Eigenvalue of Matrix over Finite Field

Question

It is known that $A\in M_{n\times n}(\mathbb{C})$ is singular if and only if $0$ is an eigenvalue of $A$. Is this still true when $A\in M_{n\times n} (F_q)$ where $F_q$ is a finite field?
Thanks.

Answer 1

Yes, let $k$ be an arbitrary field. If $A\in M_{n\times n}(k)$ has $0$ as an eigenvalue, then $A$ cannot be invertible, since if $v$ denotes the eigenvector then $$ BAv=0 $$ for all matrices $B\in M_{n\times n}(k)$, and in particular $BA$ cannot equal the identity matrix.

Conversely, if $A$ does not have $0$ as an eigenvalue, then the same holds true over any algebraic closure $\overline{k}$ of $k$, hence the determinant of $A$ (equal to the product of the eigenvalues of $A$ acting on $\overline{k}$) is non-zero, hence $A$ is invertible.