How can we geometrically prove that least norm solution is given by $$x_{LN}=A^T(AA^T)^{-1}b$$
Edit Here is what I did :
If $A$ is full rank, i.e., rank($A$)=$m$,
then every solution of the under-determined system can be represented as :
$x=x_P-x_N$
where $x_P$ is a particular solution of the system $Ax=b$
and$\quad x_N\in \mathcal{N}(A), i.e., Ax_N=0$.
Therefore least norm solution is equivalent to the solution of the problem of
finding a $x_N^\prime\in \mathcal{N}(A)$ such that $||x||$ is minimized.
Hence $x_{LN}=x_P-x_N^\prime\in (\mathcal{N}(A))^\perp$
For any $z\in\mathbb{R}^m$ and $x_N\in \mathcal{N}(A)$ we have
$$z\cdot Ax_N=z\cdot 0=0$$
Also we have
$$0=z\cdot Ax_N=z^TAx_N=(A^Tz)^Tx_N$$
Hence we see that $x_N\perp \mathcal{R}(A^T)$
Since $x_N\in \mathcal{N}(A)$ is arbitrary we can conclude that $(\mathcal{N}(A))^\perp=\mathcal{R}(A^T)$
For any particular solution $x_P\in \mathcal{R}(A^T)$, we have $x_P=A^Tw$ for some $w\in \mathbb{R}^m$
Since $Ax_P=b$, it follows that $w$ is the solution of the system $AA^Tw=b$
Now $A$ is full rank and therefore $AA^T$ is non singular.
We obtain $w=(AA^T)^{-1}b$
Hence $x_P=A^T(AA^T)^{-1}b$
Now for any $x_N\in \mathcal{N}(A), x_N $ is projection of some $y\in\mathbb{R}^n$ on $\mathcal{N}(A)$
Thus $x_N$ is least square solution of $||y-u||^2, u\in \mathcal{N}(A)$
As before $y-x_N\in (\mathcal{N}(A))^T=\mathcal{R}(A^T)$
Therefore $y-x_N=A^T\hat{u}$ for some $\hat{u}$
Since $x_N\in \mathcal{N}(A)$, $Ay=AA^T\hat{u}$
Now $A$ is full rank, therefore $AA^T$ is non singular.
We get $\hat{u}=(AA^T)^{-1}Ay$
Hence $x_N=y-A^T\hat{u}=(I_n-A^T(AA^T)^{-1}A)y$
Therefore $x=A^T(AA^T)^{-1}b-(I_n-A^T(AA^T)^{-1}A)y$ for some arbitrary $y\in\mathbb{R}^n$.
We get least norm solution by taking projection $y=0$ on $\mathcal{N}(A)$.
Hence $x_{LN}=A^T(AA^T)^{-1}b$
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Please improve the solution as you see fit.