TITLE: Prove that p divides to algebraic multiplicity of the eigenvalue
QUESTION [7 upvotes]: 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.

REPLY [5 votes]: 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 $\lambda_j\not= 0$.