Eigenvalues of a Lie bracket

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?

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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 $\lambda=0$.
Thanks for your response. Can you please elaborate on why $[\bar a,\bar b]=\bar c=\lambda\mathrm{id}_{V_\lambda}$ – Jack Oct 23 '12 at 2:18
Okay. I got that part. Could you also explain this part. $a(V_\lambda)\subseteq V_\lambda$ and $b(V_\lambda)\subseteq V_\lambda$ – Jack Oct 23 '12 at 2:32
If $f$, $g:W\to W$ 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 $\lambda$. – Mariano Suárez-Alvarez Oct 23 '12 at 2:35