Let $A \in \mathbb{R}^{n \times m}$ and $b \in \mathbb{R}^n $ with $m < n$. Moreover let $ |.|_2$ be the euclidian norm on $\mathbb{R}^n$ and $\mathbb{R}^m$. Consider $L(b)$ = {$ x \in \mathbb{R}^m : |Ax-b| = \min_{y \in \mathbb{R}^m} |Ay-b|$}.
Show there is exactly one $\hat{x} \in L(b)$ with $ | \hat{x}| = \min_{x \in L(b)} |x|$.
So this claim follows from the proof of $1)$ and from $2)$, but I don't know why. Can you please explain it to me?
here is $1)$ and his proof :
Let $ 1 \leq m < n$, $ b \in \mathbb{R}^n$, $A \in \mathbb{R}^{n \times m}$. There is a $x \in \mathbb{R}^m$, which solves $ F(x) = |Ax-b|^2 = \min_{y \in \mathbb{R}^m} |Ay-b|^2$.
Proof: Let Image($A$) = {$Ax \in \mathbb{R}^n : x \in \mathbb{R}^m$ }. Then $\mathbb{R}^n$ = Image($A$) $\otimes$ Image($A$)$^{\perp}$. Moreover $b$ has an unique decomposition: $ b = Ax_0 + v$, where $x_o \in \mathbb{R}^n$ and $ v \in $ Image($A$)$^{\perp}$. Now we have $F(x) = |Ax-b|^2 = |Ax - Ax_o -v|^2 = |A(x-x_o) -v |^2$. We know $ A(x-x_0) \in $ Image($A$). Therefore $A(x-x_o) \perp v$. Using Pythagoras: $|A(x-x_0) - v|^2 = |A(x-x_0)|^2 + |v|^2 \geq |v|^2$. The inequality above is an equality if $x-x_o \in $ Kernel($A$). So $x \in $ Kernel($A$) + $x_o$ minimize $F$.
_
here is $2)$ Let $V \subset \mathbb{R}^k $ be a vector subspace. Consider the affine space $\mathbb{A} = x_0 + V$. Regarding every norm on $\mathbb{R}^k$ the affine space $\mathbb{A}$ has exactly one $\hat{x}$ with $||\hat{x}|| = \min_{y \in \mathbb{A}}||y||$.
Thank you in Advance.