Denote $\mathcal{O}(p,r)$ be the collection of all $p$-by-$r$ matrices with orthonormal columns: $\mathcal{O}(p,r)=\{U\in\mathbb{R}^{p\times r}|U^{\intercal}U=I_p\}$. Let $A\in \mathcal{O}(p_1,r_1)$ and $B\in \mathcal{O}(p_2,r_2)$, with $p_i\ge r_i$. It is easy to see that $A\otimes B\in \mathcal{O}(p_1p_2,r_1r_2)$ by $(A\otimes B)^{\intercal}(A\otimes B)=(A^{\intercal}A)\otimes(B^{\intercal}B)=I.$
So here is my question, given a matrix $C\in\mathcal{O}(p_1p_2,r_1r_2)$, is it possible to construct $D\in\mathcal{O}(p_1,r_3)$ and $E\in\mathcal{O}(p_2,r_4)$ such that $\operatorname{col}(C)\subseteq\operatorname{col}(D\otimes E)$? (Here $\operatorname{col}(X)$ stands for the column space of $X$.)
Actually, I was trying to find $D$ and $E$ with $r_3=r_1,r_4=r_2$ s.t. $C=D\otimes E$ at the first sight. Then I notice that it may be impossible to do so. Take $r_1=r_2=1$ for instance, there are always vectors with $p_1*p_2$ length that can not be written as a Kronecker product of two vectors of length $p_1$ and $p_2$. So I relaxed the condition. Obviously when $r_3=p_1$ and $r_4=p_2$ we are able to meet the requirement. I was wondering if we can find $D\in\mathcal{O}(p_1,r_3)$ and $E\in\mathcal{O}(p_2,r_4)$ while $r_3$ and $r_4$ are only functions of $r_1$ and $r_2$?