Say you have a $m$ x $n$ matrix, $A$, with $rank(A) = m$ and let $s \in \mathbb{R}^n$. I am trying to find a formula for the point $y \in \mathbb{R}^n$ that satisfies $Ay = As$ that is closest to the origin.
I know that $y \in s + Null(A)$ since this is the solution set to the equation. From here, I've tried to find a formula for $Proj_{s + Null(A)}(0)$ but with no luck.
At this point, I've run out of ideas, and was hoping for a hint in the right direction!
On
Call $A^\dagger$ the Moore-Penrose Inverse, or pseudoinverse of $A$. It has the property you want, namely $$y = A^\dagger A s$$ is the minimum norm solution. It can be built through SVD (singular value decomposition).
You are looking for the point of $s + \ker(A)$ that lies closest to $0$. Equivalently, we are looking for the projection of $s$ onto $\ker(A)^\perp = \operatorname{range}(A^T)$.
If $B$ has full column-rank, then the projection onto the range of $B$ is given by the matrix $B(B^TB)^{-1}B^T$. So, the solution to our problem will be $$ \operatorname{Proj}_{\operatorname{range}(A^T)}(s) = [A^T(A^{TT}A^T)^{-1}A^{TT}]s = A^T(AA^T)^{-1}As. $$
Regarding $\operatorname{Proj}_{s + \ker(A)}(0) = \operatorname{Proj}_{\ker(A)^\perp}(s)$:
Let $v = \operatorname{Proj}_{\ker(A)^\perp}(s)$. We can write $s = v + (s-v)$, and we note that $s-v \in \ker(A)$. We find that $s + \ker(A) = v + \ker(A)$.
To see that $v$ is the closest to $0$ among the elements of $v + \ker(A)$, note that every element of $v + \ker(A)$ not equal to $v$ can be written in the form $v + w$, where $w \in \ker(A)$ and $w \neq 0$. Since $w$ is perpendicular to $v$, the distance of $v + w$ from $0$ is $$ \|v+w\| = \sqrt{\|v\|^2 + \|w\|^2} > \sqrt{\|v\|^2} = \|v\|. $$ So, $v$ (which has distance $\|v\|$ from $0$) is indeed the closest point.