special case of the minimax theorem

Question

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?

Answer 1

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.