Exercise 3.2.2. Use the formula (where is the projection of on ) to confirm that (where is the angle between and ).

Answer: Since we can compute the square of the length of as

Since is a scalar quantity we can bring it out of the transpose expression and then bring it to the left of the expression:

This leaves us with the scalar quantity at the right, so we can further simplify this expression as follows:

Taking the positive square root of both sides we then have

At the same time from (2) on page 146 (theorem 3G) we have

so that

Substituting into the above expression for we then have

(since is guaranteed to be positive but is not).

UPDATE: I expanded the derivation above in response to a question posted on MathHub.

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.

Pingback: Question regarding trasposes and norms - MathHub