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Unformatted text preview: 18.06 Problem Set 9 Solution Due Wednesday, 29 April 2009 at 4 pm in 2106. Total: 130 points. Problem 1: Let A = 1 s 1 3 , where s is some real number. (a) Give a value of s where A is defective; use this s in the subsequent parts. (b) Compute a set of eigenvectors and generalized eigenvectors (as defined in the handout) of A to give a complete basis for R 2 . (Use this basis in the subsequent parts.) (c) For the column vector ~u = (1 , 0) T , compute ~u ( t ) = ( I + e At ) 1 ~u (as an explicit formula with no matrix operations). (Hint: use the formula for f ( A ) from the handout; note that I = A .) (d) For the column vector ~u = (1 , 0) T , compute ~u k = A k ~u (as an explicit formula with no matrix operations). (e) Write an explicit formula for A k , for any k (as an explicit formula with no matrix operations). (Consider your answer for the previous part, and ask what matrix you would multiply by an arbitrary vector to obtain A k times that vector.) (f) Suppose we perturb the matrix slightly, changing s to s + 0 . 0001. Does k ~u k k grow more slowly or more quickly with k than when A was defective? Solution (30 points = 5+5+5+5+5+5) (a) Since we need A to be defective, its two eigenvalues must be the same. det( A λI ) = λ 2 4 λ + (3 s ) = 0. It has repeated roots when 3 s = 4, that is when s = 1. In this case, A = 1 1 1 3 . (b) When s = 1, solving det( A λI ) = λ 2 4 λ + 4 = 0 gives λ = 2. A λI = 1 1 1 1 ⇒ ~v 1 = 1 1 . 1 To find the generalized eigenvector, we solve the linear system ( A λI ) ~v 1 = ~v 1 ; 1 1 1 1 ~v 1 = 1 1 . We can find a particular solution ~v 1 = 1 . Then, we use GramSchmidt to make it perpendicular to ~v 1 as follows. ~v (2) 1 = ~v 1 ~v T 1 ~v 1 k ~v 1 k 2 ~v 1 = 1 + 1 2 1 1 = 1 2 1 2 . (c) First, we need to write ~u = c 1 ~v 1 + c 2 ~v (2) 1 . 1 1 2 1 1 2 c 1 c 2 = 1 ⇒ c 1 = 1 2 , c 2 = 1 . Let f ( X ) = (1 + e Xt ) 1 . Then f ( X ) = te Xt · (1 + e Xt ) 2 . By the formula from the handout, ~u ( t ) = ( I + e At ) 1 ~u = c 1 f ( λ ) ~v 1 + c 2 ( f ( λ ) ~v (2) 1 + f ( λ ) ~v 1 ) = 1 2 1 1 + e 2 t 1 1 1 1 + e 2 t 1 2 1 2 + te 2 t (1 + e 2 t ) 2 1 1 = 1 1 + e 2 t 1 + te 2 t (1 + e 2 t ) 2 1 1 . REMARK: One may notice that the first term is exactly 1 1+ e 2 t ~u . This is because the eigenvalues of the two eigenvectors are the same. The second term is contributed by the generalized eigenvector ~v (2) 1 , but it is a multiple of ~v 1 . (d) Continuing with the calculation above, we have ~u k = A k ~u = c 1 λ k ~v 1 + c 2 ( λ k ~v (2) 1 + kλ k 1 ~v 1 ) = 1 2 2 k 1 1 2 k 1 2 1 2 + k 2 k 1 1 1 = 2 k 1 + k · 2 k 1 1 1 = 2 k 1 2 k k . 2 (e) Since A k v 1 = λ k v 1 and A k v (2) 1 = λ k v (2) 1 + kλ k 1 v 1 , we have A k 1 1 2 1 1 2 = 2 k 1 2 2 k + k · 2 k 1 2 k 1 2 2 k k · 2 k 1 = 2 k 1 2 k 1 2 k 1 ....
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This note was uploaded on 05/06/2010 for the course 18 18.06 taught by Professor Strang during the Spring '08 term at MIT.
 Spring '08
 Strang

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