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Unformatted text preview: Solutions for Homework 1 Numerical Linear Algebra 1 Fall 2010 Problem 1.1 A matrix A C n n is nilpotent if A k = 0 for some integer k > 0. Prove that the only eigenvalue of a nilpotent matrix is 0. Solution: There are multiple ways to prove this. The simplest is to use contradiction on a matrix vector product identity. We have that a matrix A R n n is nilpotent of degree k if k is a positive integer such that A p = 0 p k A p 6 = 0 < p < k Suppose 6 = 0 is an eigenvalue corresponding to the eigenvector x 6 = 0 n . It follows that Ax = x A k x = k x However, by the nilpotent assumption A k = 0 and therefore A k x = 0 n n x = 0 n = k x Since x 6 = 0 n it follows that = 0 which is a contradiction. Therefore all must be 0. Problem 1.2 Let the matrix A C n n be unitary. Show that if is an eigenvalue of A then | | = 1. Solution: Since A is unitary we have AA H = A H A = I and therefore A is normal. It follows that A = Q Q H where is a diagonal matrix with possibly complex scalars on the diagonal and Q is a unitary matrix. We therefore have I = A H A = ( Q Q H ) H ( Q Q H ) = Q Q H Q Q H = Q | | 2 Q H where | | 2 is a diagonal matrix with elements | i | 2 on the diagonal. Q | | 2 Q H = I | | 2 = Q H IQ = I | i | 2 = 1 So | i | = 1 for 1 i n as desired. 1 Problem 1.3 Prove that a matrix A C n n is normal if and only if there exists a unitary matrix U such that U H AU is a diagonal matrix. Solution: Assume A is such that U H AU = is a diagonal matrix and U is unitary. We have AA H = U U H ( U U H ) H = U U H ( U U H ) H = U U H U U H = U U H = U U H = U U H U U H = ( U U H ) H ( U U H ) = A H A Assume that AA H = A H A . By the Schur decomposition we have A = QUQ H where U is upper triangular and Q is unitary. It follows trivially that AA H = QUU H Q H A H A = QU H UQ H UU H = U H U So U is upper triangular and normal . It can be shown that if U is upper triangular...
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- Fall '06