I am trying to prove that a mapping has a unique fixed-point by showing that its Jacobian is a P-matrix. In this particular case the Jacobian can be decomposed as the sum of two matrices and I would like to show the following result:



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> Let $M \in \mathbb R^{n \times n}$ be a *M-matrix* and $S \in \mathbb R^{n \times n}$ be a *S-matrix* (positive definite matrix) with non-negative entries. Is it the case that $M + S$ is a *P-matrix*?
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The result holds trivially for $n=1,2$, but I was wondering if it holds for arbitrary $n$. I generated millions of random matrices with $n \le 10$ and the result seems to be true. I would truly appreciate any comments, references or help!


### Some definitions


A matrix $M \in \mathbb R^{n \times n}$ is a *M-matrix* if (i) all off-diagonal entries are less than or equal to zero, and (ii) all principal minors of $M$ are positive. A matrix $P \in \mathbb R^{n \times n}$ is a *P-matrix* if all principal minors of $P$ are positive. Horn Johnson's *Topics in Matrix Analysis* (1991, pp. 112-125) discusses various equivalent definitions for these matrices.


Interestingly, all M-matrices are P-matrices and all S-matrices are P-matrices, but P-matrices are not close under addition. For example, $A = \begin{bmatrix} \frac 1 2 & -1 \\ 0 & \frac 1 2 \end{bmatrix}$ and $B = \begin{bmatrix} \frac 1 2 & 0 \\ -1 & \frac 1 2 \end{bmatrix}$ are both P-matrices but the sum $A+B$ is not. This example is taken from Parthasarathy's *On Global Univalence Theorems* (1983, p. 15).


### Some Background


In many equilibrium models one proves the uniqueness of an equilibrium by showing that an equation $F(x) = 0$ has at most one solution. In many cases the Jacobian of $F$ can be decomposed as the sum of a positive diagonal matrix and M- or P-matrix, which is again a P-matrix. Then one invokes the univalence result of Gale and Nikaido's *The Jacobian matrix and global univalence of mappings* (1965) to show that the mapping admits at most one solution. The previous result would allow one to prove uniqueness in more general mappings and may be of interest to the community.



M-matrix plus S-matrix is P-matrix?
This is not true in general.


$\left(\begin{array}{ccc} 4 & 0 & -16 \\ 2 & 4 & 0 \\ 0 & 2 & 4 \end{array}\right)= \left(\begin{array}{ccc} 1 & -2 & -16 \\ 0 & 1 & -2 \\ 0 & 0 & 1 \end{array}\right) +\left(\begin{array}{ccc} 3 & 2 & 0 \\ 2 & 3 & 2 \\ 0 & 2 & 3 \end{array}\right)$


A matrix with determinant $0$ is $M+S$. (Thanks to S. Sra for the correction. I multiplied by $4$ so everything's an integer now.)


