College Algebra Exam Review 406

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

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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 1 ²²² .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 Œx ± ± 1 Ł ˚ ²²² ˚ W Œx ± ± t Ł; as KŒxŁ –modules, and W Œx ± ± 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 Œx ± ± i Ł . A basis for W Œx ± ± 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 ² : With respect to this basis, the matrix of the restriction of T to W Œx ± ± 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 Œ 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 Kx 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 x 1 V 2 x 2 . The subspace V 2 x 1 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 x 2...
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