# special case of the minimax theorem

The minimax theorem of von Neumann says that for any payoff matrix $$A$$, we have $$\begin{equation} \max_x \min_y x^T A y = \min_y \max_x x^T A y. \end{equation}$$ In the above, $$x$$ and $$y$$ are probability distributions. In the special case when $$A$$ satisfies: $$\max_i \min_j A_{ij} = \min_j \max_i A_{i,j} (=A_{\bar{i},\bar{j}}),$$ the maximizing distribution $$x$$ and the minimizing $$y$$ in von Neumann's equality above is simply $$x = e_{\bar{i}}$$ and $$y = e_{\bar{j}}$$. (The Nash equilibrium is attained by pure strategies for both players.)

In a yet more (?) specific case when $$A$$ has a dominant column or a dominant row, i.e. there exists a $$\bar{j}$$ so that $$A_{i,\bar{j}} \leq A_{i,j}, \;\;\; \forall i, \; \forall j,$$ or there exists a $$\bar{i}$$ so that $$A_{\bar{i},{j}} \geq A_{i,j}, \;\;\; \forall i, \; \forall j,$$ it is quite easy to show that $$A$$ satisfies the second equality above.

Question: Does the second equality above imply the existence of a dominant row or column?

This is a $$3 \times 3$$ counterexample. $$\begin{bmatrix} 2 & -1 & -2 \\ 1 & 0 & 1 \\ -2 & -1 & 0 \end{bmatrix}$$ It has maxmin = minmax = 0, but it does not have a dominant column and does not have a dominant row.
Smaller $$2 \times 3$$ counterexample: $$\begin{bmatrix} 1 & 2 & 3 \\ 3 & 2 & 2 \end{bmatrix}$$
For $$2 \times 2$$ matrices the implication holds. If $$A$$ has maxmin=minmax, without loss of generality assume it appears at $$a_{11}=0$$, and $$A = \begin{bmatrix} 0 & b \\ c & d \end{bmatrix}$$ with $$b\ge 0$$ and $$c \le 0$$. Now if $$d \ge 0$$, the first column is dominant , and if $$d < 0$$, the first row is dominant.