Let $A,B$ be two $3 \times 3$ matrices with complex entries such that $$(A-B)^2=O_3$$ Prove that $$\det(AB-BA)=0$$

I tried to prove this with ranks. I denoted $X=A-B$ and thus $X^2=O_3$ which means that $\det X=0$ and $\operatorname{rank}X \leq 2$. Then, I wrote $AB-BA=(X-B)B-B(X-B)=XB-BX$ and finally I used $\operatorname{rank}(M \pm N) \leq \operatorname{rank}M+\operatorname{rank}N$ and Frobenius's inequality in order to get $$\operatorname{rank}(XB-BX) \leq \operatorname{rank}BX+\operatorname{rank}XB \leq \operatorname{rank}X+\operatorname{rank}BXB$$ and if we knew that $\operatorname{rank}BXB=0$, the problem would be solved. However, I don't quite know if the latter is true.

  • 1
    $\begingroup$ Does $O_3$ denote the zero matrix of dimension $3 \times 3$? $\endgroup$ – md2perpe Dec 26 '17 at 21:49
  • 5
    $\begingroup$ I reckon that $\text{rank }X\le1$. $\endgroup$ – Lord Shark the Unknown Dec 26 '17 at 21:50
  • $\begingroup$ @md2perpe yes, that's it! $\endgroup$ – AndrewC Dec 26 '17 at 21:51
  • $\begingroup$ Oh yes indeed, it follows from Sylvester's inequality: $\operatorname{rank}X^2=0 \geq 2\operatorname{rank}X-3$ and thus $\operatorname{rank}X \leq 1$. The problem is then solved since $\operatorname{rank}BXB \leq \operatorname{rank}X \leq 1.$ $\endgroup$ – AndrewC Dec 26 '17 at 21:59
  • $\begingroup$ From $X=A-B$ we get $A=X+B$. You wrote $AB-BA=(X-B)B-B(X-B)$, which needs to be $AB-BA=(X+B)B-B(X+B)$. $\endgroup$ – user236182 Dec 26 '17 at 22:32

Here is a more or less direct, less creative solution. Since $(A-B)^2=0$, then either $A-B=0$ (in which case $AB-BA=0$), or its Jordan form is $$J=\begin{bmatrix} 0&1&0\\0&0&0\\0&0&0\end{bmatrix}.$$ So $A-B=SJS^{-1}$ for some $S$. Let $A'=S^{-1}AS$, $B'=S^{-1}BS$. Then $A'=B'+J$, and $$A'B'-B'A'=(B'+J)B'-B'(B'+J)=JB'-B'J.$$ Now check directly that $$ JB'-B'J=\begin{bmatrix}B'_{31}&B'_{32}&B'_{33}-B'_{11}\\ 0&0&-B'_{21}\\ 0&0&-B'_{31} \end{bmatrix}. $$ Thus $\det(JB'-B'J)=0$. Finally, \begin{align} \det(AB-BA)&=\det(SA'S^{-1}SB'S^{-1}-SB'S^{-1}SA'S^{-1})\\ \ \\ &=\det(A'B'-B'A')=\det(JB'-B'J)=0. \end{align}

  • $\begingroup$ Thank you! But why has $J$ the form you mentioned? I am sorry, but I barely know about Jordan forms at the moment. Could you please provide a good source where I could learn about them? $\endgroup$ – AndrewC Dec 27 '17 at 9:36
  • $\begingroup$ I think I see now: is it because $X^2$ would be the minimal polynomial of $A-B$ and since $0$ is its only eigenvalue, it means that the size of its largest Jordanian block is $2$? $\endgroup$ – AndrewC Dec 27 '17 at 10:30
  • 1
    $\begingroup$ The largest Jordan block has size 2, because if it had size 3 you would have $J^2\ne0$. And if it had size 1 you would have $J=0$. $\endgroup$ – Martin Argerami Dec 27 '17 at 11:21

As pointed above by @Lord Shark the Unknown (whose comment struck me, pointing the right way) we have from Sylvester's inequality: $$0=\operatorname{rank}O_3=\operatorname{rank}(X\cdot X) \geq \operatorname{rank}X+\operatorname{rank}X-3 \Rightarrow \operatorname{rank}X \leq 1$$ Thus going back to my last inequality, $$\operatorname{rank}(XB-BX) \leq \operatorname{rank}X+\operatorname{rank}BXB \leq \operatorname{rank}X+\operatorname{rank}X \leq 2$$ and so $\det(AB-BA)=0$.

  • $\begingroup$ Thank you very much! :-D $\endgroup$ – AndrewC Dec 26 '17 at 22:20
  • 4
    $\begingroup$ Since $\text{rank}X \leq 1$, you do not really need the inequality in red $$\operatorname{rank}(XB-BX) \leq {\color{red}{\operatorname{rank}X+\operatorname{rank}BXB} \leq \operatorname{rank}X+\operatorname{rank}X} \leq 2.$$ Instead,$$\operatorname{rank}(XB-BX) \leq {\color{red}{\operatorname{rank}XB+\operatorname{rank}BX} \leq \operatorname{rank}X+\operatorname{rank}X} \leq 2.$$ $\endgroup$ – clark Dec 26 '17 at 22:25
  • $\begingroup$ Indeed, nice catch! $\endgroup$ – AndrewC Dec 26 '17 at 22:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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