Matrix for reflection about the line $y = \tan (\theta) \, x$
By Sebastian Wright
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}
$\endgroup$ 12 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$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?
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