Solutions to Selected Problems Chapter 1 1.7. Solution Clearly 1]_1 [2]-1=[1/2], 3 1]-1=~ [ 3-1] 1 =_1! -I5 -1 5-1 -1] [1 3 8 -I 3' 4 18 -I -I 5 and the sum of the elements in the inverse matrix is ~ in each case. We show that this is true in general. The matrix A3=nl + j, where j is the matrix every element of which is I, and in the special cases above the inverses are linear combinations of I and j. Let us see if this is true in general. Assume A;;1=CtJ+/3J. Then (al+/3})(nl+})=1 which gives nal +(a+/3n)) +/3]2= I which can be satisfied, since ]2=nj, by taking 1 a=-, n Since the inverse is unique we have and the sum of its elements is nx~+n2X ( __1_) = I_~=~. n 2n2 2 2 The answer in the case of the Hilbert matrix is n2• See e.g. D. E. Knuth, The art of computer programming, I (1968), pp. 36/7, 473/4. 1.9. Solution Since R(x}=R(rx) for any r¥O we may replace the condition x¥O by x' x = 1. We know that we can choose an orthonormal system of vectors 120 Solutions to Selected Problems c1 , C2 , ••• , Cn which span the whole space Rn and which are characteristic vectors of A, say Ac;=ex; c;, i= 1,2, ... , n. Hence we can express any x, with x'x= 1 as where .2'~; = 1. Since =.2'exj~;~jC;Cj i, j = .2' ex; ~r (by orthonormality) ; we have exn = an .2' ~r;§ R (x) = .2' ai ~r;§ a1 .2' ~r = a1 • iii Also, clearly, for any i, and so the bounds are attained. In view of the importance of the Rayleigh quotient in numerical mathe­ matics we add three remarks, the first two dealing with the two-dimensional case. (1) We show how the Rayleigh quotient varies when A=[~ ~l By homo- geneity we can restrict ourselves to vectors of unit length say x = [~~~:l Then Q (8) = a cos2 8 + 2 h cos 8 sin 8 + b sin2 8 =~ [(a-b) cos 28+2 h sin 28+ (a + b)]. To study the variation of Q(8) with 8 observe that q ( <p) = a cos <p + (3 sin <p + y = yex 2+ (32 [(exl Ya 2+ (32) cos <p+ ((31 Yex 2+ (32) sin <PJ + y = Yex 2 + (32 sin (<p + 1/1) + y where sin 1/1 = aNex 2+ (32, cos 1/1 = (3N ex2+ 132, and so q(<p) oscillates between y±yex2 +(32. Hence Q(8) oscillates between (the two real numbers) i.e., between the characteristic values of A. Chapter 1 121 (2) The fact that the characteristic vectors are involved can be seen by use of the Lagrange Multipliers. To find extrema ofax2+2hxy+by 2 subject to x 2 +y2=1, say, we compute Ex, Ey where Then Ex =2(a-A)x+2hy, Ey=2hx+2(b-A)Y and at an extremum (a-A)x+hY=O} hx+(b-A)y=O . For a non-trivial solution we must have a-A h] det [ h b-A =0, i.e., A must be a characteristic value of [~ ~]. (3) A very important general principle should be pointed out here. At an extremum Xo of y=f(x) at which f(x) is smooth, it is true that x "near" Xo implies f(x) "very near" f(xo). In the simplest case, f(x)=x2 and xo=O, we have f(x)=x2 of the "second order" in x; this is not true if we do not insist on smoothness, as in shown by the case g(x) = lxi, xo=O, in which g(x) is of the same order as x. We are just using the Taylor expansion about Xo: f(x)-f(xo) = (X_XO)2 [~f"(xo)+ ... ] in the case where f'(xo) =0. This idea can be generalized to the case where y=f(x) is a scalar function of a vector variable x in particular y=R(x). It means that from a "good" guess. at a characteristic vector of A, the Rayleigh quotient gives a "very good" estimate of the corresponding characteristic value. 1.10. Solution (1- 2 m m') (1- 2 m m')' = (1- 2 m m') (1- 2 m m') = 1- 4 m m' + 4 m m' m m' = 1- 4 m m' + 4 m (m' m) m' = 1- 4 OJ m' + 4 m m' Matrices of the form of 0 were introduced by Householder and are of great use in numerical algebra. (See e.g. Chapter 8.) 122 Solutions to Selected Problems Chapter 2 2.4. Solution Assume p:> 1, .!:.+~= 1, a:>O, {3:>O. .1'=.'