Uniqueness of minimal norm solution. $\newcommand{\R}{\operatorname{Ran}} \newcommand{\K}{\operatorname{Ker}}\newcommand{\b}{\mathbf}$
Let an equation $A\b x = \b b$ has a solution, and let $A$ has non-trivial kernel. 
Prove that there exist a unique solution $\b x_0$ of $A\b x =\b b$ minimizing the norm $ ||\b x||$.

Any solution to  $A \b x = \b b$ can be written as $\b x_1 - \b x_h$ for $\b x_1$ being a solution of $A\b x = \b b$  and $\b x_h \in \K A$. So I need to minimize $||\b x_1 -\b x_h||$ for varying $\b x_h$. 
Since $||\b x_1 -\b x_h||$ is distance from the subspace $\K A$ to $\b x_1$, therefore it will be minimum when $\b x_h = P_{\K A} \b x_1$. So the minimum occurs at $\b x_0 =\b x_1 - P_{\K A} \b x_1 = P_{(\K A)^\perp} \b x_1$, for any $x_1 : A\b x_1 = \b b$. 
How can I prove the uniqueness of this solution ?
 A: If there are any distinct and parallel solutions (regardless of norm), then $\mathbf b=0$ (*), and $0$ is the unique minimal norm solution.
If there are two distinct non-parallel solutions with a given norm, then their average has smaller norm (the triangle inequality is strict for non-parallel vectors). So if there is a solution with minimal norm, it must be unique.
Proof of (*): Say $A\mathbf x_1=A(\lambda\mathbf x_1)=\mathbf b$ and $\lambda\neq1$. Then $$\mathbf b=A(\lambda\mathbf x_1)=\lambda A\mathbf x_1=\lambda \mathbf b$$
A: Let $x_1$ be an element of the pre-image of $b$ that is perpendicular to all of $\ker A$. Then any other pre-image $x_2$ is of the form $x_1+y$ for $y\in\ker A$ and so has norm squared $\langle x_1+y,x_1+y\rangle=|x|^2+|y|^2>|x|^2$, so $x_1$ is the unique pre-image of a $b$ with minimal norm.
A: This isn’t for any $x_1\in\mathrm{Ran}(A),$ but for any $x_1$ solving $Ax_1=b.$ Now observe that $Ax_1=Ax_1’$ if and only if $x_1$ and $x_1’$ differ by an element of $\mathrm{Ker}(A)$. But this means that $P_{\mathrm{Ker}(A)^{\perp}}(x_1)=P_{\mathrm{Ker}(A)^{\perp}}(x_1’),$ which proves the point is unique.
I would consider another approach, personally. Suppose $Ax=Ay=b,$ where $x$ and $y$ are minimum-length solutions. Then $A(tx+(1-t)y)=Ay+tA(x-y)=b,$ for all $t\in[0,1].$ But $\|tx+(1-t)y\|<t\|x\|+(1-t)\|y\|$ whenever $t\in(0,1)$ and $x$ and $y$ are not linearly dependent, which is a contradiction because $\|x\|=\|y\|$ is the length of the minimum-norm solution. We need to show that $x$ and $y$ are not linearly dependent, but in this case, $y=cx,$ so $b=Ay=cAx=cb,$ which shows that $c=1$ if $b\neq0.$ It is easy to show that $x=0$ is the unique minimum-length solution when $b=0.$
