Differential Equations Solutions 105

Differential Equations Solutions 105 - If A were...

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115 CHALLENGE 19.7. The probabilities are clearly nonnegative and sum to 1. Note that the j th component of e T A is the sum of the elements in column j , and this is 1, so e T A = e T . Therefore, e T A =1 e T , and this means that the vector e T is unchanged in direction when multiplied on the right by A . This is the deFnition of a left eigenvector of A , and the eigenvalue is 1. Apply the Gerschgorin circle theorem to A T , which has the same eigenvalues as A . If the main diagonal element of A T is 0 <α< 1, then the o±-diagonal elements are nonnegative and sum to 1 α . Therefore, the Gerschgorin circle is centered at α with radius 1 α . This circle touches the unit circle at the point (1 , 0) but lies inside of it. The eigenvalues lie in the union of the Gerschgorin circles, so all eigenvalues lie inside the unit circle.
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Unformatted text preview: If A were irreducible then the eigenvalue at 1 would be simple; see, for exam-ple, [1]. 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. [1] 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.

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