Proof of Nilpotency of Matrix All Whose Powers Have Vanishing Trace Without Using Eigenvalues Let $k$ be a field and let $A\in M^{n\times n}(k)$ be such that $tr(A^m)=0$ for all $m\in \mathbb N$ (or all $m\leq n$). Prove that $A$ is nilpotent.
The canonical proof can be found in this answer. However, this answer uses the existence of a splitting field of the characteristic polynomial. I wonder if there are any more elementary proofs.
 A: I will consider only the case where $k$ has characteristic $0$.
One approach is as follows. As is noted here, the coefficients of the characteristic polynomial of $A$ can be expressed in terms of the trace of the powers of $A$, with
$$
\operatorname{tr}\left(\textstyle\bigwedge^m A\right) = \frac{1}{m!}  
\begin{vmatrix}  \operatorname{tr}A  &   m-1 &0&\cdots & \\
\operatorname{tr}A^m  &\operatorname{tr}A&  m-2 &\cdots & \\
 \vdots & \vdots & & \ddots & \vdots    \\
\operatorname{tr}A^{m-1} &\operatorname{tr}A^{m-2}& & \cdots & 1    \\ 
\operatorname{tr}A^m  &\operatorname{tr}A^{m-1}& & \cdots & \operatorname{tr}A
\end{vmatrix},
$$
where $(-1)^m \operatorname{tr}\left(\textstyle\bigwedge^m A\right)$ is the coefficient of $t^{n-m}$ in the characteristic polynomial $p(t)$ of $A$.  If all powers of $A$ have trace zero, then it follow via the above that the non-leading coefficients of the characteristic polynomial are zero, which is to say that $p(t) = t^n$.
By the Cayley-Hamilton theorem, it follows that $A$ is nilpotent.
