Schur complement like operation on a singular matrix For the classical definition of matrix inversion by Schur complement, given by:
\begin{aligned}
M^{-1}=\left[\begin{array}{ll}
A & B \\
C & D
\end{array}\right]^{-1} &=\left(\left[\begin{array}{cc}
I_{p} & B D^{-1} \\
0 & I_{q}
\end{array}\right]\left[\begin{array}{cc}
A-B D^{-1} C & 0 \\
0 & D
\end{array}\right]\left[\begin{array}{cc}
I_{p} & 0 \\
D^{-1} C & I_{q}
\end{array}\right]\right)^{-1} \\
&=\left[\begin{array}{cc}
I_{p} & 0 \\
-D^{-1} C & I_{q}
\end{array}\right]\left[\begin{array}{cc}
\left(A-B D^{-1} C\right)^{-1} & 0 \\
0 & D^{-1}
\end{array}\right]\left[\begin{array}{cc}
I_{p} & -B D^{-1} \\
0 & I_{q}
\end{array}\right] \\
&=\left[\begin{array}{cc}
\left(A-B D^{-1} C\right)^{-1} & -\left(A-B D^{-1} C\right)^{-1} B D^{-1} \\
-D^{-1} C\left(A-B D^{-1} C\right)^{-1} & D^{-1}+D^{-1} C\left(A-B D^{-1} C\right)^{-1} B D^{-1}
\end{array}\right] \\
&=\left[\begin{array}{cc}
(M / D)^{-1} & -(M / D)^{-1} B D^{-1} \\
-D^{-1} C(M / D)^{-1} & D^{-1}+D^{-1} C(M / D)^{-1} B D^{-1}
\end{array}\right]  \qquad \qquad \qquad \qquad \qquad \qquad \qquad (1)
\end{aligned}
the matrix $D$ is considered non-singular.
However, for generalised Schur complement, when $A$, $D$ and $M$ are singular, substituting the formulation for Schur complement using a Moore-Penrose inverse, i.e.,:
\begin{aligned}
M/A := A - (BD^{\dagger}C) \qquad \qquad(2)
\end{aligned}
and substituting the value in (1) results in a value of $M^{-1}$ (or should it be denoted by $M^{\dagger}$ ?) which gives $MM^{-1} \neq I$.
Granted M is ill-conditioned for this problem, but how do we go about inverting such matrices?
Any ideas/hints would be very helpful. Thanks in advance.
 A: While the Moore-Penrose inverse does indeed mean that $MM^\dagger \neq  I$, this is more a "this is not always the case" statement. See the actual statement is:
$$MM^\dagger M = M, ~~~~ M^\dagger MM^\dagger = M^\dagger$$
Which includes many use cases. For instance, this could mean that $MM^\dagger \neq  I$ but $M^\dagger M = I$ and vise versa. However, it can also mean that $M = M^\dagger$ and $MM^\dagger = M$ (take for instance an Identity matrix where some diagonal entries are 0).
With respect to the Schur complement, I think this boils down to 2 things: First, is $(A-BD^\dagger C)$ invertible? Second, is either $D^\dagger D = I$ or $DD^\dagger = I$, or the same but for $A$.
But this is when we only look for a $M^\dagger$ such that either $MM^\dagger = I$ or $M^\dagger M = I$. This does still mean that $M$ is ill-conditioned, but for singular matrices the question is never "find an unique matrix such that if multiplied with $M$ yields the identity matrix", but "find any matrix such that if multiplied with $M$ minimizes the squared difference with the Identity matrix"
