# $AB-BA$ 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.

• This is one of the facts known as Jacobson's lemma. See, e.g., Janko Bracic, On the Jacobson's lemma or Irving Kaplansky, Jacobson's lemma revisited. Note that $A$ and $B$ need to be matrices over a field of characteristic $0$ for this to hold. Otherwise, this would fail for the $2\times 2$-matrices $A = E_{1,2}$ and $B = E_{2,1}$ over the field $\mathbb{F}_2$ (where $E_{i,j}$ denotes the $2\times 2$-matrix with a $1$ in cell $\left(i, j\right)$ and $0$ elsewhere). – darij grinberg Jul 26 at 13:41

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.

• 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 – M.H 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-Álvarez 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.)