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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 $det(B)=0$.

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  • $\begingroup$ For $n=1$ there is no such matrix $B$. So we can assume that $n\ge 2$. Did you try $n=2$? $\endgroup$ Commented Jul 6, 2021 at 15:14
  • $\begingroup$ B is non-invertible right? So does it have a left or right inverse? $\endgroup$
    – Calvin Lin
    Commented Jul 6, 2021 at 15:14
  • $\begingroup$ An example for b: $$ A = \pmatrix{0&1\\0&0}, \quad B = \pmatrix{1&0\\0&0}, \quad C = \pmatrix{0&0\\1&0}. $$ $\endgroup$ Commented Jul 6, 2021 at 15:20
  • $\begingroup$ Forgot to write that n>1 $\endgroup$
    – XuUserAC
    Commented Jul 6, 2021 at 15:21
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    $\begingroup$ B cannot have a left or right inverse either as it is square matrix. $\endgroup$ Commented Jul 6, 2021 at 15:27

2 Answers 2

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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.

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(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 $KB=B$.

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