This question is hard to ask, but I'll try to be as specific as I can.
For matrices $A \in \mathbb{R}^{m \times n}$, $B \in \mathbb{R}^{p \times q}$, is there an equivalent representation (see below) for the vectorization of the Kronecker product of $A$ and $B$, i.e. $\text{Vec}(A \otimes B)$?
When I say an "equivalent representation", I am specifically asking for a quantity which is equal to $\text{Vec}(A \otimes B)$ where the $\text{Vec}$ operator only appears through $\text{Vec}(A)$ or $\text{Vec}(B)$
For example, if I were asking for an "equivalent representation" for $\text{Vec}(AB)$ (now assuming $n=p$), one could answer $(I \otimes A)\text{Vec}(B)$ or $(B^{T} \otimes I)\text{Vec}(A)$.
I hope the question is clear enough. I have tried to have a look around, and even show something, but I could not.
edit: One can write $\text{Vec}(A)=\sum\limits_{i=1}^{n}\mathbf{e}_i \otimes (A\mathbf{e}_i)$ and so $\text{Vec}(A \otimes B)=\sum\limits_{i=1}^{nq}\mathbf{e}_i \otimes ((A \otimes B)\mathbf{e}_i)$, which I guess works. Is there a nicer form for this?
Best Answer
Result: $\operatorname{vec}(A \otimes B) = [I_m \otimes K^{(q,m)} \otimes I_q][\operatorname{vec}(A)\otimes \operatorname{vec}(B)]$, where $I_k$ denotes the identity matrix of size $k$ and $K^{(q,m)}$ denotes a commutation matrix.
Proof: It suffices to consider the case that $A = uv^T$ and $B = xy^T$ (for column vectors $u,v,x,y$). We have $$ \operatorname{vec}(A \otimes B) = \operatorname{vec}[(uv^T) \otimes (xy^T)] = \operatorname{vec}[(u \otimes x)(v \otimes y)^T] = v \otimes y \otimes u \otimes x. $$ It now suffices to use the fact that $K^{(q,m)}(u \otimes y) = y \otimes u$ to show that the matrix $P = I_m \otimes K^{(q,m)} \otimes I_q$ satisfies $$ P(\operatorname{vec}(A) \otimes \operatorname{vec}(B)) = P(v \otimes u \otimes y \otimes x)= v \otimes y \otimes u \otimes x = \operatorname{vec}(A \otimes B). $$