Suppose that we have some $m \times n$ matrix $C$, and its full rank (skeleton) decomposition $$ C = AB^T, $$ where $A$ is $m\times r$ and $B$ is $n\times r$ for some $r$.
We know that frobenius norm is a matrix norm: $$ \|C\|_F \leq \|A\|_F\|B\|_F. $$
But what if I need some "reverse" of that inequality?
Imagine that we know matrix $C$, and we know that $\|A\|_F \approx \|B\|_F$. Can we then find some constant $\alpha$ using matrix $C$ such that $$ \|C\|_F \geq \alpha \|A\|_F\|B\|_F? $$
In other words, is there possible to bound norm of $A$ (and $B$, because of our restrictions) for all possible skeleton decompositions by some predefined constant just by scaling down matrix $C$ once?
This is not possible
For simplicity, just take $r=2$ and the nonzero singular values of $C$ are $1,1$. So that the F norm of C is $\sqrt2$.
Now you may put the nonzero singular values of $A$ as $1/N, N$ and those of $B$ as $N, 1/N$ aligned accordingly
Then $A,B$ have the same F-norm $\sqrt{N^2+1/N^2}$, but $N$ can be as big as you like