Suppose I have vectors $u, v \in \mathbb{R}^n$ such that $u \le v$, meaning that, component-wise, each element of $u$ is less than or equal to the corresponding component in $v$.
I am curious about under what conditions real $n \times n$ matrices $M$ necessarily satisfy
$$M u \le M v$$
Clearly, diagonal matrices with positive diagonal coefficients work (for somewhat trivial reasons), but I feel as though there is a larger class of matrices that respects the inequality structure. I conjecture that matrices with all non-negative entries respect the inequality structure.
What I am curious about is if a set of matrices even larger than that can be considered.
It shall be $$ \left\{ {\matrix{ {{\bf 0} \le {\bf M}\,\left( {{\bf v} - {\bf u}} \right) \le {\bf M}\,{\bf w}} \hfill & {\left| {\;0 \le w_{\,k} \;\left| {\;\forall k} \right.} \right.} \hfill \cr {{\bf M}\,{\bf w} < {\bf 0}} \hfill & {\left| {\;w_{\,k} < 0\;\left| {\;\forall k} \right.} \right.} \hfill \cr } } \right. $$
Then taking $$ {\bf w} = \left( {0, \ldots ,0,1,0, \ldots ,0} \right)^T $$ the conclusion follows easily, as per Stan Tendijck comment.