Let $g$ be an $n \times m$ matrix and let $q$ be an $m\times k$ matrix, with $k\leq m$ and such that $q$ has orthogonal columns: $q^{T}q = I$.
I am trying to relate $\| gq\|_F$ and $\|g\|_F$, where $\|\cdot\|_F$ is the Frobenius norm. I know that if $q$ was an $m\times m$ orthogonal matrix, then $\|gq\|_F = \|g\|_F$. Also, we can have $\|gq\|_F < \|g\|_F$ (for example, if $g = I$ and $q=(1,0,0,\dots)$.) But my question is can we have something in general like: $$\frac{\sqrt{k}}{\sqrt{n}}\|gq\|_F \leq \|g\|_F \leq \frac{\sqrt{n}}{\sqrt{k}}\|gq\|_F$$ Not sure how to prove something like this (or a related bound)
After thinking on this, I believe there can be no such bound. Let $g$ be an $n\times n$ matrix with $0$'s in the first column and $1$'s everywhere else. Let $q$ be the $n\times 1$ matrix $q = (1,0,\dots,0)^T$. Then $\|g\|_F = \sqrt{n(n-1)}$ while $\|gq\|_F= 0$. So we can not expect the inequality on the right in the question to hold.
For the inequality on the left, we can note that $q$ is made up of $k$ columns that have unit norm, and use the submultiplicativity of the Frobenius norm to see that $\|g q\|_F \leq \|g\|_F\|q\|_F = k\|g\|_F$. So the following inequality always holds: $$\frac{1}{k}\|gq\|_F \leq \|g\|_F.$$