Let $A$ be a skew-symmetric real matrix, prove that there exists a vector $x\ge0$ such that $Ax\ge0$ and $Ax + x > 0$

Question

This is an assignment that I am struggling with.

Let $A$ be a skew-symmetric real matrix, prove that there exists a vector $x\ge0$ such that $Ax\ge0$ and $Ax + x > 0$.

Not sure how to proceed here and would appreciate some pointers/solution.

My first thought is to:

1. Show there exists a vector $x\ge0$ so that $Ax\ge0$
2. Use proof by contradiction by starting with the claim: for any $x\ge0$, either $Ax<0$ or $Ax+x\le0$, and derive contradiction, but I did not see a way how.

and the second thought is to use Farkas' lemma, which states that for any given matrix A and vector b, one and only one of the following statements is true: (a) There exists a vector $x\ge0$ so that $Ax=b$; (b) There exists a vector $y$ s.t. $A^Ty\ge0$ AND $b^Ty<0$

I start by choosing $A+I$ as the "A" matrix and some positive vector $b>0$ as the "b" vector here. If I can prove that $y$ does not exist under this situation, I would have proven (a) is correct, which means there exists a $x\ge0$ so that $(A+I)x > 0$, which proves the second half of the case. But unfortunately I cannot find a way to disprove (b).

Appreciate your kind advice. Thanks.

Answer 1

This is known as “Tucker’s theorem”, “Tucker’s existence lemma” or “Tucker’s theorem of alternative” in the literature.

A proof attempt had been made by user “sourisse” on this site before, but the proof is incomplete. Below is a fix.

Let $A$ be an $n\times n$ real skew-symmetric matrix and $i\in\{1,2,\ldots,n\}$. By Farkas’ lemma, either there exist some vectors $v\ge0$ and $w\ge0$ that satisfy $$ \pmatrix{-A&I}\pmatrix{v\\ w}=-e_i $$ or there is some vector $y$ such that $$ y^T\pmatrix{-A&I}\ge0 \quad\text{and}\quad y^T(-e_i)<0. $$ In the former case, we have $Av=w+e_i$. Hence $v\ge0,\,Av\ge0,\,Av+v\ge0$ and $e_1^T(Av+v)>0$.

In the latter case, we have $-y^TA\ge0$ (i.e., $Ay\ge0$ because $A$ is skew-symmetric), $y^T\ge0$ and $y^Te_i>0$. Hence $y\ge0,\,Ay\ge0,\,Ay+y\ge0$ and $e_i^T(Ay+y)>0$.

So, in either case, we have, for each $i$, some vector $x^{(i)}\ge0$ such that $Ax^{(i)}\ge0,\,Ax^{(i)}+x^{(i)}\ge0$ and $e_i^T(Ax^{(i)}+x^{(i)})>0$. Now take $x=\sum_{i=1}^nx^{(i)}$ and we are done.