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Let $A$ and $B$ be $n \times n$ complex matrices. Then $\operatorname{rank}(AB-BA)=1$ implies $AB-BA$ is nilpotent.

I have seen proof here that $\operatorname{rank}(AB-BA)=1$ implies $A$ and $B$ are simultaneously triangularisable, which implies that $AB-BA$ is nilpotent. I was wondering if there is an easier way to show the result. (I could show $A$ and $B$ share a common eigenvector but that does not imply $AB-BA$ is nilpotent.)

luxerhia
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james black
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1 Answers1

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Since $\text{rank}(AB-BA) = 1$, we can write $AB-BA = uv^*$ for some vectors $u,v \in \mathbb{C}^n$.

Since $\text{tr}(AB) = \text{tr}(BA)$, we have $0 = \text{tr}(AB-BA) = \text{tr}(uv^*) = v^*u$.

Hence, $(AB-BA)^2 = (uv^*)^2 = u\underbrace{v^*u}_{= 0}v^* = 0_{n \times n}$, i.e. $AB-BA$ is nilpotent.

JimmyK4542
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  • that makes so much more sense thank you; how would one know to use the properties of trace here in the first place? – james black Dec 20 '20 at 04:04
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    A few years ago, I had an exam where one problem asked to prove that $n \times n$ matrices $A,B$ couldn't satisfy $AB-BA = I$. I couldn't solve the problem during the exam. The solutions to the exam said to take the trace of both sides to get $0 = n$, a contradiction. The trick that $\text{tr}(AB) = \text{tr}(BA)$ has since been added to my toolbox. – JimmyK4542 Dec 20 '20 at 04:32
  • Here is another way to look at this answer. Let $C=AB-BA$, so $C$ is a matrix of rank $1$ and trace $0$. Such a matrix is always nilpotent: by this question, the characteristic polynomial $C$ is $x^n-ax^{n-1}$, where $a$ is the trace of $C$. Thus $C$ has characteristic polynomial $x^n$.

    Even better, this question shows that the minimal polynomial of a rank one matrix is $x(x-a)$, where $a$ is the trace of the matrix. Thus the minimal polynomial of $C$ is $x^2$ and so $C^2=0$.

     – Allen Bell Dec 21 '20 at 03:54