Finding a basis for the eigenspace is the same as finding a basis for a null space of $\left[ \matrix{ -1&0&1\cr 3&0&-3\cr 1&0&1 }\right] $.
So, we find the solution set of the equation $\tag{1}\left[ \matrix{ 1&0&-1\cr 0&0&0\cr 0&0&0 }\right] {\bf x}={\bf 0} $
There are two free variables (corresponding to the columns that don't have a "leading row entry"):
$x_3$ is free, so set $x_3$ equal to an arbitrary value, say $s$. $x_2$ is free, so set $x_2$ equal to an arbitrary value, say $t$. The first row of the matrix in (1) will tell you $x_1=x_3=s$.
So the general solution to (1) is ${\bf x}=\left[ \matrix {s\cr t\cr s }\right]$, where $s$ and $t$ are arbitrary scalars.
Two free variables tells you the dimension of the eigenspace is 2. To find a basis for the eigenspace, we need to find two independent eigenvectors. We may do so by "spliting up the vector $\bf x$ into parts" (as the sum of the "$s$ part" and the "t part"): $ {\bf x}= \left[ \matrix {s\cr t\cr s }\right]=\left[ \matrix {s\cr 0\cr s }\right] +\left[ \matrix {0\cr t\cr 0}\right]= s\left[ \matrix {1\cr 0\cr 1 }\right] +t\left[ \matrix {0\cr 1\cr 0}\right]. $ The vectors $\left[ \matrix {1\cr 0\cr 1 }\right]$ and $\left[ \matrix {0\cr 1\cr 0}\right]$ form a basis for the eigenspace.