matrix kronecker product property $\det(\mathbf{A}\otimes \mathbf{B})=(\det(\mathbf{A})^m)(\det(\mathbf{B})^n)$. 
$\mathbf{A}$ is $n\times n$ matrix. $\mathbf{B}$ is $m\times m$ matrix.
$\otimes$ is Kronecker product.
How can i prove that?
 A: If $A$ and $B$ are diagonal matrices, it is easy to image that the equation is satisfied.
If $A$ and $B$ are not diagonal matrices, we can do some transformation like this:
$A^{\prime} = P A P^{-1}$
where $A^{\prime}$ is a diagonal matrix. $P$ is transformation matrix. 
we can do similar:
$B^{\prime} = Q B Q^{-1}$
Note: ${\rm det}(A^\prime) = {\rm det}(P) {\rm det}(A) {\rm det}(P^{-1}) = {\rm det}(A)$
Using the relation $(AC)\otimes(BD)=(A \otimes B)(C \otimes D)$
$A \otimes B = (P^{-1} A^\prime P) \otimes (Q^{-1} B^\prime Q) 
= [(P^{-1}A^\prime)\otimes(Q^{-1}B^\prime)][P \otimes Q]=[P^{-1} \otimes Q^{-1}][A^\prime \otimes B^\prime][P\otimes Q]=[P\otimes Q]^{-1}[A^\prime \otimes B^\prime][P \otimes Q]
$
So 
$\det(A \otimes B) = \det([P\otimes Q]^{-1}[A^\prime \otimes B^\prime][P \otimes Q]) = 
\det [A^\prime \otimes B^\prime]$
Since $A^\prime$ and $B^\prime$ are diagonal matrices, easy to image
$\det [A^\prime \otimes B^\prime] = \det(A^\prime)^m \det(B^\prime)^n = \det(A)^m \det(B)^n$
complete.
