# Show that a reflection matrix is given by $\begin{bmatrix}\cos2\theta&\sin2\theta \\ \sin2\theta&-\cos2\theta\end{bmatrix}$

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:

1. 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?

$$\text{Rotation}(2\theta) = \begin{bmatrix} \cos2\theta & -\sin2\theta \\ \sin2\theta & \cos2\theta \end{bmatrix} \text{ for reference. }$$

1. 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.

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 $$F_\theta = \pmatrix{\cos 2 \theta & \sin 2 \theta \\ \sin 2 \theta & - \cos 2 \theta}.$$

• Hey @Omnomnomnom, can you elaborate more on why entry $d$ would be wrong? I was imagining the same reflecting line for both basic vectors (at least in my head). – Darius Oct 20 '19 at 9:27
• @Darius my point against your explanation is that while $e_1$ moves through an angle $\theta$, $e_2$ moves an amount corresponding to its angle of separation from the axis, which would be $90^\circ - \theta$. Really, though, I can't be sure how exactly you got your entry $d$ because all you said is that you "did it similarly". – Ben Grossmann Oct 20 '19 at 9:35
• I see my mistake now. I was under the impression that $\text{Reflection}(\theta)$ is how much I want to "rotate" my point ($e_2$ in this case) when $\theta$ is only referring to the angle that reflecting line makes with the $x$-axis. +1 for your explanation of where I went wrong and providing a nice solution that I like! – Darius Oct 20 '19 at 9:46

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.

• +1 for pointing out my error ("a rotation is not how far a line moves when reflected, but the tilt of the line in which reflection occurs") and giving an example method to prove :) – Darius Oct 20 '19 at 9:25