Prove that if $Q^tQ = I$ and $A = QR$, then $\|Ax - b\| = \|Rx - Q^tb\|$

Question

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:

$\|Ax - b\| = \|Rx - Q^tb\|$

See here for how to format mathematics on the site using MathJax. — Zev Chonoles, May 13 '13 at 05:37
One of my doubts was that I do not know if this implication holds true for all matrices:
If Q^tQ = I, then QQ^t = I as well. — Alexander Ventura, May 13 '13 at 05:42
@AlexanderVentura Since $Q$ is a square matrix, the left inverse is the same as right inverse and hence $Q^TQ = I \implies QQ^T = I$. — , May 13 '13 at 05:50

score 2 · Accepted Answer · answered Jun 22 '13 at 22:54

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.

Prove that if $Q^tQ = I$ and $A = QR$, then $\|Ax - b\| = \|Rx - Q^tb\|$

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