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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) 2 1 Solution: Let A = 2 1 then A T A = 4 0 0 1 . The eigenvalues of A T A are λ 1 = 4 and λ 2 = 1 and A T A is already diagonal so we take V = 1 0 0 1 . The singular values of A are σ 1 = √ 4 = 2 and σ 2 = √ 1 = 1. Thus Σ = 2 0 0 1 . Next compute ~u 1 = 1 σ 1 A~v 1 = 1 2 2 1 1 = 1 ~u 2 = 1 σ 2 A~v 2 = 1 1 2 1 1 = 1 Hence we take U = 1 1 and get that A = U Σ V T as required. b) 4 2 2 1 Solution: Let A = 4 2 2 1 then A T A = 20 10 10 5 . The eigenvalues of A T A are λ 1 = 25 and λ 2 = 0 with corresponding normalized eigenvectors ~v 1 = 2 / √ 5 1 / √ 5 and ~v 2 = 1 / √ 5 2 / √ 5 . Thus we take V = 2 / √ 5 1 / √ 5 1 / √ 5 2 / √ 5 . The singular values of A are σ 1 = √ 25 = 5 and σ 2 = √ 0 = 0. Thus Σ = 5 0 0 0 0 0 . Next compute ~u 1 = 1 σ 1 A~v 1 = 1 5 4 2 2 1 2 / √ 5 1 / √ 5 = 2 / √ 5 1 / √ 5 We only have one vector, ~u 1 , in R 3 , so we must extend this to form an orthonormal basis for R 3 . Observe that we just need to pick two vectors that are orthogonal to ~u 1 and to each other. That is vectors x 1 x 2 x 3 such that 2 / √ 5 1 / √ 5 · x 1 x 2 x 3 = 0 . This gives the system 2 x 1 + x 2 = 0. Hence, we pick (1 , 2 , 0) and (0 , , 1) which are clearly orthogonal to each 2 other. Normalizing them gives ~u 2 = 1 / √ 5 2 / √ 5 , ~u 3 = 1 and so we take U = ~u 1 ~u 2 ~u 3 . Then A = U Σ V T as required. c) 1 1 1 1 1 Solution: Let A = 1 1 1 1 1 then A T A = 2 0 0 3 and the eigenvalues of A T A are seen to be (in decreasing order) λ 1 = 3 and λ 2 = 2. Corresponding normalized eigenvectors are ~v 1 = 1 for λ 1 and ~v 2 = 1 for λ 2 . Thus one choice for V is V = 0 1 1 0 . The singular values of A are σ 1 = √ 3 and σ 2 = √ 2. Thus Σ = √ 3 √ 2 Next compute ~u 1 = 1 σ 1 A~v 1 = 1 √ 3 1 1 1 1 1 1 = 1 / √ 3 1 / √ 3 1 / √ 3 ~u 2 = 1 σ...
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This note was uploaded on 04/18/2010 for the course MATH 235 taught by Professor Celmin during the Spring '08 term at Waterloo.
 Spring '08
 CELMIN
 Math, Matrices

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