# 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 $$det(B)=0$$.

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

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) 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 $$KB=B$$.