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soln7 - MATH 245 Linear Algebra 2 Solutions to Assignment 7...

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MATH 245 Linear Algebra 2, Solutions to Assignment 7 1: Find a singular value decomposition A = Q Σ P * for the matrix A = 1 2 2 0 3 1 1 1 . Solution: We have A * A = 1 2 3 1 2 0 1 1 1 2 2 0 3 1 1 1 = 15 6 6 6 . The characteristic polynomial of A * A is A * A - λI = 15 - λ 6 6 6 - λ = λ 2 - 2 λ + 54 = ( λ - 3)( λ - 18) so the eigenvalues of A * A are λ 1 = 18, λ 2 = 3, and hence the singular values of A are σ 1 = 3 2, σ 2 = 3, so we can take Σ = σ 1 0 0 σ 2 0 0 0 0 = 3 2 0 0 3 0 0 0 0 . When λ = λ 1 = 18 we have A * A - λI = - 3 6 6 - 12 - 1 2 0 0 so u 1 = 1 5 2 1 is a unit eigenvector for λ 1 . The eigenspace for λ 2 is orthogonal so u 2 = 1 5 - 1 2 is a unit eigenvector for λ 2 , and so we can take P = ( u 1 , u 2 ) = 2 5 - 1 5 1 5 2 5 ! . Next let v 1 = Au 1 σ 1 = 1 3 10 1 2 2 0 3 1 1 1 2 1 = 1 3 10 4 4 7 3 and v 2 = Au 2 σ 2 = 1 15 1 2 2 0 3 1 1 1 - 1 2 = 1 15 3 - 2 - 1 1 then extend { v 1 , v 2 } to an orthonormal basis V = { v 1 , v 2 , v 3 , v 4 } for R 4 . We have 4 4 7 3 3 - 2 - 1 1 1 6 8 2 3 - 2 - 1 1 1 6 8 2 0 20 25 5 1 6 8 2 0 1 5 4 1 4 1 0 1 2 1 2 0 1 5 4 1 4 So the orthogonal complement of Span { v 1 , v 2 } has basis (2 , 1 , 0 , - 4) t , (2 , 5 , - 4 , 0) t . We apply the Gram- Schmidt Procedure, replacing the second vector in the basis by 2 5 - 4 0 - 9 21 2 1 0 - 4 = 1 7 14 35 - 28 0 - 6 3 0 - 12 = 1 7 8 32 - 28 12 = 4 7 2 8 - 7 3 , and then we normalize to obtain v 3 = 1 21 2 1 0 - 4 and v 4 = 1 126 2 8 - 7 3 . Thus we can take Q = ( v 1 , v 2 , v 3 , v 4 ) = 4 3 10 3 15 2 21 2 126 4 3 10 - 2 15 1 21 8 126 7 3 10 - 2 15 0 21 - 7 126 3 3 10 1 15 - 4 21 3 126 .
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2: Let A M n × n ( C ) be Hermitian and positive-definite. Show that if A = Q Σ P * is a singular value decompo- sition of A , then Q = P .
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