We are still short one linearly independent solution

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We are still short one linearly-independent solution, so we’ll look for a generalized eigenvector of the repeated eigenvalue λ 1 = - 1 , with k = 2 . So we solve ( A - λ 1 I ) 2 v = 0 0 0 0 0 0 0 0 0 1 v 1 v 2 v 3 = 0 0 0 . The first and second equations give that v 1 and v 2 are undetermined, respectively, so take v 1 = a and v 2 = b . The third gives v 3 = 0 , so we have generalized eigenvectors of the form v = a b 0 and taking a = b = 1 gives v 3 = 1 1 0
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(note that we could not have chosen a = 1 and b = 0 , since that would just give back our eigenvector v 1 ). Then we have a solution y 3 = e λ 1 x I + x ( A - λ 1 I ) v 3 = e - x 1 - x 0 0 1 0 0 0 1 - x 1 1 0 = e - x 1 - x 1 0 = (1 - x ) e - x e - x 0 . Therefore the general solution is y ( x ) = c 1 y 1 + c 2 y 2 + y 3 = c 1 e - x 0 0 + c 2 0 0 e - 2 x + c 3 (1 - x ) e - x e - x 0 Remark: This example was taken from p. 352 problem 3. To compare this to the solution in the back of the text, we need to simplify by factoring out the exponentials and reassign the constants c 1 , c 2 , c 3 (we can do this because they are arbitrary) to rewrite above solution as y ( x ) = e - x c 1 1 0 0 + c 2 - x 1 0 + c 3 e - 2 x 0 0 1 . These solutions look different, but in fact they parameterize the same solution space.
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  • Fall '08
  • staff
  • Linear Algebra, eigenvector, Generalized eigenvector, linearly-independent solutions

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