Prove $BA - A^2B^2 = I_n$.

I have a problem with this. Actually, still don't have the right way to start :/

Problem :

Let $A$ and $B$ be $n \times n$ complex matrices such that $AB - B^2A^2 = I_n$. Prove that if $A^3 + B^3 = 0$, then $BA - A^2B^2 = I_n$.

Thanks for any help.

The two given conditions $AB-B^2A^2=I$ and $A^3+B^3=0$ can be rewritten as $$ \pmatrix{A&B^2\\ -B^2&A} \pmatrix{B&A^2\\ -A^2&B} = \pmatrix{I&0\\ 0&I}. $$ Since $XY=I$ implies that $YX=I$ for any two square matrices $X$ and $Y$, we have $$ \pmatrix{B&A^2\\ -A^2&B} \pmatrix{A&B^2\\ -B^2&A} = \pmatrix{I&0\\ 0&I} $$ and the assertion follows.

Another solution:

As several people noticed, $AA^{2}=A^{3}=-B^{3}$ (since $A^{3}+B^{3}=0$), so that

$AA^{2}B^{2}=-B^{3}B^{2}=-B^{5}=-B^{2}\underbrace{B^{3}}_{\substack{=-A^{3} \\\text{(since }A^{3}+B^{3}=0\text{)}}}=B^{2}A^{3}=B^{2}A^{2}A$,

and hence

$A\left( BA-A^{2}B^{2}\right) =ABA-\underbrace{AA^{2}B^{2}}_{=B^{2}A^{2} A}=ABA-B^{2}A^{2}A=\underbrace{\left( AB-B^{2}A^{2}\right) }_{=I_{n}}A=A$.

But also, $B^{2}B=B^{3}=-A^{3}$ (since $A^{3}+B^{3}=0$), thus

$B^{2}BA=-A^{3}A=-A^{4}=-A\underbrace{A^{3}}_{\substack{=-B^{3}\\\text{(since }A^{3}+B^{3}=0\text{)}}}=AB^{3}=ABB^{2}$

$B^{2}\left( BA-A^{2}B^{2}\right) =\underbrace{B^{2}BA}_{=ABB^{2}} -B^{2}A^{2}B^{2}=ABB^{2}-B^{2}A^{2}B^{2}=\underbrace{\left( AB-B^{2} A^{2}\right) }_{=I_{n}}B^{2}=B^{2}$.

Thus,

$\left( BA-A^{2}B^{2}\right) ^{2}=\left( BA-A^{2}B^{2}\right) \cdot\left( BA-A^{2}B^{2}\right) $

$=B\underbrace{A\left( BA-A^{2}B^{2}\right) }_{=A}-A^{2}\underbrace{B^{2} \left( BA-A^{2}B^{2}\right) }_{=B^{2}}=BA-A^{2}B^{2}$.

As a consequence, $BA-A^{2}B^{2}$ is a projection.

But recall a known fact which says that if an endomorphism $E$ of a finite-dimensional vector space $V$ over a field $k$ is a projection, then $\operatorname*{Tr}E=\dim\left( E\left( V\right) \right) \cdot1_{k}$. Applied to $V=\mathbb{C}^{n}$ and $E=BA-A^{2}B^{2}$, this yields $\operatorname*{Tr}\left( BA-A^{2}B^{2}\right) =\dim\left( \left( BA-A^{2}B^{2}\right) \left( V\right) \right) \cdot1_{\mathbb{C}}$. Hence,

$\dim\left( \left( BA-A^{2}B^{2}\right) \left( V\right) \right) \cdot1_{\mathbb{C}}=\operatorname*{Tr}\left( BA-A^{2}B^{2}\right) =\underbrace{\operatorname*{Tr}\left( BA\right) }_{=\operatorname*{Tr} \left( AB\right) }-\underbrace{\operatorname*{Tr}\left( A^{2}B^{2}\right) }_{=\operatorname*{Tr}\left( B^{2}A^{2}\right) }$

$=\operatorname*{Tr}\left( AB\right) -\operatorname*{Tr}\left( B^{2} A^{2}\right) =\operatorname*{Tr}\left( \underbrace{AB-B^{2}A^{2}}_{=I_{n} }\right) =\operatorname*{Tr}\left( I_{n}\right) =n$.

Since $\operatorname*{char}\mathbb{C}=0$, this yields $\dim\left( \left( BA-A^{2}B^{2}\right) \left( V\right) \right) =n$. Thus, $\left( BA-A^{2}B^{2}\right) \left( V\right) $ is an $n$-dimensional vector subspace of $V$. But since the only $n$-dimensional vector subspace of $V$ is $V$ itself, this yields $\left( BA-A^{2}B^{2}\right) \left( V\right) =V$. Hence, the image of the projection $BA-A^{2}B^{2}$ is the whole space $V$. But since the only projection of $V$ whose image is the whole space $V$ is the identity map $I_{n}:V\rightarrow V$, this yields that $BA-A^{2}B^{2}=I_{n}$, qed.