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Unformatted text preview: s is an eigenvalue of A . (I did not require this part for full credit, but it’s great if you realized that it should be checked!) This time we should use the ”other version” of eigenvalue. Since s 2 is an eigenvalue of A 2 , we know that A 2-s 2 I is not invertible. But, ( A + sI )( A-sI ) = A 2 + sIA-sAI-s 2 I 2 = A 2 + sA-sA-s 2 I = A 2-s 2 I So if that matrix on the far right is noninvertible, at least one of the two matrices in the factorization on the left is noninvertible as well (because otherwise they would both be invertible and then so would their product). One of those matrices being noninvertible shows that one of ± s is an eigenvalue of A . But, ± s are the two possible squareroots of s 2 , so we again see that the squares of the eigenvalues of A are eigenvalues of A 2 ....
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