Blockwise inversion case when $\textbf{D} - \textbf{C}\textbf{A}^{-1}\textbf{B}$ is singular What means in blockwise matrix inversion when $\textbf{D} - \textbf{C}\textbf{A}^{-1}\textbf{B}$ is singular but $\textbf{A}$ is not? is that necessary and sufficient for the whole composed matrix be singular as well? are there any cases where the whole matrix is not singular but blockwise matrix inversion fails because $\textbf{D} - \textbf{C}\textbf{A}^{-1}\textbf{B}$ is singular?
 A: It cannot happen. You are starting with a block matrix of the form
$$\mathbf{M}=\left(\begin{array}{cc}
\mathbf{A} & \mathbf{B}\\
\mathbf{C} & \mathbf{D}
\end{array}\right),$$
which we are assuming is invertible; that is, $\det(\mathbf{M})\neq 0$.
But if $\mathbf{A}$ is invertible, then we also have
$$\det(\mathbf{M}) = \det(\mathbf{A})\det(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B}),$$
so $\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B}$ cannot be singular. 
To see this, note first that
$$\det\left(\begin{array}{cc}
\mathbf{X}& \mathbf{Y}\\
\mathbf{0} & \mathbf{W}\end{array}\right) = \det(\mathbf{X})\det(\mathbf{W})$$
and
$$\det\left(\begin{array}{cc}
\mathbf{X}&\mathbf{0}\\
\mathbf{Z}&\mathbf{W}\end{array}\right) = \det(\mathbf{X})\det(\mathbf{W}).$$
If $\mathbf{A}$ is invertible, then we have
$$\mathbf{M}=\left(\begin{array}{cc}
\mathbf{A} & \mathbf{B}\\
\mathbf{C} & \mathbf{D}\end{array}\right) = \left(\begin{array}{cc}
\mathbf{A} & \mathbf{0}\\
\mathbf{C} & \mathbf{I}
\end{array}\right)\left(\begin{array}{cc}
\mathbf{I} & \mathbf{A}^{-1}\mathbf{B}\\
\mathbf{0} & \mathbf{D}-\mathbf{CA}^{-1}\mathbf{B}
\end{array}\right),$$
where $\mathbf{I}$ is a suitably sized identity matrix. Since the determinant of a product is the product of the determinants, we get the desired formula.
So if $\mathbf{M}$ and $\mathbf{A}$ are both nonsingular, then $\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B}$ is nonsingular as well.
