Textbook question:
(True/False) It is impossible to get more than one weakly dominant strategy equilibrium.
I came up with this table below, which I think it depicts one weakly dominant strategy. For player 2, picking B weakly dominates A, and, for player 1, picking A weakly dominates B. Thus, there is one weakly dominant equilibrium at $(4, -2)$. I can't seem to find a table with more than two.
However, I can see that the table has two Nash equilibriums at $(2,-2$) and $(4, -2)$.
$$ \begin{array}{cc} & \text{Player $2$} \\ \text{Player $1$} & \begin{array}{c|c|c|} & A & B \\ \hline A & (2,-2) & (4,-2) \\ \hline B & (2,3) & (3,4) \\ \hline \end{array} \end{array} $$
UPDATE:
I tried doing another table with another row with the same payoff as the first row, and I think it now represents two weakly dominant strategies. Can someone confirm this? $$ \begin{array}{cc} & \text{Player $2$} \\ \text{Player $1$} & \begin{array}{c|c|c|} & A & B \\ \hline A & (2,-2) & (4,-2) \\ \hline B & (2,3) & (3,4) \\ \hline C & (2,-2) & (4,-2) \\ \hline \end{array} \end{array} $$
Player 1 has two weakly dominant strategies $A$ and $C$.
Player 2 has one weakly dominant strategies $B$.
Thus, two weakly dominant equilibriums at $(A, B)$ and $(C, B)$
Good attempt, but by introducing strategy $C$, $A$ is no longer weakly dominant. In fact, this should suggest to you the following conjecture:
Show why this is the case and the answer to your stated question should follow immediately.