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Let $A\in M_{n}(\mathbb R)$ and $\mathrm{tr}(A^{m})=0$ for every positive integer $m$. Is $A$ nilpotent? Is $A^{2}-I$ invertible?

user26857
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Aliakbar
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    Also see: http://math.stackexchange.com/questions/159167/trace-of-powers-of-a-nilpotent-matrix – anon Dec 08 '12 at 05:19
  • This is a direct combination of https://math.stackexchange.com/questions/159167/traces-of-all-positive-powers-of-a-matrix-are-zero-implies-it-is-nilpotent and https://math.stackexchange.com/questions/637423/a1-a-1-invertible-for-nilpotent-element , so voting to close. – darij grinberg Feb 09 '19 at 22:35

1 Answers1

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Write $A$ in its Jordan form: $A=SJS^{-1}$, with $S$ invertible. As $A^m=SJ^mS^{-1}$ for all $m$ and $\mathrm{tr}(SJ^mA^{-1})=\mathrm{tr}(J^m)$, we can work directly with $J$.

Since $\mathrm{tr}(J^m)=0$ for all $m$ and the trace is linear, we conclude that $\mathrm{tr}(p(J))=0$ for every polynomial $p$ such that $p(0)=0$.

Now, since $J$ is upper triangular, the diagonal of $p(J)$ consists of $p(J_{11}),\ldots,p(J_{nn})$. If one or more of the $J_{11},\ldots, J_{nn}$ is nonzero, we can get a polynomial $p$ such that the diagonal of $p(J)$ has positive sum; but this is a contradiction, which shows that the diagonal of $J$ is zero. This implies that $J$ is nilpotent, and so is $A$.

For the last question, $A^2-I=S(J^2-I)S^{-1}$. Since $J^2-I$ is upper triangular with every entry in its diagonal equal to $-1$, its determinant is nonzero, and so it is invertible.

user26857
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Martin Argerami
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  • @aliakbar: For the second part, if $A^k=0$ then $$(A^2-I)(-A^{2k-2}-A^{2k-4}-A^{2k-6}-\cdots-A^4-A^2-I)=I.$$ – P.. Dec 08 '12 at 06:58
  • "If one or more of the $J_{11},…,J_{nn}$ is nonzero, we can get a polynomial $p$ such that ... " is not necessarily true in positive characteristic. Note how e.g. if the characteristic is $p$, this fails for the $p \times p$ identity matrix (all of whose powers have trace $0$). It would be nice if you expanded this part of the argument to show exactly what is needed about the characteristic, and how. – Torsten Schoeneberg Jul 04 '22 at 21:05
  • Of course this can fail in positive characteristic. But the context of the question is $A\in M_n(\mathbb R)$. And I have to admit that I don't even know if there is an analog result in positive characteristic. – Martin Argerami Jul 04 '22 at 21:13
  • OK, here we go: Wlog let $J_{11} = ... = J_{kk} \neq 0$. If $char(K) = 0$ or $char(K) > n$, the polynomial $p(x) = x \cdot \prod_{i=k+1}^n (x-J_{ii})$ satisfies $\sum_{i=1}^n p(J_{ii}) \neq 0$ i.e. does the job. In case $p:=char(K) \le n$, the diagonal matrix with first $p$ entries $1$, all other entries $0$ is a counterexample to the claim, and no polynomial, including the analogue of the one just considered, can work in general. – Torsten Schoeneberg Jul 07 '22 at 19:07