Suppose
$$ R = \begin{bmatrix} A & B\\ C & D\end{bmatrix} $$
is a $2 \times 2$ block matrix of real numbers, where $A$ and $D$ are squared diagonal matrices.
Is it possible that the following four conditions hold simultaneously?
$R$ is invertible
$D$ is nonsingular
the Schur complement of $D$, $A-BD^{-1}C$ is singular.
$A$ is singular.
If so, could you please provide a way to find the inverse $R$ in terms of the partitions of $R$?
On
Suppose $A$ is a $1\times 1$ equal to zero. Let $D$ be any invertible matrix and $C$ any column vector. Choose $B$ to be a row vector which is orthogonal to $D^{-1}C.\;$ These choices will meet all of the stated conditions.
Assume that the initial elements of $\{B,C,D\}$ are $\{1,2,3\},$ respectively. Then repartitioning produces $A=\pmatrix{0&1\\2&3}$ or some other invertible matrix depending on those initial elements.
The important point is that the invertibility of new $A$ means that $R^{-1}$ can be calculated using its Schur complement.
It's true in general since Volker Strassen found the following formula :
$$\begin{pmatrix}A&B\\C&D\end{pmatrix}^{-1}=\begin{pmatrix}\left(A-B D^{-1}C\right)^{-1}&-\left(A-B D^{-1}C\right)^{-1} BD^{-1}\\-D^{-1}C\left(A-B D^{-1}C\right)^{-1}&D^{-1}+D^{-1}C\left(A-BD^{-1}C \right)^{-1}BD^{-1}\end{pmatrix}$$