TITLE: Determinant of the Kronecker Product of Two Matrices
QUESTION [27 upvotes]: I'd like to know how can be shown that $\det(A \otimes B) = \det(A)^m \det(B)^n$ when $A$ and $B$ are square matrices of size $n$ and $m$ respectively and $\otimes$ represents the Kronecker product of $A$ and $B$.
I've seen some proofs using eigenvalues of $A$ and $B$, but since not every matrix has eigenvalues (at least not in all rings), I'd like a more intrinsic proof, maybe using the fact that $M_{(nm)^2} \cong M_{n^2} \otimes M_{m^2}$. 
It all comes down to showing that $\det(A \otimes I) = \det(A)^m$ and using some smart property (e.g. $(A\otimes B)(C\otimes D) = (AC) \otimes (BD)$), but I wasn't able to do even that.

REPLY [3 votes]: Use the commutation matrix; recall that Km,n(A⊗B)=(B⊗A)Km,n. It follows that det(A⊗B)=det(B⊗A). Then det(A⊗B)=det((A⊗I)(I⊗B))=det(I⊗A)det(I⊗B)=det(A)^m det(B)^n (since I⊗A is a block diagonal matrix with all blocks on the diagonal equal to A, and the determinant of a block diagonal matrix equals the product of the determinants of the blocks).