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# hw5solutions - Oct 12 2001 8:43AM IDA CCRP N0 7255 P 5/7 78...

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Unformatted text preview: 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. 6/7 79 4.2-l8 Let V = span(P) (the space that P projects onto), and let S =: V + W, where V .L W. Let x = v +w be a decomposition of x into the orthogonal spaces, v E V, w E W. Then P = max Px = max Px ll "xnﬂ l“Hulk!” = max ||Pv|l = max = 1. Ilv+wll=l llv+wll=l 4.2-19 Let d = vec(D), the vector obtained by stacking the columns of D. Furthermore, let e be a vector which selects from d those elements that were diagonal in D. For example, 4111 r112 413 T D = 421 dzz dze d = [du d21 4:1 tin (in dsz 413 423 435] dial €152 €133 and e = {1,0,0,o, 1, 0, 0,0, 1]T Then tr(D) = dTe. Now apply CS: mm)? = IdTeI’ s lldllzllell’ = dem = IIDII’m 4.2-20 Let b} be the jth column of B. I ' llABll’r = ZZIZWMV i j I: = 220;,kﬁi,mbk,jzmd i j k m = Zbg’AHAb; j where bj is the jth column of B. By the deﬁnition of the 2-nonn, b,” AHAb, 2 by,“ 3 "Aug SO IIABlllr S ||A||2 2b? bi = llAllillBll’p- 1 4.2-21 lellw = IIleI. so llAllw = "\$311 llellw = "flail llWAxll = "‘Rﬁl llWAxll. Lety = Wx,sox =W*1y. Then “An.” = Im :IWAW"yu = IIWAW-‘n. 4.2-22 If Q is unitary, then “0!”: = xHQHQx = I”): = "x"?- The converse is somewhat more tricky. Suppose that Q is isometric, so that ||Qx||z = llxllz for all x. Then in particular. when x = 6;, the ith unit vector, "ox" = ql'qr =1. Thus the diagonal elements of Q” Q are 1. We can write Q”Q = r + B where B is a hermitian matrix that is zero along the diagonal. Now let x = e; + ej, 2' 5t j. Then llQXlli = 2 + (ed + ej)TB(e€ + 8i) = HI“: = 2- Thus (a: + e; )TB(et + ej) = bij + by,- = 0. We thus have bij = "bji, which, by the symmetry of B, must be 0. ——_———.——————w——————w_v , 1% i. i e 101 __________________________..__.———————-——— 5.3.19 (a) Let y’” = £4. and let J = [1,0, . .. ,0]. We want to maximize yTe = xTAe subject ton? = 1. Setting Ar J = xTAe - -2-x x wherexisalagnngemultiplier,weﬁnd bax—J = Ae - Ax = 0 so that x = «i-Ae = ﬁat, where a; is the ﬁrst column ofA. Then xTx = :lglflt = §2IIIiII2. so that A =||a1l|- Wehave 1 T 1 T = -———A Ae = ——A 81, ” "mu "an \$01!“) = "31"- (b) Since y(1) = Ila; II. the nonzero value in the ﬁrst column of HA is the same as the value obtained by the optimization. 5.3-20 If Px = (I - 2wT/vTv) 6 span(y) then we must have v E span(x, y). Let us set therefore v = x + try for some a to be found. We have r T Px=x__ T2v(x xIax y) 2 x x+ 2m: y+a||y|l _z__ 2xxTx—2axx7y _ (_2 xxT—xTy £3: + 2ax7'y + allylP y xTx + ZaxTy + dilly2 Tx — ZanTy + alelyll2 — 2x3th — mTy xxT — xTy — _________————————————- + y(—2—-—————-—-—— — xTx + ZaxTy + ally"2 XTX + 2037'! + allyll’ To obtain a result proportional to y, the ﬁrst term must be zero. This means that we must have tnzllyll2 = llxll’, I]! lell2 v = x + -—y. llyll2 Another v which is proportional to this (and hence has the same overall result) is llxll2 Ilyll" A geometric interpretation is given in the ﬁgure below: v is chosen so that VJ” is midway between x and y (normalized). so that the reﬂection through v‘ is proportional to y. V 5.3—2l AH = A — 2Avv” /v”v = A + ﬁwv” where [3 = —2/vv” and w = Av. 53-22 A segment of code to compute the results is shown here: Algorithm 5.7 Compare least-squares solutions _____________________________.