Finding equivalent Kronecker product Let's consider $\bar{\bar{A}}$ and $\bar{b}$ known. I am looking for $\bar{\bar{C}}$ and $\bar{d}$
s.t.
$\bar{\bar{A}}\otimes \bar{b}= \bar{d}\otimes \bar{\bar{C}}$
I'm trying to define $\bar{\bar{C}}$ and $\bar{d}$ maybe using permutations as $I_{3,3}(\bar{\bar{A}}\otimes \bar{b})I_{1,3}= \bar{b} \otimes \bar{\bar{A}}$, but I can't find a solution Can you help me giving some hints ? 
Thank you :)
 A: I think this is in general not possible (trivial cases like zero vectors set aside). Your left-hand side contains repeated copies of the vector $\bar{b}$ (scaled by the elements $\bar{\bar{A}}$), whereas the right-hand side contains repeated copies of a matrix ($\bar{\bar{C}}$). These structures will not be compatible in general.
There is a way to be sure, of course. You can rearrange $\bar{d} \otimes \bar{\bar{C}}$ into $\bar{d} \cdot \bar{\bar{c}}^T$, where $\bar{\bar{c}} = {\rm vec}\{\bar{\bar{C}}\}.$ If you apply this rearrangement of elements to your given matrix $\bar{\bar{A}} \otimes \bar{b}$, you can check the rank of the resulting matrix. If it is rank one, its dominant singular values will provide a solution for $\bar{d}$ and $\bar{\bar{c}}$. If it is not, there is no (exact) solution. A truncated SVD would then give an approximate solution (best one in the least squares sense).
*edit: Here is a concrete counter-example to demonstrate that there will not always be a solution. Let's say $\bar{\bar{A}} = I_2$ and $\bar{b}=\begin{bmatrix}1\\1\end{bmatrix}.$ Then $$\bar{\bar{A}} \otimes \bar{b} = \begin{bmatrix} 1 & 0 \\ 1 & 0 \\ 0 & 1 \\ 0 & 1 \end{bmatrix}.$$ Further, let $\bar{d} = \begin{bmatrix}d_1\\d_2\end{bmatrix}$ and $\bar{\bar{C}} = \begin{bmatrix}c_1&c_2\\c_3&c_4\end{bmatrix}$  so that $$\bar{d} \otimes \bar{\bar{C}} = \begin{bmatrix}c_1 d_1 & c_2 d_1 \\ c_3 d_1& c_4 d_1 \\c_1 d_2 & c_2 d_2 \\ c_3 d_2& c_4 d_2 \end{bmatrix}.$$
Now, for these two to be equal, we have 


*

*$c_1d_1=c_3d_1=1$ which implies $c_1=c_3\neq 0$ and $d_1 \neq 0$

*$c_2d_1=c_4d_1=0$ which implies $c_2=c_4 = 0$ since $d_1 \neq 0$

*$c_1d_2=c_3d_2=0$ which implies $d_2 = 0$ since $c_1=c_3 \neq 0$

*$c_2d_2=c_4d_2=1$ which is a contradiction since $c_2=c_4 =d_2= 0$.

A: Consider the situation in which the elements of $(A,b)$ consist of distinct prime numbers, so that each element of their Kronecker product is a composite number with a unique factorization $K_{ijn}=A_{ij}b_n$.
The only way for $(C,d)$ to replicate all of these composite values is to have
$$d_n=b_n\\C_{ij}=A_{ij}$$
One way to generalize this result is to shuffle the order of the individual elements within $C$ and $d$ using permutations $\{P_k\}$.
$$d=P_1b \\C =P_2AP_3$$
This results in
$$\eqalign{
A\otimes b &= (P_1b)\otimes(P_2AP_3) \cr
 &= (P_1\otimes P_2)\;(b\otimes A)\;({\tt 1}\otimes P_3) \cr
 &= P_{12}\;(b\otimes A)\;P_3 \cr
}$$
The well-known commutation matrices are precisely the permutations which will reverse the order of a Kronecker product.
The proposed factorization is possible in a few degenerate cases:
$\quad-$ when $b$ is a scalar, i.e. a $\,{\tt 1}\times{\tt 1}$ vector, then $(d=b,\; C=A)$
$\quad-$ when $A$ is a scalar, i.e. a $\,{\tt 1}\times{\tt 1}$ matrix, then $(d=b,\; C=A)$
$\quad-$ when $A$ is a vector, then $(d=A,\; C=b)$
