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 α j u j , so, by part (b), e ( k ) = n X j =1 α j λ k j u j . Now, if all eigenvalues λ j lie within the unit circle, then λ k j → 0 as k → ∞ , so e ( k ) → 0. On the other hand, if some eigenvalue λ ± is outside the unit circle, then by choosing x (0) so that e (0) = u ± , we see that e ( k ) = λ k ± u ± does not converge to zero, since its norm | λ k ± |± u ± ± → ∞ ....
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- Fall '11
- Linear Algebra, induction hypothesis, Matrix Factorizations, induction hypothesis yields