# $AB-BA$ is a nilpotent matrix if it commutes with $A$

I saw this in a MathOverflow post and am putting it here for posterity.

Problem: Let $A$ and $B$ by square matrices and set $C=AB-BA$. If $AC=CA$, prove $C$ is nilpotent.

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What MathOverflow post? Matrices with what kind of entries? –  Jonas Meyer Feb 10 '13 at 19:54
Add the information on the question body! –  Mariano Suárez-Alvarez Feb 10 '13 at 19:58

Hint: I use this theorem: If $\forall i\ge1$ trac ${C^i}=0$, then $C$ is nilpotent.

You can easily prove by induction that trac (${C^i})=0$ for all $i\ge1$.

Edit1: Theorem :$\forall i\ge1$ trac ${C^i}=0$ iff C is nilpotent.

Proof: $C$ is a real matrix but you assume $A$ is a complex matrix and $f(x)=(x-a_1)(x-a_2)...(x-a_n)$ is its characteristic polynomial in the complex field. You can prove that trac($C^k$)=$\sum_{i=1}^n {a_i^k}$ by induction, and if $\forall k\in\mathbb N$ trac($C^k$)=$\sum_{i=1}^n {a_i^k}=0$ then $a_i=0$. Hence, $f(x)=x^n$, so $A^n=0$ and it is shown that C is nilpotent.

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Could you please use sentences with punctuation? It looks like a good approach, but I have a little trouble parsing what is meant. –  Jonas Meyer Feb 10 '13 at 20:25
@jonas-meyer:induction that use in body proof is so prolix sorry i don't write it –  Maisam Hedyelloo Feb 10 '13 at 20:53
But you can at least separate this into a couple of paragraphs and so on. Readability goes a long way into turning something into a useful and enjoyable piece of text. –  Mariano Suárez-Alvarez Feb 10 '13 at 20:54
@Maisam: You can separate into paragraphs using blank lines between them. Within a paragraph, you can separate ideas into sentences, by placing a period (dot) at the end of each sentence, and capitalizing the first letter of the subsequent sentence. (This has to do with communication rather than mathematics per se.) –  Jonas Meyer Feb 10 '13 at 20:56

Suppose $\mathcal A$ is a normed algebra and $\delta:\mathcal A\to\mathcal A$ is a bounded derivation. If $x\in A$ and $\delta(\delta(x))=0$, then $\lim\limits_{n\to\infty}\|\delta(x)^n\|^{1/n}=0$. This is proved as Theorem 2.2.1 in Sakai's Operator algebras in dynamical systems.

This applies to the case where $\mathcal A=M_n(\mathbb C)$, $\delta(X)= AX-XA$. For an $n$-by-$n$ real or complex matrix $C$, $\lim\limits_{n\to\infty}\|C^n\|^{1/n}=0$ if and only if $C$ is nilpotent.

(I also remarked on this application in an answer to a different question where the result was applicable.)

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I think you need to edit this a few more times. –  Will Jagy Feb 10 '13 at 20:07
@WillJagy: I hope you are wrong. –  Jonas Meyer Feb 10 '13 at 20:12
It's funny, if edits occur in really quick succession, the system dose not count them as separate. So, while i had left the screen on this one and saw a half dozen edits, if I click on the middle saying "edited 11 minutes ago" it says just two total. –  Will Jagy Feb 10 '13 at 20:16
@WillJagy: If I recall correctly (it was over 10 minutes ago now and getting hazy) those edits were all getting the links straightened out. It would be a rather banal revision history if it were all recorded. –  Jonas Meyer Feb 10 '13 at 20:18