<s>[INST] 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 
 [/INST] Edit. For any $3\times3$ rotation matrix $K$, if it is not symmetric, you can read off the rotation axis directly from its skew-symmetric part $W=K-K^T$. More specifically, the rotation axis is parallel to $(w_{32}, w_{13}, w_{21})^T=-(w_{23}, w_{31}, w_{12})^T$.
Reason: any anticlockwise rotation for an angle $\theta$ about a unit vector $\mathbf u=(x,y,z)^T$ can be put into axis-angle form:
$$
K=\begin{bmatrix}
\cos\theta + x^2(1-\cos\theta) &xy(1-\cos\theta) - z\sin\theta &xz(1-\cos\theta) + y\sin\theta\\
yx(1-\cos\theta) + z\sin\theta &\cos\theta + y^2(1-\cos\theta) & yz(1-\cos\theta) - x\sin\theta\\
zx(1-\cos\theta) - y\sin\theta &zy(1-\cos\theta) + x\sin\theta & \cos\theta + z^2(1-\cos\theta)
\end{bmatrix}.
$$
When it is not symmetric, $\sin\theta\ne0$. Hence the skew-symmetric part of $K$ is given by
$$
W=K-K^T=2\sin\theta\,\begin{bmatrix}
0&-z&y\\ 
z&0&-x\\ 
-y&x&0
\end{bmatrix}.
$$
Thus $(w_{32}, w_{13}, w_{21})=2(\sin\theta)(x,y,z)=-(w_{23}, w_{31}, w_{12})$.
In your case,
$$
W = K-K^T = 
\begin{pmatrix}
\ast & \ast & 1\\ 
-1 & \ast & \ast\\ 
\ast & -1 & \ast
\end{pmatrix}.
$$
Therefore the rotation axis is parallel to $(-1,1,-1)^T$.
</s>