Linear Algebra and Its Applications, Exercise 3.4.4

Exercise 3.4.4. Given two orthogonal matrices Q_1 and Q_2, show that their product Q_1Q_2 is also orthogonal. If Q_1 represents rotation through the angle \theta and Q_1 represents rotation through the angle \phi, what does Q_1Q_2 represent? What trigonometric identities for \sin (\theta + \phi) and \cos (\theta + \phi) can be found in multiplying Q_1 and Q_2?

Answer: For any orthogonal matrix Q we have Q^TQ = I. We then have

(Q_1Q_2)^T(Q_1Q2) = (Q_2^TQ_1^T)(Q_1Q2) = Q_2^T(Q_1^TQ_1)Q_2 = Q_2^TQ_2 = I

Since (Q_1Q_2)^T(Q_1Q2) = I the matrix Q_1Q_2 is orthogonal.

If the matrices Q_1 and Q_2 represent rotations through \theta and \phi respectively, then Q_1Q_2 represents rotation through \theta + \phi, and will contain elements including \sin (\theta + \phi) and \cos (\theta + \phi). These will be produced in the matrix multiplication through the following trigonometric identities:

\sin (\theta + \phi) = \sin \theta \cos \phi + \cos \theta \sin \phi

\cos (\theta + \phi) = \cos \theta \cos \phi - \sin \theta \sin \phi

where \sin \theta and \cos \theta come from Q_1  and \sin \phi and \cos \phi come from Q_2.

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, Fifth Edition and the accompanying free online course, and Dr Strang’s other books.

This entry was posted in linear algebra and tagged , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

w

Connecting to %s