Monotone matrix

A real matrix $A$ is called monotone if $Ax\geq 0$ implies $x \geq 0$. If inverse of $A$ exists and is real, then prove that $A$ is monotone if and only if inverse of $A \geq 0$.

($x\geq 0$ means $x$ is a column vector whose all entries are non-negative.

$A \geq 0$ means $A$ is a square matrix whose all entries are non-negative.)

Solution 1:

Suppose that $A^{-1}\ge 0$ and $Ax \ge 0$. We have $0\le A^{-1}(Ax)=(A^{-1}A)x=x$.

On the other hand, suppose the $i^\textrm{th}$ column of $A^{-1}$ (which we call $x$) contains a negative entry. We have $Ax$ a column of $I$, hence $Ax\ge 0$. But $x\not \ge 0$, which contradicts $A$ monotone. Hence $A^{-1}\ge 0$.