space IRn under the inner product given by (x1y) = f (x, y). Let x = E 1 xiu1 and y = ni_1 yiuiThen n
n
n
/
f (x) y) _ (xIy) = (E xiui I > yiui) = E xiyi i=1
i=1
i=1
and, for some real symmetric matrix B = [bij[nxn, n
n
g(x, y) = > > bi? xi yj = xt By i=1 j=1
where xl
Yi
x=
y= yn
Lxn
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 (relative to the inner product determined by f) and consists of eigenvectors of B. Let x = EE ;vi and y = 1 pivi. Then, relative to this orthonormal basis, we have n
f (x, y) _
e£fli; ti=1
n
eirli
g(x,y) _ i=1
Consequently,
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
Solutions to Chapter 2 Since, from the above expressions for f (x, y) and g(x, y) computed relative 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 respect to some ordered basis of IRn. If xl
y l
Y=
x 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 the equation 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
B= 0
A= 1 101, 0
1
0
0
1
0
0
-1 0.
1
Since 1
-A
1
-A
0
-1 -A =A2+1
det(B - AA) = det -A 1
the equation det(B-AA) = 0 has no solutions. But, as observed above, if a simultaneous reduction to sums of squares were possible, such solutions would be the coefficients in one of these sums of squares. Since neither of the given forms is positive definite, the conclusion follows. 2.45
The exponent is -xeAx where 1
A=
1
1
2
2
2
1
2
1
1
93
1
Book 4
Linear algebra
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 to ir3/2/ 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 matrix 1
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 Jordan normal form.
S
Let rP1, ... , rPn E (IRn)d. Prove that the solution set C of the linear inequalities
rP1(x)>0, rai(x)>0, ... ,rpn(x)>0 satisfies
(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)x show, by considering yty, that y = 0. Hence prove that an orthogonal matrix satisfies an equation without repeated roots. Prove that a real orthogonal matrix with all its eigenvalues real is necessarily symmetric.
5
Prove that if a real quadratic form in n variables is reduced by a real non-singular linear transformation to a form
having p positive, q negative, and n - p - q zero coefficients then p and q 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 + an_1xn_1xn + anxnxl
a1x1x2 + a2x2x3 +
in which each a;
0, show that
Ip - ql = I
if n is even; if n is odd.
0 1
97
Test paper 2
Time allowed : 3 hours (Allocate 20 marks for each question)
Let V be a finite-dimensional vector space and let e E £(V, V) be a projection. 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 that Im 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
ifs=r+
_ J1
ura = l 0
otherwise,
and J = [jre] E Matnxn(C) is given by
jr° -
ifr+s= n+1;
11
otherwise,
0
show that Ut = JUJ. Deduce that if A E Matnxn(() then there is an invertible 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 W is 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, show that 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 f(x) = 0
f2(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 and only if there are complex scalars Al, ... , Ak and self-adjoint projections
el,...,ek on V such that (1) f = A1e1 + (2) ides = e1 + b
+ Akek; + ek;
(3) (i34 j) eioej=0. (a) Show that the quadratic form xtAx is positive definite if and only if there exists a real non-singular matrix P such that A = PP. Show also that if Ei j_1 bijxixj > 0 for all non-zero vectors x then bijpixipjxj > 0 for all x. Hence show that if xtAx and xtBx are both positive definite then so is n
E at,bijxtx,'
i,j=1
(b) For what values of k is the quadratic form n
xr+k>xixj i<j
r=1
positive definite? 99
Test paper 3
Time allowed : 3 hours (Allocate 20 marks for each question) 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
n
dim ®(U1 n S). i=1
2
For the matrix 0
1
A= -1
0
1
1
-1
0
2
find a non-singular matrix P such that P-'AP is in Jordan normal form.
If A is positive, obtain real values of x;j such that C2 = D where X101
X12
x13
0 0
x22 0
x23 x33
,
U
A
I
D=0
A
1
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 be subspaces of V. Prove that
(W+X)-L=W1nXl and (WnX)1=W1+Xl. Given gi, ... , gn E V d, prove that the following conditions concerning f 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
A=
1
1
1
1
2
2
2
2
1
1
1
1
2
2
2
2
12
-2i
12
-2
1
1
1
1
2
2
2
2
1
is orthogonal. Find its eigenvalues and show that the matrix A2 - I4 has 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
Time allowed : 3 hours (Allocate 20 marks for each question)
Let F be the vector space of infinitely differentiable complex functions and let Pn be the subspace of complex polynomial functions of degree less than n. For every A E d define Pn,a = {eazp I p E Pn}. Show that Pn,.\ is a subspace of F and that zk
B = {eaz
I
k = 0,...,n - 1}
k!
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 characteristic polynomial of Dn,A. If µ iA A show, by considering the characteristic polynomial of Dn,A + (µ - A) id, that (Dn,A - µ id)n is invertible. 2
Given A = I
a
b
use the Cayley-Hamilton theorem and euclidean
division to show that every positive power of A can be written in the form
An=th12+Q2A.
If the eigenvalues of A are Al, A2 show that n
A
-
\n \n A2A1 - A1A2 lA I2 + A2 - Al AZ - Al
if Al # A2,
(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 Jordan normal form of f. Let f, g E C((6, Q8) be nilpotent with the same minimum polynomial and dim Ker f = dim Ker g. Show that f, g have the same Jordan normal form. By means of an example show that this fails in general for f, g E
C(d', (7). Deduce that if s, t E C(Cn, Qn) have the same characteristic polynomial
(X - a1)k'(X - a2)k' ... (X - ar)kr and the same minimum polynomial, and if
dim Ker(s - at id) = dim Ker(t - ai id)
4
for 1 < i < r, then s and t have the same Jordan normal form provided ki < 6 for 1 < i < r. 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
in. a be such that p' = 0 and pn-1 # 0. (ii) Let p E C(VV)
which case dim Im s <
Show
that there is a basis B = {z1,...,xn} of V such that p(xj) = xj+1 for
j = 1,...,n - 1 and p(xn) = 0. Show that if t = > 1 Aipi-1 where each Ai E F then t commutes with p. Conversely, suppose that t E C (V, V) commutes with p and is represented relative to the basis B by the matrix [a=j]nxn Prove by induction 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, ... , 103
pn-1
b
Show that each of the quadratic forms
6x?+5x2+7x3-4fx2x3i 7yi + 6y2 + 5y3 + 4yi y2 + 4y2Y3
can be reduced by an orthogonal transformation to the same form 2 2 aIzl+a2z2+a3z3.
Obtain an orthogonal transformation which will convert the first of the above forms into the second.
104
