Let $V$ be the space of $n\times n$ matrices, $B$ a fix matrix, then what is matrix of operator $L_B(A) = BA$ on $V$? I guess the answer is the $n^2\times n^2$ block matrix $diag(B, B, ..., B)$. Where shall I start to prove it? I tried to find the matrix by figuring out the image of the standard basis of $V$, but it gets complicated.
Any idea would be appreciated.
One more question

$\bullet$ Any idea about how to compute $det(L_B)$? I am sure that the answer is $det(B)^n$ if $B$ is invertible.

I found the answer of question $\bullet$ here :)
 A: If we use a standard basis in which the matrix
$$
A=\begin{pmatrix}
a_{11}&a_{12}&\cdots &a_{1n}\\
a_{21}&a_{22}&\cdots &a_{2n}\\
\cdots\\
a_{n1}&a_{n2}&\cdots &a_{nn}\\
\end{pmatrix}
$$
is represented as a vector of components
$$
A=\begin{pmatrix}
a_{11}\\\cdot\\\cdot \\a_{1n}\\
a_{21}\\\cdot\\\cdot \\a_{2n}\\
\cdots\\\cdots\\
a_{n1}\\\cdot \\\cdot\\a_{nn}\\
\end{pmatrix}
$$
than the matrix that represents the transformation $L_B(A)=BA$ is a block matrix of the form
$$
\begin{pmatrix}
B_{11}&B_{12}&\cdots &B_{1n}\\
B_{21}&B_{22}&\cdots &B_{2n}\\
\cdots\\
B_{n1}&B_{n2}&\cdots &B_{nn}\\
\end{pmatrix}
$$
where $B_{ij}$ is a diagonal matrix with as diagonal  value the element $b_{ij}$ of the matrix $B$
You can prove this result starting from the case $n=2$ and using induction.
A: Hint If you use the standard basis $E_{ij}$m, then 
$$L_B(E_{ij})=BE_{ij}=\begin{bmatrix} 0&0 &...& b_{1i} & 0 & ..0 \\
0&0 &...& b_{1i} & 0 & ..&0 \\
0&0 &...& b_{2i} & 0 & ..&0 \\
...& ...&...&...&...&...&...\\
0&0 &...& b_{ni} & 0 & ..0 \\
\end{bmatrix}$$
where the non-zero elements appear in column $j$. Write this as a linear combination of elementary matrices.
