Finding the span for an eigen space I have matrix $A=\begin{bmatrix} 1&&-3&&3 \\ 3&& -5&&3 \\ 3&&-3&&1 \end{bmatrix}$
I got $x^3 +3x^2-4 =0$ for the characteristic polynomial for which I got $x=-2,-2,1$ 
I have to write the eigenspace associated with each eigenvalue as span.
My solution: For the first I put $x=-2$ in the characteristic polynomial and performed row reduction to get $\begin{bmatrix} 1&&-1&&1 \\ 0&&0&&0 \\ 0&&0&&0 \end{bmatrix}$. How do I proceed from here?
I know that the eigenspace is simply the eigenvectors associated with a particular eigenvalue. 
 A: $\begin{bmatrix} 1&&-1&&1 \\ 0&&0&&0 \\ 0&&0&&0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}$
Find a basis to the solution of linear system above. 
Method $1$:
You can do it as follows:
Let the $x_2=s, x_3=t$.
Then we have $x_1=s-t$
Hence $\begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix}=sv_1 + tv_2 $ for some vector $v_1$ and $v_2$.
Can you find vector $v_1$ and $v_2$?
Method $2$: 
Suppose $x_2=0$ and $x_3=1$, solve for $x_1$. Suppose $x_3=0$ and $x_2=1$ solve for $x_1$. Check whether the two solutions are linearly independent.
A: Like for any linear system: the eigenspace  $E_2$ for the eigenvalue $2$ is defined by the sole equation $\; x-y+z=0$, hence it has dimension $2$. As the  geometric multiplicities are equal to the algebraic multiplicities, the matrix is diagonalisable in a basis of eigenvectors.
One has to find two linearly independent eigenvectors in $E_2$. We  deduce  from the equation of $E_2$ an isomorphism :
$\begin{aligned}[t]
\mathbf R^2&\longrightarrow E_2\\
(x,y)&\longmapsto (x, y, y-x)
\end{aligned}$
Starting from the canonical basis of $\mathbf R^2$: $e_1=(1,0)$, $e_2=(0,1)$, we obtain the following basis of $E_2$:
$$e'_1=(1,0,-1), \quad e'_2=(0,1,1).$$
