27_hwc1SolnsODDA

27_hwc1SolnsODDA - (b) Things are easy now because we have...

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Unformatted text preview: (b) Things are easy now because we have already computed A2 and B 2 . We find: A2 − 2I = and 0 −3 2 3 B − 2I = 4 1 5 −1 1 − 20 5 10 0 0 1 0 0 −2 −3 −1 0 = 3 2 5 . 1 1 5 8 −1 −4 8 1 −2 7 0 0 −3 = 1 −4 8 5 , 3.31 Let A be an n × n matrix such that Ak+1 = 0 for some k. Show I − A is invertible, by checking that the matrix B defined by B = I + A + A2 + . . . + Ak is the inverse of I − A. (This shows that if Ak+1 = 0, the inverse of I − A is a polynomial in A of degree (at most) k. SOLUTION We compute (I − A)B = (I − A) I + A + A2 + . . . + Ak−1 + Ak = (I − A) + (A − A2 ) + (A2 − A3 ) + . . . + (Ak−1 − Ak ) + (Ak − Ak+1 ) = (I − A) + (A − A2 ) + (A2 − A3 ) + . . . + (Ak−1 − Ak ) + Ak = I + (−A + A) + (−A2 + A2 ) + (−A3 + A3 ) + . . . + (−Ak + Ak ) where in the next to the last last line (penultimate line) we have used the fact that Ak+1 = 0, and where in the last line we have regrouped the terms. Now notice that A enters once with a plus sign, and once with a minus sign. Also, A2 enters once with a plus sign, and once with a minus sign. In fact this is true for Aj for each power j = 1, 2, . . . , k . Therefore, all of these terms cancel out, leaving only I . That is, (I − A)B = I . This shows that (I − A) is a left inverse of B . But the same sort of computation also shows that B (I − A) = I . Hence, (I − A) is a right inverse also, and therefore it is the inverse. 3.33 In fact, it turns out that whenever an n × n matrix A is invertible, there is a polynomial p of degree n − 1 at most such that A−1 = p(A). Why this is true, and why this is useful will be explained later on, but we can see some examples now. (a) Suppose that A is an n × n matrix, and q (x) is a polynomial of degree k such that q (A) = 0 but q (0) = 0. Define a function p(x) by q (0) − q (x) . p(x) = xq (0) Show that p(x) is a polynomial of degree k − 1, that A is invertible, and that A−1 = p(A). 21/september/2005; 22:09 28 ...
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This note was uploaded on 10/03/2010 for the course MATH 380 taught by Professor Gladue during the Fall '08 term at Roger Williams.

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