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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.

Or Kedar
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  • Perhaps you would prefer the second proof given in that same link; arguably, using Newton's identities is still "elementary". – Ben Grossmann Sep 08 '21 at 14:53
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    Note that this result in general fails to hold in the case that $k$ is of finite characteristic. For instance, all powers of $$ \pmatrix{1&1\0&1} $$ have trace zero when $k$ has characteristic $2$. – Ben Grossmann Sep 08 '21 at 14:55
  • @BenGrossmann This solution still assume the existence of $n$ generalized eigenvalues, which I disallow. – Or Kedar Sep 08 '21 at 16:18
  • you really don't need a splitting field of an arbitrary char poly over $\mathbb K$. One sufficient approach: prove Newton's Identities as a trace identity over $\mathbb C$. Then consider the polynomial ring $R:=\mathbb Z[\mathbf x]$ and $A' \in R^{n\times n}$ -- trace form of Newton's Identities holds here (why?) and then do a substitution homomorphism into your field $\mathbb K$. This last step requires care when you have positive characteristic-- as evidenced by the fact that your opening two sentences are still innacurate. No field theory required -- just ring theory and FTA. – user8675309 Sep 08 '21 at 17:40

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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.

Ben Grossmann
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