Inversion of a matrix in a system of linear inequalities

Question

I would like to know if someone knows sufficient conditions on $A\in\mathbb{R}^{n\times n}$ and $b\in\mathbb{R}^{n}$ such that for all $x\in\mathbb{R}^{n}$: $$Ax\leq b \Rightarrow x\leq A^{-1}b \text{ or } x\geq A^{-1}b.$$ $A$ is assumed to be positive definite.

Answer 1

If $A$ is an M-matrix (all symmetric M-matrices are SPD but not all SPD matrices are M-matrices), then $Az\geq 0$ implies $z\geq 0$. So if $y:=A^{-1}b$, then $b-Ax=A(y-x)\geq 0$ implies $y-x\geq 0$ and hence $x\leq A^{-1}b$. This is simply due to the fact that M-matrices have nonnegative inverses.

If $A$ is SPD, $Ax\leq b$ and $x\geq A^{-1}b$ for some $x$ means that $Az\geq 0$ and $z\leq 0$ for some $z$. But then $z^TAz$ is a sum of nonpositive numbers so $z^TAz\leq 0$. Since $A$ is SPD, this is possible if only if $z=0$, that is, $Ax=b$.