Proving a matrix has Kronecker product form Is there a property unique to Kronecker product of two matrices, so that one could use it to prove that a certain matrix has Kronecker product form? Is there a proof technique to this type of objective?
An example problem is the following:
Prove that if $P = \{P_i\}$ is the group of all permutation matrices that commute with a particular matrix $A$: $A P_i = P_i A$, and similarly $Q = \{Q_j\}$ has all permutations commuting with $B$, $Q_j B = B Q_j$, then ALL permutations that commute with $A \otimes B$ have Kronecker product form: 
$$W (A \otimes B) = (A \otimes B) W \quad \Rightarrow  \quad W = P_i \otimes Q_j$$
In this example, our goal is to show that $W$ has a Kronecker form. Is there any technique to use for this type of proof?
 A: Let us consider the case of a square matrix $n \times n$ matrix $W$.
You have to consider all ways of writing $n$ as the product of two numbers $n=pq$.
Let us take for simplifying the formulas to come that $n=6$ with $p=2$ and $q=3$.
For each such writings consider a partition of $W$ in $q$ times $p \times p$ blocks under the form :
$$W=\left[\begin{array}{c|c}
    W_{11}& W_{12}& W_{13}\\
\hline
W_{21}& W_{22}& W_{23}\\
\hline
W_{31}& W_{32}& W_{33}
\end{array}\right]
$$ 
Identify one of the blocks having at least one nonzero entry. Assume for the sake of simplicity that it is the upper-left block $W_{11}$; call it $B$. 
Then, under the condition that all other blocks are proportional to $B$, i.e., giving $W$ the following aspect :
$$W=\left[\begin{array}{c|c}
    B&aB&bB\\
\hline
cB&dB&eB\\
\hline
fB&gB&hB
\end{array}\right]
=A \otimes B \ \ \text{with} \ \ A:=\left[\begin{array}{cc}
    1&a&b\\
    c&d&e\\
    f&g&h
\end{array}\right]
$$
Please note the lack of unicity : $A \otimes B = kA \otimes \tfrac{1}{k}B $ for any non zero real number $k$.
Remark : the cases where $W$ is a rectangular matrix $n_1 \times n_2$ can be treated in a very similar manner, this time by considering the different ways to write $n_1=p_1q_1$ and $n_2=p_2q_2$.
