Consider the linear transformation $\ltdefn{T}{M_{22}}{M_{22}}$ defined by \begin{equation*} \lt{T}{\begin{bmatrix}a&b\\c&d\end{bmatrix}} = \begin{bmatrix} -17a+11b+8c-11d & -57a+35b+24c-33d \\ -14a+10b+6c-10d & -41a+25b+16c-23d \end{bmatrix} \end{equation*} and the vectors \begin{align*} \vect{x}_1 &= \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix} & \vect{x}_2 &= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} & \vect{x}_3 &= \begin{bmatrix} 1 & 3 \\ 2 & 3 \end{bmatrix} & \vect{x}_4 &= \begin{bmatrix} 2 & 6 \\ 1 & 4 \end{bmatrix} \end{align*}
Then compute \begin{align*} \lt{T}{\vect{x}_1} &= \lt{T}{\begin{bmatrix} 0 & 1 \\ 0 & 1\end{bmatrix}} = \begin{bmatrix} 0 & 2 \\ 0 & 2 \end{bmatrix} = 2\vect{x}_1\\ \lt{T}{\vect{x}_2} &= \lt{T}{\begin{bmatrix} 1 & 1 \\ 1 & 0\end{bmatrix}} = \begin{bmatrix} 2 & 2 \\ 2 & 0 \end{bmatrix} = 2\vect{x}_2\\ \lt{T}{\vect{x}_3} &= \lt{T}{\begin{bmatrix} 1 & 3 \\ 2 & 3\end{bmatrix}} = \begin{bmatrix} -1 & -3 \\ -2 & -3 \end{bmatrix} = (-1)\vect{x}_3\\ \lt{T}{\vect{x}_4} &= \lt{T}{\begin{bmatrix} 2 & 6 \\ 1 & 4\end{bmatrix}} = \begin{bmatrix} -4 & -12 \\ -2 & -8 \end{bmatrix} = (-2)\vect{x}_4\\ \end{align*}
So $\vect{x}_1$, $\vect{x}_2$, $\vect{x}_3$, $\vect{x}_4$ are eigenvectors of $T$ with eigenvalues (respectively) $\lambda_1=2$, $\lambda_2=2$, $\lambda_3=-1$, $\lambda_4=-2$.