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Unformatted text preview: Solutions for Homework 4 Numerical Linear Algebra 1 Fall 2010 Problem 4.1 Given that we know the SVD exists for any complex matrix A C m n , assume that A R m n has rank k with k n , i.e., A is real and it may be rank deficient, and show that the SVD of A is all real and has the form A = U S V T = U k k V T k where S R n n is diagonal with nonnegative entries, U = ( U k U m- k ) , U T U = I m V = ( V k V n- k ) , V T V = I n U k R m k , and V k R n k Solution: The result follows simply from the relationship of the SVD to the symmetric eigenvalue decomposition. A T A and AA T are both real symmetric positive semidefinite and therefore have real orthogonal eigenvectors that define orthogonal U and V as needed. The rank is reflected in the nonzero eigenvalues which are the squares of the nonzero singular values and therefore define k and S . U k and V k follow immediately from the appropriate partitionings of U and V . Problem 4.2 Let T R n n be a symmetric tridiagonal matrix, i.e., e T i Te j = e T j Te i and e T i Te j = 0 if j < i- 1 or j > i + 1. Consider T = QR where R R n n is an upper triangular matrix and Q R n n is an orthogonal matrix. Recall, the nonzero structure of R was derived in class and shown to be e T i Re j = 0 if j < i (upper triangular assumption) or if j > i + 2, i.e, nonzeros are restricted to the main diagonal and the first two superdiagonals. (4.2.a) Show that Q has nonzero structure such that e T i Qe j = 0 if j < i- 1, i.e., Q is upper Hessenberg. (4.2.b) Show that T + = RQ is a symmetric triagonal matrix. (4.2.c) Prove the Lemma in the class notes that states that choosing the shift = , where is an eigenvalue of T , results in a reduced T + with known eigenvector and eigenvalue....
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- Fall '06