__——————-——-— 102 Someimpomntmatrixfactoriutions % A = [10000 10001; 10001 10002; 10002 10003; 10003 10004; 10004 10005]; b = [20001; 20003; 20005; 20007; 20009]; cond(A) cond(A'*A) x1 = i’nv(A"A) *A'*b % QR method [V,R] = qrhouse(A),- Q = ququ); C Q'*b; c c(1:2); R1 = R(1:2.1:2); x2 = backsub(R1.c) A"A: choleskwx); A'*b; forsub(L,p) ; x3 = backsub(L’,y) _____________________________————————-—-—- 5.3-23 inf. The matrix (a) Using the Mom cond command. we ﬁnd that cond (A) = 1.41 x 10‘, and com! (A’ «11) is poorly conditioned! The rest of the results follow: x1 = 1.6258 1.4550 x2 = 1.0000 1.0000 x3 = 1.3334 0.6667 We see that the QR method worked the best! (a) G6 __ c039 -sin9 cos9 sin9 _ c0329+sin29 cosOsinO—cosOsinO __I 9 ‘0'— sin9 c059 —sin9 cos9 _ cosOsinO-cosasina c0529+sin29 —_ ' Interpretation: ﬁrst rotating an angle of -9, then an angle of 9 is the same as no rotation at all. (b) ca _ c059 —sin0 cosd: —sin¢ _ cosOcos¢—sin95in¢ -cos€sin¢—cos¢sin9 a ‘- sin0 c039 sin¢ c054: _ cos¢sin0+cosesin¢ cochosda-sinﬂsind) = cos(9+¢) —sin(¢+0) sin(¢+9) cos(9+¢) ' Since there is an identity operation and each element has an inverse operation. and the operation is associative. the set of rotations forms a group. 107 62-8 LetA‘ beaneigenvalue ofA.witheol-lespondingeigenvectorx. Then Ax = Xx. Then det(A‘I - A) = 0. NowdetQI- (A+rI)) =det((A-r)I—A).whichiszerowhenA—r= A‘,orA = A‘ +r. Forthisvalue,letybe the eigenvector: (A + rI)y = (A‘ + r)y implies that Ay = X)! That is. y is the eigenveetor of A corresponding to A', so y = x. 6.2-9 lf Ax = Ax. then, multiplying both sides by A we have A’x = AAx = MAX) = A’x. Thus A“ is an eigenvalue of A”. and x is an eigenvector. The process repeats for arbitrary powers of A. 6.2-10 Ifo = Ax. multiply both sides by A": A'le = AA"): x = AA"1x i—x = A-lx so 1/) is an eigenvalue of A", and x is the eigenvector. 62-]! Let g(:l:) = z" +,<),.—1z""l + - - - + go Then 9(A) = A" + sin—1.4”" + - -- +goA Then i i g(A)x = A”): + g.._1A""1x + - - - + goAx But ij = ij, so we have g(A)x = X'x + g..-1A”"x + - - - + goAx = g(A)x. So if A is an eigenvalue of A. then 9(A) is an eigenvalue of g(A). 6.2-12 With P2 = P, we must have A2 = A, so ,\ = 0 or 1. 6.243 (a) Let AB): = Ax. Then (mull. both sides by B) we have (BA) (Bx) = ABx “owl...” i Let y = Bx. Then we have BAy = Ay so A is an eigenvalue of BA. (b) bet Ax.- = xix.- fori = 1,2, . .. ,n, and let By.- = My; fori = 1,2,... ,n. If, as the problem suggests. there are n ,wwsswm . ,, § common eigenvectors, order them so that x.- = y.- fori = 1, 2, . . . , 11. With n independent eigenvectors. we can write A = TAIT" _ where T = [x1 , . . . ,xnl and A; is diagonal. We can also write B = TAzT" Then AB = TAiT‘1TA2T" = Timur" = BA. 110 Eigenvalues and Eigeuvectors stW 63-25 In the Schur lemma, the triangularization T = UHAU has the same eigenvalues as A. Similarity preserves tank. so T has 6.4 the same rank as A. When rank(T) = r < m, only the ﬁrst 1' columns of T have new information in them. so the lower 7 right (m — r) x (m — r) comer of T must be zero. But since the eigenvalues of T are on the diagonal. we must have at least at — r zero eigenvalues. 