If $A$ is invertible, then $A^{-1} = V \sum ^{-1}U^*$

Question

The singular value decomposition of a complex-values $n\times m$ matrix $A$ is defined as

$$A = U \sum V^*$$

where $U$ and $V$ are unitary $n \times n$ matrices and $m \times m$ matrices respectively, and $\sum $ is an $n \times m$ rectangular diagonal matrix, whose diagonal elements are the singular values $\sigma _I$ of $A$. $V^*$ denotes the conjugate transpose of $V$. The singular value decomposition s a generalization of the eigen decomposition to arbitrary $n\times m$ matrices

I am trying to prove the following $2$ things:

1) The rank of $A$ equals the number of nonzero entries in $\sum $

2) If $A$ is invertible, then $A^{-1} = V \sum ^{-1}U^*$

Below is my attempt for $1$. I am not sure if I am correct or not. I found this proof online! Iam not sure how to proceed with $2$ looking for some help with this, thanks!

If we have the singular value decomposition $A=VDU$, with $V,U$ orthogonal and $D$ diagonal, you have $$ A^TA=U^TDV^TVDU=U^TD^2U,$$ and since conjugation with an orthogonal preserves eigenvalues, the eigenvalues of $A^TA$ are precisely the squares of the diagonal entries of $D$. As $D$ is diagonal, its rank is equal to the number of nonzero diagonal entries. And since $U,V$ are invertible, the rank of $A$ is equal to the rank of $D$.

Answer 1

Guide:

To show that $A^{-1} = V \Sigma^{-1}U^*$, you just have to verify that

$$A(V\Sigma^{-1} U^*)=I.$$

Note that $U$ and $V$ are unitary, hence we have $UU^*=I=VV^*$

Alternatively, you can use the fact that $(CD)^{-1}=D^{-1}C^{-1}$.

Edit:

\begin{align}(U \Sigma V^*)(V \Sigma^{-1} U^*) &=U \Sigma(V^*V)\Sigma^{-1} U^*\\ &= U\Sigma I\Sigma^{-1}U^*\\ &= U (\Sigma \Sigma^{-1})U^*\\ &= UIU^*\\ &= UU^* \\ &= I \end{align}

Edit:

If we have the singular value decomposition $A=U\Sigma V^*$, with $V,U$ unitary and $\Sigma$ diagonal, you have $$ A^*A=V\Sigma^* U^*U\Sigma V^*=V (\Sigma^*\Sigma)V^*,$$ and since conjugation with an orthogonal preserves eigenvalues, the eigenvalues of $A^*A$ are precisely the squares of the diagonal entries of $\Sigma^*\Sigma$. As $\Sigma^*\Sigma$ is diagonal, its rank is equal to the number of nonzero diagonal entries. And since $U,V$ are invertible, the rank of $A$ is equal to the rank of $\Sigma^*\Sigma$.