Q:

general inverse of a matrix

If we have a matrix $A=\begin{bmatrix} C_{r \times r} & D \\ E & F\end{bmatrix}_{m \times n}$ where the rank of $A$ is $r$ and $C$ is full rank, i.e. $r(C)=r$. So the general inverse of $A$ is defined as $A^g$ where $AA^gA=A$.
The general inverse is not unique and one of the $A^g$ can be found as
$$A=\begin{bmatrix} C^{-1}_{r \times r} & 0 \\ 0 & 0\end{bmatrix}_{n \times m}$$
I want to check we indeed have $AA^gA=A$, but I got
$$AA^gA=\begin{bmatrix} C & D \\ E & F\end{bmatrix} \begin{bmatrix} C^{-1} & 0 \\ 0 & 0\end{bmatrix} \begin{bmatrix} C & D \\ E & F\end{bmatrix}=\begin{bmatrix} I & 0 \\ EC^{-1} & 0\end{bmatrix} \begin{bmatrix} C & D \\ E & F\end{bmatrix}= \begin{bmatrix} C & D \\ E & EC^{-1}D\end{bmatrix}$$
So how could we justify $EC^{-1}D=F$ here?
Thanks!

A:

One approach is to notice that $F - EC^{-1}D$ is the Schur complement $A/C$.  It follows that
$$
r = \operatorname{rank}(A) = \operatorname{rank}(C) + \operatorname{rank}(A/C) = r + \operatorname{rank}(A/C)
$$
it follows that $\operatorname{rank}(A/C) = 0$, which is to say that $A/C =F - EC^{-1}D = 0$, so that $F = EC^{-1}D$ as desired.

