Let $\sigma>0$ be a singular value of $A$, then it is also a singular value of $A^T$. Prove or disprove.
I think this should be true because if we consider the singular value decomposition of $A$, we will have:
$$A=U\Sigma V^T,$$
where $\Sigma$ contains $\sigma$.
Thus $$A^T=V\Sigma^TU^T.$$
Now let $U':=V$, $\Sigma':=\Sigma^T$, $V' :=U$, then
$$A^T=U' \Sigma' V'^T$$
is a singular value decomposition of $A^T$, and $\Sigma'$ contains $\sigma$. Moreover, if $\sigma=0$ then if $A$ is not a square matrix, then $\sigma$ is either a singular value of $A$ or $A^T$, but not both.
Please let me know if this is correct, or if there's a fallacy in my proof.
Assuming that you are working with real matrices. I agree up to the point where you address the case $\sigma >0$. (which is actually what the question is asking).
Consider an example where $\sigma=0$, Let $A=(0,0)$.
$0$ is a singular value for both $A$ and $A^T$.
Your initial argument should work regardless of whether $\sigma =0$ or $\sigma >0$.