Let $A, B\in Mat_{n,n}(\Bbb R)$. Suppose $AB=BA=0$, $\mathrm{rank}(A^2)=\mathrm{rank}(A)$. Show $\mathrm{rank}(A)+\mathrm{rank}(B)=\mathrm{rank}(A+B)$

Question

The question is:

If $A$ and $B$ are two $n\times n$ matrices，$AB=BA=0$ and $\mathrm{rank}(A^2)=\mathrm{rank}(A)$ then show that $\mathrm{rank}(A+B)=\mathrm{rank}(A)+\mathrm{rank}(B)$.

I only manage to prove that $\mathrm{rank}(A^n)=\mathrm{rank}(A)$ and some common conclusions. I don't know how to go further. Any help would be appreciated.

Answer 1

Certainly $\text{Image}(A+B)\subseteq\text{Image}(A)+\text{Image}(B)$. One has to show that reverse inclusion holds and that $\text{Image}(A)\cap\text{Image}(B)=\{0\}$. If $u\in \text{Image}(A)$ then $u\in\text{Image}(A^2)$, so $u=A^2w=(A+B)Aw\in\text{Image}(A+B)$. If $u=Bx\in \text{Image}(B)$ then $u=(A+B)x-Ax$ so $u\in\text{Image}(A+B)$ also.

Let $u\in\text{Image}(A)\cap\text{Image}(B)$. Then $Au=0$. But $A$ takes $\text{Image}(A)$ bijectively to itself. Therefore $u=0$.