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 
 A: $\newcommand{\e}{\mathbf{e}}$Your matrix $K$ cyclically permutes the vectors
$$
\e_{1} = (1, 0, 0),\quad
-\e_{2} = (0, -1, 0),\quad
\e_{3} = (0, 0, 1).
$$
It should be visually apparent where the axis lies. ;)

A: 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 $+1$.
A: 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 $(-1,1,-1)^T$.
A: I can offer a remark on how you could simplify in some sense the calculation, but I'm not sure whether you would call the final result "inspection" and not "calculation".
Sometimes it easier to do mental calculations with matrices by thinking of the result of matrix multiplication by a vector as a linear combination of the matrix rows. That is, 
$$ \begin{pmatrix}
a_{11} & a_{12} & a_{13} \\ 
a_{21} & a_{22} & a_{23} \\ 
a_{31} & a_{32} & a_{33}
\end{pmatrix} \cdot 
\begin{pmatrix}
x \\ 
y \\ 
z 
\end{pmatrix} = x \begin{pmatrix} a_{11} \\ a_{21} \\ a_{31} \end{pmatrix} + y \begin{pmatrix} a_{12} \\ a_{22} \\ a_{32} \end{pmatrix} + z \begin{pmatrix} a_{13} \\ a_{23} \\ a_{33} \end{pmatrix}. $$ 
The result of the multiplication is what you get if you multiply $x$ by the first row, add to it $y$ multiplied by the second row and then $z$ by the third row.
In your case, we have
$$ \begin{pmatrix}
0 & 0 & 1 \\ 
-1 & 0 & 0 \\ 
0 & -1 & 0
\end{pmatrix} \cdot 
\begin{pmatrix}
x \\ 
y \\ 
z 
\end{pmatrix} = x \begin{pmatrix} 0 \\ -1 \\ 0 \end{pmatrix} + y \begin{pmatrix} 0 \\ 0 \\ -1 \end{pmatrix} + z \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} z \\ -x \\ -y \end{pmatrix} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}. $$ 
After some practice, you can do this in your head and then note that the first coordinate of the result is going to be $z$ and you want $z = x$. So you can take $z = x = 1$. Then, the second coordinate should be $-x = y$, so you can take $y = -1$.
