Sum of Difference of anti-diagonal matrix elements

Question

Let $A \in \mathbb{R}^{n \times n}$, with elements $a_{ij}$

What conditions on $A$ are required for the following to be true?

There exists some vector $x \in \mathbb{R}^n_+$, $x \neq 0$ such that for all $i=1\dots n$,

$$\sum_{j=1}^n x_j(a_{ij} - a_{ji}) \geq 0$$

Obviously if $A$ is symmetric this is true. I believe it may be true for all $A$ but I haven't been able to prove it.

(sorry really wasn't sure what to title this post)

I've tried the usual Farkas' Lemma-style trick but the problem is self-dual

Answer 1

Lawrence and Spingarn, On fixed points of non-expansive piecewise isometric mappings, Proc. London Math. Soc. (3) 55 (1987), no. 3, 605–624, MR0907234 (89d:58063), "obtain a constructive and simple new proof of Tucker's Theorem which states that for any antisymmetric real $n\times n$ matrix $A$, there exists $x\ge0$ such that $Ax\ge0$ and $x+Ax\gt0$." They also write, "This theorem is equivalent to the Farkas Lemma and the linear programming duality theorem in the sense that each may be easily derived from the other."

Answer 2

The original question asks if we can find a nonzero vector $x \ge 0$ such that $(A-A^T)x \ge 0$, of equivalently, for a skew-symmetric (aka antisymmetric) matrix $M$, the claim is that there always exists an $0 \neq x \ge 0$ such that $Mx \ge 0$. Chasing Gerry Myerson's suggestion above, here is a link to a proof of this claim by invoking Farkas' Lemma.

Lemma 4.5 here.