TITLE: Pseudoinverse matrix and SVD
QUESTION [7 upvotes]: I'm trying to solve an homework question but I got stuck.
Let A be a m x n matrix with the SVD $A = U \Sigma V^*$ and $A^+ = (A^* A)^{-1} A^*$ its pseudoinverse.
I'm trying to get $A^+ = V \Sigma^{-1} U^*$, but I'm missing something.
Can anyone help me with this please?
Thanks!

REPLY [3 votes]: If the dimensions of A are m x n and $m\not\equiv n$ then there isn't any way of deriving $A^+ = U\Sigma^{-1}V$. The reason is because $\Sigma$ has the same dimensions as $A$ therefore it is not invertible. If you see any source about SVD you will see that the equation is $A = U_{mxm} \Sigma_{mxn}V^T_{nxn}$. 
If A is rectangular maybe the possible derivation you're looking for is $$\begin{align}
A^+ &=(A^TA)^{-1}A^T\\
&=(V\Sigma^TU^T U\Sigma V^T)^{-1}V\Sigma^TU^T\\
&=(V\Sigma^T\Sigma V^T)^{-1}V\Sigma^TU^T  \\
&=(V^T)^{-1}(\Sigma^T \Sigma)^{-1}V^{-1}V\Sigma^TU^T \\
&=V(\Sigma^T \Sigma)^{-1}\Sigma^TU^T
\end{align}$$