(,-1 p q B a P Then area OA'A= IxP-ldx =~ o p P {3q and area OBB'= J yl/(P-l)dy=_. o q x Clearly the area of the rectangle OA' C B' is not greater than the sum of the areas of the curvilinear triangles OA' A and OBB' and equal to it only if A, B and C coalesce. Hence with strict inequality unless {3q=aP. This inequality, when written in the form Al/p B1/q;§ (Alp) + (Blq) can be recognized as a generalization of the Arithmetic-Geometric Mean inequality from which it can be deduced, first when the weights p, q are rational and then by a limiting process for general p, q. If we write a-H {3-JbL in this inequality we find - Ilxllp' - Ilyllq (1) Adding the last inequalities for i=l, 2, ... , n we find so that (H) Chapter 2 123 This is the Holder inequality. There is equality in the last inequality if and only if there is equality in all the inequalities (1) which means that the Ixil P are proportional to the IYil q. Observe that when p = q = 2 the inequality (H) reduces to the Schwarz inequality (S) Observe also that the limiting case of (H), when p = 1, q = =, is also valid. In order to establish the Minkowski inequality (M) we write and sum, applying (H) twice on the right to get L: (IXi! + !YiJ)P;:§ Ilxll p[L: (lXi! + !Yil)(P-ljqJ/q + lIyllp [L: (lxd + !YiJ)(P-ljqJ/q. Observe that (p-1)q=p, so that the terms in [ ] on the right are identical with that on the left. Hence, dividing through, [L: (lxd + !YiI)P]l-(l/qj;:§ Ilxll p+ Ilyllp, i.e., since 1-(l/q) = lip, The equality cases can easily be distinguished. We have therefore shown that the p-norm satisfies Axiom 3, the triangle­ inequality. The proofs that Axioms 1, 2 are satisfied are trivial. To complete the solution we observe that which we can write as Ilxll!.;:§ Ilxll~;:§ n Ilxll!.· Taking p-th roots we get IIxll=;:§ Ilxllp;:§ n1 / P Ilxll= and, since as p -+ =, we have llxll=;:§ lim II xll p;:§ Ilxll=· P-= 124 Solutions to Selected Problems 2.5. Solution See sketch. For simplicity we have only drawn the part in the first quadrant. Each set is bounded, closed, convex and symmetrical about the origin ("equilibrated") and has a not-empty interior. 2.6. Solution See sketch. For simplicity we have only drawn the part in the first quadrant. This set Ilxll ~ I is not convex but has the other properties of those in Problem 2.5. The triangle inequality is not satisfied: e.g., x=[O, I)" y=[l,O)" x+y=[I, I)' Ilx+ yll =23/2>2= Ilxll + Ilyll· 2.7. Solution See sketch. For simplicity we have only drawn the part in the first quadrant. The set Ilxll ~ 1 has the properties listed in Problem 2.5 and the axioms are satisfied. Chapter 2 125 .,, 2.B. Solution If PI and P2 are equivalent and if P2 and P3 are equivalent then PI and P3 are equivalent for we have -c PI (x) P2 (x) PI (x) -c O<P12P23=-(-) .-(-) =-P( ) =P32P21 <=. P2 X P3 X 3 X It will be enough, therefore, to prove that any norm p(x) is equivalent, e.g., to the Chebyshev norm ll(x)=llxL. The set S={x: ll(x) = I}, the surface of the appropriate cube, is closed and bounded. Any norm p(x) is continuous everywhere. Let m, M be its lower and upper bounds on S, so that m~p(x)~M, xE S. Now, by continuity there are vectors x m , XM in S such thatp(xm)=m,p(xM)=M and, since Ilxmll = 1, m>O and we have O<m~M<=. For any vector x;;:<,O there is a k such that x=k;, where ll(;) = 1. We have, therefore, p(x) p(k;) Iklpm p@. q(x) ll(k;) Iklll(;) Hence, for x;;:<,O, p(x) O<m:s--:sM<=- ll(x) - . It is instructive to deal with the two-dimensional case of the second part geometrically, drawing the contour lines of the norm surfaces z=p(xI , x 2). By homogeneity, the ratio IIXI12/11XIII is equal to its value at the vectors Xl' X2 where these are chosen to make II XliiI = 1, II X2 111 = Vi. Since Xl is inside the 126 Solutions to Selected Problems circle IIx11 2= 1 we have IIXI 1l 2:2I so that IIXI I1 2:2I = IIXIII I and then IIXI12:2IIXIII' Since X2 is outside the circle Ilx1l 2=I we have IIX2112/IIX2111~I/y'2 and so V21IXI12~IIXIIII' x 2 We deal with the general case analytically.