8.3 - Math 2940 Solutions, Fall 2011 Section 8 . 3 3 ) A =...

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Unformatted text preview: Math 2940 Solutions, Fall 2011 Section 8 . 3 3 ) A = 1 . 2 . 8 has characteristic polynomial ( λ- 1)( λ- . 8) which has roots 1 , . 8. The steady state eigenvector has eigenvalue 1 and is (1 , 0). A = . 2 1 . 8 0 has characteristic polynomial λ 2- . 2 λ- . 8 which has roots 1 ,- . 8. The steady state eigenvector has eigenvalue 1 and is (5 / 9 , 4 / 9). Consider A = 1 / 2 1 / 4 1 / 4 1 / 4 1 / 2 1 / 4 1 / 4 1 / 4 1 / 2 . We could do the calulation to find the eigenvalues. On the other hand we can see that (1 / 3 , 1 / 3 , 1 / 3) is an eigenvector with eigenvalue 1 and that the first two rows of A- I = - 1 / 2 1 / 4 1 / 4 1 / 4- 1 / 2 1 / 4 1 / 4 1 / 4- 1 / 2 are independent so this last matrix has rank 2 so A- I has only one independent vector in its solution space and then A has only one independent eigenvector with eigenvalue 1. We have the steady state solution....
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This note was uploaded on 01/02/2012 for the course MATH 2940 at Cornell.

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