Let $F \in \mathbb{R}^{m \times p}$ and $X \in \mathbb{R}^{p \times n}$ be matrices, and let $r \leq p \land n$ be the rank of $X$. Suppose that $\sigma_{i}(FX) = \sigma_{i}(X)$ for all $i \in [r]$, i.e., the singular values are the same. Is it true that $F$ is an isometry on $\operatorname{Col}(X)$ (say with respect to the Euclidean norm)?
Stuff I have tried:
- Looking into SVD of $X$ and seeing what happens when you multiply it by $F$ -- in general, nothing useful happens, since left-multiplication may change the right-singular vectors.
- Using Courant-Fischer minimax theorem -- nothing happens because the dimension counts don't quite work, at least without some clever argument I can't come up with.
Nonetheless, it seems intuitively true to me.