Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
5  
Please specify where the question is from, and what you've tried so far. Welcome! –  nbubis Oct 30 '12 at 6:50
1  
This is a very common problem in linear algebra that can be found in many places (with solution). See here. –  user26857 Nov 5 '12 at 10:51

1 Answer 1

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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.