Provided $A$ is a full rank matrix $\in\mathbb{C^{m\times n}}$, then $A^{*}A$ is of full rank. Suppose $m\gt n$.
There is a solution to this problem: solution link, and the top solution makes sense to me.
But when I use SVD to solve this problem:
since $A=U\Sigma V^{*}$
then $A^{*}A =V\Sigma^{*}U^{*}U\Sigma V^{*}= V\Sigma^{*}\Sigma V^{*}$
$rank(A^{*}A)=rank(\Sigma^{*}\Sigma)$
because $rank(\Sigma)=n$, $rank(A^{*}A)=n$
I got a quite different answer. Where is error in my SVD proof?