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College Algebra Exam Review 406

# College Algebra Exam Review 406 - A on Q 5 We found that Q...

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416 8. MODULES To simplify notation, consider one of these submodules. Call the sub- module W , the generator w 0 , and the period a.x/ . Write a.x/ D .x 1 / m 1 .x t / m t : Now we compute the primary decomposition of the cyclic submodule W exactly as in the discussion preceding Lemma 8.5.11 . We have W D W OEx 1 Ł ˚ ˚ W OEx t Ł; as KOExŁ –modules, and W OEx i Ł is cyclic of period .x i / m i . Set let r i .x/ D Q k ¤ i .x k / m k . Then r i .A/w 0 D Q k ¤ i .A k E/ m k w 0 is a generator of the module W OEx i Ł . A basis for W OEx i Ł is r i .A/w 0 ; .A i E/r i .A/w 0 ; : : : ; .A i E/ m i 1 r i .A/w 0 : With respect to this basis, the matrix of the restriction of T to W OEx i Ł is the Jordan block J m i . i / . Example 8.7.16. Consider the matrix A D 2 6 6 6 6 4 1 0 0 0 3 1 2 0 4 0 3 1 2 4 3 0 0 0 1 0 2 0 0 0 4 3 7 7 7 7 5 from Example 8.6.11 . We computed that the characteristic polynomial of A is .x 1/ 2 .x 2/ 3 , so A has a Jordan canonical form in Mat 5 OE Q Ł . Let T be the linear transformation determined by left multiplication by
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Unformatted text preview: A on Q 5 . We found that Q 5 is the direct sum of two T –invariant subspaces V 1 and V 2 . The subspaces V 1 is one dimensional, spanned by v 1 D 2 6 6 6 6 4 4 1 8=3 3 7 7 7 7 5 , and Av 1 D v 1 . The subspace V 2 is four dimensional, and generated as a KŒxŁ –module by v 2 D 2 6 6 6 6 4 3 3 1 3 7 7 7 7 5 . The period of V 2 is .x ± 1/.x ± 2/ 3 . We get the Jordan canonical form by computing the primary decomposition of V 2 , V 2 D V 2 Œx ± 1Ł ˚ V 2 Œx ± 2Ł . The subspace V 2 Œx ± 1Ł is one dimensional, and spanned by w 1 D .A ± 2E/ 3 v 2 D 2 6 6 6 6 4 3 ± 3 2 3 7 7 7 7 5 : The subspace V 2 Œx ± 2Ł...
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