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Unformatted text preview: 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 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 , ( A I ) k ( v ) = 0 for some positive integer k , then v is a generalized eigenvector of A associated to the eigen value . The smallest k with ( A I ) 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 I )( v ) = 0 ( A ) v = ( I ) v Av = v and so v is an (ordinary) eigenvector. Recall that, for an eigenvalue of A , E is the eigenspace of , E = { v  Av = v } = { v  ( A I ) v = 0 } . We let E denote the generalized eigenspace of , E = { v  ( A I ) k ( v ) = 0 for some k } . It is easy to check that E is a subspace. Since every eigenvector is a generalized eigenvector, we see that E E . The following result (which we shall not prove) is an important fact about generalized eigenspaces. Proposition A.2. Let be an eigenvalue of A of multiplicity m. Then E is a subspace of dimension m. 2 Example A.3. Let A be the matrix A = bracketleftbigg 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 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 1dimensional. 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 , and set v k 1 = ( A I ) v k . We claim that v k 1 is a generalized eigenvector of index k 1 associated to the eigenvalue . To see this, note that ( A I ) k 1 v k 1 = ( A I ) k 1 ( A I ) v k = ( A I ) k v k = 0 but ( A I ) k 2 v k 1 = ( A I ) k 2 ( A I ) v k = ( A I ) k 1 v k negationslash = 0 ....
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This note was uploaded on 02/29/2008 for the course MATH 22 taught by Professor Dodson during the Summer '05 term at Lehigh University .
 Summer '05
 Dodson
 Differential Equations, Equations

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