matrix kronecker product property

Question

$\det(\mathbf{A}\otimes \mathbf{B})=(\det(\mathbf{A})^m)(\det(\mathbf{B})^n)$. $\mathbf{A}$ is $n\times n$ matrix. $\mathbf{B}$ is $m\times m$ matrix. $\otimes$ is Kronecker product. How can i prove that?

Answer 1

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.