Let $A\in\mathbb{M}_n(\mathbb{R})$ be a singular matrix. If $AXA=A,\ XAX=X$ and $AX=XA$, then what are the conditions on $A$ so that we obtain a solution?
Specifically, can we say that $A$ is not nilpotent or a projection? What if we restrict $n$ to $2$. I easily obtained that $AX$ is idempotent. But, I am unable to proceed further. Any hints? Thanks beforehand.
If $AXA=A,\ XAX=X\ $ and $AX=XA$, then $X$ is called the group inverse of $A$.
A singular matrix $A\in\mathbb{M}_n(\mathbb{R})$ has a group inverse $ \iff rank(A)=rank(A^2)$.
In this case, the group inverse of $A$ is unique.