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Differential Equations Solutions 18

# Differential Equations Solutions 18 - k 1 = Ae k by part(a...

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28 Chapter 5. Solutions: Matrix Factorizations where D is a diagonal matrix containing the values α 1 ,...,α r ,and W = 1 λ 1 λ 2 1 ... λ r 1 1 1 λ 2 λ 2 2 ... λ r 1 2 . . . . . . . . . ... . . . 1 λ r λ 2 r ... λ r 1 r . Now W is a Vandermonde matrix and has full rank r , and so does the matrix formed by the u vectors. Therefore the vectors x , Ax ,..., A r 1 x must form a matrix of rank r and therefore are linearly independent, which contradicts the statement that S has dimension k<r . Therefore, every vector in S must have components of at most k diFerent eigenvectors, and we can take them as a basis. CHALLENGE 5.15. (a) Subtracting the relations x ( k +1) = Ax ( k ) + b and x true = Ax true + b , we obtain e ( k +1) = x ( k +1) x true = A ( x ( k ) x true )= Ae ( k ) . (b) If k = 1, then the result holds by part (a). As an induction hypothesis, suppose e ( k ) = A k e (0) . Then e (
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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 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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