## 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.

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