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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 A2 = 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
4216
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)2iAxi2tr(xx )
2 1
= “All?  ﬁrzAxlli + Ell4345
_ 2 _ AK 2
— llAllr xﬂx . since (st(xxHAHA) = tr(x”AHAx) = KHAH Ax. 4217 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 4.2l8 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 Pvl = max = 1.
Ilv+wll=l llv+wll=l 4.219 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
4.220 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 2nonn, b,” AHAb, 2
by,“ 3 "Aug SO IIABlllr S A2 2b? bi = llAllillBll’p 1
4.221 lellw = IIleI. so llAllw = "$311 llellw = "flail llWAxll = "‘Rﬁl llWAxll. Lety = Wx,sox =W*1y. Then “An.” = Im :IWAW"yu = IIWAW‘n. 4.222 If Q is unitary, then
“0!”: = xHQHQx = I”): = "x"? The converse is somewhat more tricky. Suppose that Q is isometric, so that Qxz = 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  2x x
wherexisalagnngemultiplier,weﬁnd bax—J = Ae  Ax = 0
so that x = «iAe = 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. 5.320 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+ayl
_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. 5322 A segment of code to compute the results is shown here: Algorithm 5.7 Compare leastsquares 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) _____________________________—————————— 5.323 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 _ cosOsinOcosasina 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¢ cochosdasinﬂsind)
= 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. 107 628 LetA‘ beaneigenvalue ofA.witheollespondingeigenvectorx. Then
Ax = Xx.
Then det(A‘I  A) = 0.
NowdetQI (A+rI)) =det((Ar)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.
6.29 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.
6.210 Ifo = Ax. multiply both sides by A":
A'le = AA"): x = AA"1x
i—x = Alx 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). 6.212 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 6325 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.
6326 (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. 6.327 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. 6328 (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 __
Nl
Z e—j21r(m—n)k/N = N —
Is=o
and when m  n aé 0 we have
N24 —j21r(mn)k/N ej2'(m_")k " 1
e = —.———— = 0.
e—121r(mn) __ 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
Nl N—l
_ _ —j21rl(iP+P)/N _ —’21rlp/N '21rl(i—p)/N
 E :Ca—pe —e ’ C((imue 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 Nl
_' 1'
I" = Z C((apmve ’2” (' “M, i=0 which is the DFI' of the sequence {ci}. lfx = Uzy. then (mu? + 2,1152: = 0, so
N04") = span(Uz) _______________________________————————————————
7.34 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. 7.35 "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 Uf' 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.
7.57 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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This note was uploaded on 01/24/2012 for the course EECS 250 taught by Professor Swindlehurst during the Fall '09 term at UC Irvine.
 Fall '09
 Swindlehurst
 Digital Signal Processing, Signal Processing

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