Pseudoinverse matrix and SVD 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!
 A: First you need to assume that the matrix $A^*A$ is invertible. For which you need $n \leq m$ and rank($A$) is $n$.
So when $n \leq m$ and when rank($A$) is $n$, then the reduced SVD of $A$ is $A = U \Sigma V^*$ where $U \in \mathbb{R}^{m \times n}$, $\Sigma \in \mathbb{R}^{n \times n}$ and $V \in \mathbb{R}^{n \times n}$ such that $U^* U = I_{n \times n}$, $V^* V = I_{n \times n}$, $V V^* = I_{n \times n}$ and $\Sigma$ is a square diagonal matrix and has only (positive) real entries.
Note that $V^{-1}=V^*$.
Also note that $A^* = V \Sigma^* U^* = V \Sigma U^*$ since $\Sigma^* = \Sigma$.
Further note that if $M_1,M_2 \text{and} M_3$ are invertible matrices then $(M_1 M_2 M_3)^{-1} = M_3^{-1} M_2^{-1} M_1^{-1}$.
Use these to get the final answer.
A: 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 $$\begin{align}
A^+ &=(A^TA)^{-1}A^T\\
&=(V\Sigma^TU^T U\Sigma V^T)^{-1}V\Sigma^TU^T\\
&=(V\Sigma^T\Sigma V^T)^{-1}V\Sigma^TU^T  \\
&=(V^T)^{-1}(\Sigma^T \Sigma)^{-1}V^{-1}V\Sigma^TU^T \\
&=V(\Sigma^T \Sigma)^{-1}\Sigma^TU^T
\end{align}$$
A: $$
\begin{align}
A^+ &= (A^*A)^{-1}A^* \\
    &=(V\Sigma U^*U\Sigma V^*)^{-1} V\Sigma U^* \\
    &=(V\Sigma^2 V^*)^{-1} V\Sigma U^* \\
    &=(V^*)^{-1} \Sigma^{-2} V^{-1} V\Sigma U^* \\
    &= V \Sigma^{-2}\Sigma U^* \\
    &= V\Sigma^{-1}U^*
\end{align}
$$
using the properties of the matrices $U,V,\Sigma$ in the Singular_value_decomposition.
