What's the reason to use Singular Value Decomposition instead io $(A^TA)^{-1}A^T$ for pseudo inverse?

Question

I wonder what's the reason to use this formula from Singular Value Decomposition

$$ A = U\Sigma V $$ $$ A^{\dagger} = V\Sigma^{-1}U^T $$

Instead of $$ A^{\dagger} = (A^TA)^{-1}A^T $$

Both give the same result.

Answer 1

What is $A$ isn't full-rank? Then $A^T A$ will be singular and its inverse will not exist. However, you can still do singular value decomposition and will see the $0$'s along the diagonal of the $\Sigma$ matrix. In some applications, it is okay to ignore the zeros and simply invert the non-zero entries of the $\Sigma$ matrix. Also, it is often the case that $A$ is close to being singular which can lead to numerical problems when you invert $A^TA$. This can be mitigated again by ignoring the really small eigen values in $\Sigma$ if appropriate.