Step 1 :Given the matrix $A=\left[\begin{array}{ll}3 & 1 \\ 2 & 2\end{array}\right]$ and the eigenvector $v=\left[\begin{array}{c}1 \\ -2\end{array}\right]$, we can find the associated eigenvalue by using the property of eigenvectors and eigenvalues that $Av = \lambda v$.
Step 2 :Multiplying the matrix $A$ by the eigenvector $v$ gives $Av = \left[\begin{array}{c}1 \\ -2\end{array}\right]$.
Step 3 :Dividing $Av$ by $v$ gives $\lambda_v = \left[\begin{array}{c}1 \\ 1\end{array}\right]$.
Step 4 :Since the result of the division is $\left[\begin{array}{c}1 \\ 1\end{array}\right]$, this means that the eigenvalue associated with the eigenvector $v$ is 1.
Step 5 :Final Answer: The eigenvalue associated with the eigenvector $v$ is \(\boxed{1}\).