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A5_soln - Math 235 Assignment 5 Solutions 1 Find a basis...

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Unformatted text preview: Math 235 Assignment 5 Solutions 1. Find a basis for the orthogonal complement of S = Span { E 8] , [f :31] } under the standard inner product for M'zxgflR). CL1 02 Solution: Let A = [ a3 a4 __ a1 (Lg 2 —-1 __ _ -—-<[a3 (14],[1 3]>—2a1 a2+a3+3a4 Row reducing the coefficient matrix of the homogeneous system gives 1010~1010 2—113 011—3 a1 ‘1 0 . . a2 "1 3 Hence, we get the general solution is z: 3 +t (L3 1 0 (14 0 1 . L —1 —1 0 3 Therefore, the orthogonal complement of S IS S = Span 1 0 , O 1 . 2. On P2 define the inner product (p,q) = p(—1)q(-—1) +p(0)q(0) + p(1)q(1) and let S = Span{1, a: — $2}. (a) Use the Gram—Schmidt procedure to determine an orthogonal basis for S. Solution: We let 171 :2: 1. Then <w—w2,1> 2 ——2 2 2 —————1= -—. -—-—-—-1=— ‘— “W2 a: r 3 3 +t 2: Thus, an orthogonal basis for S is {1, 2 + 3:1: -— 3:52}. —; vgzx—x2— (b) Use the orthogonal basis you found in part (a) to determine projS(1 + at + :32). Solution: We have 1+$+x21) (1+x+x22+3x~3x2) ' 1 2 =< ’ 1 ’ PTOJs( PM”) 1111!2 + llZ+393—3932H 5 4 =—1 —2 3 -—3 2 3 +24( + a: :13) 1 1 =2+~x~~x2 2 2 1 3. Define the inner product (51?, g) 2 25mm + $2y2 + 3m3y3 on 1R3. Extend the set 0 ~1 to an orthogonal basis for R3. Solution: We first extend the set to a basis for R3. We take 1 1 0 B={w1,w2,w3}= 0 , o , 1 —1 o o 1 We then apply the Gram-Schmidt procedure. Take 171 = 0 ] . Then —1 4 (“2,173.1 __ 1 2 1 — 3/5 Uig— _, 2 ’Ul -- 0 —— 0 -- 0 “”1” 0 5 _1 2/5 3 Hence, we take 172 = 0 . 2 .1 _, .. .1 0 1 3 0 Ug=w3—M51"<wi’v:>62= 1 “0 0 “0 0 z 1 ””1” ”'02” 0 _1 2 0 Hence, an orthogonal basis is {51, 172, 173}. 4. Suppose that W is a k—dimensional subspace of an inner product Space V. Prove that projw is a linear operator on V with kernel Wi. Solution: Let {171, . . . ,fi’k} be an orthonormal basis for W. Then, for any 735,17 6 V and s,t E R, we have —; 4—: —; pr01w(317+ “7) 3 (317+ 75771 6:1) v1 + ' 1 ~ + (817+ to, 11k) ,3 :3 (11mm +---+(1?,17k)17,,) +t((17,171)6’1+~-+ (17,17k)17,,) = s projw 1? + tprojW 6 Hence, projW is linear. -—o —; -—o—;>—; Let 113 E Wi. Then (11167,) = 0 for 1 g 1' g k. Hence proqu? = (117,111) 111+~ . -+(w,vk 0, so 13 E ker(projw). Therefore, Wi C; ker(projw). Let f E ker(projW), then 6 : projwf :2 (5,171)171 + ‘ - ' + (55,17,951, Since {171, . . . ,17k} is linearly independent, we get that (:3, 6}) z: 0 for 1 S i _<_ k. Therefore, a? E V17i by theorem. Hence, ker(projw) g Wi. Consequently, V17“L = ker(projW) as required. 3 5. Let V be an n dimensional inner product space and let S be a subspace of V. Prove that projSi, f = perpS a? for all :3 E V. Solution: Let {171,. . . ,Uk} be an orthonormal basis for S and let {17k+1, . . . ,Un} be an or— thonormal basis for 8““. Then {171, . . . , 77”} is an orthonormal basis for V and for any 55 E V we have if : <57€1>61 + + <51?) 17n>fin Hence, perpg a? = a? -— projs :E’ = ( 53’,"1>171 +~ +(i’,?7n>17n) —((f,171>171 + + ($692719) 2 <1 77k+1>77k+1 + + (55,77an 2 projgr 51? 