Basis of eigenspace For eigenvalue $1$ I did $(A-I)x=0$ which is 
$$\left(\begin{array}{ccc}
0 & 1 & 0 & 0\\
0 & 0 & 0 & 0\\
2 & 2 & 1 & 0\\
-1 & 1 & -1 & 1
\end{array}\right).$$
I got row reduced echelon form 
$$\left(\begin{array}{ccc}
1 & 0 & 0 & 1/3\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & -2/3\\
0 & 0 & 0 & 0
\end{array}\right).$$  
I'm a little confused with the free variables that I have and how to find a basis for this eigenspace. 
If I set first column $x$, second $y$, third $z$ and fourth $w$:
$x = -1/3t$
$y = s$
$z = 2/3 t$
$w = t$
Is this correct?
 A: No, I don't think that is correct. Reading the third line you get
$$1z-\frac23w=0\implies z=\frac23w$$
And then second line:
$$y=0$$
And finally first line
$$x+\frac13w=0\implies x=-\frac13w$$
The general solution is the line
$$\left\{\left(-\frac13w\,,\,\,0\,,\,\,\frac23w\,,\,w\right)\;:\;\;w\in\Bbb R\right\}=\left\{\left(-w\,,\,0\,,\,2w\,,\,3w\right)\;;\;w\in\Bbb R\right\}=$$
In general, after reducing, we better read from bottom to top.
A: Your reduced row echelon matrix is not correct.  It should be
$$RREF(A-I) = \left[\begin{array}{cccc}1 & 0 & 0 & 1\\0 & 1 & 0 & 0\\0 & 0 & 1 & -2\\0 & 0 & 0 & 0\end{array}\right]$$
Now to see how to find the eigenspace from this let's rewrite this matrix as a set of linear equations:
$$\left[\begin{array}{cccc}
1 & 0 & 0 & 1\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & -2\\
0 & 0 & 0 & 0
\end{array}\right] \implies \begin{cases} x+w=0 \\ y=0 \\ z-2w=0 \\ 0=0\end{cases}$$
The only column without a pivot position is the $4^{th}$ column, so there's only $1$ free variable.  If we call the $4^{th}$ variable $w$, then we first start by setting $w=t$.  Then we see that the solutions are of the form $$\begin{bmatrix} x \\ y \\ z \\ w\end{bmatrix} = \begin{bmatrix}-t \\0 \\2t \\ t\end{bmatrix} = t\begin{bmatrix}-1 \\0 \\2 \\ 1\end{bmatrix}$$
for any $t\in \Bbb R$.
