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July 5

Eigenvalues of A^2

Hello,

It's straightforward to show that if λ is an eigenvalue of A then λ^2 is an eigenvalue of A^2. I'm trying to prove that opposite, that if λ^2 an eigenvalue of A^2 (over the complex numbers C) then λ or -λ is an eigenvalue of A. I would appreciate any direction. — Preceding unsigned comment added by 77.127.22.176 (talk) 15:32, 5 July 2017 (UTC)

$\det [A^2-{\lambda }^{2}I]=\det \left[(A-\lambda I)(A+\lambda I)\right]=\det [A-\lambda I]\det [A+\lambda I]$ Count Iblis (talk) 20:55, 5 July 2017 (UTC)

Could you pleas elaborate? why it cannot be the case that (x-λ^2) appears in the characteristic polynomial of A-λI or A+λI? — Preceding unsigned comment added by 77.127.22.176 (talk) 06:08, 6 July 2017 (UTC)

never mind, I understand. Thanks! — Preceding unsigned comment added by 77.127.22.176 (talk) 06:42, 6 July 2017 (UTC)

By a similar argument you get a generalization: If p is a polynomial and μ is an eigenvalue of p(A), then every eigenvalue of A is a solution of p(λ)=μ.--RDBury (talk) 12:58, 7 July 2017 (UTC)

Shouldn't that be, instead of "every eigenvalue of A is", "there exists an eigenvalue of A which is"? -- Meni Rosenfeld (talk) 23:50, 8 July 2017 (UTC)

Yes. Good catch. --RDBury (talk) 10:04, 10 July 2017 (UTC)