Find the Axis of rotation of rotation matrix $K$ after solving $(K-I)v=0$ $$K=\
\begin{pmatrix}
0 & 0 & 1\\ 
-1 & 0 & 0\\ 
0 & -1 & 0
\end{pmatrix}$$
Find the axis of rotation for the rotation matrix $K$. 
This is from my previous thread click here to view it
Noting that the axis of rotation consists of vectors that remain unmoved. That is a vector $v$ satisfying  $Kv = v$. Or, $Kv - Iv=0$ where $I$ is the $3\times3$ identity matrix. For matrix $K$ after solving the homogeneous equations given by $(K-I)v=0$ and showing the working: 
$(K-I)v=0$
So
$$K-I=\
\begin{pmatrix}
0 & 0 & 1\\ 
-1 & 0 & 0\\ 
0 & -1 & 0
\end{pmatrix}-\begin{pmatrix}
1 & 0 & 0\\ 
0 & 1 & 0\\ 
0 & 0 & 1
\end{pmatrix}=\
\begin{pmatrix}
-1 & 0 & 1\\ 
-1 & -1 & 0\\ 
0 & -1 & -1
\end{pmatrix}$$ 
therefore 
$$\begin{pmatrix}
-1 & 0 & 1\\ 
-1 & -1 & 0\\ 
0 & -1 & -1
\end{pmatrix}v=0$$
writing out the components for $v$ gives
$$\begin{pmatrix}
-1 & 0 & 1\\ 
-1 & -1 & 0\\ 
0 & -1 & -1
\end{pmatrix}\begin{pmatrix}
x \\ 
y \\ 
z 
\end{pmatrix}=0$$
Multiplying out gives three equations
$-x+z=0$
$-x-y=0$
$-y-z=0$
Now that i've done everything P Vanchinathan told me to do in the last thread, the question is how do i get the rotation axis from those 3 equations? As this method is time consuming, more importantly how would I get the answer by inspection?
Thank you to all that help and especially to P Vanchinathan in the last thread as if it wasn't for him I wouldn't have made it this far.
Regards, BLAZE

You found that
$$
v=\begin{bmatrix}x\\y\\z\end{bmatrix}
$$
where
\begin{align*}
z&= x\\
y&=-x
\end{align*}
Hence the axis of rotation is given by the line
$$
\begin{bmatrix}
x\\-x\\x
\end{bmatrix}=x\begin{bmatrix}1\\-1\\1\end{bmatrix}\quad x\in\Bbb R
$$
That is, the axis of rotation is
$$
\operatorname{Span}\left\{\begin{bmatrix}1\\-1\\1\end{bmatrix}\right\}
$$
