If $t$ is an eigenvalue of $A^{n}$, then must an $n$th root of $t$ be an eigenvalue of $A$?

Question

If $t$ is an eigenvalue of $A^{n}$, then must an $n$th root of $t$ be an eigenvalue of $A$? Also, does doing it over the real or complex numbers matter?

Answer 1

Consider $A= \begin{bmatrix} 1 & -2 \\ 1 & -1 \end{bmatrix}$ as a matrix over the reals.

Then $A^2=-I$ has repeated eigenvalue $-1$.

Since the square root of $-1$ is not real, $A$ has no real eigenvalues.

If, on the other hand $A$ is considered to be a matrix over the complex numbers them it will have eigenvalues of $j$ and $-j$. So, it does matter what field you are over.