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# supplementABC-new - JORDAN CANONICAL FORM WITH APPLICATION...

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JORDAN CANONICAL FORM WITH APPLICATION TO SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS by STEVEN H. WEINTRAUB c circlecopyrt 2006, Steven H. Weintraub Dept. of Mathematics Lehigh University Bethlehem, PA 18015 USA

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1 JCF and DEs 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 negationslash = 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 = bracketleftbigg 0 1 4 4 bracketrightbigg . Then, as you can check, if u= bracketleftbigg 1 2 bracketrightbigg , 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 = bracketleftbigg 1 0 bracketrightbigg , then ( A 2 I ) 2 v = 0 but ( A 2 I ) v negationslash = 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 negationslash = 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 .

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