Eigen values of a nilpotent matrix over a finite field?

Question

If $A$ is a nilpotent matrix over a field of characteristics $0$, then all eigen values of $A$ are $0$. Does the same result holds even for the nilpotent matrices over a finite field?

Answer 1

This is obviously true over any field. Say $A^n=0$ and $Ax=\lambda x$, $x\ne0$. Then $\lambda^n x=A^nx=0$, so $\lambda^n=0$, hence $\lambda=0$. I can't imagine a proof that depends on the field having characteristic zero...

What about the converse? The converse is less trivial, but it doesn't require Cayley-Hamilton.

Of course the converse is false in general: The matrix $\begin{bmatrix}0&-1&0\\1&0&0\\0&0&0\end{bmatrix}$ has $0$ as its only real eigenvalue but it's not nilpotent.

Assume below we're talking about operators on a finite-dimensional vector space over an algebraically closed field. And let $\sigma(T)$ denote the set of eigenvalues of $T$.

Lemma If $p$ is a polynomial and $\lambda\in \sigma(p(T))$ then there exists $\alpha\in\sigma(T)$ with $p(\alpha)=\lambda$.

Proof: There exist $\alpha_j$ with $p(x)-\lambda=\prod(x-\alpha_j)$. So $p(T)-\lambda I=\prod(T-\alpha_jI)$; hence there exists $j$ such that $T-\alpha_jI$ is not invertible.

Theorem If $0$ is the only eigenvalue of $T$ then $T$ is nilpotent.

Proof: Say $T:V\to V$. Say $N$ is the kernel of $T$. If $N=Y$ we're done; suppose not and let $W=V/N$. Now $T$ induces an operator $\tilde T:W\to W$ and $\tilde T$ has an eigenvalue $\lambda$. This says that there exists $x\in V$ with $x\ne 0$ and $y\in N$ such that $$Tx=\lambda x+y.$$ It follows that $\lambda=0$, because if $\lambda\ne0$ then $$T(x+\lambda^{-1}y)=\lambda(x+\lambda^{-1}y),$$so $\lambda$ is an eigenvalue of $T$.

So $Tx=y$, which says that the nullspace of $T^2$ is a proper superspace of the nullspace of $T$. The lemma shows that $0$ is the only eigenvalue of $T^2$. Repeating the argument shows that there exists $n$ such that $T^{2^n}=0$.