supplement

# supplement - Math 205 Supplement c 2006 Steven H Weintraub...

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Math 205 Supplement c 2006, Steven H. Weintraub

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1 Mathematics 205 Supplement A. Jordan Canonical Form In order to understand and be able to use Jordan Canonical Form, we must introduce a new concept, that of a generalized eigenvector. Definition A.1. If v = 0 is a vector such that, for some λ , ( A - λI ) k ( v ) = 0 for some positive integer k , then v is a generalized eigenvector of A associated to the eigen- value λ . The smallest k with ( A - λI ) k ( v ) = 0 is the index of the generalized eigenvector v . Let us note that if v is a generalized eigenvector of index 1, then ( A - λI )( v ) = 0 ( A ) v = ( λI ) v Av = λv and so v is an (ordinary) eigenvector. Recall that, for an eigenvalue λ of A , E λ is the eigenspace of λ , E λ = { v | Av = λv } = { v | ( A - λI ) v = 0 } . We let ˜ E λ denote the generalized eigenspace of λ , ˜ E λ = { v | ( A - λI ) k ( v ) = 0 for some k } . It is easy to check that ˜ E λ is a subspace. Since every eigenvector is a generalized eigenvector, we see that E λ ˜ E λ . The following result (which we shall not prove) is an important fact about generalized eigenspaces. Proposition A.2. Let λ be an eigenvalue of A of multiplicity m. Then ˜ E λ is a subspace of dimension m.
2 Example A.3. Let A be the matrix A = 0 1 - 4 4 . Then, as you can check, if u= 1 2 , then ( A - 2 I ) u = 0 , so u is an eigenvector of A with associated eigenvalue 2 (and hence a generalized eigenvector of index 1 of A with associated eigenvalue 2). On the other hand, if v = 1 0 , then ( A - 2 I ) 2 v = 0 but ( A - 2 I ) v = 0 , so v is a generalized eigenvector of index 2 of A with associated eigenvalue 2. In this case, as you can check, the vector u is a basis for the eigenspace E 2 , so E 2 = { cu | c R } is 1-dimensional. On the other hand, u and v are both generalized eigenvectors associated to the eigenvalue 2, and are linearly independent (the equation c 1 u + c 2 v = 0 only has the solution c 1 = c 2 = 0 , as you can readily check), so ˜ E 2 has dimension at least 2. Since ˜ E 2 is a subspace of R 2 , it must have dimension exactly 2, and ˜ E 2 = R 2 (and { u, v } is indeed a basis for R 2 ). Let us next consider a generalized eigenvector v k of index k associated to an eigenvalue λ , and set v k - 1 = ( A - λI ) v k . We claim that v k - 1 is a generalized eigenvector of index k - 1 associated to the eigenvalue λ . To see this, note that ( A - λI ) k - 1 v k - 1 = ( A - λI ) k - 1 ( A - λI ) v k = ( A - λI ) k v k = 0 but ( A - λI ) k - 2 v k - 1 = ( A - λI ) k - 2 ( A - λI ) v k = ( A - λI ) k - 1 v k = 0 . Proceeding in this way, we may set v k - 2 = ( A - λI ) v k - 1 = ( A - λI ) 2 v k v k - 3 = ( A - λI ) v k - 2 = ( A - λI ) 2 v k - 1 = ( A - λI ) 3 v k . . . v 1 = ( A - λI ) v 2 = · · · = ( A - λI ) k - 1 v k and note that each v i is a generalized eigenvector of index i associated to the eigenvalue λ . A collection of generalized eigenvectors obtained in this way gets a special name.

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3 Definition A.4. If { v 1 , . . . , v k } is a set of generalized eigenvectors associated to the eigenvalue λ of A , such that v k is a generalized eigenvector of index k and also v k - 1 =( A - λI ) v k , v k - 2 = ( A - λI ) v k - 1 , v k - 3 = ( A - λI ) v k - 2 , · · · , v 2 = ( A - λI ) v 3 , v 1 = ( A - λI ) v 2 , then { v 1 , . . . , v k } is called a chain of generalized eigenvectors of length k . The vector v k is called the top
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