Funny fact in Linear Algebra

from Hoffman & Kunze Linear Algebra 5.4.12

A funny fact came from this problem.

Consider the two ordered basis of $M_{n}(F)$ which are

\[\alpha=\{E_{11},E_{12},\cdots,E_{1n},\cdots,E_{n1},\cdots,E_{nn}\}\]

\[\beta=\{E_{11},E_{21},\cdots,E_{n1},\cdots,E_{1n},\cdots,E_{nn}\}\]

We have $[L_{B}]=[L_{B}(E_{11})\ L_{B}(E_{12})\ \cdots\  L_{B}(E_{nn})]$. Consider $BE_{ij}$. We have ${\rm col}_{j}(BE_{ij})=Be_{i}={\rm col}_{i}(B)$. Thus, first column is form of $B_{11},B_{21},\cdots,B_{n1},\cdots,0$. From that point we get

For basis $\beta$

\[    [L_{B}]=\begin{bmatrix}
            B_{11} & B_{12} & \cdots & B_{1n}\\
            & & \cdots\\
            B_{n1} & B_{n2} & \cdots & B_{nn}\\
            & & & & \ddots\\
            & & & & & B_{11} & B_{12} & \cdots & B_{1n} \\
            & & & & & & & \cdots\\
            & & & & & B_{n1} & B_{n2} & \cdots & B_{nn} \\
        \end{bmatrix} = \begin{bmatrix} B \\ & B \\  & & \ddots \\ & & & B \\ & & & & B \end{bmatrix} \]

For basis $\alpha$

\[ [L_{B}]=\begin{bmatrix}
            B_{11}I_{n} & B_{12}I_{n} & \cdots & B_{1n}I_{n}\\
            B_{21}I_{n} & B_{22}I_{n} & \cdots & B_{2n}I_{n}\\
            \vdots & \vdots & \cdots & \vdots \\
            B_{n1}I_{n} & B_{n2}I_{n} & \cdots & B_{nn}I_{n}
        \end{bmatrix} \]

Then the $\det([L_{B}])=\det(B)^{n}$ from basis $\beta$. If you are working with a basis $\alpha$ we will have difficulty in finding the determinant of that matrix. One idea you can get from here is a change of basis may help you to find the determinant easier!