Suppose $A$ and $B$ are real or complex $n \times n$ matrices and $C = [A,B]$ is their commutator. If $C$ commutes with $A$, show that $C$ is nilpotent.

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    $\begingroup$ Please specify where the question is from, and what you've tried so far. Welcome! $\endgroup$ – nbubis Oct 30 '12 at 6:50
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    $\begingroup$ This is a very common problem in linear algebra that can be found in many places (with solution). See here. $\endgroup$ – user26857 Nov 5 '12 at 10:51

I guess there are elementary proofs for this result but the only one I can think of now is the following result for bounded derivations, which applies to many other cases. You can see it in Murphy's book(Problem 12-14 to Chapter 1), I just sketch the key steps here.

Let $\mathcal{A}$ be an algebra, a linear map $d:\mathcal{A}\to \mathcal{A}$ is a derivation if \begin{equation} d(ab)=ad(b)-bd(a) \end{equation} for all $a,b\in \mathcal{A}$. Derivations satisfy the Leibniz formular \begin{equation} d^n(ab)=\operatorname{\sum_{k=0}^n}\frac{n!}{(n-k)!k!}d^k(a)d^{n-k}(b). \end{equation} (Think about derivatives.)

Now let $\mathcal{A}$ be a unital Banach algebra and $d$ be bounded. If we have $a\in \mathcal{A}$ such that $d(a)=\lambda a$ for some $\lambda\neq 0$, then you can apply Leibniz to $d(a^n)$ to see $a^n=0$ for large $n$. (Note that $\lambda$ is in the spectrum of $d$, which is a bounded set.)

Now if we have $d^2(a)=0$, then we can show $d^n(a^n)=n!(d(a))^n$, which then gives $d(a)$ is quasi-nilpotent. Because \begin{equation} \|d^n\|\ge n!\frac{\|(da)^n\|}{\|a^n\|} \end{equation} and then \begin{equation} \|d\|=\operatorname{lim}\|d^n\|^{1/n}\ge\operatorname{lim}(n!)^{1/n}\frac{\|(da)^n\|^{1/n}}{\|a^n\|^{1/n}}, \end{equation} so the only possibility that $\|d\|$ can stay bounded is when $\operatorname{lim}\|(da)^n\|^{1/n}=0$, which says $da$ is quasi-nilpotent.

To sum up, what we have shown is that in a unital banach algebra $\mathcal{A}$, if $d^2(a)=0$ for some derivation $d$, then $d(a)$ is quasi-nilpotent. Apply this to $\mathcal{A}=M_n(\mathbb{C})$, and the derivation defined by $B\mapsto AB-BA$, then you can see $[A,B]=d(B)$.

If $[A,B]$ commutes with $A$, then $d^2(B)=d([A,B])=0$, so $[A,B]=d(B)$ is quasi-nilpotent. But in finite dimensional spaces like $M_n{\mathbb{C}}$, this is the same as nilpotent.

By the way, this is the Kleinecke-Shirokov theorem.

  • $\begingroup$ Can you explain why $d^n(a^n) = n! (da)^n$ ($n \geq 1$) implies $da$ nilpotent ? $\endgroup$ – user10676 Nov 3 '12 at 16:51
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    $\begingroup$ @user10676 I made a mistake. It should be quasi-nilpotent. I edited the answer. $\endgroup$ – Hui Yu Nov 4 '12 at 1:51

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