## A reflection followed by a rotation is a reflection

In preparation for answering exercise 2.6.3 in Gilbert Strang’s Linear Algebra and Its Applications, Third Edition, I wanted to derive in detail the effect of a rotation followed by a rotation, a reflection followed by a reflection, a rotation followed by a reflection, and a reflection followed by a rotation. This post demonstrates that a reflection followed by a rotation is equivalent to a reflection.

In the last post we considered a matrix $Q_{\theta}$ that rotates vectors through an angle $\theta$ and a second matrix $H_{\varphi}$ that reflects vectors in the line through the origin with angle $\varphi$ (the $\varphi$-line). We showed that $H_\varphi Q_\theta = H_{\varphi-\frac{1}{2}\theta}$ so that the effect of applying a rotation of angle $\theta$ followed by a reflection through the $\varphi$-line is equivalent to a reflection through the $(\varphi - \frac{1}{2}\theta)$-line:

$H_\varphi Q_\theta= \begin{bmatrix} \cos 2\varphi&\sin 2\varphi \\ \sin 2\varphi&-\cos 2\varphi \end{bmatrix} \begin{bmatrix} \cos \theta&-\sin \theta \\ \sin \theta&\cos \theta \end{bmatrix}$

$= \begin{bmatrix} \cos 2(\varphi - \frac{1}{2}\theta)&\sin 2(\varphi - \frac{1}{2}\theta) \\ \sin 2(\varphi - \frac{1}{2}\theta)&-\cos 2(\varphi - \frac{1}{2}\theta)\end{bmatrix} = H_{\varphi-\frac{1}{2}\theta}$

Consider the reverse operation represented by the matrix $Q_\theta H_\varphi$ in which we first reflect vectors in the line through the origin with angle $\varphi$ (the $\varphi$-line) and then rotate vectors through an angle $\theta$. We have

$Q_\theta H_\varphi = \begin{bmatrix} \cos \theta&-\sin \theta \\ \sin \theta&\cos \theta \end{bmatrix} \begin{bmatrix} \cos 2\varphi&\sin 2\varphi \\ \sin 2\varphi&-\cos 2\varphi \end{bmatrix}$

$= \begin{bmatrix} \cos\theta\cos 2\varphi - \sin\theta\sin 2\varphi&\cos\theta\sin 2\varphi + \sin\theta\cos 2\varphi \\ \sin\theta\cos 2\varphi + \cos\theta\sin 2\varphi&\sin\theta\sin 2\varphi - \cos\theta\cos 2\varphi \end{bmatrix}$

We can simplify this using the trigonometric identities

$\begin{array}{rcl} \cos (\alpha + \beta)&=&\cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \sin (\alpha + \beta)&=&\sin \alpha \cos \beta + \cos \alpha \sin \beta \end{array}$

We then have

$Q_\theta H_\varphi = \begin{bmatrix} \cos\theta\cos 2\varphi-\sin\theta\sin 2\varphi&\sin\theta\cos 2\varphi+\cos\theta\sin 2\varphi \\ \sin\theta\cos 2\varphi+\cos\theta\sin 2\varphi&-(\cos\theta\cos 2\varphi-\sin\theta\sin 2\varphi) \end{bmatrix}$

$= \begin{bmatrix} \cos (\theta+2\varphi)&\sin (\theta+2\varphi) \\ \sin (\theta+2\varphi)&-\cos (\theta+2\varphi) \end{bmatrix}$

$= \begin{bmatrix}\cos 2(\varphi+\frac{1}{2}\theta)&\sin 2(\varphi + \frac{1}{2} \theta) \\ \sin 2(\varphi + \frac{1}{2} \theta)&-\cos 2(\varphi + \frac{1}{2} \theta \end{bmatrix} = H_{\varphi+\frac{1}{2}\theta}$

We have thus shown that $Q_\theta H_\varphi = H_{\varphi+\frac{1}{2}\theta}$ so that the effect of applying a reflection through the $\varphi$-line followed by a rotation of angle $\theta$ is equivalent to a reflection through the $(\varphi + \frac{1}{2}\theta)$-line.

If $\theta= 0$ then this equation reduces to $Q_0 H_\varphi = H_{\varphi+\frac{1}{2} \cdot 0}$ or $H_\varphi = H_\varphi$ since $Q_0 = I$.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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