Linear Algebra and Its Applications, Exercise 2.6.20

Exercise 2.6.20. A nonlinear transformation f from a vector space V to a vector space W is invertible a) if for any b in W there exists some x in V such that f(x) = b and b) if x and y are in V then f(x) = f(y) implies that x = y. Describe which of the following transformations from \mathbb{R} to \mathbb{R} are invertible, and explain why or why not:

a) f(x) = x^3

b) f(x) = e^x

c) f(x) = x+11

d) f(x) = \cos x

Answer: a) If b is any real number then x = \sqrt[3]{b} exists and is a unique solution to x^3 = b. The transformation f(x) = x^3 is therefore invertible.

b) We have e^x > 0 for all x in \mathbb{R}, so if b \le 0 then there is no x for which e^x = b. The transformation f(x) = e^x is therefore not invertible.

c) If b is any real number then b-11 exists and is a unique solution to x+11 = b. The transformation f(x) = x+11 is therefore invertible.

d) We have -1 \le \cos x \le 1 for all x in \mathbb{R} so if b < -1 or b > 1 then there is no x for which \cos x = b. Also note even for -1 \le b \le 1 solutions to \cos x = b are not unique, since (for example) \cos 0 = \cos 2\pi = 1. The transformation f(x) = \cos x is therefore not invertible.

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.

 Buy me a snack to sponsor more posts like this!

This entry was posted in linear algebra. 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 )

Twitter picture

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

Facebook photo

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

Google+ photo

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

Connecting to %s