By solving the equation a λi v 0 2 from step 1 we

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by solving the equation ( A - λI ) v = 0 . (2) From step (1) we now have at least m linearly-independent solutions. If m < n , for each of the repeated eigenvalues start looking for generalized eigenvectors with k = 2 by solving ( A - λI ) 2 v = 0 . For each generalized eigenvector (that isn’t also an eigenvector), you now have a new solution y = e λx I + x ( A - λI ) v . (3) If step (2) didn’t produce enough solutions, look at repeated eigenvalues with multiplicity 3 and try to find generalized eigenvectors with k = 3 by solving ( A - λI ) 3 v = 0 . For each generalized eigenvector (that isn’t also an generalized eigenvector with k = 1 , 2 ), you now have a new solution y = e λx I + x ( A - λI ) + x 2 ( A - λI ) 2 2! v . (4) If step (3) didn’t produce enough solutions, proceeding in this manner for higher values of k will always eventually produce enough solutions, with the number of steps determined by multiplicity of the most repeated eigenvalue. Once you have n linearly-independent solutions, the general solution is as always y ( x ) = c 1 y 1 + c 2 y 2 + · · · + c n y n .
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I Example Find the general solution to the system ˙ y = - 1 - 1 0 0 - 1 0 0 0 - 2 y . Solution To find the eigenvalues we compute det - 1 - λ - 1 0 0 - 1 - λ 0 0 0 - 2 - λ = ( - 2 - λ )( - 1 - λ )( - 1 - λ ) = - ( λ + 2)( λ + 1) 2 so A has eigenvalues λ 1 = - 1 and λ 2 = - 2 . For λ 1 = - 1 we solve ( A - λ 1 I ) v = 0 0 - 1 0 0 0 0 0 0 - 1 v 1 v 2 v 3 = 0 0 0 . The first equation gives v 2 = 0 and the third gives v 3 = 0 , but v 1 is undetermined so we have eigenvectors of the form v = c 0 0 and taking c = 1 gives v 1 = 1 0 0 so we have one solution y 1 = e λ 1 x v 1 = e - x 0 0 .
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For λ 2 = - 2 we solve ( A - λ 2 I ) v = 0 1 - 1 0 0 1 0 0 0 0 v 1 v 2 v 3 = 0 0 0 . The third equation means v 3 is undetermined so take v 3 = c . The second equation gives v 2 = 0 and the first gives v 1 - v 2 = 0 , so v 1 = 0 . Then we have eigenvectors of the form v = 0 0 c and taking c = 1 gives v 2 = 0 0 1 so we have one solution y 2 = e λ 2 x v 2 = 0 0 e - 2 x .
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  • Fall '08
  • staff
  • Linear Algebra, eigenvector, Generalized eigenvector, linearly-independent solutions

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