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.

  • $\begingroup$ Polar Decomposition is really for square matrices and with $\Sigma\succ \mathbf 0$, interpreting this through polar decomposition means $\det\big(M\big) \neq 0$ $\endgroup$ Oct 1, 2021 at 19:07
  • $\begingroup$ @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) $\endgroup$ Oct 1, 2021 at 19:22
  • $\begingroup$ 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. $\endgroup$ Oct 1, 2021 at 19:28

1 Answer 1


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.


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