This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1. SOLUTIONS (1)(20 points) ( a ) If the sequence a n satisfies lim n a n +1 a n = 0 then lim n a n exists and is finite. F Take a n = n. ( b ) If q is a polynomial of degree 5 and q (0) = q 00 (0) = q (4) (0) = 0 and q (0) = 1 , q 000 (0) = 1 6 , q (5) (0) = 1 120 then q is the fifth degree Taylor polynomial for f ( x ) = sin x at c = 0 . T Since f (0) = f 00 (0) = f (4) (0) = 0 and f (0) = 1 , f 000 (0) = 1 6 , f (5) (0) = 1 120 . ( c ) The matrix A = 0 1 0 0 0 1 0 0 0 is diagonalizable. F One way to see this is that the only eigenvalue is 0 , it has multiplicity 3 , and the eigenspace N ( A I ) = { ( t, , 0) T : t R } so there are not three linearly independent eigenvectors. Another way to see this is that if A were diagonalizable, then for some invertible matrix X we would have A = XDX 1 and D is the diagonal matrix with the eigenvalues of A on its diagonal. But this would be the 0 matrix and as a result A is also the 0 matrix, but it is not, so A can not be diagonalizable. 1 2 ( d ) If an eigenvalue of the matrix A has multiplicity greater than one, then A is defective....
View
Full
Document
 Spring '10
 DONALDSON,NEIL

Click to edit the document details