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anonymous
 5 years ago
suppose that y is an eigenvalue of matrix A with corresponding eigenvector X.Then if K is a +integer Y^k is an eigenvalue of the matrix A^k with corresponding eigenvector X.
anonymous
 5 years ago
suppose that y is an eigenvalue of matrix A with corresponding eigenvector X.Then if K is a +integer Y^k is an eigenvalue of the matrix A^k with corresponding eigenvector X.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0recall: AX=YX (A^K)X=A^(K1)(AX) " =A^(K1)(YX) =YA^(K2)(AX) =YA^(K2)(YX) =Y^(K1)(AX) =Y^(K1)(YX) =Y^K(X)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What were you trying to do/ prove?
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