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

$\newcommand{\e}{\mathbf{e}}$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. ;)

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 $+1$.