Producing a $G$-invariant form from the standard Hermitian product using the averaging process 
Problem statement: Let $G$ be a cyclic group of order $3$. The matrix $$A=\begin{bmatrix} -1 && -1 \\ 1 && 0 \end{bmatrix}$$ has order $3$, so it defines a matrix representation of $G$. Use the averaging process to produce a $G$-invariant form from the standard Hermitian product $\mathbf{x}^*\mathbf{y}$ on $\mathbb{C}^2$.

So I am working on this proof and I am still confused what exactly the problem statement means when it says "$G$-invariant form from the standard Hermitian product $\mathbf{x}^*\mathbf{y}$ on $\mathbb{C}^2$".
I understand that the form is defined by the inner product, but am I just supposed to find the $G$-invariant vectors in $\mathbb{C}^2$?
I know that $$A=\begin{bmatrix} -1 && -1 \\ 1 && 0 \end{bmatrix}$$
$$A^2=\begin{bmatrix} 0 && 1 \\ -1 && -1 \end{bmatrix}$$
and then we have eigenvectors for $A$ and $A^2$ respectively
$$\left\{\begin{bmatrix} -1 +\sqrt{3}i \\ 2\end{bmatrix}, \begin{bmatrix} -1 -\sqrt{3}i \\ 2\end{bmatrix}\right\}$$
$$\left\{\begin{bmatrix} 1 +\sqrt{3}i \\ -2\end{bmatrix}, \begin{bmatrix} 1 -\sqrt{3}i \\ -2\end{bmatrix}\right\}$$
So, clearly the space 
$$\mathrm{span}\left\{\begin{bmatrix} 1 +\sqrt{3}i \\ -2\end{bmatrix}\right\}$$ 
is $G$-invariant. I am confused how I use the averaging process to find my form? I don't understand how the sum
$$\langle \mathbf{x}, \mathbf{y}\rangle = \frac{1}{|G|}\sum_{g \in G}\{g\mathbf{x}, g\mathbf{y}\}$$ 
is interpreted in this formula. My book says $\{\ , \}$ represents a positive definite form, so $\mathbf{x}^TA\mathbf{y} > 0$, but I am just unsure where to go from here. In this case, we would have $$\langle \mathbf{x}, \mathbf{y}\rangle = \frac{1}{3}\sum_{i=1, A^i \in G}^{3}\{A^i\mathbf{x}, A^i\mathbf{y}\},$$ but I don't understand how to interpret this. Any suggestions are appreciated!
 A: Finding $G$-invariant vectors in not what you are asked to do. You are given a pairing $\{\cdot, \cdot\}$ on $\mathbb{C}^2$ (the standard hermitian form on $\mathbb{C}^2$) and you want to find a $G$-invariant pairing $\langle\cdot, \cdot\rangle$. In order to do this, you use the formula towards the end of your post.
We have a faithful two-dimensional complex representation $\rho : \mathbb{Z}_3 \to GL_2(\mathbb{C})$ generated by $1 \mapsto A$. So $\rho(0) = I$, $\rho(1) = A$, and $\rho(2) = A^2$; i.e. $\rho(i) = A^i$. With this representation at hand, we obtain $\varphi : \mathbb{Z}_3\times\mathbb{C}^2 \to \mathbb{C}^2$, a $\mathbb{Z}_3$ action on $\mathbb{C}^2$ given by $\varphi(i, v) = \rho(i)v$.
The correction interpretation of the formula is then
$$\langle x, y\rangle = \frac{1}{3}\sum_{i \in \mathbb{Z}_3}\{\rho(i)x, \rho(i)y\} = \frac{1}{3}\sum_{i \in \mathbb{Z}_3}\{A^ix, A^iy\} = \frac{1}{3}\sum_{i=0}^2\{A^ix, A^iy\}.$$
Now note that $\{x, y\} = x^*y = \bar{x}^Ty$, so 
$$\langle x, y\rangle = \frac{1}{3}\sum_{i=0}^2\{A^ix, A^iy\} = \frac{1}{3}\sum_{i=0}^2\overline{A^ix}^TA^iy = \frac{1}{3}\sum_{i=0}^2\bar{x}^T(A^i)^TA^iy = \bar{x}^T\left(\frac{1}{3}\sum_{i=0}^2 (A^i)^TA^i\right)y.$$
So $\langle x, y\rangle = \bar{x}^TBy$ where $B = \frac{1}{3}(I + A^TA + (A^2)^TA^2)$.
