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Unformatted text preview: L. Vandenberghe 11/04/10 EE103 Homework 5 solutions 1. Exercise 8.2 (e,f,g). We first recall some useful properties of matrix norms. Orthogonal matrices have unit norm (if A T A = I then bardbl A bardbl = 1). This follows from the fact that a multiplication with an orthogonal matrix preserves the norm of a vector: bardbl Ax bardbl = bardbl x bardbl if A is orthogonal. The norm of a diagonal matrix is the maximum absolute value of its diagonal elements. The norm of an outer product A = uv T , where u and v are vectors, is given by bardbl A bardbl = bardbl u bardblbardbl v bardbl . This can be shown as follows: bardbl A bardbl = max x negationslash =0 bardbl uv T x bardbl bardbl x bardbl = max x negationslash =0 bardbl u bardbl| v T x | bardbl x bardbl = bardbl u bardblbardbl v bardbl . The last step follows from the Cauchy-Schwarz inequality: | v T x | bardbl v bardblbardbl x bardbl , with equality if v is parallel to x . (e) The matrix A can be written as a product of a permutation matrix and a diagonal matrix: A = QD with Q = 0 1 0 1 0 0 0 0 1 , D = 2 1 3 . This gives bardbl A bardbl = max x negationslash =0 bardbl QDx bardbl bardbl x bardbl = max x negationslash =0 bardbl Dx bardbl bardbl x bardbl = bardbl D bardbl = 3 . The second step follows because bardbl Qy bardbl = bardbl y bardbl if Q is a permutation matrix. 1 The inverse is A 1 = D 1 Q T = 1 / 2 1 1 / 3 . The norm is bardbl A 1 bardbl = max x negationslash =0 bardbl D 1 Q T x bardbl bardbl x bardbl = max x negationslash =0 bardbl D 1 Q T x bardbl bardbl Q T x bardbl = max y negationslash =0 bardbl...
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