Find all idempotent matrices such that $(A-B)^2 = 0$

Find all idempotent matrices such that $$(A-B)^2 = 0$$

We can see that the hypotheses imply that $$A+B=AB+BA$$, and if we multiply by $$AB$$ on the right, we get $$AB+BAB=(AB)^2+BAB$$, which also implies that $$AB$$ is idempotent (and we can also see that $$BA$$ is idempotent).

After that, I am not sure what to do next.

I add a necessary condition that $$\operatorname{rank} (A) = \operatorname{rank} (B)$$, since $$A-B$$ is nilpotent, then $$\operatorname{Trace} (A-B) = 0$$, so $$\operatorname{Trace} (A) = \operatorname{Trace} (B)$$. But $$A$$ and $$B$$ are projectors, so $$\operatorname{rank} (A) = \operatorname{rank} (B)$$.

additional sufficient condition: if $$\text{im}(A) = \text{im}(B)$$, then $$AB=B$$ and $$BA=A$$. This implies that $$A+B=AB+BA$$, which further implies that $$(A-B)^2=0$$.

• I work in characteristic zero or, to simplify, I consider the field $\mathbb{R}$ or $\mathbb{C}$. Jun 4, 2023 at 21:12
• Where does this come from? Jul 14, 2023 at 9:51

Let $$N=B-A$$. Then $$N^2=0$$ and from $$A+N=(A+N)^2$$, we obtain $$AN+NA=N$$. By a change of basis, we may assume that $$A=\pmatrix{I_r&0\\ 0&0}$$. Let $$N=\pmatrix{X_{r\times r}&Y_{r\times(n-r)}\\ Z_{(n-r)\times r}&W_{(n-r)\times(n-r)}}$$. The two conditions $$AN+NA=N$$ and $$N^2=0$$ imply that $$X,W,YZ$$ and $$ZY$$ are zero. That is, the general solution is given by $$A=P\pmatrix{I_r&0\\ 0&0}P^{-1} \quad\text{and}\quad B=P\pmatrix{I_r&Y\\ Z&0}P^{-1}$$ for some $$r\in\{0,1,\ldots,n\}$$ and for some matrices $$Y,Z$$ and $$P$$ such that $$YZ=0$$ and $$ZY=0$$.
If you want the roles of $$A$$ and $$B$$ to look more symmetric, observe that \begin{aligned} \pmatrix{I&Y\\ 0&I}\pmatrix{I&0\\ 0&0}\pmatrix{I&-Y\\ 0&I} &=\pmatrix{I&-Y\\ 0&0},\\ \pmatrix{I&Y\\ 0&I}\pmatrix{I&Y\\ Z&0}\pmatrix{I&-Y\\ 0&I} &=\pmatrix{I&0\\ Z&0}. \end{aligned} So, by negating $$Y$$, the general solution can also be expressed as $$A=P\pmatrix{I_r&Y\\ 0&0}P^{-1} \quad\text{and}\quad B=P\pmatrix{I_r&0\\ Z&0}P^{-1}$$ with $$YZ=0$$ and $$ZY=0$$.