College Algebra Exam Review 404

College Algebra Exam Review 404 - J 3.1 ˚ J 1.1 J 2.1 ˚ J...

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414 8. MODULES that the equation .T ± ±/x D x ± r has no solutions. Then the dimension of V is r C 1 , and x ± r is a generator of the cyclic KŒxŁ –module V . Example 8.7.14. Let A D 2 6 6 4 3 2 ± 4 4 ± 6 ± 3 8 ± 12 ± 3 ± 2 5 ± 6 ± 1 ± 1 2 ± 1 3 7 7 5 . Let T be the linear transformation determined by left multiplication by A on Q 4 . We can compute that the characteristic polynomial of A is ² A .x/ D x 4 ± 4x 3 C 6x 2 ± 4x C 1 D .x ± 1/ 4 : The possible Jordan forms for A correspond to partitions of 4 : They are J 4 .1/ , J 3 .1/ ˚ J 1 .1/ , J 2 .1/ ˚ J 2 .1/ , J 2 .1/ ˚ diag .1;1/ , and diag .1;1;1;1/ . The number of Jordan blocks is the number of linearly independent eigen- vectors of A , since each Jordan block has one eigenvector. Solving the linear equation .A ± E/ x D 0 , we find two linearly independent solutions, a D 2 6 6 4 ± 2 0 0 1 3 7 7 5 and b D 2 6 6 4 0 2 1 0 3 7 7 5 . Therefore, there are two Jordan blocks in the Jordan canonical form of A . The possible Jordan forms are thus
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Unformatted text preview: J 3 .1/ ˚ J 1 .1/ , J 2 .1/ ˚ J 2 .1/ . We find that .A ± E/ a ± 1 D a has solution a ± 1 D 2 6 6 4 2 ± 3 3 7 7 5 , but .A ± E/ x D a ± 1 has no solution. Likewise, the equa-tion .A ± E/ b ± 1 D b has solution b ± 1 D 2 6 6 4 ± 1 1 3 7 7 5 , but .A ± E/ x D b ± 1 has no solution. Therefore, the Jordan form of A is J 2 .1/ ˚ J 2 .1/ , and the matrix of T with respect to the ordered basis B D . a ± 1 ; a ; b ± 1 ; b / is in Jordan canonical form. Let S be the matrix whose columns are the elements of the ordered basis B , S D 2 6 6 4 2 ± 2 ± 1 0 ± 3 1 2 0 1 1 0 0 3 7 7 5 Then ŒT Ł B D S ± 1 AS D J 2 .1/ ˚ J 2 .1/...
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This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.

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