Unformatted text preview: Math 2940 Solutions, Fall 2011 Section 8 . 3 3 ) A = 1 . 2 . 8 has characteristic polynomial ( Î» 1)( Î» . 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 Î» 2 . 2 Î» . 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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 '05
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 Math, Linear Algebra, Algebra, Steady State, steady state solution

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