matricesreflection
I have been looking into matrix transformations and found the following matrix to reflect about the line $y=(\tan\theta)x$.
$$R = \begin{bmatrix}
\cos(2\theta)& \sin(2\theta)\\
\sin(2\theta)& -\cos(2\theta)\\
\end{bmatrix}$$
However, is there a general matrix to reflect about the line $y=mx+c$?
On the part about $uu^T$, your error is that $u$ needs to be considered a column vector, not a row vector.
As for the part about $2P - I$, you seem to have generally the right idea, but you're not executing it properly. If you start with a vector $v$, its reflection $Rv$ is symmetric to $v$ with respect to the projection $Pv$. This means that the vector from $v$ to $Rv$, which is $Rv - v$, is twice the vector from $v$ to $Pv$, which is $Pv - v$. Can you finish the calculation from there?
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?
Best Answer
Matrices use for linear transformation. Any linear transformation keep the origin fixed. So only the reflection in lines that pass the origin make linear transformations and has matrix representation. Therefore $c$ must be $0$ and $y=mx$. Hence just set $m=\tan\theta$ to get answer.