$Av>0, v\ge 0$, $Aw<0, w\ge 0$ for a square matrix $A$ of certain type Consider an $n\times n$ matrix $A = B - D$, where $B$ has positive coefficients and $D$ is diagonal with positive coefficients.
I want to prove that for such $A$ there don't exist vectors $v, w \in\mathbb{R_{\ge 0}^n}$ such that $Av$ has strictly positive coefficients and $Aw$ has strictly negative coefficients.
Geometric interpretation: hyperplanes $x_i = \pm 1$, $x_i=0$ define $2^n$ cubes. Consider a cube $C$, pick points $P_1, P_2, ..., P_n$ from inside each neighbour of $C$ (one point from each). Let $O$ be the origin. Then convex hull of $OP_1P_2...P_n$ intersects either $C$ or the cube symmetric to $C$ trivially (the intersection is $O$).
I proved the claim for $n=2$, and drew a picture in Geogebra for $n=3$, which made me believe that the fact should be true.
Can it be proven using Farkas' lemma? Seems like very similar setting, but the lemma involves $A^T$.
If not, then what to do? I see that it is enough to prove that the plane $P_1P_2...P_n$ doesn't intersect one of $C$, $-C$, because then I could separate $C$, $-C$ by drawing parallel plane through $O$, but this is where I'm stuck.
 A: Here's a solution due to Pavel Myaktinov.
We work by induction on n. We will show the step for $n=3$, it obviously generalises for $n>3$.
If $A$ has a non-negative row, then obviously there is no vector $v$ such that $v\ge 0$, $Av<0$. So, the diagonal entries of $A$ are negative. We may scale any row by a non-negative number, so, WLOG the diagonal entries of $A$ are $-1$.
Then we have a system of inequalities (where $\vee$ stands either for $>$ or for $<$ and all letters are positive)
$$    \begin{cases}
      x \vee ay + cz\\
      y \vee cx + dz \\
      z \vee ex + fy
    \end{cases}\,.
 $$
If this system has a solution, then the following system does, too:
$$    \begin{cases}
      y \vee c(ay+cz) + dz \\
      z \vee e(ay+cz) + fy
    \end{cases}\,.
 $$
Note that the coefficient of $z$ in the first inequality is positive and so is the coefficient of $y$  in the second inequality. That means that the new matrix $2\times 2$ is again of the form $B - D$ and we've succesfully reduced the case $n=3$ to the case $n=2$.
