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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