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Let $A,B$ be two matrices with $A,B \in \mathcal{M}_{n}(\mathbb{R})$ and $ABA-BAB=I$ and $A^2B+B^2A=0$. Prove that $A,B$ are invertible matrices.

Git Gud
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Iuli
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  • Why (-1) dear downvoter? thanks! is it to easy? – Iuli Mar 15 '14 at 18:37
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    I think it's because you haven't given any context. Please show your working so far :) – Shaun Mar 15 '14 at 18:39
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    what work? I tried some methods, some particularization... but nothing... it's not very useful to write 2-3 A4 pages with blank work – Iuli Mar 15 '14 at 18:41
  • Well, show what you've tried since you've tried some methods. I only know that you've got nothing now that I've been annoyed that you didn't tell me you've got nothing. – Shaun Mar 15 '14 at 18:45
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    @ Shaun: in re. your use of the words "we": you and who else? – Robert Lewis Mar 15 '14 at 18:47
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    @RobertLewis I've wondered the same when mathematicians write 'we', so I mostly stick to 'one' and 'I'. – Git Gud Mar 15 '14 at 18:48
  • That's a fair point: I've edited the comment :) – Shaun Mar 15 '14 at 18:48
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    @ Shaun: launds for that. And thanks. I don't mind saying that I sometimes get annoyed when individual users speak/write as if for the totality of MSE participants. I mean, who asked me what I think? Regards, RKL. – Robert Lewis Mar 15 '14 at 18:50
  • @RobertLewis It was a bit presumptuous. Sorry :) – Shaun Mar 15 '14 at 18:52
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    @ Git Gud: well, the "we" here is a somewhat different usage than the mathematical "we have" und so weiter. Anyway, I tend to use both "we" and "I" in writing math, depending on the context, my mood, etc. ;-)! – Robert Lewis Mar 15 '14 at 18:52
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    @ Shaun: no need to apologize. And thanks again! ;-)! – Robert Lewis Mar 15 '14 at 18:56
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    @Iuli Sorry to ask: Does this statement hold for sure and you want a proof, or do you want to see if it is possible to prove it (and you are not sure whether it holds true or not)? – Jimmy R. Mar 15 '14 at 19:25
  • Why somebody wants to close this topic? – Iuli Mar 16 '14 at 07:29

2 Answers2

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Set $X:=\begin{bmatrix} A & B\\ -B & A\end{bmatrix}$ and $Y:=\begin{bmatrix} BA & AB\\ -AB & BA\end{bmatrix}$.

Note that $XY=\begin{bmatrix} ABA-BAB & A^2B+B^2A\\ -B^2A-A^2B & -BAB+ABA\end{bmatrix}=\begin{bmatrix} I & \bf 0\\ \bf 0 & I \end{bmatrix}$.

So $XY=YX=\begin{bmatrix} BA^2-AB^2 & BAB+ABA\\ -ABA-BAB & -AB^2+BA^2\end{bmatrix}$.

You can now use any entry of the last matrix above to conclude. Taking, for instance, the entry $(1,2)$ you can find $BAB+ABA=\bf 0$ and one of the hypothesis is $ABA-BAB=I$. Summing yields $2ABA=I$ and so both $B$ and $A$ are invertible.

Git Gud
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We can do better. Indeed, according to Git Gud post, $BA=1/2A^{-1}$ and consequently $AB=BA$. Then $BAB=-1/2 I$ implies $BA=-1/2B^{-1}$. Finally, the relations are $A=-B$ and $2A^3+I=0$.

EDIT: Some have ridiculous habit of removing a point without saying why, and without giving their name. This gives a very bad image of (pseudo ?) mathematicians.