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solhw4

# solhw4 - Solutions for Homework 4 Numerical Linear Algebra...

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Solutions for Homework 4 Numerical Linear Algebra 1 Fall 2011 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 parenleftbigg S 0 parenrightbigg V T = U k Σ 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 Σ 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 μ = λ , where λ is an eigenvalue of T , results in a reduced T + with known eigenvector and eigenvalue. 1

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Solutions: To show that Q is upper Hessenberg we can analyze the structure of the product of Given’s rotations used to produce it, Q = G 1 G 2 · · · G n - 1 . We have the column structure G i e i = γ i e i σ i e i +1 G i e i +1 = σ i e i + γ i e i +1 G i e j = e j , j < i, j > i + 1 and the row structure e T
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solhw4 - Solutions for Homework 4 Numerical Linear Algebra...

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