Let $A\in\mathbb{R}^{m\times n}$, $m<n$, rank($A$) $= m$. Given $b\in\mathbb{R}^m$ we want to find $x\in\mathbb{R}^n$ such that $Ax = b$.
We can solve by SVD, but I've also looked at two other ways:
- By setting $x = A^Ty$ so that we now have to solve $AA^Ty=b$, where the matrix $AA^T$ is square and full-rank so it's better, and
- By using the QR factorization of $A^T$, $A^T = QR$, so then we solve $(QR)^Tx = b$.
I'm trying to show that the solution obtained by any of these methods are the same. For SVD $\Leftrightarrow$ $AA^Ty=b$, we can simply plug in the SVD of A on the right to get $y = V\Sigma^{-2}U^Tb$, x = $V\Sigma^{-1}U^Tb$.
Now I'm having trouble showing that SVD $\Leftrightarrow$ QR or $AA^Ty=b$ $\Leftrightarrow$ QR. My friend suggested using the SVD of R but I couldn't get that to work. Any suggestions would be much appreciated!