Least Squares Solution of $Ax = b$ using SVD of $A$ or SVD of $A^\top A$ For $m > n$ and a matrix $A \in \mathbb R^{m \times n}$ of rank $r \leq n$, we want to
\begin{equation}
\text{minimize } \| A\mathbf x - \mathbf b\|
\end{equation}
over all $\mathbf x \in \mathbb R^n$. This can be done using the SVD of $A$. Solving the above problem is equivalent to solving the linear system
\begin{equation}
A^\top A\mathbf x = A^\top \mathbf b,
\end{equation}
which, again, can be done using SVD, however, the one of $A^\top A$, which has less entries than $A$ as $n < m$.
My question is why would one want to compute the SVD of $A$ instead of the SVD of $A^\top A$?

As said in the comment, the condition number of $A^TA$ is the condition number of $A$ squared, so is usually a lot higher.
This leads to imprecisions in the computation of the SVD, or in inverting the matrix $A^TA$ in the system $A^TAx = A^Tb$.
