TITLE: how to compute the determinant of a linear map
QUESTION [5 upvotes]: 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?

REPLY [5 votes]: 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 $\det \phi=(\det A)^n(\det C)^m$.