Show that an entirely real, non-negative covariance matrix has a precision matrix with non-positive off-diagonal elements

Question

I was told in lecture that if a precision matrix $J$ has non-positive off-diagonal elements i.e. $\forall i\neq j, J_{ij} \leq 0 $, then the corresponding covariance matrix $J^{-1}$ has all non-negative off diagonal elements i.e $\forall i \neq j, (J^{-1})_{ij} \geq 0$. Is the converse true? How can I show this?

Answer 1

The answer is no. A counter example is:

$A=\left[ \begin{array}{ccc} 4 & 0 & 2 \\ 0 & 4 & 3 \\ 2 & 3 & 4 \\ \end{array} \right] $

with

$A^{-1}=\left[ \begin{array}{rrr} 0.5833 & 0.5000 & -0.6667 \\ 0.5000 & 1.0000 & -1.0000 \\ -0.6667 & -1.0000 & 1.3333 \\ \end{array} \right] $.

This counterexample can be found in

Nabben, Reinhard, and Richard S. Varga. "A linear algebra proof that the inverse of a strictly ultrametric matrix is a strictly diagonally dominant Stieltjes matrix." SIAM Journal on Matrix Analysis and Applications 15.1 (1994): 107-113. http://www.math.kent.edu/~varga/pub/paper_205.pdf