pset9-s09-soln

# pset9-s09-soln - 18.06 Problem Set 9 Solution Due Wednesday...

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18.06 Problem Set 9 Solution Due Wednesday, 29 April 2009 at 4 pm in 2-106. 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 0 = (1 , 0) T , compute ~u ( t ) = ( I + e At ) - 1 ~u 0 (as an explicit formula with no matrix operations). (Hint: use the formula for f ( A ) from the handout; note that I = A 0 .) (d) For the column vector ~u 0 = (1 , 0) T , compute ~u k = A k ~u 0 (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

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To find the generalized eigenvector, we solve the linear system ( A - λI ) ~v 0 1 = ~v 1 ; - 1 - 1 1 1 ~v 0 1 = 1 - 1 . We can find a particular solution ~v 0 1 = - 1 0 . Then, we use Gram-Schmidt to make it perpendicular to ~v 1 as follows. ~v (2) 1 = ~v 0 1 - ~v 0 T 1 ~v 1 k ~v 1 k 2 ~v 1 = - 1 0 + 1 2 1 - 1 = - 1 2 - 1 2 . (c) First, we need to write ~u 0 = c 1 ~v 1 + c 2 ~v (2) 1 . 1 - 1 2 - 1 - 1 2 c 1 c 2 = 1 0 c 1 = 1 2 , c 2 = - 1 . Let f ( X ) = (1 + e Xt ) - 1 . Then f 0 ( X ) = te Xt · (1 + e Xt ) - 2 . By the formula from the handout, ~u ( t ) = ( I + e At ) - 1 ~u 0 = c 1 f ( λ ) ~v 1 + c 2 ( f ( λ ) ~v (2) 1 + f 0 ( 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 0 + 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 0 . 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 0 = c 1 λ k ~v 1 + c 2 ( λ k ~v (2) 1 + 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 0 + 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 + 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 . Hence, A k = 2 k - 1 2 k - 1 - 2 - k - 1 1 - 1 2 - 1 - 1 2 - 1 = 2 k - 1 2 k - 1 - 2 - k - 1 1 2 - 1 2 - 1 - 1 = 2 k - 1 2 - k - k k k + 2 . (f) If we change s to s + 0 . 0001, then to get the eigenvalue, we need to solve λ 2 - 4 λ + 4 - 0 . 0001 = 0, that is ( λ - 2) 2 = 0 . 0001. We get λ 1 = 2 . 01 and λ 2 = 1 . 99. Since λ 1 > λ , k ~u k k is going to grow more quickly with k than when A was defective. More precisely, since an exponential always grows faster than any polynomial, 2 . 01 k = 1 . 05 k 2 k grows faster than k 2 k .

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