If the singular value decomposition $U\Sigma V^T$ has rank $r$ and $Q$ is semi-orthogonal, then $Q\Sigma V^T$ has rank $r$ as well

Question

Let $m,n\in\mathbb N$ and $A\in\mathbb R^{m\times n}$ with reduced singular value decomposition $U\Sigma V^T$, \begin{align}U&\in\mathbb R^{m\times r},\\\Sigma&\in\mathbb R^{r\times r},\\V&\in\mathbb R^{n\times r}.\end{align} Now let $Q\in\mathbb R^{l\times r}$ be semi-orthogonal, $$QQ^T=I_l\tag1.$$

Let $B:=Q\Sigma V^T$. How can we conclude that $\operatorname{rank}B=r$?

Clearly, $r=\operatorname{rank}A$.

Answer 1

I think it is not true.

Let $Q=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \end{bmatrix}$, and $\Sigma=V=I_3$.

Then $\mathrm{rank}\Sigma=r=3$ and $\mathrm{rank}Q\Sigma V^T=2\neq r$.