hw3_2009_10_08_01_solutions

hw3_2009_10_08_01_solutions - EE263 S. Lall 2009.10.08.01...

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EE263 S. Lall 2009.10.08.01 Homework 3 Solutions Due Thursday 10/15. 1. Some basic properties of eigenvalues. 50045 Show that (a) the eigenvalues of A and A T are the same (b) A is invertible if and only if A does not have a zero eigenvalue (c) if the eigenvalues of A are λ 1 ,...,λ n and A is invertible, then the eigenvalues of A - 1 are 1 1 ,..., 1 n , (d) the eigenvalues of A and T - 1 AT are the same. Hint: you’ll need to use the facts that det A = det( A T ), det( AB ) = det A det B , and, if A is invertible, det A - 1 = 1 / det A . Solution. (a) The eigenvalues of a matrix A are given by the roots of the polynomial det( sI A ). From determinant properties we know that det( sI A ) = det( sI A ) T = det( sI A T ). We conclude that the eigenvalues of A and A T are the same. (b) First we recall that A is invertible if and only if det( A ) n = 0. But det( A ) n = 0 ⇐⇒ det( A ) n = 0. i. If 0 is an eigenvalue of A , then det( sI A ) = 0 when s = 0. It follows that det( A ) = 0 and thus det( A ) = 0, and A is not invertible. From this fact we conclude that if A is invertible, then 0 is not an eigenvalue of A . ii. If A is not invertible, then det( A ) = det( A ) = 0. This means that, for s = 0, det( sI A ) = 0, and we conclude that in this case 0 must be an eigenvalue of A . From this fact it follows that if 0 is not an eigenvalue of A , then A is invertible. (c) From the results of the last item we see that 0 is not an eigenvalue of A . Now consider the eigenvalue/eigenvector pair ( λ i ,x i ) of A . This pair satis±es Ax i = λ i x i . Now, since A is invertible, λ i is invertible. Multiplying both sides by A - 1 and λ - 1 i we have λ - 1 i x i = A - 1 x i , and from this we conclude that the eigenvalues of the inverse are the inverse of the eigenvalues. (d) First we note that det( sI A ) = det( I ( sI A )) = det( T - 1 T ( sI A )). Now, from determinant properties, we have det( T - 1 T ( sI A )) = det( T - 1 ( sI A ) T ). But this is also equal to det( sI T - 1 AT ), and the conclusion is that the eigenvalues of A and T - 1 AT are the same. 2. Tridiagonal Toeplitz matrices 0660 Some matrices have simple formulae for eigenvalues and eigenvectors. An example we have seen is the circulant matrices. Another example is given by tridiagonal Toeplitz matrices, as follows. Suppose G = b a c b a c b a . . . a c b is an n × n real matrix, and a n = 0 and c n = 0. 1
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EE263 S. Lall 2009.10.08.01 (a) Suppose x = x 1 x 2 . . . x n is an eigenvector of G . Show that cx k - 1 + bx k + ax k +1 = λx k for all k = 1 ,...,n , where we deFne for convenience x 0 = 0 and x n +1 = 0. Hence show that the solution x must be of the form x k = αr k 1 + βr k 2 for some choice of α,β,r 1 and r 2 .
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hw3_2009_10_08_01_solutions - EE263 S. Lall 2009.10.08.01...

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