College Algebra Exam Review 405

# College Algebra Exam Review 405 - ical form of a matrix...

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8.7. JORDAN CANONICAL FORM 415 Example 8.7.15. Let A D 2 6 6 4 1 2 ± 4 4 2 ± 1 4 ± 8 1 0 1 ± 2 1 1 ± 2 3 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 ± A .x/ D x 4 ± 4x 3 C 2x 2 C 4x ± 3 D .x ± 3/.x ± 1/ 2 .x C 1/: There are two possible Jordan canonical forms: diag . ± 1;3/ ˚ J 2 .1/ and diag . ± 1;3;1;1/ . By solving linear equations .A ± ²E/ x D 0 for ² D ± 1;3;1 , we ﬁnd that there is one eigenvector for each of the three eigenval- ues, respectively a D 2 6 6 4 ± 2 4 1 0 3 7 7 5 , b D 2 6 6 4 2 ± 4 ± 1 2 3 7 7 5 , and c D 2 6 6 4 0 2 1 0 3 7 7 5 . There- fore, the Jordan form is diag . ± 1;3/ ˚ J 2 .1/ . We ﬁnd that the equation .A ± E/ c ± 1 D c has solution c ± 1 D 2 6 6 4 0 1 0 ± 1 2 3 7 7 5 . (As a check, we can compute that the equation .A ± E/ x D c ± 1 has no solution.) The matrix of T with respect to the basis B D . a ; b ; c ± 1 ; c / is in Jordan form. Let S be the matrix whose columns are the elements of the ordered basis B , S D 2 6 6 4 ± 2 2 0 0 4 ± 4 1 2 1 ± 1 0 1 0 2 ± 1 2 0 3 7 7 5 Then ŒT Ł B D S ± 1 AS D diag . ± 1;3/ ˚ J 2 .1/ Second method. Our second method for computing the Jordan canon-
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Unformatted text preview: ical form of a matrix proceeds by ﬁrst computing the rational canonical form and then applying the primary decomposition to the generator of each cyclic submodule. Let A be a matrix in Mat n .K/ whose characteristic polynomial factors into linear factors in KŒxŁ . Let T be the linear operator of left multiplica-tion by A on K n . Suppose we have computed the rational canonical form of A (by the method of the previous section). In particular, suppose we have a direct sum decomposition .T;K n / D .T 1 ;V 1 / ˚ ²²² ˚ .T s ;V s /; where V j is a cyclic KŒxŁ –submodule with generator v .j/ and period a j .x/ , where a 1 .x/;:::;a s .x/ are the invariant factors of A ....
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