Coming from a problem in game theory, I arose at some dubious monotonicity like property for matrices of the following art. Let $H=\lbrace h\in\mathbb{R}^{n}\colon h\_{1}+\dots+h\_{n}=0\rbrace$. I'm interested in matrices $A\in\mathbb{R}^{n\times n}$ which satisfy for all $h\in H$ and $i\ne j$
\begin{equation}
(h\_{i}-h\_{j})((Ah)\_{i}-(Ah)\_{j})\le0.
\end{equation} 
Does this property have a common name and has it been studied anywhere? What are sufficient conditions on $A$ for this property to hold? What else can be said about such matrices? For instance, one can easily check, that they must be conditionally negative semi-definite. I'm thankful for any remark, literature advice, etc.



Dubious matrix monotonicity
Such a matrix has the form $\theta I\_n+ew^T+ve^T$ where $T$ denotes transposition and $e$ is the vector $(1,\ldots,1)^T$. The parameter $\theta$ is $\le0$.


Here is the proof when $n\ge4$. By continuity, your assumption implies that
$$(e\cdot h\quad\hbox{and}\quad h\_i=h\_j)\Longrightarrow((Ah)\_i=(Ah)\_j).$$
Denoting $B=A^T$, this means that $h\in(e,e\_i-e\_j)^\bot$ implies $h\in(B(e\_i-e\_j))^\bot$. Therefore there exists scalars such that
$$B(e\_i-e\_j)=\lambda\_{ij}e+\theta\_{ij}(e\_i-e\_j).$$
Writing $B(e\_i-e\_k)=B(e\_i-e\_j)+B(e\_j-e\_k)$ and using the fact that $e,e\_i-e\_j,e\_i-e\_k$ and $e\_j-e\_k$ are linearly independent when $i,j,k$ are pairwise distinct, you find that $\theta\_{ij}=\theta$ does not depend upon the indices, and $\lambda\_{ij}=w\_i-w\_j$ for some vector $w$. Finally, we obtain that the vector $v:=Be\_i-\theta e\_i-w\_ie$ does not depend on the index $i$. Thus $B=\theta I\_n+ew^T+ve^T$.


