College Algebra Exam Review 398

College Algebra Exam Review 398 - 1 in the.1;d position In...

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408 8. MODULES that is, T applied to each of these basis vectors is the next basis vector. However, T.T d ± 1 p.T / k v 0 / D T d p.T / k v 0 D p.T / k C 1 v 0 ± .a d ± 1 T d ± 1 C ²²² C a 1 T C a 0 /p.T / k v 0 : That is, T applied to the last basis vector in a block is the sum of the first vector of the next block and a linear combination of elements of the current block. (If k D m ± 1 , then p.T / k C 1 v 0 D p.T / m v 0 D 0 .) For example, if p.x/ D x 3 C x 2 C 3x C 5 , and m D 3 , then the matrix ŒT 1 Ł B is 2 6 6 6 6 6 6 6 6 6 6 6 6 4 0 0 ± 5 0 0 0 0 0 0 1 0 ± 3 0 0 0 0 0 0 0 1 ± 1 0 0 0 0 0 0 0 0 1 0 0 ± 5 0 0 0 0 0 0 1 0 ± 3 0 0 0 0 0 0 0 1 ± 1 0 0 0 0 0 0 0 0 1 0 0 ± 5 0 0 0 0 0 0 1 0 ± 3 0 0 0 0 0 0 0 1 ± 1 3 7 7 7 7 7 7 7 7 7 7 7 7 5 : Let C p be the d –by– d companion matrix of p.x/ . Let N be the d –by– d matrix with all zero entries except for a
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Unformatted text preview: 1 in the .1;d/ position. In general, the matrix ŒT 1 Ł B is J m .p.x// D 2 6 6 6 6 6 6 6 4 C p ²²² N C p ²²² N C p ²²² : : : : : : : : : : : : : : : : : : : : : C p ²²² N C p 3 7 7 7 7 7 7 7 5 : Now consider the special case that p.x/ is linear, p.x/ D x ± ± . Since d D 1 , the ordered basis B reduces to .v ;.T ± ±/v ;:::;.T ± ±/ m ± 1 v /...
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