Let $V$ be a complex vector space. Suppose that $a,b \in gl(V )$ (the set of all linear maps from $V$ to $V$) satisfies $[a,[a,b]] = [b,[a,b]] = 0.$ how do I show that all eigenvalues of $[a,b]$ are zero?
Eigenvalues of a Lie bracket
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lie-algebras
1 Answers
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Your hypothesis is that $c=[a,b]$ commutes with $a$ and with $b$.
Let $\lambda$ be an eigenvalue of $c$, and let $V_\lambda$ be the corresponding eigenspace. The hypothesis implies that $a(V_\lambda)\subseteq V_\lambda$ and $b(V_\lambda)\subseteq V_\lambda$, so we can consider the maps $\bar a$, $\bar b$ and $\bar c:V_\lambda\to V_\lambda$ obtained by restricting $a$, $b$ and $c$, respectively, to $V_\lambda$.
Notice that $[\bar a,\bar b]=\bar c=\lambda\mathrm{id}_{V_\lambda}$, so taking traces we see that $0=\operatorname{tr}[\bar a,\bar b]=\operatorname{tr}\lambda\mathrm{id}=\lambda\dim V_\lambda.$ Since $\dim V\neq0$ in $\mathbb C$, this implies that $\lambda=0$.
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0If $f$, $g:W\to W$ are two endomorphisms of a vector space which commute and $v$ is an eigenvector of $f$ with respect to the eigenvalue $\lambda$, then $g(v)$ is also an eigenvector of $f$ with respect to $\lambda$. – 2012-10-23