Exercise 2.6.20. A nonlinear transformation from a vector space to a vector space is invertible a) if for any in there exists some in such that and b) if and are in then implies that . Describe which of the following transformations from to are invertible, and explain why or why not:

a)

b)

c)

d)

Answer: a) If is any real number then exists and is a unique solution to . The transformation is therefore invertible.

b) We have for all in , so if then there is no for which . The transformation is therefore not invertible.

c) If is any real number then exists and is a unique solution to . The transformation is therefore invertible.

d) We have for all in so if or then there is no for which . Also note even for solutions to are not unique, since (for example) . The transformation 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.