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).

Answer: (a) We have

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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