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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

Assuming $A$ is invertible, the first relation implies

$B = A^{-1}(I+B^2A^2)$, then:

$BA -A^2 B^2= A^{-1}(I+B^2A^2) A -A^2 B^2= I + A^{-1} B^2A^3 -A^2B^2= I - A^{-1} B^2B^3 -A^2B^2 = I - A^{-1} B^3B^2 -A^2B^2 = I + A^{-1} A^3B^2 -A^2B^2 = I + A^2B^2 -A^2B^2 = I$.

Answer 5

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:

$BA - A^2B^2 = I_n - A^3A^2B^{-1} -A^2B^2 = I_n + A^2 B^3 B^{-1} - A^2B^2 = I_n + A^2B^2 - A^2B^2 = I_n$.