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 equation .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.
 Fall '08
 EVERAGE
 Algebra

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