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It's clear that $k$-th powers of EV of $A$ are all EV of $A^k$, but the reverse inclusion isn't obvious to me.

For reference, this question has been posted before (Eigenvalues and power of a matrix), but I couldn't find an actually valid proof for the general case in that thread. For $\Bbb C$ there's a theorem that states for $f$ analytic that $f(\text{Spec}(A))=\text{Spec}(f(A))$, but this doesn't work for an arbitrary field $\Bbb K$.

Hilbert Jr.
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  • I think the argument based on Jordan form in that post works for any algebraically closed field. – podiki Nov 11 '21 at 14:39
  • This can be false for non-algebraically-closed fields. You can pass to an algebraic closure and make the argument work. – Randall Nov 11 '21 at 14:40
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    Assuming that the field is algebraically closed, triangularise $A$ and the rest is straightforward. – user1551 Nov 11 '21 at 14:48
  • It's not true for an arbitrary field. Not even for $\Bbb R$: Let $A=\begin{bmatrix}0&1\-1&0\end{bmatrix}$, $k=2$. $A^2=-I$ but $A$ has no eigenvalues... – David C. Ullrich Nov 11 '21 at 14:48

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It's not true for an arbitrary field. Let $\Bbb K=\Bbb R$, $A=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$, $k=2$; then $A^k=-I$ but $A$ has no eigenvalues.

A simple self-contained proof for an algebraically closed field: Say $a$ is an eigenvalue of $A^k$. There exist $\lambda_1,\dots,\lambda_k\in\Bbb K$ with $$z^k-a=\prod_{j=1}^k(z-\lambda_j).$$Hence $$A^k-aI=\prod_{j=1}^k(A-\lambda_jI).$$Since $A^k-aI$ is singular this shows that at least one of the $A-\lambda_jI$ must be singular. (Note that $\lambda_j^k=a$, so this shows that $a$ is the $k$-th power of an eigenvalue of $A$.)

David C. Ullrich
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  • I completely forgot to consider the case of when the matrix has no EV in the ground field, thanks. Also, I think you forgot to include the power in "say $a$ is an EV of $A$" (you meant to say $A^k$, no?). – Hilbert Jr. Nov 11 '21 at 19:28
  • @V.Ch. yes of course that was supposed to be $A^k$, thanks. – David C. Ullrich Nov 11 '21 at 21:53