I need help in the following exercise of a qualifying exam:
Let $A$ be a matrix of size $m$ by $m$ over the finite field $\mathbb{F}_p$ such that $\operatorname{trace}\left(A^n\right)=0$ for all $n$. If $\lambda$ is a nonzero eigenvalue of $A$, prove that the algebraic multiplicity of $\lambda$ is divisible by $p$.
Thank you by some hints.
Darij's argument is undoubtedly much smarter (I don't understand it), but the following simple approach works too. In the algebraic closure, we have the Jordan normal form available, so if $m_j$ denotes the algebraic multiplicity of $\lambda_j$, then your assumption now says that $\sum m_j \lambda_j^n=0$ for all $n\ge 1$, or, equivalently, $\sum m_jp(\lambda_j)=0$ for all polynomials with $p(0)=0$. We can now take $p=\lambda\prod_{k\not= j} (\lambda-\lambda_k)$ to see that $m_j\equiv 0\mod p$ if $\lambda_j\not= 0$.