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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
4.11L=%,—2%,—{t. 4.12 With the given solution. t
if}: = zoe‘" + b“) + / “BaaﬂbUWT = “(0 + W),
0 so that %% — b = a1:(t). as required.
4.13 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.
4.24 “A1H1 = 9 "Alnz = 8.1623 "Alllp = 8.3666 “A1llw = 9
“Ath = 4 "Ath = 3.2566 [IAzHF = 3.7417 “11sz = 3 1A31 = 3 A32 = 2.4142 A3F = 2.4495 “Ash” = 3 Oct.12.2001 8:43AM lDA CCRP N0. 7255 P. 3/7 76 Linear operators and matrix lnverses
4.25  ‘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]. IakjmJl = Ej lat} i. We thus obtain the largest row sum. 4.26 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 + =$1Zlaj1+$2l2012+"'+[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. 4.27 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.
4,28 (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 ﬁnitedimensional subspace M , by the previous result, is complete. Every convergent sequence has its limit in M ,
which must therefore be closed. 4.29 Let “All = maxlaul, and let A = B = H] . Then IlAII = IIBII = 1 IIABII = H [é i] II = 2 > llAll llBll
4.210 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.
4.212 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]=fF(I — F)“1ISIIFIIH(I  FYI" S 1 _ HF“ where the last result follows from exercise 4210. 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 4214 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 A2 = 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 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. 7/? so Linear operators and matrix lnverses 4.223 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.
4324 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. 4.325 By deﬁnition, (A‘)"A‘ = I.
Now let a: = A2. Then (144%?!) = (141142. 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‘)".
4326
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/Lofj9'f» * m n, w m
‘w M—fO EIIUii'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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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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