Does there exist $n\times n$ matrices such that $BABCB-BCBAB=B$? Found this past paper exercise and I can't figure out how to go with it.
Suppose $n \times n$ matrix $B$, non invertible, non zero.
Does there exist $n\times n$ matrices $A,C$ such that:
a) $BABCB - BCBAB = B$
b) $ACB - CAB = B$
My try:
a) $BABCB - BCBAB = B$
$B(ABC-CBA)B=B$
Then, $ABC-CBA$ is the left inverse or right inverse of $B$. But I can not continue from here. Also, tried the trace cycle property, which leads nowhere. The determinant properties give no important results, since $det(B)=0$.
 A: Here is a solution only for b), in the case $n \ge 2$, assuming your matrices are defined over the field of complex numbers $\mathbb{C}$:
Lemma 1: If $T \in \mathcal{M}_{n}(\mathbb{C})$ with $\text{Tr}(T)=0$ and $n \ge 2$ there exist $A,B \in \mathcal{M}_{n}(\mathbb{C})$ such that $T = [A,B] = AB-BA$. 
Proof: Induction on $n$. See this post.
Lemma 2: There exists some $M \in \mathcal{M}_n(\mathbb{C})$ such that $\text{Tr}(M) \neq 0$ and $MB = O_n$. 
Proof: This is quite easy to prove. Since $B$ is not invertible, there exists some $x \in \mathcal{M}_{1,n}(\mathbb{C})$ nonzero, $x = [x_1, \ldots, x_n]$ such that that $xB = O_{1,n}$. Then we may take $M$ to be the matrix whose lines are all $x$. If $\text{Tr}(M) \neq 0$ we are done. If not, consider $i \in \{1, \ldots, n \}$ such that $x_i \neq 0$ and replace the $i$-th line of $M$ with $2x$.
Now, let us prove b):
Note that $ACB-CAB = B$ is equivalent to $$(AC-CA-I_n)B = O_n$$ Let $M$ be as in Lemma 2 and denote by $\lambda = \text{Tr}(M)$. Take $T = (-n)\cdot \lambda^{-1}M + I_n$. Note that $\text{Tr}(T) = 0$. Therefore, by Lemma 1, there exists some $A,C \in \mathcal{M}_n(\mathbb{C})$ such that $AC-CA = T$. Now, just notice that
$$(AC-CA-I_n) B= (T-I_n)B = (-n)\lambda^{-1}MB = O_n$$ as desired.
A: (a) Solution may not always exist. Consider, for example, a rank-one matrix $B=uv^T\ne0$. We have
$$
BABCB=u(v^TAu)(v^TCu)v^T=(v^TAu)(v^TCu)B
$$
and by a similar argument, $BCBAB$ is also equal to $(v^TAu)(v^TCu)B$. Therefore $BABCB-BCBAB=0\ne B$.
(b) Solution always exists. Let $B^g$ be a generalised inverse of $B$, so that $BB^gB=B$. Let $K=BB^g+X^T(I-BB^g)$ where the matrix $X$ is to be determined. Then
\begin{aligned}
\operatorname{tr}(K)
&=\operatorname{tr}(BB^g)+\operatorname{tr}(X^T(I-BB^g))\\
&=\operatorname{tr}(BB^g)+\left\langle\operatorname{vec}(X),\,\operatorname{vec}(I-BB^g)\right\rangle.\\
\end{aligned}
Since $B$ is singular, $I-BB^g\ne0$. Therefore, we can always pick an $X$ such that $\operatorname{tr}(K)=0$. Since every traceless matrix is a commutator, $K$ is equal to $AC-CA$ for some square matrices $A$ and $C$. The result now follows because $KB=B$.
