Find the axis of rotation of a rotation matrix by inspection (NOT by solving $Kv=v$) $$K=\
\begin{pmatrix}
0 & 0 & 1\\ 
-1 & 0 & 0\\ 
0 & -1 & 0
\end{pmatrix}$$
Find the axis of rotation for the rotation matrix $K$ by INSPECTION. 
This is from my other thread click here to view it
Everything you see below is me finding the axis of rotation by solving $Kv=v$. Just to show you how much working it requires:
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
$-x+z=0$
$-x-y=0$
$-y-z=0$
Since
$$
v=\begin{bmatrix}x\\y\\z\end{bmatrix}
$$
Here's the solution parametrically in terms of $x$
\begin{align*}
z&= x\\
y&=-x\\
x&=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
$$
\operatorname{Span}\left\{\begin{bmatrix}1\\-1\\1\end{bmatrix}\right\}
$$
As you can see this was a lot of work so i would be so grateful if someone could please explain in simple english how to get the answer: $$
\operatorname{Span}\left\{\begin{bmatrix}1\\-1\\1\end{bmatrix}\right\}
$$ 
by using Inspection? 
Many thanks to all that helped so far particularly Brian Fitzpatrick in the last thread 

Neglect the sign for the moment and think of 
$$
K=\
\begin{pmatrix}
0 & 0 & 1\\ 
-1 & 0 & 0\\ 
0 & -1 & 0
\end{pmatrix}
$$
as a permutation matrix: $(1\to 3), (2\to 1), (3\to 2)$ or shorter $(312)$. So you permute your coordinate axes. This points to a rotation axis along one of the vectors $(\pm 1,\pm 1,\pm 1)^T$. Apply this set of vectors on $K$ to get
$$
\begin{pmatrix}
0 & 0 & 1\\ 
-1 & 0 & 0\\ 
0 & -1 & 0
\end{pmatrix}
\begin{pmatrix}
\color{red}{\pm 1} \\ \color{blue}{\pm 1} \\ \pm 1 
\end{pmatrix}=
\begin{pmatrix}
\pm 1 \\ \color{red}{\mp 1} \\ \color{blue}{\mp 1}
\end{pmatrix}
$$
So you choose $\color{red}{+ 1},\color{blue}{- 1}$ and $+1$.
