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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

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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: $$ \begin{pmatrix} (P_1+P_2)^{-1} & (P_1+P_2)^{-1}-P_2^{-1} \\ (P_1+P_2)^{-1}-P_1^{-1} & (P_1+P_2)^{-1} \end{pmatrix} \begin{pmatrix} P_1 \\ P_2 \end{pmatrix} \\ = \begin{pmatrix} (P_1+P_2)^{-1}P_1 + (P_1+P_2)^{-1}P_2-I \\ (P_1+P_2)^{-1}P_1-I + (P_1+P_2)^{-1}P_2 \end{pmatrix} \\ = \begin{pmatrix} (P_1+P_2)^{-1}(P_1+P_2)-I \\ (P_1+P_2)^{-1}(P_1+P_2)-I \end{pmatrix} \\ = \begin{pmatrix} I-I \\ I-I \end{pmatrix} \\ = 0 $$

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  • $\begingroup$ Thank you very much, it also solves a stronger version of the problem (see edited question) ! $\endgroup$
    – M. Boyet
    Commented Mar 17, 2017 at 14:48

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