Linear Algebra with Applications (3rd Edition)

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Math 417 homework 6 solutions 1 Section 5.2 problem 14 ~u 1 = 1 10 1 7 1 7 , ~u 2 = 1 2 - 1 0 1 0 , ~u 3 = 1 2 0 1 0 - 1 . 2 Section 5.2 problem 18 Q = 1 5 - 4 - 3 0 0 0 1 - 3 4 0 , R = - 5 - 5 0 0 - 35 0 0 0 - 2 . 3 Section 5.3 problem 6,8,10 - B is orthogonal as well: ( - B ) T ( - B ) = B T B = I. A + B is not always orthogonal. For example A = I and B = I are orthog- onal, but A + B = 2 I is not: (2 I ) T (2 I ) = 4 I 6 = I. B - 1 AB is orthogonal too: note that B - 1 = B T for orthogonal matrices, so ( B - 1 AB ) T ( B - 1 AB ) = ( B T AB ) T ( B T AB ) = B T A T ( B T ) T B T AB = B T A T BB T AB. BB T = I as well, so after three cancellations we get = I . 4 Section 5.3 problem 28 An orthogonal transformation L has a matrix A that is orthogonal: L ( ~v ) = A~v. Then L ( ~v ) · L ( ~w ) = ( A~v ) · ( A~w ) = ( A~v ) T ( A~w ) = ~v T A T A~w. 1
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Here we exploit that A is orthogonal: = ~v T ~w = ~v · ~w. 5 Least squares problem: We have to find A so that ~ f = A~x. To this end, calculate the entries of ~ f . For example f ( - 1) = x 2 ( - 1) 2 + x 1 ( - 1) + x 0 = x 0 - x 1 + x 2 . Continuing in this way, we see that A = 1 - 1 1 1 0 0 1 1 1 1 1 1 . The rest can be done in Matlab:
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