TITLE: Prove that if $Q^tQ = I$ and $A = QR$, then $\|Ax - b\| = \|Rx - Q^tb\|$
QUESTION [6 upvotes]: I have a linear algebra final tomorrow and was practicing a few proofs. I want to make sure this proof is correct.

Prove that: If $Q^tQ = I$ and $A = QR$, then $\|Ax - b\| = \|Rx - Q^tb\|$ 
$$\begin{align*}
A &= QR\\[0.1cm]
Ax &= QRx\\[0.1cm]
Ax - b &= QRx - b\\[0.1cm]
\qquad Ax - b &= QRx - QQ^tb \quad\text{(since $QQ^t = I$)}\\[0.1cm]
Ax - b &= Q(Rx - Q^tb)\\[0.1cm]
\|Ax - b\| &= \|Q(Rx - Q^tb)\|\\[0.1cm]
\end{align*}$$
  Since the orthogonal transformation preserves length, $\|Q(Rx - Q^tb)\| = \|(Rx - Q^tb)\|$.
This completes the proof:
$\|Ax - b\| = \|Rx - Q^tb\|$

REPLY [2 votes]: This CW answer intends to remove the question from the unanswered queue.

As copper.hat already noted in the comments, your proof looks good. Also $QQ^T=I$ follows from $Q^TQ=I$ since $Q$ is a square matrix (and hence left and right inverses agree). This was noted by user17762 in the comments.