Matrix for reflection about the line $y = \tan (\theta) \, x$ How would I show that a reflection about the line $y = \tan (\theta) \, x$ is the following?
\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}
 A: Show that$$\begin{bmatrix}\cos(2\theta)&\sin(2\theta)\\ \sin(2\theta)&-\cos(2\theta)\end{bmatrix}.\begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix}=\begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix}$$and that$$\begin{bmatrix}\cos(2\theta)&\sin(2\theta)\\ \sin(2\theta)&-\cos(2\theta)\end{bmatrix}.\begin{bmatrix}-\sin\theta\\\cos\theta\end{bmatrix}=\begin{bmatrix}\sin\theta\\-\cos\theta\end{bmatrix}=-\begin{bmatrix}-\sin\theta\\\cos\theta\end{bmatrix}.$$Besides, note that $\left[\begin{smallmatrix}\cos\theta\\\sin\theta\end{smallmatrix}\right]$ belongs to the line $y=\tan(\theta)x$ and that $\left[\begin{smallmatrix}\cos\theta\\\sin\theta\end{smallmatrix}\right]$ and $\left[\begin{smallmatrix}-\sin\theta\\\cos\theta\end{smallmatrix}\right]$ are orthogonal.
A: HINT
Consider the line $y = mx$. Since reflexions are linear transformations, we can proceed as follows.
Given the basis $\mathcal{B} = \{(1,m),(-m,1)\}$, the matricial representation of such reflection is given by:
\begin{align*}
[T]_{\mathcal{B}} =
\begin{bmatrix}
1 & 0\\
0 & -1
\end{bmatrix}
\end{align*}
which can be rewrriten in terms of the standard basis $\mathcal{B}' = \{(1,0),(0,1)\}$ as
\begin{align*}
[T]_{\mathcal{B}'} = [I]_{\mathcal{B}}^{\mathcal{B}'}[T]_{\mathcal{B}}[I]_{\mathcal{B}'}^{\mathcal{B}}
\end{align*}
Having said that, can you take it from here?
A: The transformation\begin{pmatrix} \cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{pmatrix}
maps any arbitrary vector $P=\begin{pmatrix}\alpha\\\beta\end{pmatrix}\in\mathbb R^2$ to the vector
$P'=\begin{pmatrix}\alpha\cos2\theta+\beta\sin2\theta\\\alpha\sin2\theta-\beta\cos2\theta\end{pmatrix}$.
It is easy to verify that the mid-point of line joining vectors $P$ and $P'$ i.e.
$M=\begin{pmatrix}\alpha\cos^2\theta+\beta\sin\theta\cos\theta\\\alpha\sin\theta\cos\theta+\beta\sin^2\theta\end{pmatrix}$
always lie on the line $y\cos\theta=x\sin\theta.$
