Reflection matrix:
$$ \text{Reflection}(\theta) = \begin{bmatrix} \cos2\theta & \sin2\theta \\ \sin2\theta & -\cos2\theta \end{bmatrix}$$
Attempt:
Inspiration:Speaking non-rigorously, it seems like the angle between the reflected vector and the original vector will be $2\theta$. Armed with this, let's consider how $e_1 = \begin{bmatrix}1\\0\end{bmatrix}$ and $e_2 = \begin{bmatrix}0\\1\end{bmatrix}$ change when we reflect them across an arbitrary line.
Let $$ \text{Reflection}(\theta) = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$
Then,
$$\begin{align} \text{Reflection}(\theta) \cdot e_1 &= \begin{bmatrix} a & b \\ c & d \end{bmatrix} \cdot \begin{bmatrix}1\\0\end{bmatrix} \\ &= \begin{bmatrix}a\\c\end{bmatrix} \end{align}$$
Using my assumption that reflected vectors have angle of $2\theta$ between itself and the original vector, $(\text{Reflection}(\theta)\cdot e_1) \cdot e_1 = \begin{bmatrix}a\\c\end{bmatrix}\cdot \begin{bmatrix}1\\0\end{bmatrix} = a = ||\text{Reflection}(\theta)\cdot e_1||\cdot||e_1|| \cos(2\theta)$ (dot product).
Simplifying the other side of the equation, we get:$$a = 1\cdot 1 \cos(2\theta) = \cos(2\theta)$$
Doing it similarly for $e_2$ yields $d=\cos(2\theta)$ which means our reflection matrix currently looks like this:
$$ \text{Reflection}(\theta) = \begin{bmatrix} \cos2\theta & b \\ c & \cos2\theta \end{bmatrix}$$
which is not correct.
Questions:
- I suspect my implicit assumption that a reflection can be represented as a rotation of $2\theta$, where $\theta$ is the angle between the original vector and reflection line, is where I went wrong. Why is this wrong?
$$ \text{Rotation}(2\theta) = \begin{bmatrix} \cos2\theta & -\sin2\theta \\ \sin2\theta & \cos2\theta \end{bmatrix} \text{ for reference. }$$
- What's the correct way to do this?
I would normally ask one question per SE question but I think my two questions are tightly coupled.
$\endgroup$2 Answers
$\begingroup$The problem with your attempted derivation is that you keep changing the definition of $\theta$ since "the original vector" can be any input to the linear transformation.
Presumably, the $\theta$ in the definition of the reflection is such that the axis of reflection is along the vector $(\cos \theta, \sin \theta)$, which is to say that $\theta$ is the angle between the axis of rotation and the vector $e_1$. With that in mind, your derivation of the entry $a$ is fine, but your derivation of the entry $d$ is unjustified and ultimately incorrect.
Here's a geometric approach involving rotations that you might like. Let $F_\theta$ denote the matrix of the reflection through the axis at angle $\theta$, and let $R_\theta$ denote the matrix of the counterclockwise rotation by $\theta$. Note that the matrix of the reflection through the axis at $\theta = 0$ is given by$$ F_0 = \pmatrix{1&0\\0&-1}. $$On the other hand, we can implement a reflection through the axis at angle $\theta$ by first rotating that axis to $\theta = 0$, then applying $F_0$, then rotating the axis back. In other words, we have$$ F_\theta = R_{\theta} F_0 R_{-\theta} = \pmatrix{\cos \theta & - \sin \theta\\ \sin \theta & \cos \theta} \pmatrix{1&0\\0&-1} \pmatrix{\cos \theta & \sin \theta \\ - \sin \theta & \cos \theta}. $$After multiplying and simplifying with the double-angle laws, you should indeed find that$$ F_\theta = \pmatrix{\cos 2 \theta & \sin 2 \theta \\ \sin 2 \theta & - \cos 2 \theta}. $$
$\endgroup$ 3 $\begingroup$The angle parameterising a rotation is not how far a line moves when reflected, but the tilt of the line in which reflection occurs. The line $\ell$ of equation $y=x\tan\theta$ sends $p:=\binom{a}{b}$ to a point $p^\prime:=\binom{a^\prime}{b^\prime}$ for which $\vec{pp^\prime}$ is orthogonal to $\ell$ and hence of gradient $-\cot\theta$, so $\vec{pp^\prime}$ has equation $y-b=(a-x)\cot\theta$. The intersection with $\ell$ solves $x(\tan\theta+\cot\theta)=a\cot\theta+b$, so has $x$-coordinate $\cos^2\theta(a+b\tan\theta)$ and $y$-coordinate $\sin\theta\cos\theta(a+b\tan\theta)$. So $$a^\prime=2\cos^2\theta(a+b\tan\theta)-a=a\cos2\theta+b\sin2\theta,\\b^\prime=2\sin\theta\cos\theta(a+b\tan\theta)-b=a\sin2\theta-b\cos2\theta.$$This agrees with the reflection matrix here.
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