Check that a Nash equilibrium point is given by $\left(0,\frac {1} {2}, \frac {1} {2}\right)$ $\left(0,\frac {1} {2}, \frac {1} {2}\right)$ Given the game matrix 
\begin{bmatrix}
    1 & 1 & 1 \\ 1 & 2 & 0 \\
    1 & 0 & 2
  \end{bmatrix}
I already see a Nash equilibrium in pure strategies, which is $a_{11}$, where $v^+ = v^- =1$. So how do I verify that  $\left(0,\frac {1} {2}, \frac {1} {2}\right), \left(0,\frac {1} {2}, \frac {1} {2}\right)$ is a Nash equilibrium point? Is it with the formula $xAy^T$?
I'm not completely sure how that works.
Thanks!
 A: You need to check that $x^*=(0,1/2,1/2)$ is a best response to $y^*=(0,1/2,1/2)$ i.e. that $$\max_{x\in\Bbb P^3}xA(y^*)^T=x^*A(y^*)^T$$ (where $\Bbb P^3$ is the simplex in $\Bbb R^3$, or simply the set of probability vectors in $\Bbb R^3$) and vice versa, or if you know that equalizing strategies are optimal then do the latter. In any case you need to calculate the product $A(y^{*})^T$ and $x^*A$ so lets start with: $$A(y^*)^T=
\begin{bmatrix}
    1 & 1 & 1 \\ 1 & 2 & 0 \\
    1 & 0 & 2
  \end{bmatrix}
\begin{pmatrix}
    0 \\ 
    1/2 \\
    1/2
  \end{pmatrix}=
\begin{pmatrix}
    1 \\ 
    1 \\
    1
  \end{pmatrix}$$ Now maximize $xAy^T$ by choosing an appropriate $x \in \mathbb P^3$, i.e. solve $$\max_{x\in \Bbb P^3}xA(y^*)^T=\max_{x\in \Bbb P^3}(x_1,x_2,x_3)
\begin{pmatrix}
    1 \\ 
    1 \\
    1
  \end{pmatrix}$$ But since $Ay^T$ is constant (i.e. $y$ is equalizing) then any $x$ would do, so in particular $x^*$. To see this computationally, note that for any $x\in \Bbb P^3$ $$x
\begin{pmatrix}
    1 \\ 
    1 \\
    1
  \end{pmatrix}=(x_1,x_2,x_3)
\begin{pmatrix}
    1 \\ 
    1 \\
    1
  \end{pmatrix}=x_1+x_2+x_3=1$$
Hence $x^*$ is a best response to $y^*$. Due to symmetry (or by a similar calculation) you can also establish the converse.
