Let $M = \sum_{k} \mathbf u_k \mathbf v_k^{\top}$, where both sets of vectors $\{\mathbf u_k\}$, $\{\mathbf v_k\}$ are linearly independent.
Can we conclude that $\mathbf u_k \in \operatorname{colspan}(M)$?
Some comments: Maybe this is trivial (it's just late here). I guess since the $\mathbf v_k$ are independent, we can multiply on the right by $V^{-1}$ (pseudo-inverse) to obtain $U$ as linear combinations of columns of $M$? Maybe that's all that's needed here. In any case me asking here is just to confirm my intuition on this?
For context, I'm trying to show that a certain maximum likelihood problem has a certain solution, which is actually a stable local maximum. The local stability condition I have derived depends on this "lemma", but I don't think it makes sense to go deeper into that here.
Choose a vector $\mathbf w$ orthogonal to all the $\mathbf v_k$ for $k\ne i$ and not orthogonal to $\mathbf v_i$. (This is possible precisely because the $\mathbf v_k$ are linearly independent.) Then $M\mathbf w$ is a nonzero multiple of $\mathbf u_i$. This is a less fancy way of stating what you were already thinking. Linear independence of the $\mathbf u_k$ is not required.