For instance 1 e a b e a e b if ab ba 2 d dx e ax ae

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While difficult to compute in general, the matrix exponential has several nice properties. For instance: (1) e A + B = e A e B if AB = BA (2) d dx e Ax = Ae Ax (3) e λIx v = e λx v for any scalar λ and vector v . Now the reason the matrix exponential is so interesting to us, is that given any constant vector v (that is, a vector that doesn’t depend on x ), suppose y = e Ax v . Then we have ˙ y = d dx y = d dx e Ax v = Ae Ax v = A y . Thus we have the surprising result that y = e Ax v is a solution to the homogeneous equation for any constant vector v ! The problem however, is that e Ax v is still defined by an infinite series and is thus very difficult to compute except for a few very special vectors v . This brings us to the next definition. Definition Let A be an n × n matrix. We say that v is a generalized eigenvector of A with eigenvalue λ if ( A - λI ) k v = 0 for some integer k 1 . I Example The usual eigenvectors we’ve been dealing with so far are in fact generalized eigenvectors with k = 1 . The reason generalized eigenvectors appeal to us, is that for a generalized eigenvector v , we can actually compute the vector e Ax v . Suppose v is a generalized eigenvector of A with k = 2 . We calculate: e Ax v = e λIx e ( A - λI ) x v by algebra and property (1) above = e λIx I + x ( A - λI ) + x 2 ( A - λI ) 2 2! + · · · v by definition of e X = e λIx I v + x ( A - λI ) v by the fact v is a generalized eigenvector = e λx I + x ( A - λI ) v by property (3) above .
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Thus for every generalized eigenvector v of A with eigenvalue λ and with k = 2 we get a solution of the form y = e λx I + x ( A - λI ) v For a generalized eigenvector with k = 3 we can perform the same calculation and since ( A - λI ) 3 v = 0 we must retain the quadratic term in the Taylor series, but higher degree terms will be zero and we get the solution y = e λx I + x ( A - λI ) + x 2 ( A - λI ) 2 2! v This can be easily be extended to any value of k by simply keeping more terms of the Taylor series. We now summarize these results: EV/ev Method Case 3: Repeated Eigenvalues To solve the homogeneous equation ˙ y = A y in the case where we have only m eigenvalues λ 1 , λ 2 , · · · , λ m , do the following: (1) Find as many linearly independent eigenvectors as possible for each eigenvalue λ i
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  • Fall '08
  • staff
  • Linear Algebra, eigenvector, Generalized eigenvector, linearly-independent solutions

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