TITLE: Determinant of block matrix with singular blocks on the diagonal
QUESTION [9 upvotes]: Let $A$ and $D$ be square matrices, and let $B$ and $C$ be matrices of valid shapes to allow the formation of
$$
M =
\begin{bmatrix}
    A & B \\
    C & D
\end{bmatrix}.
$$
If $\det{A}\neq0$, we may use the Schur complement to express $\det{M}$ in terms of its constituent blocks as
$$
\det{M} = \det{A}\cdot\det(D-CA^{-1}B),
$$
and if $\det{D}\neq0$ we have in a similar fashion that
$$
\det{M} = \det(A-BD^{-1}C)\cdot\det{D}.
$$
My question: Does there exist a similar formula expressing $\det{M}$ in terms of its constituent blocks, that is valid in case $\det{A}=\det{D}=0$?

REPLY [6 votes]: For $M=\begin{bmatrix}A & B\\C & D\end{bmatrix}$ let $N=M^TM$. So that
$$N:=\begin{bmatrix}E & F\\F^T & G\end{bmatrix}$$
where
$$\begin{align}
E&=A^T A+C^T C\\F&=A^T B+C^T D\\G&=B^T B+D^T D
\end{align}$$
Now, if $M$ is non-singular then $N$ is a positive definite matrix. Hence according to this other post, $E$ is a positive definite (i.e. non-singular) block matrix.
Since $E$ is non-singular, we can use Schur complement to obtain $\det N$.
$$\det{N} = \det{E}\cdot\det(G-F^T E^{-1}F)=(\det M)^2$$
The only remaining part in this case is to determine the sign of $\det M$, which apparently, there is no easy way to do that in general.
We might be able to get the sign of $\det M$ by a trick similar to the one discussed at the end of @hypernova's answer. I'll update this answer if anything comes up.