Linear Algebra and Its Applications, Exercise 3.2.15

Exercise 3.2.15. For a matrix $A$ show that if $AA^T = A^TA$ then the length of $Ax$ is equal to the length of $A^Tx$ for all $x$ .

Answer: We have $\|A^Tx\|^2 = \left(A^Tx\right)^T\left(A^Tx\right)$ . By the rule for transposes of products we have $\left( A^Tx \right)^T = x^T \left(A^T \right)^T =x^TA$ so that $\|A^Tx\|^2 = x^TAA^Tx$ . But since we assumed that $AA^T = A^TA$ we then have

$\|A^Tx\|^2 = x^TAA^Tx = x^TA^TAx$

Applying the rule for transposes of products once more we have $x^TA^T = \left(Ax\right)^T$ so that

$\|A^Tx\|^2 = \left(Ax\right)^T\left(Ax\right) = \|Ax\|^2$

So the length of $Ax$ is equal to the length of $A^Tx$ .

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!

Leave a Reply Cancel reply

Enter your comment here...

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

Email (required) (Address never made public)

Name (required)

Website

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

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

Cancel

Connecting to %s