$$K=\ \begin{pmatrix} 0 & 0 & 1\\ -1 & 0 & 0\\ 0 & -1 & 0 \end{pmatrix}$$
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 $$ \operatorname{Span}\left\{\begin{bmatrix}1\\-1\\1\end{bmatrix}\right\} $$