Why is this determinant zero ? (block matrix) I have two non-singular matrices $P_1$ and $P_2$ such that their sum $P_1+P_2$ is also non-singular.
The calculations I need to do lead me to the following block matrix:
$$\begin{pmatrix} (P_1+P_2)^{-1} & (P_1+P_2)^{-1}-P_2^{-1} \\ (P_1+P_2)^{-1}-P_1^{-1} & (P_1+P_2)^{-1}\end{pmatrix}$$
Its determinant appears to be always null (i tried with random-generated $P_1$ and $P_2$) but I do not find any reason why, even though it is easy to prove it when $P_1$ and $P_2$ are $1\times 1$.
To follow up with this question, can we find a linear combination of these matrix that is zero ?
Edit : this weaker version of my problem has been solved. It also solves this stronger version : let's consider the whole problem then. We have $P_1$, $P_2$, ... $P_n$ non-singular such that every sum of these is also non-singular. The block matrix is now
$$
\begin{pmatrix}
\Sigma^{-1} & \Sigma^{-1}-A_1 & \dots & \Sigma^{-1}-A_1 \\
\Sigma^{-1}-A_2 & \Sigma^{-1} & \dots & \Sigma^{-1}-A_2 \\
\vdots & \vdots & \ddots & \vdots \\
\Sigma^{-1}-A_n & \Sigma^{-1}-A_n & \dots & \Sigma^{-1}
\end{pmatrix}$$
Where $\Sigma = \displaystyle \sum_{i}P_i$ and $A_k = \displaystyle \left(\sum_{i\neq k}P_i\right)^{-1}$.
Multypling by $(P_1, P_2,\dots,P_n)$ gives zero
 A: It can easily be shown (by multiplying both sides from the left with $P_1+P_2$) that
$$
0= (P_1+P_2)^{-1}x +\left((P_1+P_2)^{-1}-P_2^{-1}\right)y
$$
as well as
$$
0= \left((P_1+P_2)^{-1}-P_1^{-1}\right)x +(P_1+P_2)^{-1}y
$$
are equivalent with 
$$
P_2^{-1}y = P_1^{-1}x
$$
So simply take an arbitrary vector $x\neq 0$, calculate $y = P_2P_1^{-1}x$ and you will get
$$
\begin{pmatrix}
(P_1+P_2)^{-1}  &  (P_1+P_2)^{-1}-P_2^{-1} \\
(P_1+P_2)^{-1}-P_1^{-1}  & (P_1+P_2)^{-1}
\end{pmatrix}
\begin{pmatrix}
x \\
y
\end{pmatrix}
=0
$$
which means that the matrix is singular.
An even simpler variant:
$$
\begin{pmatrix}
(P_1+P_2)^{-1}  &  (P_1+P_2)^{-1}-P_2^{-1} \\
(P_1+P_2)^{-1}-P_1^{-1}  & (P_1+P_2)^{-1}
\end{pmatrix}
\begin{pmatrix}
P_1 \\
P_2
\end{pmatrix}
\\
=
\begin{pmatrix}
(P_1+P_2)^{-1}P_1  +  (P_1+P_2)^{-1}P_2-I \\
(P_1+P_2)^{-1}P_1-I  + (P_1+P_2)^{-1}P_2
\end{pmatrix}
\\
=
\begin{pmatrix}
(P_1+P_2)^{-1}(P_1+P_2)-I \\
(P_1+P_2)^{-1}(P_1+P_2)-I
\end{pmatrix}
\\
=
\begin{pmatrix}
I-I \\
I-I
\end{pmatrix}
\\
= 0
$$
