Find the eigenvectors of $ A = \begin{bmatrix} 0 & 2 \\ 1 & 1 \end{bmatrix} $. Find the eigenvectors of
$$
A = \begin{bmatrix} 0 & 2 \\ 1 & 1 \end{bmatrix}.
$$
I know you can solve $ \det(A - \lambda I) = 0 $ to find the eigenvalues of $ A $, but I keep getting no free variables. However, I thought this was impossible, but I know this problem works.
 A: $$
A-\lambda I=  \left[\begin{matrix}0 - \lambda &2  \\1 & 1-\lambda\end{matrix}\right]
$$
so your equation is
$$
(0-\lambda)(1-\lambda) - 2\times 1 = 0
$$
giving
$$
\lambda^2 - \lambda -2 =0
$$
or
$$
(\lambda +1)(\lambda-2)=0
$$
(thanks to @graydad for spotting an error)
A: If you solve for the eigenvalues you should get $\lambda = -1,2$. Now use these values to solve the equation $A\mathbf{v} = \lambda \mathbf{v}$ for $\mathbf{v}=\left[\begin{matrix}v_1  \\v_2 \end{matrix}\right]$. For example in the case of $\lambda = -1$ we want to solve $$A\mathbf{v} = - \mathbf{v} \\ \implies  \left[\begin{matrix}0&2  \\1 & 1\end{matrix}\right] \left[\begin{matrix}v_1  \\v_2 \end{matrix}\right] = \left[\begin{matrix}-v_1  \\-v_2 \end{matrix}\right]$$ Completing the matrix multiplication on the left yields a set of two equations. Namely, $$\begin{align}2v_2 = -v_1 \\ v_1+v_2 = -v_2\end{align}$$ Since any scalar multiple of a vector is still the same vector, we have some free reign in choosing $v_1$ and $v_2$, so long as the two equations above are satisfied for the pair of numbers chosen. I recommend sticking with integers for eigenvector entries. It makes the vector easier to plot ($v_2=1$ makes for an integer $v_1$) Once you have your pair you have one eigenvector. Repeat the process using $\lambda = 2$ to get your second eigenvector.