6. If {171, . . .,17k} is an orthonormal basis for a subspace S of IR", verify that the standard matrix of projg can be written lprOJsl- — 771711 + 7725; + -- + WW Solution: Using the fact that "T31: 55 d4“ 41‘ _. a—qx. _...T... (v1v1+ +1711) 2: 1v1w+~-+ kvka: for all a? E R”. Thus, [projS]: 17113? + ' « - + 77,166,? MATH 235 Assignment 5 MATLAB Orthogonality and Gram-Schmidt Question 1(a) A = [~3 2 3 -- 2 2 -2 -l] A= —3 2 -3 2 l O 0 ”S 2 0 0 3 l 1 x1 =A(:,l) x1 = -3 —3 1 O 2 O 1 x2 =A(:,2) x2 = 2 2 O —5 O 3 1 X3 = A( ,3) x3 = 3 —6 4 -7 5 —l 1 7 l; '3 2 —6 2 0 5; NMNNaNi—J 1 wwwwmmw 0 4 4 0 -5; 0 -5 —7 2 —4 3; x4 x4 x5 x5 x6 x6 H N I! A(:,4) NNNNbNH A(:,5) 2 0 5 2 —4 3; 0 3 '1 2 l 3; l l v1 v1 v2 V2 v3 v3 N M I HWOU'IOOO H x1 X2 - .6250 .6250 .4583 .0000 .9167 .0000 .4583 x3 - .4753 .5247 .4819 .8024 .9638 .1186 .5576 (X2'*v1)/(v1'*v1)*v1 (X3'*Vl)/(Vl'*vl)*vl (x3'*v2)/(v2'*v2)*v2 v4 v4 v5 v5 v6 v6 (I HI—‘I—‘NDNO M H x4 - .6949 .2603 .0652 .1353 .6522 .0899 .4962 x5 - .4375 .7026 .3697 .3883 .2391 .7194 .6866 x6 - .8666 .0981 .1073 .7919 .8861 .9273 .3594 (x4'*v1)/(v1‘*v1)*v1 (x4'*v2)/(v2‘*v2)*v2 (x4'*v3)/(v3'*v3)*v3 (X5'*v1)/(v1'*vl)*v1 (x5'*v2)/(v2‘*v2)*v2 (x5'*v3)/(v3'*v3)*v3 (x5‘*v4)/(v4'*v4)*v4 (x6'*v1)/(v1'*v1)*v1 (x6‘*v2)/(v2'*v2)*v2 (x6'*v3)/(v3‘*v3)*v3 (x6'*v4)/(v4‘*v4)*v4 (x6'*v5)/(v5'*v5)*v5 H n1 v1/norm(v1) n1 H -0.6124 ~0.6124 0.2041 0 0.4082 0 0.2041 n2 = v2/norm(v2) ll n2 .1014 .1014 .0744 .8116 .1488 .4869 .2367 | OOOOOOO n3 : v3/norm(v3) n3 ll 0.6411 -0.4127 0.2906 -0.2110 0.2299 ~0.4822 —0.0653 Question 1(b) N = [n1 n2 n3 n4 n5 n6} N = -0.6124 0.1014 ~0.6124 0.1014 0.2041 0.0744 0 —0.8116 0.4082 0.1488 0 0.4869 0.2041 0.2367 N'*N ans = 1.0000 0.0000 0.0000 1.0000 -0.0000 -0.0000 ~0.0000 0 0.0000 0.0000 0.0000 0.0000 .6411 .4127 .2906 .2110 .2299 .4822 .0653 .0000 .0000 .0000 .0000 .0000 .0000 0000000 .1213 .3945 .7095 .3726 .2883 .1902 .2611 .0000 .0000 .0000 .0000 .0000 OF—‘OOOO n4 = 0000000 n5 = n5 H n6 = n6 = .1188 .4624 .3720 .1055 .6081 .4670 .1865 .0000 .0000 .0000 .0000 .0000 .0000 v4/norm(v4) .1213 .3945 .7095 .3726 .2883 .1902 .2611 v5/norm(v5) .1188 .4624 .3720 .1055 .6081 .4670 .1865 v6/norm(v6) .3348 .1969 .3779 .3214 .5176 .5250 .2438 0.3348 -0.1969 —0.3779 0.3214 0.5176 0.5250 ~0.2438 .0000 .0000 .0000 .0000 .0000 .0000 I HOOOOO Question 2(a) N = orth(A) N: —0.2984 0.6528 0.1081 ~0.0273 ’0.6393 —0.0294 0.5190 0.3435 0.3744 0.3671 0.1063 —0.5387 v0.3890 —0.3074 *0.0296 0.7675 -0.2826 ~0.0218 0.6719 -0.2999 -0.1650 0.0855 ~0.5794 0.2115 -0.1158 -0.4123 0.7283 —0.3819 -0.2979 —0.1189 0.1033 0.2051 0.4847 0.2589 0.2050 0.7764 ~0.1213 -0.2436 0.2337 0.2350 0.1836 -0.2163 Question 2(b) N'*N 1.0000 -0.0000 ’0.0000 ~0.0000 -0.0000 -0.0000 —0.0000 1.0000 —0.0000 0.0000 0.0000 0.0000 ~0.0000 —0.0000 1.0000 ~0.0000 -0.0000 ~0.0000 ~0.0000 0.0000 ~0.0000 1.0000 0.0000 -0.0000 ~0.0000 0.0000 —0.0000 0.0000 1.0000 0.0000 ~0.0000 0.0000 -0.0000 -0.0000 0.0000 1.0000 ...
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