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Let $A=QR$ be the full $QR$ factorization of $Ainmathbb{C}^{mtimes n}$, $mge n$, where

$$Q=begin{bmatrix}
hat{Q}_1 & hat{Q}_2
end{bmatrix}, R=begin{bmatrix}
hat{R}_1\
0
end{bmatrix},$$
where $hat{Q}_1$ and $hat{R}_1$ are reduced $QR$-factorization such that $A=hat{Q}_1hat{R}_1$. I want to prove that the norm of the least squares residual $$r=|b-Ay|$$
(where $y$ is the minimizing vector for the least squares problem) is equal to the norm of $hat{Q}_2b$.

So I proceeded directly and showed that

$$r=b-Ay=b-AA^+b=b-QRR^+Q^*b=left(I-Qbegin{bmatrix}
I_n & 0\
0 & 0
end{bmatrix}Q^*right)b$$
$$=left(I-begin{bmatrix}
hat{Q}_1 & hat{Q}_2
end{bmatrix}begin{bmatrix}
I_n & 0\
0 & 0
end{bmatrix}begin{bmatrix}
hat{Q}_1^*\hat{Q}_2^*
end{bmatrix}right)b=begin{bmatrix}
0&0\0&I_{m-n}
end{bmatrix}b.$$

So that $|r|$ is the norm of the last $m-n$ entries of $b$.

We can also multiply $r$ by $Q^*$ on the left to get that $|r|=|Qr|=|hat{Q}_2r|$, which is the desired result.

But is it then true that $|hat{Q}_2r|=left|begin{bmatrix}
0&0\0&I_{m-n}
end{bmatrix}bright|$ or am I missing something? If if it is true then why not just evaluate the norm of the last $m-n$ entries of $b$ rather than the norm of $hat{Q}_2r$?