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Unformatted text preview: k +1) = Ae ( k ) by part (a), and substituting the induction hypothesis yields e ( k +1) = AA k e (0) = A k +1 e (0) . Therefore, the result holds for all k = 1 , 2 , . . . , by mathematical induction. (c) ollowing the hint, we express e (0) = n X j =1 j u j , so, by part (b), e ( k ) = n X j =1 j k j u j . Now, if all eigenvalues j lie within the unit circle, then k j 0 as k , so e ( k ) 0. On the other hand, if some eigenvalue is outside the unit circle, then by choosing x (0) so that e (0) = u , we see that e ( k ) = k u does not converge to zero, since its norm | k | u ....
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.
- Fall '11