Consider a real symmetric matrix $P$ that is partitioned as
$$
P = \begin{bmatrix}
A & AB \\
B^\top A & C+B^\top A B
\end{bmatrix},
$$
where P is singular, while $A$ is positive definite, and $C$ is a non-zero matrix.
I am wondering if the following relation is true:
\begin{equation}
Det(P) = \det(A) Det(C), \quad (1)
\end{equation}
where $Det$ denotes the pseudo determinant, i.e. $Det(P)$ is the product of all the non-zero eigenvalues of $P$.
The above relation is straightforward when only the determinant is concerned, in which case we can apply the following formula:
$$
\det(P) = \det(A) \det[(C+B^\top A B) – (B^\top A) A^{-1} (AB) ] = \det(A) \det(C).
$$
However, I do not know if $(1)$ holds.
Best Answer
This doesn't hold. E.g. when $A=B=1$ and $C=0$, we have $$ P=\pmatrix{1&1\\ 1&1} $$ and hence $$ Det(P)=2\ne 1=\det(A)Det(C).\tag{2} $$
This counterexample can be easily extended to the case where $C$ is nonzero or even entrywise nonzero. First, by appending $1$s to $A,B$ and $C$, i.e. when $A=B=I_2$ and $C=\operatorname{diag}(0,1)$, the new $P$ is (by a simultaneous permutation of rows and columns) similar to $\pmatrix{1&1\\ 1&1}\oplus\pmatrix{1&1\\ 1&2}$. Therefore $Det(P)=2$ and the inequation $(2)$ still holds. Next, to make $C$ entrywise nonzero, we can simply replace $C$ by $QCQ^T$ for some appropriate real orthogonal matrix $Q$. In particular, the inequation $(2)$ is still true when $A=B=I_2$ and $$ C=\left[\frac{1}{\sqrt{2}}\pmatrix{1&1\\ 1&-1}\right]\pmatrix{0&0\\ 0&1}\left[\frac{1}{\sqrt{2}}\pmatrix{1&1\\ 1&-1}\right]=\frac12\pmatrix{1&-1\\ -1&1}. $$