how to compute the determinant of the linear map $f(X)=AXC$ Let $V$ be the vector space of $m\times n$ matrices over a field $F$. Fix an $m\times m$ matrix $A$ and an $n\times n$ matrix $C$, and consider the map $\phi: V\longrightarrow V$ defined by $\phi(B)=ABC$ for $B\in V$. Compute the determinant of $\phi$ in terms of $A$ and $C$. How to compute? What is the result?
 A: Branimir's answer is smarter than what follows, but I thought I'd give a basic linear algebra class argument. I'll use the canonical basis $\{E_{ij}\}$ of $M_{m\times n}(F)$, where  $E_{ij}$ is the matrix with $1$ in $(i,j)$ position and $0$ elsewhere.
Note that $\phi$ is the composition of the (commuting) linear maps $L_A:B\longmapsto AB$ and $R_C:B\longmapsto BC$. In particular, $\det \Phi=\det L_A\det R_C$.
Now just note that the matrix representation of $L_A$ in the canonical basis $\{E_{11},E_{21}\ldots,E_{m1},E_{12},\ldots,E_{mn}\}$ is block-diagonal with $n$ copies of $A$ on the diagonal. Therefore $\det L_A=(\det A)^n$.
Likewise, but reordering the canonical basis as follows: $\{E_{11},E_{12}\ldots,E_{1n},E_{21},\ldots,E_{mn}\}$, the matrix representation of $R_C$ is block-diagonal with $m$ copies of $C$ down the diagonal. So $\det R_C=(\det C)^m$.
Hence $\det \phi=(\det A)^n(\det C)^m$. 
A: The point is that
$$
 V = M_{m \times n}(F) \cong L(F^n,F^m) \cong F^m \otimes_F (F^n)^\ast \cong F^m \otimes_F F^n,
$$
where the composite isomorphism $S : F^m \otimes_F F^n \cong V$ is given by $S : v \otimes w \mapsto vw^T$. In particular, then, for any $v \otimes w \in F^m \otimes F^n$,
$$
 \phi(S(v \otimes w)) = \phi(vw^T) = Avw^TC = (Av)(C^T w)^T = S(Av \otimes C^T w) = S((A \otimes C^T)(v \otimes w))
$$
so that $S^{-1}\phi S = A \otimes C^T$. Hence, using the general fact that $\det(X \otimes Y) = \det(X)^n \det(Y)^m$ for $X \in M_m(F)$ and $Y \in M_n(F)$, it therefore follows that
$$
 \det(\phi) = \det(S^{-1}\phi S) = \det(A \otimes C^T) = \det(A)^n \det(C^T)^m = \det(A)^n \det(C)^m.
$$
