Rank of $UV^\top$ in SVD decomposition

Let $$M\in\mathbb{R}^{m\times n}$$ be a matrix with $$\text{rank}(M) = r$$ and let $$M = U\Sigma V^\top$$ be its compact SVD decomposition. In other words

• $$U\in\mathbb{R}^{m\times r}$$ is semi-orthogonal i.e. $$U^\top U = I_r$$
• $$\Sigma\in\mathbb{R}^{r\times r}$$ is diagonal with strictly positive entries
• $$V\in\mathbb{R}^{n\times r}$$ is semi-orthogonal i.e. $$V^\top V = I_r$$

What is the rank of $$UV^\top$$?

According to this property we have $$\text{rank}(UV^\top) = \text{rank}(U)$$ as long as $$V^\top$$ has rank $$r$$, which I think it does. I sense that somehow the rank of $$UV^\top$$ must be computable somehow but not sure if this is the right way.

One thing that might help is that I think $$UV^\top$$ is one of the matrices in the polar decomposition.

• Polar Decomposition is really for square matrices and with $\Sigma\succ \mathbf 0$, interpreting this through polar decomposition means $\det\big(M\big) \neq 0$ Oct 1, 2021 at 19:07
• @user8675309 actually here it says this can easily be extended to rectangular matrices by requiring the orthogonal matrix to become semi-orthogonal (or semi-unitary) Oct 1, 2021 at 19:22
• A closer read: It says you may (generalize /) extend the definition of Polar Decomposition to non-square matrices... that means the definition Polar Decomposition is constrained to square matrices. If you want to call $UV^T$ a matrix in generalized polar decomposition, ok fine. Oct 1, 2021 at 19:28

If $$y\in \text{Col}(UV^T)$$ then $$y=UV^Tx$$ for some $$x \in \mathbb{R}^n$$ and so $$y=UV^Tx=U\big(V^Tx\big)\in\text{Col}(U)$$ On the other hand, if $$y\in \text{Col}(U),$$ then $$y=Ux$$ for some $$x\in \mathbb{R}^r$$ and so $$y=Ux=UI_{r}x=UV^T(Vx)\in \text{Col}(UV^T)$$ This shows $$\text{Col}(U)=\text{Col}(UV^T)$$ and since $$\text{rank}(U)=r$$ (columns of $$U$$ are orthogonal) we must also have $$\text{rank}(UV^T)$$ equaling $$r$$ as well.