I am having trouble understanding how we would derive a potential function for a bimatrix game.
For example, the matrix $G$ shown below is a variant of the Prisoner's dilemma game, and $P$ represents its Potential Function:
$$G=\left[\begin{array}{cc} (1,1)&(9,0)\\(0,9)&(6,6) \end{array}\right],\quad P=\left[\begin{array}{cc} 4&3\\3&0 \end{array}\right].$$
However, I am having trouble understanding how the Potential Function is derived. In a resource online, I saw the definition:
A function $\Phi: S \to \mathbb{R}$ is called an (exact) potential function for the game $G$ if for each $i \in \mathscr{I}$ and all $s_{-i} \in S_{-i}$, $$u_i(x,s_{-i}) -u_i(z,s_{-i}) = \Phi(x,s_{-i}) - \Phi(z,s_{-i}), \text{ for all } x,z \in S_i.$$ $G$ is called an (exact) potential game if it admits a potential.
I am having trouble with how to break the definition down to actually get the Potential Function/Matrix.
Any help?
According to the definite, denote by $P=\left[\begin{array}{cc} a&b\\c&d \end{array}\right]$ the potential function.
For player one, $(1,-1)Aq=(1,-1)Pq\,$ holds for any mixed strategy $q$, from which you can get $a-c=1, b-d=3$. Similarly for player 2, $p^tB(1,-1)^t=p^tP(1,-1)^t$, and you can get $a-b=1,c-d=3$.
Together it admits a solution for $P$ as written in the question, and notice that $P+\lambda \left[\begin{array}{cc} 1&1\\1&1 \end{array}\right]$ is also a solution for any $\lambda$.