# Notations:

Let $$M = (m_{ij})\in \mathbb R^{n\times n}$$ be a square real matrix, such that $$m_{ij}\leq 0$$ for all $$i\neq j$$. We say that a matrix is monotone when $$M$$ is invertible and $$M^{-1}$$ has all its coefficients positive. We call "leading principal minors" the $$\det (m_{ij})_{1\leq i,j\leq k}$$ for $$1\leq k \leq n$$.

# The problem:

Show that $$M$$ is monotone $$\iff$$ all the leading principal minors of $$M$$ are $$>0$$.

# What I've done so far:

I'm really stuck with this one. Tried lots of things but nothing seems to work yet.

• Yes, you are right. I fixed it. However, I didn't find anything on the internet to show this equivalence simply (without showing a lot of other equivalence). Do you have any idea? May 18 '20 at 17:27

These are actually two equivalent characterisations of nonsingular $$M$$-matrices. In standard terminologies for nonnegative matrices, it can be stated as follows: let $$M\in\mathbb R^{n\times n}$$ be a nonsingular $$Z$$-matrix. Then $$M$$ is inverse-positive if and only if all leading principal minors of $$M$$ are positive.
A proof of $$50$$ equivalent characterisations of nonsingular $$M$$-matrices can be found in theorem 2.3, chapter 6 of Nonnegative Matrices in the Mathematical Sciences written by Berman and Plemmons. The equivalence of the two particular characterisations in your question can be proved as follows.
Suppose $$M$$ is inverse-positive. Let $$\alpha=\max_im_{ii}$$. Then $$P=\alpha I-M\ge0$$. Let $$x\ge0$$ be a Perron vector for $$P$$. Then $$(\alpha-\rho(P))M^{-1}x=x\ge0$$. Since $$M^{-1}x$$ is also nonnegative, we must have $$\alpha>\rho(P)$$. Now denote by $$P_k$$ the leading principal $$k\times k$$ submatrix of $$P$$ and define $$M_k$$ analogously. Since $$P\ge0$$, by using Gelfand's formula with the induced $$\infty$$-norm, we have $$\rho(P_k)\le\rho(P)$$. It follows that $$\alpha-\rho(P_k)\ge\alpha-\rho(P)>0$$. Consequently, all eigenvalues of $$M_k=\alpha I_k-P_k$$ lie on the open right half plane. Therefore $$\det(M_k)>0$$ for each $$k$$, i.e. $$M$$ has positive leading principal minors.
Conversely, suppose $$M$$ has positive leading principal minors. By mathematical induction on $$n$$, one can show that $$M$$ has an LU-decomposition $$M=LU$$ such that all diagonal elements of $$L$$ are equal to $$1$$ and $$L,U$$ are inverse-positive. More specifically, in the inductive step, we have \begin{aligned} \pmatrix{L_0U_0&x\\ y^T&a} &=\pmatrix{L_0&0\\ y^TU_0^{-1}&1}\pmatrix{U_0&L_0^{-1}x\\ 0&b} \quad(b=a-y^TU_0^{-1}L_0^{-1}x),\\ \pmatrix{L_0&0\\ y^TU_0^{-1}&1}^{-1}&=\pmatrix{L_0^{-1}&0\\ -y^TU_0^{-1}L_0^{-1}&1},\\ \pmatrix{U_0&L_0^{-1}x\\ 0&b}^{-1}&=\pmatrix{U_0^{-1}&-b^{-1}U_0^{-1}L_0^{-1}x\\ 0&b^{-1}}. \end{aligned} It follows that $$M^{-1}=U^{-1}L^{-1}$$ is nonnegative, i.e. $$M$$ is inverse-positive.