## 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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