A question about generalized inverse matrix

Question

Suppose a matrix $A$, and $AGA=A$. We know $G$ is not unique, but my question is that: Is $AG$ unique? Formally,

If $AG_1A=AG_2A=A$, then $AG_1=AG_2$?

Answer 1

In general, no. Take $A$ such that $A^2\not =0$, and $A^3=0$ . If $G$ is such that $AGA=A$, then for any $\lambda$ $G_1=G+\lambda A$ is such that $AG_1A=AGA=A$, but $AG_1=AG+\lambda A^2$ for any $\lambda$.

Answer 2

No, not even for $2\times2$ matrices. Let $A=\left(\begin{matrix}0&0\\0&1\end{matrix}\right)$. Any matrix $G$ for which $AGA=A$ is of the form $G=\left(\begin{matrix}a&b\\c&1\end{matrix}\right)$. However, $AG= \left(\begin{matrix}0&0\\c&1\end{matrix}\right)$. So just pick $G_1$ and $G_2$ so that their $c$-values differ.

(Proof by Maple.)