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Unformatted text preview: If A were irreducible then the eigenvalue at 1 would be simple; see, for exam-ple, . Let the eigensystem of A be deFned by Au j = j u j , and let e 1 = n X j =1 j u j , where u 1 , . . . , u 4 are a basis for the eigenspace corresponding to the eigenvalue 1. Then verify that A k e 1 = n X j =1 j k j u j . Since k j 0 as k except for the eigenvalue 1, we see that A k e 1 1 u 1 + 2 u 2 + 3 u 3 + 4 u 4 . Therefore, we converge to a multiple of the stationary vector.  Richard Varga, Matrix Iterative Analysis , Prentice Hall, Englewood Clis, NJ, 1962....
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.
- Fall '11