determinant of crossing product of two matrix If $V$ is the vector space of $n\times n$ matrices over $F$, and $B$ is a fixed $n\times n$ matrix over $F$. Let $L_B$ and $R_B$ be the linear operators on $V$ defined by $L_B(A) = BA$ and $R_B(A) = AB$ show that:
a) $\det(L_B) = (\det(B))^n$ 
b) $\det(R_B) = (\det(B))^n$
I was wondering if I should use the representation matrix of the operators or something else. Is there any formula for cross product of two matrices?
please help me on this problem.
 A: Let 
\begin{align}
\beta
&=\{E^{ij}\in V:1\le i,j\le n\}\\
&=\{E^{11},E^{21},\ldots,E^{n1},E^{12},E^{22},\ldots,E^{n2},\ldots,E^{1n},E^{2n},\ldots,E^{nn}\}
\end{align}
be the standard ordered basis
for $V$, where $E^{ij}$ is the matrix whose $ij$-entry is one and other
entries are zero. Observe that
\begin{align}
L_B(E^{ij})=BE^{ij}=\sum_{k=1}^nB_{ki}E^{kj},
\end{align}
where $B_{ki}$ denotes the $ki$-entry of $B$. Then the
matrix representation of $L_B$ with respect to $\beta$,
denoting $[L_B]_\beta$, is of the form
\begin{align}
[L_B]_\beta
&=
\begin{pmatrix}
B_{11}&B_{12}&\cdots&B_{1n}\\
B_{21}&B_{22}&\cdots&B_{2n}\\
\vdots&\vdots&\ddots&\vdots\\
B_{n1}&B_{n2}&\cdots&B_{nn}\\
&&&&B_{11}&B_{12}&\cdots&B_{1n}\\
&&&&B_{21}&B_{22}&\cdots&B_{2n}\\
&&&&\vdots&\vdots&\ddots&\vdots\\
&&&&B_{n1}&B_{n2}&\cdots&B_{nn}\\
&&&&&&&&\ddots\\
&&&&&&&&&B_{11}&B_{12}&\cdots&B_{1n}\\
&&&&&&&&&B_{21}&B_{22}&\cdots&B_{2n}\\
&&&&&&&&&\vdots&\vdots&\ddots&\vdots\\
&&&&&&&&&B_{n1}&B_{n2}&\cdots&B_{nn}
\end{pmatrix}\\
&=
\begin{pmatrix}
B&&&\\
&B&&\\
&&\ddots&\\
&&&B
\end{pmatrix}.
\end{align}
That is, $[L_B]_\beta$ is the diagonal block matrix consisting of 
$n$ times $B$. Finally, because the determinant of a linear operator
is independent of the choice of an ordered basis for $V$, we conclude that
$$\det(L_B)=\det([L_B]_\beta)=[\det(B)]^n,$$
where the second equality follows by Appendix. This completes 
the proof of (a), and the proof of (b) is similar.

Appendix.
We show that given a $k$-block matrix
$$M=\begin{pmatrix}B_1&&&\\&B_2&&\\&&\ddots&\\&&&&B_k\end{pmatrix},$$
where $B_1,B_2,\ldots,B_k$ are square matrices (arbitrary size). Then
$$\det(M)=\det(B_1)\cdot\det(B_2)\cdots\det(B_k).$$
However, if we show the case of an $n\times n$ two-block matrix
$$M=\begin{pmatrix}A&\\&B\end{pmatrix},$$
then it can be easily extended to $k$-block matrix inductively. Fix
$B$ be $m\times m$, we use induction on $n$. Begin with $n=m+1$, then
$A=(A_{11})$ and it is clear that
$$\det(M)=A_{11}\cdot\det(B)=\det(A)\cdot\det(B).$$
Suppose that the result holds for some $n=m+l$, where $l\ge 1$. Then for
$n=m+(l+1)$, we apply the cofactor of the $ij$-entry of $A$, defined by
$$(-1)^{i+j}\cdot\det(\tilde{A}_{ij}),$$
where $\tilde{A}_{ij}$ is the $l\times l$ matrix obtained from $A$ by
deleting row $i$ and column $j$. Then
\begin{align}
\det(M)
&=\sum_{j=1}^{l+1}(-1)^{1+j}\cdot A_{1j}\cdot\det(\tilde{M}_{1j})\\
&=\sum_{j=1}^{l+1}(-1)^{1+j}\cdot A_{1j}\cdot\det(\tilde{A}_{1j})\cdot\det(B)\tag{1}\\
&=\det(A)\cdot\det(B)\tag{2},
\end{align}
where $(1)$ follows by the assumption and $(2)$ follows by the definition of
$\det(A)$.
