5.2 - 5.2 Characteristic Equation Recall: If RE form for...

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5.2 Characteristic Equation Recall: If RE form for (square) A is A U with r row interchanges, Then det A = ( - 1) r det U = ( - 1) r × product of pivots A is invertible there are no free variables det U 6 = 0 det A 6 = 0 A~x = 0 has only trivial solution ~x = ~ 0 A has an ev 0 there is ~x 6 = ~ 0 such that A~x = 0 · ~x = ~ 0 Invertible Matrix Theorem Equivalences Continued: r) A invertible s) det A 6 = 0 t) 0 is not an eigenvalue of A Another viewpoint: λ is an ev. of A A~x = λ~x for some ~x 6 = ~ 0 ( A - λI ) ~x = ~ 0 for ~x 6 = 0 p ( λ ) = det( A - λI ) = 0 Characteristic equation . 1
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For triangular A = a 11 ··· ··· a 1 n 0 . . . . . . . . . . . . . . . 0 ··· 0 a nn det( A - λI ) = ± ± ± ± ± ± ± ± ± ± ± a 11 - λ ··· ··· a 1 n 0 . . . . . . . . . . . . . . . 0 ··· 0 a nn - λ ± ± ± ± ± ± ± ± ± ± ± = ( a 11 - λ )( a 22 - λ ) ··· ( a nn - λ ) = 0 λ = a ii some i evs of A are just diagonal entries! EX:
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This note was uploaded on 04/09/2011 for the course MATH 232 taught by Professor Russel during the Spring '10 term at Simon Fraser.

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5.2 - 5.2 Characteristic Equation Recall: If RE form for...

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