Show lambda is an eigenvalue of A and find one eigenvector Show that $\lambda$ is an eigenvalue of A and find one eigenvector corresponding to this eigenvalue. 
$\lambda$ = -1
A = \begin{pmatrix}
    1    & 0  & 2 \\
    -1    & 1  & 1 \\  
    2    & 0  & 1
\end{pmatrix}
I subtract \begin{pmatrix}
    1    & 0  & 2 \\
    -1    & 1  & 1 \\  
    2    & 0  & 1
\end{pmatrix} - \begin{pmatrix}
    -1    & 0  & 0 \\
    0    & -1  & 0 \\  
    0    & 0  & -1
\end{pmatrix}
To get \begin{pmatrix}
    2    & 0  & 2 \\
    -1    & 2  & 1 \\  
    2    & 0  & 2
\end{pmatrix}
Finding the null space of that results in:
\begin{pmatrix}
    1    & 0  & 1 \\
    0    & 1  & 1 \\  
    0    & 0  & 0
\end{pmatrix}
So
x1 +  x3 = 0
x2 +  x3 = 0

x1  = - x3
x2  = - x3

I'm confused on the value of x3 here, I understand that the zero row means it's a free variable but what does that do in terms of our eigenvector?
\begin{pmatrix}
-x3 \\
-x3 \\  
x3
\end{pmatrix}

Any (nonzero!) value of $x_3$ will give you an acceptable eigenvector; so $(-1,-1,1)$ is valid, for example. In general, if $v$ is an eigenvector, then so too is $\lambda v$, $\lambda \in \mathbb{R}$ (try prove it!).  
