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Unformatted text preview: Solutions for Homework 1 Numerical Linear Algebra 1 Fall 2011 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 λ 6 = 0 is an eigenvalue corresponding to the eigenvector x 6 = 0 n . It follows that Ax = λx A k x = λ k x However, by the nilpotent assumption A k = 0 and therefore A k x = 0 n × n x = 0 n = λ k x Since x 6 = 0 n it follows that λ = 0 which is a contradiction. Therefore all λ must be 0. Problem 1.2 Problem 7.1.1 Golub and Van Loan p. 318 Solution: Lemma: If U ∈ C n × n is upper triangular and normal then it is a diagonal matrix. Proof: Partition U as follows U = U n 1 u n 1 H n 1 μ nn where U n 1 ∈ [ C ] n 1 × n 1 is upper triangular and u n 1 ∈ [ C ] n 1 . Since U is normal we have UU H = U H U e H n UU H e n = e H n U H Ue n  μ nn  2 = u H n 1 u n 1 +  μ nn  2 1 Therefore, u n 1 = 0 n 1 and we have U = U n 1 n 1 H n 1 μ nn Given this structure and since U is normal, it follows that U n 1 is a normal upper triangular matrix. Therefore, the last column of U n 1 also has 0 in all elements above the diagonal and we have U = U n 2 D 2 where D 2 is a 2 × 2 diagonal matrix and U n 2 is an upper triangular n 2 × n 2 matrix. Once again, since U is normal this structure implies U n 2 is normal. This repeats until all columns above the diagonal are shown to be 0, i.e., U is diagonal, as desired. 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....
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 Fall '06
 gallivan
 Matrices, Diagonalizable matrix, Diagonal matrix, Normal matrix, 7.1.1 Golub

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