Minimum of the 2-norm Let $A = QR$, where $Q$ is an orthogonal ($m\times m$)−matrix and $R$ is an upper ($m\times n$)-triangular matrix of rang $n$ ($m>n$). 
I want to show that $$\min_{x\in \mathbb{R}^n}\|Ax-y\|_2=\|(Q^Ty)_{n+1}^m\|_2, \ \forall y\in \mathbb{R}^m$$ 
It is $(a)_k^l=(a_k, \ldots , a_l)^T$ if $a=(a_1, \ldots , a_l)^T\in \mathbb{R}^l$. 
$$$$ 
I have done the following: 
$$\min_{x\in \mathbb{R}^n}\|Ax-y\|_2=\min_{x\in \mathbb{R}^n}\|QRx-y\|_2=\min_{x\in \mathbb{R}^n}\|QRx-QQ^{-1}y\|_2=\min_{x\in \mathbb{R}^n}\|Q(Rx-Q^{T}y)\|_2=\min_{x\in \mathbb{R}^n}\|Rx-Q^{T}y\|_2$$ 
(We have used here the properties of an orthogonal matrix.) 
How could we continue? Do we have to show that this expression is minimum if $x=0$ ? 
 A: We shouldn't get that the minimum is achieved when $x=0$ or else we'll have that the min is $\| Q^{T}y \|_{2}$ and not $\|(Q^{T}y)_{n+1}^{m} \|_{2}$.
$R$ is upper triangular of rank $n$, so we can see that $Rx$ will have a bunch of zeros for its $(n+1)$-th through $m$-th entries. This means we can't do anything to minimize $\|(Rx-Q^{T}y)_{n+1}^{m} \|_{2}$.
Let $\widehat{R}$ be the matrix consisting of the first $n$ rows of $R$ (just chopping the zeros off) and let $\widehat{Q^{T}}$ be the matrix consisting of the first $n$ rows of $Q^{T}$. Using our observation above, we can show that $x$ solves
$$ \min_{x \in \mathbb{R}^{n}} \| Rx - Q^{T}y \|_{2}$$
if and only if $x$ solves 
$$\min_{x \in \mathbb{R}^{n}} \| \widehat{R}x-\widehat{Q^{T}}y \|_{2}.$$
Since $R$ is rank $n$, we get that $\widehat{R}$ is rank $n$. From here, you can use the fact that $\widehat{R}$ is $n$-by-$n$ to get that
$$\min_{x \in \mathbb{R}^{n}} \| \widehat{R}x-\widehat{Q^{T}}y \|_{2} = 0$$
and you can use norm properties to deduce that the solution to that least-squares problem solves the system $\widehat{R}x-\widehat{Q^{T}}y = 0$. The result you want follows pretty quickly from there.
