hw4solutions

# hw4solutions - Qct.12.2001 8:42AM IDA CCRP No 7255 P 2/7...

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Unformatted text preview: Qct.12.2001 8:42AM IDA CCRP No. 7255 P. 2/7 Chapter 4 Linear operators and matrix inverses 3 2 4.1-1L=%,—2%,—{-t. 4.1-2 With the given solution. t if}: = zoe‘" + b“) + / “Baa-ﬂbUWT = “(0 + W), 0 so that %% — b = a1:(t). as required. 4.1-3 We will ﬁrst ﬁnd 9. m) = j: a.(t) f0 2 11mm) ds dt, Let 0 otherwise. W’s) = {11(3) 5 <t Then 1 1 ~ 1 1 _ f(z) =/o (1(0/0 b(t1 s)m(s)dsdt=/O 11(3) [/0 a(t)b(t,s)dt] ds Let y(9) = f; a(t)5(t,s) dt, or 1 3/(t)=/t a(a)b(t)ds. To show bounded, we have “in” = max "mug = max I mama“? ||=l|=1 ||=1||=1 By the CS inequality, 1 1 1 1 1 2 2 a a 2 '2 < max/O z (Unit/0 y (t)dt < max :1: Writ/D a (Gait/0 b (t)dt - lllll=1 _ llwl|=l 0 Since a and b are in L2, they are each bounded, and hence I; is also. Hence f is bounded. 4.2-4 “A1H1 = 9 "Alnz = 8.1623 "Alllp = 8.3666 “A1llw = 9 “Ath = 4 "Ath = 3.2566 [IAzHF = 3.7417 “11-sz = 3 |1A3||1 = 3 ||A3||2 = 2.4142 ||A3||F = 2.4495 “Ash” = 3 Oct.12.2001 8:43AM lDA CCRP N0. 7255 P. 3/7 76 Linear operators and matrix lnverses 4.2-5 | ‘24 011'ij | 2,- “Mil A = max Ax = max H “m uxu..=1" “°° nxuu=1 lEjamjwjl By the triangle inequality, the ith sum appearing in the vector can be written I :1“;sz S 2 laiﬂjly J' 1' with equality ifall (liij 2 0,j = 1,2, . .. ,m. For each row i, then, we consider a vector x having elements :l:1 (since at has ||x|]°° = 1) as necessary to make 04ij 2 0. Then for this row EU 2:,- = |a.-,-|. Then we consider each absolute row sum in turn, selecting the row k that has the largest Z]. IakjmJ-l = Ej lat} i. We thus obtain the largest row sum. 4.2-6 Let y = Ax. Then IIYIli = |y1|+|y21+"-+lym| =|a11111 + 0.12m + ' ’ ' + armalml + lazrwi + Q2222 + ‘ " + aszlv'ml + - ' - +Iam1w1 + amzwz + - ' ' + ammwm S lan + lazillzll + ' ' ‘ + lamillfb‘rl + lar2llwzl + lazzllm‘zl + ‘ H + lamzllrzl + lalmllwml 'l" la‘zmllzml ‘l‘ ‘ ' ' + Iammllzml + =|\$1|Zlaj1|+|\$2l2|012|+"'+[mmIZlaml- j j :i where the inequality follows from the triangle inequality. Now, since ||xl| 1 = 1, the latter term represents an average of the sum of the columns of A. Since the convex average of a set of numbers cannot be larger than its largest element, we can obtain the largest sum by selecting |xl~| = 1 corresponding to the largest column sum. 4.2-7 Letus write ﬂax, . .. ,an) = xTa. Then lf(011) * f(02)| = lxrlal — a2)l Let M = max; Then lf(01) - flaw” S M(l0!1.1 - 011,2l+ "- Ham» — (12ml) = MAa Thus for sufﬁciently small A0,. | f (on) —— f (a2)| is small. so f is continuous. 4,2-8 (a) Let x1, . . .xn be a basis for X , and let 2;; be aCauchy sequence in X, with representation m = mean + - - - + aimx» Then 2;, —z. = (am — a|,1)x1 +- - -+ (am -— 0;”wa By lemma 4.1, it follows that lot," —ou,,.| 5 Mllzk — z.” for some constant M. Hence, since {at} is Cauchy, {aid} is Cauchy (over the real numbers) and hence convergent. Let a‘j be the limiting value, and let 2" = aixr + - ' ' + afar”. Then "We - Zoll =lllak.1 — 011-)?“ + (ww- - all)an 5 law - ailllmll + ‘ ' ' + lam - ailllxu Thus zk -» 1;). (b) A ﬁnite-dimensional subspace M , by the previous result, is complete. Every convergent sequence has its limit in M , which must therefore be closed. 4.2-9 Let “All = maxlaul, and let A = B = H] . Then IlAII = IIBII = 1 IIABII = H [é i] II = 2 > llAll llBll- 4.2-10 Since (I w)“ = i=0 nu _ m-‘n = IIZF‘II S SW = 1 anu' i=0 i=0 Oct.12.2001 8:43AM IDA CCRP N0. 7255 P. 4/7 77 __________________________._._...—.——~———-———————— 4.2—1] If I - F is singular, there is a vector x such that (I — F)x = 0. This means that leH = II“!!- But “Fat” 3 HFIIHxII < “x”, which is acontradiction. 