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Unformatted text preview: 18.06 Problem Set 9 Due Friday, 9 May 2008 at 4 pm in 2106. Problem 1: Do problem 4 in section 6.7 (pg. 360) in the book. Solution (10 points) a) We have A T A = AA T = 2 1 1 1 This matrix has eigenvalues satisfying 2 3 + 1 = 0, so it has eigenvalues 1 = 3 2 + 5 2 and 2 = 3 2 5 2 . Its eigenvectors form the nullspace of A T A 3 + 5 2 I = (1 5) / 2 1 1 (1 + 5) / 2 This has nullspace generated by (2 , 5 1). Since the eigenvectors of A T A must be perpendicular, we know that another eigenvector is ( 5 1 , 2) (which we could also find directly). The normalized eigenvector matrix is S = 1 p 10 2 5 2 5 1 5 1 2 b) We construct the singular value decomposition A = U V H . First, we choose the matrix V to be the eigenvector matrix for A T A ; that is, it is just the S we found in part a. The matrix is the 2x2 matrix with the square roots of the eigenvalues of A T A on the diagonal: = q 3+ 5 2 q 3 5 2 Finally, we find U via the equation AV = U . We cant skip directly to U = S . It is true that U will be an eigenvector matrix for AA T , but we must pick the eigenvectors correctly! In this case the only choice in unit eigenvectors of AA T is the sign. Even so, we must have the relationship A = U V H , and if we get the sign of the vectors of U backwards this will not be true. 1 Let v i and u i be the ith columns of V and U . We know u 1 is either v 1 or v 1 , and similarly for u 2 . The question is just which way around it is. We start with v 1 : Av 1 = s 3 + 5 2 u 1 1 p 10 2 5 5 + 1 2 = s 3 + 5 2 u 1 so u 1 = s 1 (10 + 2 5) 5 + 1 2 = s 1 (10 2 5) 2 5 1 This is the same vector as v 1 . Here Av 1 = 1 u 1 is an eigenvector equation for A , since 1 is an eigenvalue of A . So v 1 keeps the same sign. For v 2 we find: Av 2 = s 3 5 2 u 2 1 p 10 2 5 5 3 5 1 = s 3 5 2 u 2 We already know that u 2 is either v 2 or v 2 . However v 2 has negative second component, and u 2 has negative first component, meaning that the sign has switched. Here Av 2 = 2 u 2 is not an eigenvector equation, since 2 = 2 . So we need to switch the sign of u 2 as well. In the end, we get the SVD: U = 1 p 10 2 5 2 ( 5 1) 5 1 2 = q 3+ 5 2 q 3 5 2 V = 1 p 10 2 5 2 5 1 5 1 2 It is almost the diagonalization of A , but not quite. Since one of the eigenvalues of A is negative, it cant appear in . We must switch its sign, and we compensate by switching the sign of the eigenvector in U . As you might guess from this problem, 2 the SVD for a positive definite matrix is its diagonalization see the last problem of this pset....
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This note was uploaded on 08/10/2008 for the course MATH 250 taught by Professor Chanillo during the Spring '08 term at Rutgers.
 Spring '08
 CHANILLO

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