Antisymmetric Matrix Games Have Value
Let $B$ be a real square matrix satisfying $B=-B^T$. Then the matrix game $B$ has a value. Equivalently, there exists $x\in\Delta(I)$ such that $$ Bx\le 0. $$
Proof
The finite minimax theorem gives a mixed value for the square matrix game. Since $B=-B^T$, every mixed strategy $z$ satisfies $$ zBz=0. $$ For any row mixed strategy $x$, the average of the column payoffs $xBe_j$ under $x$ is $xBx=0$, so some column gives payoff at most $0$ to player $1$ and the row player cannot guarantee more than $0$. Dually, for any column mixed strategy $y$, the average of the row payoffs $e_iBy$ under $y$ is $yBy=0$, so some row gives payoff at least $0$ and the column player cannot hold the payoff below $0$. Equality of maxmin and minmax therefore forces the value to be $0$.
If $x$ is an optimal column strategy for player $2$, then it holds every row payoff to at most $0$, i.e. $(Bx)_i\le0$ for all $i$. This is the displayed equivalent form.
References
- [MFoGT, Section 2.8, Exercise 10(1)] Laraki, Renault, and Sorin, Mathematical Foundations of Game Theory. Brown-von Neumann theorem for antisymmetric games.