= Trace orthogonality nilpotence lemma
{title2=$\operatorname{tr}(\alpha M)=0\ \Longrightarrow\ \alpha\text{ nilpotent}$}
Let $U\subseteq W\subseteq\operatorname{End}_{\mathbb C}(V)$ be <vector subspaces> and $M=\{T:[T,W]\subseteq U\}$. If $\alpha\in M$ and $\operatorname{tr}(\alpha\beta)=0$ for all $\beta\in M$, then $\alpha$ is a <nilpotent endomorphism>.
To prove it, let $\beta$ act by $\overline\lambda$ on the <generalized eigenspace> of $\alpha$ of <eigenvalue> $\lambda$. <Polynomial interpolation> on eigenvalue differences, together with <adjoint compatibility of additive Jordan decomposition>, expresses $\operatorname{ad}\beta$ as a <polynomial> in $\operatorname{ad}\alpha$ with zero constant term. Since $\operatorname{ad}\alpha$ maps $W$ into $U$ and preserves $U$, this gives $\beta\in M$. Taking the <matrix trace> on each <generalized eigenspace> yields $0=\sum_\lambda\dim(V_\lambda)|\lambda|^2$, so every <eigenvalue> is zero. This is the linear-algebra step behind the <Cartan solvability criterion>; $U,W$ need not be Lie subalgebras.
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