College Algebra Exam Review 407

College Algebra Exam Review 407 - ˚ J 2 .1/ . We obtain...

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8.7. JORDAN CANONICAL FORM 417 is three dimensional, and generated by x 1 D .A ± E/v 2 D 2 6 6 6 6 4 3 3 3 0 3 3 7 7 7 7 5 : The remaining vectors in a basis for V 2 Œx ± are x 2 D .A ± 2E/x 1 D 2 6 6 6 6 4 0 3 3 0 0 3 7 7 7 7 5 ; and x 3 D .A ± 2E/ 2 x 1 D 2 6 6 6 6 4 0 0 3 0 0 3 7 7 7 7 5 : Let B be the basis .v 1 ;w 1 ;x 1 ;x 2 ;x 3 / and let S be the matrix whose columns are the elements of B . Then ŒT Ł B D S ± 1 AS D J 1 .1/ ˚ J 1 .1/ ˚ J 3 .2/: Example 8.7.17. We treat the matrix 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 of Example 8.7.14 again by our second method. First we compute the Smith normal form of xE ± A in Q ŒxŁ . That is we compute invertible matrices P;Q 2 Mat 4 . Q ŒxŁ/ such that P.xE ± A/Q D D.x/ is diagonal with the monic invariant factors of A on the diagonal The result is D.x/ D 2 6 6 4 1 0 0 0 0 1 0 0 0 0 .x ± 1/ 2 0 0 0 0 .x ± 1/ 2 3 7 7 5 ; so the invariant factors of A are .x ± 1/ 2 ;.x ± 1/ 2 . From this we can al- ready see that the Jordan form of A is J 2 .1/
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Unformatted text preview: ˚ J 2 .1/ . We obtain cyclic vectors for two invariant subspaces using the last two columns of the ma-trix P ± 1 , which are 2 6 6 4 1 ± 1=2 1 3 7 7 5 : Since the entries of these columns are constant polynomials, these two columnn are already the cyclic vectors we are looking for. A second basis vector for each for the two invariant subspaces is obtained by applying A ± E to the cyclic vector. The result is the basis v 1 D 2 6 6 4 1 ± 1=2 3 7 7 5 , v 2 D .A ± E/v 1 D 2 6 6 4 2 1 3 7 7 5 , w 1 D 2 6 6 4 1 3 7 7 5 , and...
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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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