Determinant of the Kronecker Product of Two Matrices 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.
 A: HINT:
It's enough to show the identity holds for matrices with entries from $\mathbb{C}$. Moreover, it's enough to show it for complex diagonalizable matrices, a dense subset. So show that if $A$ is diagonalizable with eigenvalues $\lambda_i$ and $B$ diagonalizable with eigenvalues $\mu_j$, then $A\otimes B$ is diagonalizable with eigenvalues $\lambda_i \cdot \mu_j$. While here, you can also show that the eigenvalues of $A\otimes I_m + I_n \otimes B$ are $\lambda_i + \mu_j$. 
A: The most intrinsic proof is probably showing $\det(V\otimes W) \cong \det(V)^{\otimes\text{rk}(W)} \otimes \det(W)^{\otimes\text{rk}(V)}$, using the universal properties involved, and that under this (natural) isomorphism, $\det(f\otimes g)$ becomes $\det(f)^{\otimes\text{rk}(W)} \otimes \det(g)^{\otimes\text{rk}(V)}$. Here, $f\!: V \to V$ and $g\!: W \to W$ are endomorphisms (represented by $A$ resp. $B$). This also works over rings.
A: 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).
A: Here's another approach. Consider $A \otimes {\bf 1}_m$, we will show that this matrix can always be brought to the block form 
$$\left(
\begin{array}{cccc} A & 0 & \cdots & 0 \\ 
0 & A & \cdots & 0 \\
0 & \cdots & \cdots & A \\
\end{array}\right)
$$
To this end consider the matrix $A$ with components $(a_{ij})$ in some basis, say $\{u_i\}$ of a vector space $V$ with dimension $n$ over a ring $R$. Consider also the identity ${\bf 1}_m$ over the vector space $W$ of dimension $m$ over $R$ also. We will use the basis $\{ u_i \otimes e_a \}$ for the space $V \otimes W$, being $i,j=1,\cdots,n$ and $a,b=1, \cdots, m$. Let us further choose an ordering for the basis, this ordering will be 
$$
\{u_1 \otimes e_1, u_1\otimes e_2, \cdots ,u_1 \otimes e_m, u_2 \otimes e_1, \cdots ,u_n \otimes e_m \}
$$
Let us look at the form of the operator $A \otimes {\bf 1}_m$ in this basis, we shall see that it is the block form given above. Consider the action:
$$
(A \otimes {\bf 1}_m)u_{i}\otimes e_a = Au_i \otimes {\bf 1}_m e_a = \sum_{j,b} a_{ij}\delta_{ab} u_j \otimes e_b
$$
This means that the matrix element in this basis is $(A \otimes {\bf 1}_m)_{ia,jb} = a_{ij}\delta_{ab}$. This is the block form we are aiming to get, for notice that this matrix element is only distinct of zero when $a=b$, that is along the diagonal of a $m \times m$ block matrix, and in each block one has the matrix $(a_{ij})$, which is the operator $A$ in the basis $\{u_i\}$. The determinant is independent of the basis chosen.
Now take the determinant of this block matrix, it is easy to show that this determinant is $\det(A)^m$. Finally, as you pointed out yourself write
$$
A \otimes B = (A \otimes {\bf 1})({\bf 1}\otimes B)
$$
and use $\det(MN) = \det M \cdot \det N$. This works with any ring.
