Linear Algebra - seemingly incorrect result when looking for a basis 
For the following matrix $$ A = \begin{pmatrix}
    3 & 1 & 0 & 0 \\
    -2 & 0 & 0 & 0 \\
    -2 & -2 & 1 & 0 \\
    -9 & -9 & 0 & -3
\end{pmatrix} $$
Find a basis for the eigenspace $E_{\lambda}(A)$ of each eigen value. 

First step is to find the characteristic polynomial to find the eigenvalues. $$\det(A- I\lambda) = 0  \implies \lambda ^4 - \lambda^3 - 7 \lambda^2 + 13 \lambda -6  = 0 $$ $$\lambda = 1 \lor \lambda = 2 $$ Now that we have the eigen values, to find a basis for $E_{\lambda}(A)$ for the eigen values $\lambda$, we find vectors $\mathbf{v}$ that satisfy $(A-\lambda I)\mathbf{v} = 0$:
$$(A - 2I)\mathbf{v} = \begin{pmatrix}
    1 & 1 & 0 & 0 \\
    -2 & -2 & 0 & 0 \\
    -2 & -2 & -1 & 0 \\
    -9 & -9 & 0 & -5
\end{pmatrix}  \begin{pmatrix}
    v_1 \\
    v_2 \\
    v_3 \\
    v_4
\end{pmatrix} = \begin{pmatrix}
    0 \\
    0 \\
    0 \\
    0
\end{pmatrix} \implies
  \begin{cases}
    v_1 + v_2 = 0\\
    -2v_1 -2v_2  = 0 \\
    -2v_1 - 2v_2 -v_3 = 0 \\
    -9v_1 -9v_2 - 5v_4 = 0
  \end{cases}$$ $\implies v_2 = -v_1 \land v_3 = 0 \land v_4 = 0$. So $(v_1, v_2, v_3, v_4) = (v_1, -v_1, 0, 0) = v_1(1, -1, 0, 0)$ so $(1, -1, 0, 0)$ is the basis. 
However, this is apparently incorrect. What have I done wrong here? 
Also, for the eigenvalue $\lambda = 1$ I do get the correct answer, so only the part above gives me problems. 

If you calculate $\det(A-\lambda I)$ by the first row (and all the minors too) you get
$$
\det(A-\lambda I)=(3-\lambda)\lambda(1-\lambda)(-3-\lambda)+2(1-\lambda)(-3-\lambda)=(3+\lambda)(1-\lambda)(\lambda^2-3\lambda+2)
=(3+\lambda)(1-\lambda)(\lambda-1)(\lambda-2).
$$
So the eigenvalues are $1$ (with multiplicity $2$), $2$, and $-3$. 
The eigenvectors (and so the eigenspace) for $2$ you found correctly. You still need to find the eigenspaces for $1$ and for $-3$. 
