I'm trying to understand the conditions for which $$ A x \leq b \Leftrightarrow x \leq A^{-1}b$$ is true.
Let's assume that $A$ is a non-singular, $x$,$b\in \mathbb{R}^n$.
I have come across an optimization problem where this does not seem to hold, my matrix is positive definite, non symmetric, does not have any particular structure and the off-diagonal elements can have different signs.
I would like to understand what might be the conditions on the matrix A which are necessary for this to hold.
M-matrices do not work. E.g. $$ \pmatrix{-1\\ 0}\le\pmatrix{2&-1\\ -1&1}^{-1}\pmatrix{0\\ 0} \ \text{ but }\ \pmatrix{2&-1\\ -1&1}\pmatrix{-1\\ 0}\not\le\pmatrix{0\\ 0}. $$ The implication $\forall (x,b),\,Ax\le b\Rightarrow x\le A^{-1}b$ is actually equivalent to $0\le A^{-1}$, while the implication $\forall (x,b),\, x\le A^{-1}b\Rightarrow Ax\le b$ is equivalent to $A\ge0$. Therefore you need a matrix $A$ such that both $A$ and $A^{-1}$ are nonnegative. The only M-matrices that satisfy this requirement are the positive diagonal matrices.