## Linear Algebra and Its Applications, exercise 1.4.22

Exercise 1.4.22. The x-y plane can be rotated through an angle $\theta$ by the following matrix: $A(\theta) = \begin{bmatrix} \cos \theta&-\sin \theta \\ \sin \theta&\cos \theta \end{bmatrix}$

1. Show that $A(\theta_1) A(\theta_2) = A(\theta_1 + \theta_2)$. Hint: use the identities for $\cos(\theta_1 + \theta_2)$ and $\sin(\theta_1 + \theta_2)$.
2. Compute $A(\theta) A(-\theta)$. $A(\theta_1) A(\theta_2) = \begin{bmatrix} \cos \theta_1&-\sin \theta_1 \\ \sin \theta_1&\cos \theta_1 \end{bmatrix} \begin{bmatrix} \cos \theta_2&-\sin \theta_2 \\ \sin \theta_2 &\cos \theta_2 \end{bmatrix}$ $\quad = \begin{bmatrix} \cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2&-\cos \theta_1 \sin \theta_2 - \sin \theta_1 \cos \theta_2 \\ \sin \theta_1 \cos \theta_2 + \cos \theta_1 \sin \theta_2&-\sin \theta_1 \sin \theta_2 + \cos \theta_1 \cos \theta_2 \end{bmatrix}$

We can simplify the final matrix using the following formulas for the sine and cosine: $\sin (x + y) = \sin x \cos y + \cos x \sin y$ $\sin (x - y) = \sin x \cos y - \cos x \sin y$ $\cos (x + y) = \cos x \cos y - \sin x \sin y$ $\cos (x - y) = \cos x \cos y + \sin x \sin y$

(Due to lack of space I’ve omitted a proof of the above identities. See Will Garner’s pre-calculus textbook for a relatively simple geometric proof.)

Using the above identities we have $A(\theta_1) A(\theta_2) = \begin{bmatrix} \cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2&-\cos \theta_1 \sin \theta_2 - \sin \theta_1 \cos \theta_2 \\ \sin \theta_1 \cos \theta_2 + \cos \theta_1 \sin \theta_2&-\sin \theta_1 \sin \theta_2 + \cos \theta_1 \cos \theta_2 \end{bmatrix}$ $\quad = \begin{bmatrix} \cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2&-(\sin \theta_1 \cos \theta_2 + \cos \theta_1 \sin \theta_2) \\ \sin \theta_1 \cos \theta_2 + \cos \theta_1 \sin \theta_2&\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2 \end{bmatrix}$ $\quad = \begin{bmatrix} \cos (\theta_1 + \theta_2)&-\sin (\theta_1 + \theta_2) \\ \sin (\theta_1 + \theta_2)&\cos (\theta_1 + \theta_2) \end{bmatrix} = A(\theta_1 + \theta_2)$

So $A(\theta_1)A(\theta_2) = A(\theta_1 + \theta_2)$ as hypothesized. In other words, rotating the x-y plane through an angle $\theta_1$ followed by a second rotation through an angle $\theta_2$ is equivalent to a single rotation through the angle $\theta_1 + \theta_2$.

(b) Using the formula derived above we have $A(\theta) A(-\theta) = A(\theta - \theta) = A(0) = \begin{bmatrix} \cos 0&-\sin 0 \\ \sin 0&\cos 0 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = I$

In other words, rotating the x-y plane through an angle $\theta$ followed by a second rotation through the angle $-\theta$ (i.e., through the angle $\theta$ in the reverse direction) takes us back to the original state corresponding to no rotation at all.

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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