Let $A$ be a square matrix over some field $\Bbb{F}$. It can be shown that if $\lambda$ is an eigenvalue of $A$ then $\lambda^n$ is an eigenvalue of $A^n$. In my linear algebra class we were asked if all eigenvalues of $A^n$ can be written in this form.
In general, the answer is that they cannot. For example, if $\Bbb{F} = \Bbb{R}$ then the matrix $$A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \\ \end{bmatrix}$$ has no eigenvalues but $A^2$ has $-1$ as an eigenvalue. In fact, this argument can be generalized to show that if $\Bbb{F}$ is not closed under $n$th roots then the result does not hold over that field.
However, if $\Bbb{F}$ is algebraically closed and $A$ is as above, then it can actually be shown that all eigenvalues of $A^n$ are of the form $\lambda^n$ for some eigenvalue $\lambda$ of $A$.
Unfortunately, there are fields which are not algebraically closed but which are closed under $n$th roots. Because of this, I wanted to ask the following question:
For which fields $\Bbb{F}$ does it hold that if $A$ is a square matrix over $\Bbb{F}$ then all eigenvalues of $A^n$ are of the form $\lambda^n$ for some eigenvalue $\lambda$ of $A$?
Let $F$ be a field that contains all the $n$th roots of each of its elements. Let $A$ be a square matrix with entries in $F$. Let $\mu$ be an eigenvalue of $A^n$ with corresponding eigenvector $v\ne0$, so $(A^n-\mu I)v=0$. Then we have $$(A-\lambda_1I)(A-\lambda_2I)\cdots(A-\lambda_nI)v=0$$ where, for each $i$, $\lambda_i^n=\mu$. By hypothesis, each $\lambda_i$ is in $F$. Then there exists $j$ such that $$(A-\lambda_jI)(A-\lambda_{j+1}I)\cdots(A-\lambda_n)v=0,\quad(A-\lambda_{j+1}I)\cdots(A-\lambda_nI)v\ne0$$ Then $(A-\lambda_jI)w=0$, where $$w=(A-\lambda_{j+1}I)\cdots(A-\lambda_nI)v\ne0$$ so $\lambda_j$ is an eigenvalue of $A$. That is, $A$ has an eigenvalue $\lambda=\lambda_j$ such that $\lambda^n=\mu$.