4.2-12 First the hint: Starting from I + (F —- I)(F — I)‘1 = 0, we see that I+F(I—F)" —(1—1!«‘)‘1 =0 or I — (I — F)” = —F(I — F)". Now NF” "1 _(1 _ F)'1]|=f|F(I — F)“1ISIIFIIH(I - FYI" S 1 _ HF“ where the last result follows from exercise 42-10. 42—13 (a) A+E= A(I+A"E) Since ".4419" < 1, by exercise 4.2—1]. (1 + A'IE) is nonsingular. (b) (I — F)'1A"‘ = (I + A'IE)‘1A“ = [A(I + A“E)]“ = (A+ E)". (c) Wehavc I- I — A‘IE = —A“A 80 I— A“(A+E) = -A"1E Multiplying through on the right by (A + E)‘1 we obtain (A+ E)‘1 — A“ = —A“1E(A + E)‘1 (d) "(A +E7)'1 - A’1ll = II - A‘lElA + E)”‘|l = ll — :11'111'70 — F)"A“|l _ _ - E 5 HA ‘uzuEnua — F) ‘u s 42-14 The following illustrate typical results: IfA = [5 3] then "All: = 1 and IIAHF = 1, 50 "Allz = “AHF- lfA = [(1,3)] then lle = 1 and liAllF = ‘5, 50 ||A||2 = ﬁll/1N!”- IfA = [3, 8] then Iggxlaéyj|=1 and HAH2=L so mam-,3- lai,jl = “Anz- Oct. 12. 2001 8:43AM IDA CCRP N0. 7255 P. 5/7 78 Linear operators and matrix inverses IfA=[}}]then "Allz = 2, and maxlaiJl = 1 ‘1] so “Aug = Zmaxa'j |a.-,j[. IfA== 31] then IlAllz = V5, and “Allow = 2 5° ﬁll/1N2 = "Alloc- lfA = {3] then ||A||2 = V5, and "Allow = 1 3° "Allz = ﬁll/“loo- lfA = {‘1’} then llAllz = V5, and IIAIII = 2 5° ﬁll/H12 = “All:- IfA = (1, (1,] then “Alb = ‘5. and “Alli = 1 SO "A": = ﬁll/lili- 4.2—15 The (j, k)th element of A” A is (AHAL‘Jc = 25.70in 'i so that (AHAM = 2560M = Z laijl2 Then “(AH/1) = ZMHAM = ZZI‘MIZ = “Alli?- j j a 42-16 xx" xx" xx" “Au — min: = u [(I— F; AMI — mﬂ - tr(A”A) — 2tr(5§iA”A) + tr( 1 xxHAHAxxH) _ xHx (it”x)2 _ 2 __ _2_ H H 1 2 H _ xHx til(x A Ax) + (xgx)2|iAxi|2tr(xx ) 2 1 = “All? - ﬁrzAxlli + Ell-4345 _ 2 _ AK 2 — llAllr xﬂx . since (st-(xxHAHA) = tr(x”AHAx) = KHAH Ax. 42-17 A submatrlx B can be obtained by pre- and post- multiplying A by matrices which select the desired rows and columns of B'. i.e.. that are identity matrices in the desired rows/columns, and zero elsewhere. Then B = 0114in where 01 selects the desired columns of A and Q2 selects the desired rows of A. That is, Q; and Q; contain identity matrices (possibly permuted), with zero rows and columns in columns and/or rows not selected. Hence “01" = 1 and "02" = 1. Then "B"? = lIQlAQallp s iIQxllp “All? "02",: s HAMP- ‘Oct. 12. 2001 8:44AM IDA CCRP No. 7255 P. 7/? so Linear operators and matrix lnverses 4.2-23 lf rank(A) is n, then A” A is invertible. Let B = A(AHA)‘1A". Then B is a projection matrix, and its eigenvalues are either 0 or 1. Let x = Ay, that is, any element from 92(A). Then A(A"A)“A”x = A(A”A)-1A”Ay = Ay = x, so A : 1 is an eigenvalue. 43-24 Finite dimensional adjoints. (a) (Ax,y)w = yHWAx Using A‘ = W’IAHW we have (x, A'y)w = yHWHAW'IWx = yHWAx, so the adjoint works. 0)) Using the adjoint above, (A‘A)"A‘b =(W‘1AHWA)‘1W“1A”Wb = (A"WA)"A”Wb. 4.3-25 By deﬁnition, (A‘)"A‘ = I. Now let a: = A2. Then (144%?!) = (14-1142. 9) = (Ml)- Also (A_1Ily)=(zl(A_x)'y) = (AMA—1Y3!) =(Z.A'(A'l)‘y)- We must therefore have A“(A“)' = I, or (A")" = (A‘)". 43-26 1 t I t (A\$,y) = j; j; k(t,r)z(r)dry(t) dt = A; [o y(t)k(t,7)z(r)drdt. The region of integration is as shown here: An equivalent integration can be obtained using [013(1') [1‘ y(t)k(¢,,)dt] ch as shown here: :4“ A: ' -- — ~_....._.‘;.. .«Ml...m M‘wm_.... 1‘) Jim—v {MIAH+5A‘Vj/f)=(oe h ‘ 9:5:(hE‘c/A +v(kC*/fol)= «(3] 6061“) + JIM/f) , I ‘ HSML‘:MF mm); = xétp’lm [rt/51) * s, (srwiqwAla/Lofj-9'f»- * m n, w m ‘w M—f-O EIIU-ii'w (p h m.“ M M) = fw M» tel/v _ u . J! Sm : ,th ma u 344/ WWWMW: W}. m ez)u«-/e)= ﬁdwwg/wwm “as? MK I. lfifis . ...
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hw4solutions - Qct.12.2001 8:42AM IDA CCRP No 7255 P 2/7...

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