MATH 323 Final: MATH 323 Final Exam
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Note that the following provides answers to the sample problems and suggestions on their solutions. 1. (a) the characteristic polynomial of a turns out to be pa( ) = x x3, and so the eigenvalues are = 0, 1, 1 . (b) to nd the eigenspace e , we calculate the null space of a i. E 1 = span{(0, 1, 1)t} , e0 = span{(1, 0, 0)t} , e1 = span{(1, 2, 1)t} . (c) we let q be a matrix whose columns are linearly independent eigenvectors. 0 and we can check that q 1aq is a diagonal matrix whose entries are the eigenvalues of a. (d) since q 1aq = (cid:16) 1 0 0. 0 0 1(cid:17) from part (c), we let d be the diagonal matrix on the right. From this we see that a5 = qd5q 1 = (cid:16) 1 1 as a. 0 1 2(cid:17) , which turns out to be the same.