63-26 (a) letAbeunitary.ThenA”A =AA” =1. LetAbesymmetric. ThenATA = AAT. LetAbe Hermitian. Then AHA = AA”. Let A be skew symmetric. Then ATA = -—AA = -(—AAT) = AAT. LetAbeskewHermitian. Then A”A= -AA = -(—AA”) = AA”. (b) Let A be normal. Then the triangulan'zation UHAU = T satisﬁes (UH/1U)” = UAHU = T”. Multiplying these we have (U" AU)(U" AU)” = U” AA”U = v” A" AU so TT" = T” T. But the only way this can happen is if T is diagonal. 6.3-27 If A2 = A, then the eigenvalues are either 1 or 0. Then tr(A) is the number of nonzero eigenvalues, which is the rank of the matrix. 63-28 (a) The (111, n)th element of FF” is N—l N—l N_1 (FFHLM' = Z fijnk = z e—ermkejzrrnk = Z e—j2r(m—n)k/N k=0 k=0 k=0 6 .4- Whenm—n.=0,wehave __ N-l Z e—j21r(m—n)k/N = N — Is=o and when m - n aé 0 we have N24 —j21r(m-n)k/N e-j2'(m_")k " 1 e = —--.———— = 0. e—121r(m-n) __ 1 k=0 We also observe that this behavior is periodic with period N. Hence (FFH)m,. = N6...,,., so l/NFFH = I. (b) Let (m, n)th element of C is cm... = c((,._,,.))~. (The notatiorr ((z))N here means :1: modulo N.) Let f. be the lth column of F: n =[e'1'2*“‘/”], k = 0,1,... ,N — 1. Let us consider the product of C with f}. The pth component of the product is "'1 N—l = Z cp‘ffl'i = Z C((i_p))Ne—)21rh/N i=0 i=0 N-l N—l _ _ —j21rl(i-P+P)/N _ —’21rlp/N -'21rl(i—p)/N - E :Ca—pe —e ’ C((i-mue J - i=0 i=0 The latter sum is the same for all values of 1). Hence we see that the pth component of Cf; is proportional to the pth component of f1 , so f; is an eigenvector. The eigenvalue is N-l _' 1'- I" = Z C((a-pmve ’2” (' “M, i=0 which is the DFI' of the sequence {ci}. lfx = Uzy. then (mu? + 2,1152: = 0, so N04") = span(Uz) _______________________________————————-———————-— 7.3-4 Here is some MATLAB code to ﬁnd the solution: [U.S.V] = svd(A); xhat = v*pinv(S)'U'*b giving 0.544248 2.40265 * = 3.09292 3.73009 The norm of this solution is "it": = 5.4359. while ||x|| = 5.4772. The solution of smallest norm is selected. 7.3-5 "Air — bu: = IIUEV" vz'b - bu: = "(22' - I)U"bu§ NowEEl -I=diag(0,0,... ,0,1,1,...1).so (22* — 1w” = U," and the result is established. The quantity U-f' b is that portion of b which is orthogonal to the range of A, and the norm simply ﬁnds the length of that vector. _, _ 8.97059 7.4.5 (11) A 1:. [4.67647]. -9.85 condition of A is 100.) 0.0294118 0.117647 . i... W A " [0.0294118 0.117647]"“‘““°s°'““°“w° A, 1 _ 0.147059 A, 1 _ 0.15 1 ’ 0.147059 1.1 ‘ 0.15 Now there are only slight changes in the solution. corresponding to slight changes in the RHS. The solution vector is a multiple of [1, l]T. which points nearly orthogonal to the sensitive direction. 7.5-7 Let 2 = 71w. where (b) A‘lb = [10'15]. Even though an element of b only changed by .l, the solution changed by essentially 1. (The II .S s ‘75 Then, since ||z|| = 1, we must have k+l S/JI’WI2 = 1- i=1 k+l "(A — 3):": = Zaﬂvfz)’ i=1 k+l k+1 = Zan‘zl‘ﬁlz 2 “H‘ 2 I‘Y-‘I2 = n+1, i=1 i=1 where the inequality follows since the singular values are ordered 01 2 oz . . . 2 0,... Equalin can be achieved if z = V“. 1, which occurs when B = [v1, . . . ,vk]. ...
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