Does every real skew symmetric matrix map some non-zero vector with non-negative entries into a vector with non-negative entries?

Question

Find a matrix $A$ (or prove no such matrix exists) such that $A=-A^T$ and if $v≥0$ (pointwise) and $Av≥0$ (pointwise) then $v=0$.

Although no solution has yet been given we have currently established no such $2×2$ or $3×3$ matrix exists. Ideally a proof would be given for all $n×n$ matrices (or a single counter example of any dimension). Thanks to all of you who have contributed.

Answer 1

No such matrix exists. Tucker's existence lemma asserts that if $A$ is skew-symmetric, then there exists a vector $x$ such that $x\ge0,\,Ax\ge0$ and $Ax+x>0$ (so that $x$ must be nonzero). See Giorgio Giorgi (2014), Again on the Farkas Theorem and the Tucker Key Theorem Proved Easily, p.15, lemma 3.

Answer 2

Here is a proof of the question as it was in the original title, approximatively : "Existence of an antisymmetric matrix such that the image by it of the positive orthant $O_+$ has a $\{0\}$ intersection with $O_+$ ?".

My answer is that no such matrix exist for the $2 \times 2$ case, and that an infinity of solutions exist for the $3 \times 3$ case.

$2 \times 2$ case :

The general form of an antisymmetric matrix is (for any real $a$)

$$A=\begin{pmatrix}0&-a\\a& \ \ 0\end{pmatrix}$$

which corresponds to a $\pi/2$ rotation composed with a homothety with ratio $a$. In the $a>0$ case the image of the first quadrant is the second quadrant, sharing the half of the ordinate axis with the first quadrant. A similar conclusion for the case $a<0$, this time with the positive $x$-axis.

$3 \times 3$ case :

The general form of an antisymmetric matrix is (for any real numbers $x_0,y_0,z_0$)

$$A=\begin{pmatrix}\ \ 0&-z_0& \ \ y_0\\ \ \ z_0& \ \ 0&-x_0\\-y_0& \ \ x_0& \ \ 0\end{pmatrix}$$

with the following geometrical interpretation : $AV= U_0 \times V$ where

$$U_0=\begin{pmatrix}x_0\\y_0\\z_0\end{pmatrix}$$

Therefore the image $R(A)$ of $A$ is the plane (2 dimensional vector space) orthogonal to $U_0$.

It is visually intuitive that $R(A) \cap O_+=\{0\}$ (where $O_+$ = closed positive orthant) if we take any $V_0$ in the open positive orthant ($x_0,y_0,z_0 > 0)$.

Proof of this fact : Let us assume, as said upwards, that $x_0,y_0,z_0 > 0$. Then

$$V(x,y,z) \in R(A)\cap O_+ \ \iff \ xx_0+yy_0+zz_0=0 \ \ \text{with} \ \ x \geq 0, y \geq 0, z \geq 0$$

which is impossible unless $x=y=z=0$ (consider for example that $x >0$ ; we would have a contradiction with $xx_0+yy_0+zz_0 > 0$).