Suppose that all payoff matrix $A$ elements are non negative and that each column and each row contains at leat one positive element. To find each opponents optimal mixed strategy, a Linear Programming problem is used: $$f = \sum^k_{j=1}z_j \rightarrow \max \\ \sum^k_{j=1}a_{ij}z_j \le 1, i = 1,...,m \\ z_j\ge 0, j=1,...,k$$Corresponding dual Linear Programming problem: $$g = \sum^m_{i=1}y_i \rightarrow \min \\ \sum^m_{i=1}a_{ij}y_i \ge 1, j = 1,...,k \\ y_i \ge0, i = 1,...,m$$
1) First, does all sums over $k$ for $a_{ij}z_j$ must be less or equal to one, because the probabilities of choosing any strategy should be less or equal to 1? If so, how does that translate to opponents strategies? For dual problem the constrain is that all sums over $m$ of $a_{ij}y_i$ should be equal or greater that one. How to make sense of that?
Let's notate the solution of Linear Programming Problem $z^*$ and the solution for dual linear problem - $y^*$. Then $f(z^*)>0$. Why? The prize for the game can then be computed as $V = \frac{1}{f(z^*)}$. Optimal strategies for the first player: $p^* = Vy^* = \frac{y^*}{f(z^*)}$ and for the second player: $q^* = Vz^* = \frac{z^*}{f(z^*)}$.
Proof: It will be proved that both linear programming problems have a solution and that $f_{max} > 0$. It will also be proved, that for both of the linear programming problems set of plans is not empty (I don't know whether plans is a valid notation in English, but it's set of points that satisfy the constrains). Let's use notation: $a_j > 0 ,j = 1,...,k$ and $a_j = \max\limits_{i} a_{ij}, j = 1,...,k$. Vector $\bar{z}$ with components $z_j = \frac{1}{ka_j}, j = 1,...,k$. Then $$\sum^k_{j=1}a_{ij}\bar{z}_j \le \sum^k_{j=1}a_{j}\bar{z}_j = 1, i = 1,...,m$$. From there it follows that vector $\bar{z}$ belongs to set of plans. (Because it satisfies the constrains? Or is there something else to it?)
Now, let's suppose that $a_{j} = \max\limits_{i}a_{ij}=a_{i_jj}, j = 1,...,k$. Assuming that $$1)\ \ \bar{y}_i = 0, \text{ if there isn't such j where } i = i_j \\ 2) \ \ \bar{y}_i = \min\left[\left.\frac{1}{a_j}\right|j:i_j = i\right] \text{, if there exists such } j \text{ that } i = i_j$$
At this point, I don't understand why there was a need to introduce an element $a_{i_jj}$ in place of before defined $a_j$? Also, $a_j$ is the biggest element in each columnt, that is the biggest loss for all the strategies the 2nd player can play. What is then meaning of the $i = i_j$ and $y_i = 0$ in case of there not being $j$ column such that $i=i_j$? Why $y_i$ is chosen to be minimum of $\frac{1}{a_j}$, what is the meaning behind it?
From 1) and 2) : $\sum^m_{i=1} a_{i j}\bar{y}_i \ge a_{i_jj}\bar{y}_{i_j} \ge 1, j=1,...,k$. From there it follows that vector $\bar{y}$ belongs to set of plans. So, the set of plans for dual programming problem is non empty as well.