Can a weakly dominated strategy be optimal in a zerosum game? I saw an earlier question posted which was using the game
$$A = \left(\begin{array}{} 1& 1 \\ 1 &0 \end{array}\right)$$
But, in thise case removing the second row only removes a nash equilibrium and not an optimal strategy (and the value is also 1). Could anyone provide an example of where the weakly dominated strategy can't be deleted or is can it be deleted and why? Thanks in advance!
Suppose the value is given by a pure strategy. Let $A_{i, j}$ be a Nash equilibrium. Let's suppose it belongs to a weakly dominated strategy, $A_i$ WLOG. Lets $A_k$ be a strategy dominating $A_i$ ($A_k \geq A_i$). Because $A_{i, j}$ is a Nash equilibrium, $A_{k, j} \leq A_{i, j}$. But because $A_k \geq A_i$, $A_{k, j} = A_{i, j}$.
$A_{i, j}$ is the minimum of $A_i$, and $A_k \geq A_i$, $A_{k, j}$ is the minimum of $A_k$. $A_{i, j}$ is the maximum of $A_{\bullet, i}$, so is $A_{k, j}$. Hence $A_{k,j}$ is also a Nash equilibrium, so we can delete the weakly dominated strategy.
Let now consider mixed strategies: Let $(p,q)$ be a Nash equilibrium. Let $A_i$ be weakly dominated by $A_k$. We consider the new strategy $(p', q)$ with $p'_k = p_k + p_i$, $p'i=0$ and $p'j = p_j$ for all others $j$.
Because $(p, q)$ is a Nash equilibrium, we must have $p_{k, j} \leq p_{i, j}$ for all $j$ such that $q_j > 0$, so $p_{k, j} = p_{i, j}$ for all $j$ such that $q_j > 0$. As before, we easily show that the new strategy is still a Nash equilibrium with the same value.
In conclusion, we can always remove a weakly dominated strategy in a zero-sum game. However, this is no longer true if the game is not zero-sum, deleting weakly dominated strategies can remove all existing Nash equilibria.