Find such anti-symmetric matrix $W$ that $A^T WP \geq 0$ $P$ and $A$ are both n-dimensional vectors with non-negative components. $W$ is an $n\times n$ matrix with $W_{ij}=w_i-w_j$, where all $w_k\geq 0$. So $W$ is an anti-symmetric matrix with some stronger properties. For a specific $A$, is there such $W$ that for all $P$, $A^TWP\geq0$?
I noticed $A^TWA=0$, and I can't go any further.
 A: Let $e=(1,\dots,1)$ be the vector of all ones. Then $W = we^T -ew^T$. Then the inequality becomes
$$
a^TWp = a^T(we^T -ew^T)p = (a^Tw)(e^Tp) - (a^Te)(w^Tp) \ge0
$$
Setting $w=e$ we obtain $W=0$ and
$$
a^TWp=0.
$$ 
for all $p$. This is maybe not the answer you are looking for ;)
In the case that $a_i=0$ for some indices but $a\ne0$ there are choices of $w$ such that $a^TW\ne0$. 
Set 
$$
w_i = \begin{cases} 1 & \mbox{if } a_i>0\\0 & \mbox{if } a_i=0.
\end{cases}
$$
Then it holds $a^Tw=a^Te$ and $e^Tp \ge w^Tp$ for all $p\ge 0$. If $p_i>0$ for all $i$ then  $e^Tp > e^Tw$. This means
$$
a^TWp = a^T(we^T -ew^T)p = (a^Te)(e-w)^Tp\ge0
$$
and moreover $a^TWe\ne0$.
A: It depends. $A^TWP\ge0$ for all entrywise nonnegative $P$ if and only if $A^TW\ge0$. For examples,


*

*if $W=\pmatrix{0&-1\\ 1&0}$ and $A^T=(1,1)$, then $A^TWP<0$ when $P=\pmatrix{0\\ 1}$;

*if $W=\pmatrix{0&-1\\ 1&0}$ and $A^T=(0,1)$, then $A^TWP=(1,0)P\ge0$ for all $P\ge0$.

A: Let $w_j =0$ for all $j$, then clearly $W=0$. And this clearly automatically satisfies the equation.  In fact this is even stronger, this is a fixed $W$ such that this works for any $A$ or $P$.
