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Unformatted text preview: Math 235 Assignment 8 Solutions 1. Find a singular value decomposition of each of the following matrices. a) 1 1 1 1 Solution: Let A = 1 1 1 1 then A T A = 2 0 0 2 . Thus the eigenvalue of A T A is λ 1 = 2 with multiplicity 2 and A is orthogonally diagonalized by V = 1 0 0 1 . The singular values of A are σ 1 = √ 2 and σ 2 = √ 2. Thus the matrix Σ is Σ = √ 2 √ 2 . Next compute ~u 1 = 1 σ 1 A~v 1 = 1 √ 2 1 1 1 1 1 = 1 1 ~u 2 = 1 σ 2 A~v 2 = 1 √ 2 1 1 1 1 1 = 1 1 Thus we have U = 1 / √ 2 1 / √ 2 1 / √ 2 1 / √ 2 . Then A = U Σ V T as required. b) 1 4 2 2 2 4 Solution: Let A = 1 4 2 2 2 4 then A T A = 9 0 36 and the eigenvalues of A T A are λ 1 = 36 and λ 2 = 9. Corresponding normalized eigenvectors are ~v 1 = 1 for λ 1 and ~v 2 = 1 for λ 2 . Thus we let V = 0 1 1 0 . The singular values of A are σ 1 = √ 36 = 6 and σ 2 = √ 9 = 3. Thus the matrix Σ is Σ = 6 0 0 3 0 0 . Next compute ~u 1 = 1 σ 1 A~v 1 = 1 6 1 4 2 2 2 4 1 =  4 2 4 ~u 2 = 1 σ 2 A~v 2 = 1 3 1 4 2 2 2 4 1 = 1 2 2 We only have two orthogonal vectors in R 3 , so we must extend this to form an orthonormal basis for R 3 . So, we need to pick a vector ~u 3 = x 1 ,x 2 ,x 3 such that ~u 1 · ~u 3 = 0 and ~u 2 · ~u 3 = 0. This gives the system of equation 4 x 1 + 2 x 2 + 4 x 3 = 0, x 1 2 x 2 + 2 x 3 = 0. We find that a solution is x 1 = 2, x 2 = 2, x 3 = 1. Thus we pick ~u 3 = 2 2 1 and let U =  2 / 3 1 / 3 2 / 3 1 / 3 2 / 3 2 / 3 2 / 3 2 / 3 1 / 3 . Then A = U Σ V T as required. 2 c) 1 1 1 1 1 1 1 1 1 Solution: Let A = 1 1 1 1 1 1 1 1 1 then A T A = 3 0 0 0 3 0 0 0 3 and the eigenvalues of A T A are λ 1 = 3 with multiplicity 3 and A is orthogonally diagonalized by V = 1 0 0 0 1 0 0 0 1 ....
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This note was uploaded on 10/12/2010 for the course MATH 235 taught by Professor Celmin during the Fall '08 term at Waterloo.
 Fall '08
 CELMIN
 Matrices

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