What is the transformation matrix that I multiply a point by if I want to reflect that point across a line that goes through the origin in terms of the angle between the line and the x-axis?
In other words,
$$y = mx$$
$\theta$ is the angle between the $x$-axis and the line.
The position vector $P=\begin{bmatrix}a\\b\\\end{bmatrix}$
is a point on the same plane as the line.
The vector $P'$ is $P$ reflected across $y=mx$.
What matrix do I multiply $P$ by to get $P'$, in terms of $\theta$?
I've looked online and found
$$\begin{bmatrix}
\cos(2\theta) & \sin(2\theta) \\
\sin(2\theta) & -\cos(2\theta) \\
\end{bmatrix}
$$but I tested it and it doesn't work for me.
Also, examples with the vector
$$\begin{bmatrix}2\\1\\\end{bmatrix}$$
and the line $y = \frac 43x$
would help to stay consistent with my tests.
Best Answer
The correct answer is the one for which you said "I tested it and it doesn't work for me," namely
$$\begin{bmatrix} \cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta) \\ \end{bmatrix} $$
Here is a diagram for your example.
The line is $y=\frac 43x$, which has the angle of inclination $\theta\approx 53.1301023542°$ with $\cos(2\theta)=-0.28$ and $\sin(2\theta)=0.96$, and the points before and after reflection are
$$P=\begin{bmatrix}2\\1\\\end{bmatrix}, \quad P'=\begin{bmatrix}0.4\\2.2\\\end{bmatrix} $$
(The line and point $P$ were entered in Geogebra: the angle, cosine, sine, and point $P'$ were calculated by Geogebra.) The matrix calculation is then
$$ \begin{bmatrix}\cos(2\theta)&\sin(2\theta)\\\sin(2\theta)&-\cos(2\theta)\\\end{bmatrix}\cdot P =\begin{bmatrix}-0.28&0.96\\0.96&0.28\\\end{bmatrix}\cdot \begin{bmatrix}2\\1\\\end{bmatrix} =\begin{bmatrix}0.4\\2.2\\\end{bmatrix} =P' $$
so it all works out.
Do you want a derivation of that transformation matrix? And what tests did you try that did not work for you?