A6_soln - Math 235 Assignment 6 Solutions 1. Find a and b...

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Unformatted text preview: Math 235 Assignment 6 Solutions 1. Find a and b to obtain the best fitting equation of the form y = a + bt for the given data. t 1 2 3 4 5 y 9 6 5 31 9 11 6 12 Solution: We take fl 2: 5 andX = 1 3 .Then 3 1 4 1 1 5 9 ~1 6 a_T-1T,,__515 11111 __1~_55~1524*10.5 [61‘(XX) X “[1555] 1123451§“50 ~15 5 53”~1.9 1 Hence, the best fitting line is y = 10.5 ~ 1.916. 2. Verify that the following system A2? =2 5 is inconsistent, then determine for the vector :3 that minimizes [[1455 ~ £131 ‘1‘ 21'2 Z 5 2961 ~ 3332 2: 6 £191 "‘ 12112 I: “*4 1 2 5 1 2 5 Solution: Row reducing gives 2 ~13 6 N 0 ~7 ~41 Hence, the system is 1 ~12 «4 0 0 ~1 inconsistent. 1 2 .. We have A = 2 ~23 . Hence, the vector :5 that minimizes “As? —— [9H is 1 ~12 f:(ATA)“1A5 “ 6 ~16 "1 1 2 1 g “ ~16 157 2 ~3 ~12 __4 ____1__ 157 16 13 “685 16 6 40 2: [3385/4998] 2 3. Prove that if U and W are subspaces of a vector space V such that U 0 W = {6}, then U 63 W is a subspace of V. Moreover, if {171, . . . ,flk} is a basis for U and {2171,- -- ,w’g} is a basis for W, then {271, . . . £19,131, . . . ,u‘ig} is a basis for U EB W. Solution: We Will prove that {171, . . . ,17k+g} is a linearly independent spanning set for UEEW. Consider C1111 + - - - +Ck’17k + 0194,1131 + ' ' ‘ + Ck+gtvg == 0 Then, we have 61771 + ' ' ' + Ck'Uk : “CMflBi '* ' ' ' ‘“ Ck+€117£ The vector on the right is in U and the vector on the left is in W. But, the only vector that is both in U and W is the zero vector. Therefore, each oz- 2 O and hence {171, . . . , 17k, 21517 . . . , 117g} is linearly independent. For any 17 E U EB W we have that 17 =2 {i + ’u')’ by definition of U G9 W. We can write 11’: 01171 + ' ' ‘ + Ckflk and U7: d1131+~~+ dew} and hence 17201771+"'+Ck’17k+d1231+"‘+dg’tvg and so {171,...,17k,161,...,tb’g} also spans UEBW. 4. Let U be a subspace of a finite dimensional vector space V and suppose that there are subspaces W1 and W2 such that V = U @ W1 and V = U EB W2. (a) Prove that dim W1 = dim W2. Solution: By Theorem 3.5.1 (problem 3), we have that dimV 2: dimU + dile and dimV :2 dimU + dim W2. Thus, dim W1 :2 dimV —— dimU =2 dim W2 (b) Give an example of a vector space V With subspaces U, W1, and W2 such that V = UEBW1 andV—TUEBWQ, but W1 Solution: In R2, let U = Span { }, W1 :2 Span{ } and W2 2 Span{ Then, by Theorem 3.5.1, we have V 2 U EB W1 and V = U 89 W2, but clearly W1 % W2. 5. Let A be an m x 71 real matrix. Prove that the nullspace of AT is the orthogonal complement of the column space of A. 17? Solution: Let A = [171 17”], so AT :2 _ 5:5 Ufa: HfEQflAHJMnQdHJLEmmmATH= ; zfihmmfeNmmfly 173% On the other hand, let :3 E Null(AT), then AT” :2 6 so 13} - a? 2: 0 for all 2'. Now, pick any 56 Col(A). Then 5: 01171 + - ~ - + cuff”. But then szkfifi~~+%%yfzm hence a? E Col(A)i. MATH 235 Assignment 6 MATLAB Solutions Least Squares (a) X = [0; 10; 20; 30; 40; X = 0 10 20 30 40 50 60 70 80 90 y = [4.77; 16.47; 19.76; y = 4.7700 16.4700 19.7600 22.3500 23.6700 25.0300 27.0100 30.9300 40.9800 53.8600 plOt(XI Y! H”); (b) X = [x X.“2 x.“3] X 7:. 0 O 10 100 20 400 30 900 40 1600 50 2500 60 3600 70 4900 80 6400 90 8100 50; 60; 70; 22.35; 23.67; 0 1000 8000 27000 64000 125000 216000 343000 512000 729000 80; 90] 25.03; 27.01; 30.93; 40.98; 53.86] rref(X'*X) ans = l O O 0 1 0 0 0 1 beta = inV(X'*X)*X'*y beta = 1.6288 —0.0366 0.0003 (d) xnrange = [0: l: 901'; ymfitnline = beta(l,l)*x_range + beta(2,l)*x_range.“2 + beta(3,1)*x_range.“3; hold on plot(x_range, y_fit_line, 'r—') MATH 235 Mm fa: Assignment 5 £933: Sqaams: Safes v3. Costs costs (méltimg) wk 5313 minis WWW i235! sqeareg {it {I E} 2:3 33 413 Sit 80 m 833 238 sales (méifitms) ...
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This note was uploaded on 09/30/2011 for the course MATH 235 taught by Professor Celmin during the Spring '08 term at Waterloo.

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A6_soln - Math 235 Assignment 6 Solutions 1. Find a and b...

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