Prove that $\operatorname{adj}(AB-BA) = \operatorname{adj}(AB) - \operatorname{adj}(BA)$ then $A, B \in M_{22}$

Question

Prove or Disprove :

$$\operatorname{adj}(AB-BA) = \operatorname{adj}(AB) - \operatorname{adj}(BA)$$

where $A$ and $B$ are arbitrary $n \times n$ matrices, and $\operatorname{adj}$ is the adjugate(classical adjoint) operator.

The statement is false, of course, since $\operatorname{adj}(A) = I = \begin{bmatrix} 1 \end{bmatrix}$ for any $1 \times 1$ matrix $A$.

I am now trying to show that the identity holds only for $n = 2$, as the title says. I have checked it for $2 \times 2$ matrices by brute force calculation($A = [a_{ij}]$, $B = [b_{ij}]$) and it does hold. Also, I found a counterexample for $n = 3$, which is just a random $3 \times 3$ matrix because I was pretty sure that it would not hold for $n > 2$.

Counterexample for $M_{33}$ :

$$A = \begin{bmatrix} 1 & 2 & 3 \\\ 4 & 5 & 6 \\\ 7 & 8 & 9 \end{bmatrix} \quad \text{and} \quad B = \begin{bmatrix} 3 & 6 & 1 \\\ 2 & 5 & -1 \\\ -3 & 4 & -7 \end{bmatrix}$$

How can I show that the identity holds only for $n = 2$? The following is my approach.

We know that

$$\begin{align} (AB-BA) \operatorname{adj}(AB-BA) &= \det(AB-BA) I \\\ AB \operatorname{adj}(AB) &= \det(AB) I \\\ BA \operatorname{adj}(BA) &= \det(BA) I \end{align}$$

for all square matrices $A, B$. If the given identity is satisfied, then

$$\begin{align} \det(AB-BA)I &= (AB-BA) \operatorname{adj}(AB-BA) \\\ &= (AB-BA) (\operatorname{adj}(AB) - \operatorname{adj}(BA)) \end{align}$$

Hence, we have

$$\left( 2 \det(AB) - \det(AB-BA) \right) I = BA \operatorname{adj}(B) \operatorname{adj}(A) + AB \operatorname{adj}(A) \operatorname{adj}(B)$$

I thought about evaluating the trace of both sides since the trace is a linear operator - so you can separate two terms using the additivity property. However, I couldn't proceed further after the trace.

Answer 1

Note that for $X=\pmatrix{a&b\\ c&d}$, its adjugate $\operatorname{adj}(X)=\pmatrix{d&-b\\ -c&a}$ is a linear function in $X$. In particular, it is additive. It follows that $\operatorname{adj}(AB-BA)=\operatorname{adj}(AB)-\operatorname{adj}(BA)$.

For $n\ge3$, let $$ X=\pmatrix{0&1&0\\ 0&0&1\\ 0&0&0}. $$ Then $$ XX^T=\pmatrix{1\\ &1\\ &&0}, \quad X^TX=\pmatrix{0\\ &1\\ &&1}, \quad XX^T-X^TX=\pmatrix{1\\ &0\\ &&-1}. $$ Therefore $$ \begin{align} \operatorname{adj}(XX^T-X^TX) &=\pmatrix{0\\ &-1\\ &&0}\tag{$\ast$}\\ \ne\operatorname{adj}(XX^T)-\operatorname{adj}(X^TX) &=\pmatrix{0\\ &0\\ &&1}-\pmatrix{1\\ &0\\ &&0} =\pmatrix{-1\\ &0\\ &&1}. \end{align} $$ Thus $A=X$ and $B=X^T$ constitute a counterexample when $n=3$.

When $n\ge4$, consider $$ A=\pmatrix{X\\ &I_{n-3}},\quad B=A^T. $$ Then $$ \begin{aligned} \operatorname{adj}(AB-BA) &=\operatorname{adj}\operatorname{diag}(1,0,-1,0,\ldots,0)\\ &=0,\\ \operatorname{adj}(AB)-\operatorname{adj}(BA) &=\operatorname{adj}\operatorname{diag}(1,1,0,1,\ldots,1) -\operatorname{adj}\operatorname{diag}(0,1,1,1,\ldots,1)\\ &=\operatorname{diag}(0,0,1,0,\ldots,0) -\operatorname{diag}(1,0,0,0,\ldots,0)\\ &\ne0.\\ \end{aligned} $$