Question about Eigenspaces 
The last part is something I don't understand. 
The equation should be $x_1-x_3=0$ right? Therefore, $x_1=x_3$
I know how to write the eigenspace and find the basis if there is only one free variable in a 3x3 matrix. I don't understand what happened here though. Some help required :)
 A: 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.
