I can see that if $c$ is an eigenvalue of a matrix $A$, then $c^k$ will be an eigenvalue for the matrix $A^k$, but I'm curious about the reverse case. I checked for a simple rotation matrix counterclockwise by $180$ with single eigenvalue $-1$ that the rotation matrix counterclockwise by $\pi/2$ has eigenvalues $\pm i$.
I don't expect it to be the case, though, that if $c$ is an eigenvalue for $A^k$ ($k > 1$), that $c^{1/k}$ are eigenvalues of $A$. The example I think of to contradict this is taking a $2 \times 2$ rotation matrix by $\pi / 3$, $A$, whose cube will have eigenvalue $-1$, which has three cube roots, which exceeds the possible number of eigenvalues of $A$.
Is there, however, a more intuitive way to see this, or a weaker result which holds relating the eigenvalues in the reverse direction?