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Let $V$ a finite dimensional vector space over a characteristic zero algebraically closed field. Let $x,y\in\mathfrak{gl}(V)$ and $[x,y]=z$ such that $z$ commutes with $x$ and $y$. Prove $z$ is a nilpotent endomorphism.

Of course the Lie algebra generated by $x,y,z$ is nilpotent. Since the field is algebraically closed, $z$ admits an eigenvalue and I tried to show this eigenvalue is zero, without success. Some hints?

roob
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  • @TheSilverDoe I'd say no... – roob Jun 06 '22 at 16:30
  • A more canonical duplicate seems to be https://math.stackexchange.com/q/299640/96384. Note you only need $z$ to commute with one of either $x$ or $y$, and you do not need the field to be algebraically closed (you do need it to have good characteristic though). – Torsten Schoeneberg Jun 06 '22 at 16:34
  • The general proof idea is to show in a first step that all powers of $z$ have trace $0$. One can show this e.g. by inductively writing each power of $z$ as a commutator (the case of power $1$ is given as hypothesis), or by eigenvalue/characteristic polynomial consideration, or possibly other linear algebra methods. Anyway, once one has that, in a second step one applies the fact that (in characteristic zero) such an operator must be zero, a helpful lemma for which the canonical reference on this site seems to be https://math.stackexchange.com/q/159167/96384 – Torsten Schoeneberg Jun 06 '22 at 16:47

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