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pset9-soln

# pset9-soln - 18.06 Problem Set 9 Due Friday 9 May 2008 at 4...

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Unformatted text preview: 18.06 Problem Set 9 Due Friday, 9 May 2008 at 4 pm in 2-106. 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 λ 2- 3 λ + 1 = 0, so it has eigenvalues λ 1 = 3 2 + √ 5 2 and λ 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 Σ 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 Σ is the 2x2 matrix with the square roots of the eigenvalues of A T A on the diagonal: Σ = q 3+ √ 5 2 q 3- √ 5 2 Finally, we find U via the equation AV = U Σ. We can’t 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 Σ 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 = σ 1 u 1 is an eigenvector equation for A , since σ 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 = σ 2 u 2 is not an eigenvector equation, since σ 2 =- λ 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 Σ = 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 can’t appear in Σ. 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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pset9-soln - 18.06 Problem Set 9 Due Friday 9 May 2008 at 4...

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