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$\mathcal{M}_{n}(K) $: Set of all $n×n$ matrices over the field $K$.

$A\in \mathcal{M}_{n}(K) $ is called nilpotent if $A^k=\textbf{0}$ for some $k\in \Bbb{Z}^+$

It is clear that if $A$ is nilpotent then $\operatorname{tr}(A^k) =0$ for all $k\in \Bbb{Z}^+$

The converse is also true if $\operatorname{char}(K) =0$

My question: Suppose for a matrix $A\in\mathcal{M}_n(K)$, $\operatorname{tr}(A^k)=0 \space \forall k\in \Bbb{Z}^+ $ implies $A$ is nilpotent. Can we conclude that $\operatorname{char}(K) =0$?

Gary
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Sourav Ghosh
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    Have you considered the identity matrix? – Hanno Jul 03 '22 at 18:51
  • I'm sorry but what's $char(K)$ again ? – Hamdiken Jul 03 '22 at 20:19
  • https://en.m.wikipedia.org/wiki/Characteristic_(algebra)#:~:text=As%20mentioned%20above%2C%20the%20characteristic,positive%20characteristic%20or%20prime%20characteristic.&text=of%20prime%20order.,and%20this%20isomorphism%20is%20unique.
    Characteristic of the field $K$ which is either $0$ or prime.     – Sourav Ghosh Jul 03 '22 at 20:24
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    No. Take $n=1$. You have that $A$ is a scalar, and the trace is the identity.

    Then the property that $A^k=0\implies A=0$ is true for every field.

     – Exodd Jul 03 '22 at 20:46
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    If you mean that it must hold also for every $n$, then yes, since the identity matrix of size $p$ (being the char of the field) has trace zero with all its powers – Exodd Jul 03 '22 at 21:00
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    in general you can only apply the converse for (i) characteristic zero and (ii) when $n\lt p$ for prime $p$ – user8675309 Jul 03 '22 at 21:30
  • Are we assuming the condition is true for one specific $n$ or simultaneously for all $n$? – Torsten Schoeneberg Jul 04 '22 at 21:12

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Let $P(n)$ be the statement "If $A \in M_n(K)$ is such that for all $k \ge 1$, $\operatorname{tr}(A^k)=0$, then $A$ is nilpotent."

Claim: $P(n) \Leftrightarrow (\operatorname{char}(K) = 0 \text{ or } n < \operatorname{char}(K) )$

Proof of "$\Rightarrow$": Assume $n \ge \operatorname{char}(K) =: p$. Consider the matrix whose first $p$ diagonal entries are $1$, all other entries $0$.

Proof of "$\Leftarrow$": See e.g. the answer to Trace Criterion. Notice for the case of positive characteristic $p$ that under our assumption, $\operatorname{tr}(Id_{m\times m}) = m \neq 0 \in K$ for all $1 \le m \le n < p$, which makes the argument in that answer go through.

(It would be interesting to see which of the proofs in Traces of all positive powers of a matrix are zero implies it is nilpotent, $\operatorname{tr}(A)=\operatorname{tr}(A^{2})= \ldots = \operatorname{tr}(A^{n})=0$ implies $A$ is nilpotent, The characteristic polynomial of $A$ is $x^n$ if and only if $\text{Tr}(A^i)=0$ for all $1\le i \le n$. also go through assuming $n < p$.)

So of course

$$(\forall n: P(n)) \Leftrightarrow \operatorname{char}(K)=0$$

Torsten Schoeneberg
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