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Mathematics 20 Online
OpenStudy (anonymous):

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.

OpenStudy (anonymous):

recall: AX=YX (A^K)X=A^(K-1)(AX) " =A^(K-1)(YX) =YA^(K-2)(AX) =YA^(K-2)(YX) =Y^(K-1)(AX) =Y^(K-1)(YX) =Y^K(X)

OpenStudy (anonymous):

What were you trying to do/ prove?

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