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.
 A: By the two given conditions, we see that
\begin{align*}
A(BA-A^2B^2) &= ABA-A^3B^2\\
&=(I+B^2A^2)A-A^3B^2\\
&=A+B^2A^3-A^3B^2\\
&=A-B^5+B^5\\
&=A.
\end{align*}
Therefore, if we can prove that $A$ is invertible, we are done.
Suppose the contrary. Then there exists a nonzero vector $v$ such that $Av=0$. We now prove by mathematical induction that for $k\ge1$, $AB^kv = a_kB^{k-1}v$ for some $a_k\neq0$. The base case is easy: $ABv=(I+B^2A^2)v=v$. Now, for $k\ge1$,
\begin{align*}
AB^{k+1}v
&= (AB)B^kv\\
&= (B^2A^2+I)B^kv\\
&= B^2A(AB^kv) + B^kv\\
&= B^2A(a_kB^{k-1}v) + B^kv\\
&= a_kB^2(AB^{k-1}v) + B^kv\\
&= \begin{cases}
0+B^kv &\text{ when } k=1\\
a_k B^2 (a_{k-1}B^{k-2}v) + B^kv &\text{ when } k\ge2
\end{cases}\\
&= \begin{cases}
B^kv &\text{ when } k=1\\
(a_{k-1}a_k+1) B^kv &\text{ when } k\ge2
\end{cases}.
\end{align*}
In short, if we define $a_0=0,\ a_1=1$ and $a_k=a_{k-2}a_{k-1}+1$ for $k\ge2$, we have
\begin{align*}
Av &= 0,\tag{1}\\
AB^kv &= a_kB^{k-1}v\ \text{ for }\ k\ge1\tag{2}
\end{align*}
with $a_k\neq0$ for $k\ge1$.
Let $m\ge0$ be the largest integer such that $v,Bv,B^2v,\ldots,B^mv$ are linearly independent, and let $B^{m+1}v=\sum_{k=0}^m c_kB^kv$. By $(1)$ and $(2)$,
$$
a_{m+1}B^mv = AB^{m+1}v
= A \sum_{k=0}^m c_k B^k v
= \sum_{k=\color{red}{1}}^m c_k AB^k v
= \sum_{k=1}^m c_ka_k B^{k-1} v.\tag{3}
$$
Yet this is impossible because $v,Bv,B^2v,\ldots,B^mv$ are linearly independent and $a_{m+1}\neq0$. Hence the assumption that $v$ is nonzero cannot be true and $A$ is invertible.
A: 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.
A: Assuming $A$ is invertible, the first relation implies
$B = A^{-1}(I+B^2A^2)$, then:
$BA -A^2 B^2=  A^{-1}(I+B^2A^2) A -A^2 B^2= I + A^{-1} B^2A^3  -A^2B^2= I - A^{-1} B^2B^3  -A^2B^2  =  I - A^{-1} B^3B^2  -A^2B^2 = I + A^{-1} A^3B^2  -A^2B^2 =  I + A^2B^2  -A^2B^2 = I$.
A: First, I try to demonstrate that $A$ and $B$ are invertible.
Let $(\lambda_j, v_j)$ be a couple of eigenvalue-eigenvector of $A$. Then, $(\lambda_j^3, v_j)$ is a couple of eigenvalue-eigenvector of $A^3$. Since $A^3 = -B^3$, then we can say that $(-\lambda_j^3, v_j)$ is a couple of eigenvalue-eigenvector of $B^3$. Finally, we can state that $(-\lambda_j, v_j)$ is is a couple of eigenvalue-eigenvector of $B$.
Consider now the following equations:
$$ABv_j - B^2A^2v_j = I_n v_j$$
$$A (-\lambda_j)v_j - B^2 (\lambda_j^2)v_j = v_j$$
$$(\lambda)_j (-\lambda_i)v_j - (\lambda_j^2) (\lambda_j^2)v_j = v_j$$
$$(\lambda_j^4 + \lambda_j^2 + 1)v_j = 0$$
Posing $(\lambda_j^4 + \lambda_j^2 + 1) = 0$, we get that eigenvalues are $\frac{1 \pm i \sqrt{3}}{2}$ and $\frac{-1 \pm i \sqrt{3}}{2}$. This means that neither $A$ nor $B$ has a null eigenvalues and hence $A$ and $B$ are both invertible.
At this point, we have that:
$AB - B^2A^2 = I_n \Rightarrow A = (I_n + B^2A^2)B^{-1}$
Then:
$BA - A^2B^2 = B(I_n + B^2A^2)B^{-1} - A^2B^2 = I_n + B^3A^2B^{-1} -A^2B^2$
We know that $A^3 = -B^3$, and then:
$BA - A^2B^2 = I_n - A^3A^2B^{-1} -A^2B^2 = I_n + A^2 B^3 B^{-1} - A^2B^2 = I_n + A^2B^2 - A^2B^2 = I_n$.
A: 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.
