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

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.

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}
• Can you please elaborate? I'm just learning linear algebra and I have no idea (a) how you evaluated the adjugate of the block matrix, and (b) why that becomes a counterexample for $n > 2$.