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College Algebra Exam Review 403

College Algebra Exam Review 403 - J is similar to J t So by...

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8.7. JORDAN CANONICAL FORM 413 Lemma 8.7.11. J m . / is similar to its transpose. Proof. Write J t m . / for the transpose of J m . / . We have J m . / D E C J m .0/ , and J t m . / D E C J t m .0/ . Therefore PJ m . /P D J t m . / . n Lemma 8.7.12. Let A D A 1 ˚ ˚ A s and B D A 1 ˚ ˚ B s be block diagonal matrices. (a) A t D A t 1 ˚ ˚ A t s . (b) If A i is similar to B i for each i , then A is similar to B . Proof. Exercise. n Lemma 8.7.13. A matrix in Jordan canonical form is similar to its trans- pose. Proof. Follows from Lemmas 8.7.11 and 8.7.12 . n Proof of Proposition 8.7.9 : Let A 2 Mat n .K/ . Note that the characteristic polynomial of A is the same as the characteristic polynomial of A t . Let F be a field containing K such that A .x/ factors into linear factors in F OExŁ . By xxx, it suffices to show that A and A t are similar in Mat n .F / . In the following, similarity means similarity in Mat n .F / . A is similar to its Jordan form J , and this implies that A t is similar to J t . By Lemma 8.7.13
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Unformatted text preview: , J is similar to J t . So, by transitivity of similarity, A is similar to A t . Computing the Jordan canonical form We will present two methods for computing the Jordan canonical form of a matrix. First method. The first method is the easier one for small matrices, for which computations can be done by hand. The method is based on the following observation. Suppose that T is linear operator on a vector space V and that V is a cyclic KŒxŁ –module with period a power of .x ² ±/ . Then V has (up to scalar multiples) a unique eigenvector x with eigenvalue ± . We can successively solve for vectors x ± 1 ;x ± 2 ;::: satisfying .T ² ±/x ± 1 D x , .T ² ±/x ± 2 D x ± 1 , etc. We finally come to a vector x ± r such...
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