## A rotation followed by a reflection 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 reflection followed by a rotation, and a rotation followed by a reflection. This post demonstrates that a rotation followed by a reflection is equivalent to a reflection.

Assume that we have a matrix that rotates vectors through an angle $\theta$ and a second matrix that reflects vectors in the line through the origin with angle $\varphi$ (the $\varphi$-line). What is the effect of applying both matrices in succession?

As we did with a reflection followed by a reflection, we can try to gain an intuition about the problem by playing with a cut-out paper triangle on a piece of paper marked with axes. After trying this you should find that a rotation followed by a reflection will apparently cause the triangle to be reflected from its original position. (One hint here is that the triangle will end up as a mirror image of the original triangle.)

If this is indeed the case then the effect of a rotation through the angle $\theta$ and a reflection in the $\varphi$-line can be represented by a matrix $Q_\omega = \begin{bmatrix} \cos 2\omega&\sin 2\omega \\ \sin 2\omega&-\cos 2\omega \end{bmatrix}$

for a line of reflection through the origin with some angle $\omega$. (Here we use the expression for a reflection matrix derived in the last post.)

How does the angle $\omega$ relate to the angles $\theta$ and $\varphi$? Since we are dealing with linear transformations the simplest assumption is that $\omega = a\theta + b\varphi$ for some $a$ and $b$ where $a$ and $b$ are the same for all $\theta$ and $\varphi$.

To determine the values of $a$ and $b$ let’s look at the unit vector $(1, 0)$ and rotate it and reflect it. In the first case we rotate the vector $90^{\circ}$ and then reflect it in the $135^{\circ}$-line (the line for which $y = -x$). The rotation takes $(1,0)$ to $(0,1)$ and the reflection takes $(0,1)$ to $(-1, 0)$. The corresponding line of reflection is  the $90^{\circ}$-line (the $y$-axis).

In the second case we rotate vector $(1,0)$ $180^{\circ}$ and then reflect it in the $45^{\circ}$-line (the line for which $y = x$). The rotation takes $(1,0)$ to $(-1,0)$ and the reflection takes $(-1,0)$ to $(0,-1)$. The corresponding line of reflection is  the $325^{\circ}$-line (equivalent to the $-45^{\circ}$-line).

If the angle of rotation $\omega = a\theta + b\varphi$ then in the first case we would have $90 = 90a + 135b$ and in the second case we would have $-45 = 180a + 45b$. Subtracting 2 times the first equation from the second we have $-225 = -225b$ or $b = 1$. Substituting the value of $b$ into the first equation we have $90 = 90a + 135$ or $a = -\frac{1}{2}$.

Our hypothesis is therefore that doing a rotation through the angle $\theta$ and then a reflection through the $\varphi$-line would produce a reflection through the line of angle $\omega = -\frac{1}{2}\theta + \varphi = \varphi -\frac{1}{2}\theta$. If this is the case then the matrix representing the reflection would be $H_{\varphi-\frac{1}{2}\theta} = \begin{bmatrix} \cos 2(\varphi-\frac{1}{2}\theta)&\sin 2(\varphi-\frac{1}{2}\theta) \\ \sin 2(\varphi-\frac{1}{2}\theta)&-\cos 2(\varphi-\frac{1}{2}\theta) \end{bmatrix}$ $= \begin{bmatrix} \cos (2\varphi-\theta)&\sin (2\varphi-\theta) \\ \sin (2\varphi-\theta)&-\cos (2\varphi-\theta) \end{bmatrix}$

and this matrix should be equal to the product of the matrices corresponding to the rotation and the reflection: $H_\varphi Q_\theta = \begin{bmatrix} \cos 2\varphi&\sin 2\varphi \\ \sin 2\varphi&-\cos 2\varphi \end{bmatrix} \begin{bmatrix} \cos \theta&-\sin \theta \\ \sin \theta&\cos \theta \end{bmatrix}$ $= \begin{bmatrix} \cos 2\varphi \cos \theta + \sin 2\varphi \sin \theta&-\cos 2\varphi \sin \theta + \sin 2\varphi \cos \theta \\ \sin 2\varphi \cos \theta - \cos 2\varphi \sin \theta&-\sin 2\varphi \sin \theta - \cos 2\varphi \cos \theta \end{bmatrix}$

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

We have therefore proved what we set out to prove, 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.

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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