I'm a little confused about the various explanations for using Singular Value Decomposition (SVD) to solve the Linear Least Squares (LLS) problem. I understand that LLS attempts fit $Ax=b$ by minimizing $\|A\hat{x}-b\|$, then calculating the vector $\hat{x}$ such that $\hat{x}=(A^{\top}A)^{-1}A^{\top}b$
But my question(s) are in relation to the two explanations given at SVD and least squares proof and Why does SVD provide the least squares solution to $Ax=b$? :
Why do we need (or care to) to calculate $\hat{x}=V{\Sigma}^{-1}U^{\top}b$ where $SVD(A)=U\Sigma V^{\top}$ when $\hat{x}$ can be calculated vie at the pseudo-inverse mentioned above ($\hat{x}=(A^{\top}A)^{-1}A^{\top}b$)
The first post mentioned that we are subject to the constraint that $\|\hat{x}\|=1$? What happens when the least squares solution does not have $\|\hat{x}\|=1$? Does this invalidate using SVD for the solution of $\hat{x}$ or is there a "back-door" approach?
How do the answers to the questions above (as well as our approach) change when we are minimizing $Ax=0$ versus a generic $Ax=b$? Example: When the SVD of A is $U$, $\Sigma$, and $V^{\top}$ (that is $A\hat{x}=U\Sigma V^{\top}\hat{x}$), I would think we only care about the smallest singular value $\sigma_i$ in $\Sigma$ when solving $Ax=0$, since using the smallest $\sigma_i$ does not necessarily give the best fit to $u_i \sigma_i v^{\top}_i \hat{x} = b$?
Much thanks, Jeff