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Let $A \in \mathbb R^{n \times n}$ be a real matrix. I want to prove that there exists a vector $v \in \mathbb R_{\geq 0}^n$, $v \neq 0$ such that $Av \geq A^Tv$ in every coordinate.

This comes for free from an "unrelated" claim, and I would like to know what is a more natural approach to attack this problem from the first place.

(I think it might be better not to say what is this claim, for now, so as not to fixate similar directions to possible answers).

What might be an algebraic approach to this question? (or is it not "algebraic in nature", and if so why?)

Emolga
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  • Note that your question can be rephrased as: prove that $A-A^T$ sends at least one element of the positive orthant to the positive orthant. Maybe it's of interest that $A-A^T$ is skew-symmetric. – Gabriel Romon Apr 09 '18 at 16:28
  • @GabrielRomon Under this formulation, I know that it is enough if the property "there is such a $v$ with $Bv \geq 0$" is preserved under orthogonal similarity, by a classification of skew-symmetric matrices. Do you perhaps know if this is the case? – Emolga Apr 09 '18 at 16:31
  • Now that somebody has answered, maybe you can disclose the way you solved the problem ? – Gabriel Romon Apr 10 '18 at 19:39
  • @GabrielRomon In a symmetric zero-sum game, player A has a strategy that regardless of the actions of player B will guarantee player A a nonnegative expected outcome, so the game has zero value. The matrix of the game gives the claim. This follows from the minimax theorem. – Emolga Apr 11 '18 at 05:53

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As noted in a comment, your conjecture is equivalent to the statement that if $K$ is real skew-symmetric, then there is a nonzero and nonnegative vector $v$ such that $Kv\ge0$. This is a consequence of a result known as "Tucker existence lemma". See my other answer on this site.

user1551
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