If $A$ has orthonormal columns then $||Ax||^2_2 = ||x||^2_2$, why?

Question

In the lecture notes we have a fact:

If $A$ has orthonormal columns then $||Ax||^2_2 = ||x||^2_2$

Why is it the case? What properties of matrix-vector multiplication should I know to reason about this?

Thank you

Answer 1

The fact that $A$ has orthonormal columns is expressed concisely by the statement that $A^T A = I$. It follows from this fact that \begin{align} \| Ax \|^2 &= (Ax)^T Ax \\ &= x^T A^T A x \\ &= x^T x \\ &= \| x \|^2. \end{align}

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Here's an alternative proof. Let $u_i$ be the $i$th column of $A$ and let $x_i$ be the $i$th component of a vector $x$. If $y = Ax = \sum_i x_i u_i$, then \begin{align} \|y\|^2&= \sum_i \| x_i u_i \|^2 \qquad \text{(by Pythagorean theorem)} \\ &= \sum_i x_i^2 \| u_i \|^2 \\ &= \sum_i x_i^2 \\ &= \| x \|^2. \end{align}

Answer 2

Since $A$ has orthonormal columns, $A^TA=I$.