## 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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