I was absent from class and I am reading a classmate's notes. On them, it says:
Notice that $A = U \Sigma V^* \implies \Sigma = U^*AV$, so $A$ and $\Sigma$ has the same rank.
My question is, why would the LHS imply that the rank of $\Sigma$ and the rank of $A$ are the same?
The following is my idea; could you tell me please if it is correct?
I think I was able to prove that in general, if $B$ (square) has full rank, then the rank of $AB$ and of $BA$ are the same as the rank of $A$. Therefore, we can write
$$U \Sigma = AV$$
The rank of $U \Sigma$ is the same as the rank of $\Sigma$, and the rank of $AV$ is the same as the rank of $A$. Therefore the rank of $A$ is the same as the rank of $\Sigma$.
$U$ and $V$ are unitary, so they are isomorphisms/invertible (or determinant non-zero, if you like).