SVD and least square solution

Question

Let $K \in \mathbb{R}^{m,n}$, $u \in \mathbb{R}^n$, and $f \in \mathbb{R}^m$. Assume that $m < n$ and $K$ have full rank so a solution exists but is not unique. I want to understand why this smallest solution is spanned by the first $m$ right singular vectors, denoted by $V_m$ = $(v_1, v_2, \dots, v_m)$: $$ K V_m z = f $$ .

Answer 1

If $K=U\Sigma V^T$, then the minimization task $$ \text{minimize}~~\|Kz-f\|_2 $$ is equivalent to $$ \text{minimize}~~\|\Sigma y-b\|_2,~~y=(V^Tz),~~b=(U^Tf) $$ This last system is completely decoupled, the influence of each row on the residual is independent of all the other rows and can be considered in isolation.

---

One does not need the full SVD for such a solution, a QR decomposition allows a similar simplification.