Identity relating scalar triple product of basis

Question

$\newcommand{\vek}[1]{\boldsymbol{#1}}$ $\det( \vek{e}_i ~\vek{e}_j ~\vek{e}_k) \det \begin{pmatrix} \vek{e}_i^T \\ \vek{e}_l^T \\ \vek{e}_m^T \\ \end{pmatrix} = \det \begin{pmatrix} \vek{e}_i^T \vek{e}_i & \vek{e}_i^T \vek{e}_j & \vek{e}_i^T \vek{e}_k \\ \vek{e}_l^T \vek{e}_i & \vek{e}_l^T \vek{e}_j & \vek{e}_l^T \vek{e}_k \\ \vek{e}_m^T \vek{e}_i & \vek{e}_m^T \vek{e}_j & \vek{e}_m^T \vek{e}_k \\ \end{pmatrix}$

From

I have two questions on the above equation:

1. How do you represent $\det(e_i,e_j,e_k)$ as a matrix?
2. How did the equality follow?

Answer 1

Each $e_i$ itself is a vector, so this is nothing else than the matrix with three column vectors $e_i,e_j,e_k$, so in total we have $A\in k^{3\times 3}$ with $a_{1i}=a_{2j}=a_{3k}=1$ and zeros everwhere else. The other matrix is the same just flipped, so instead of column vectors you have row vectors with $b_{i1}=b_{j2}=b_{k3}=1$ and zeros elsewhere.

Example for $i=3,j=l=1,k=m=1$:

$$ \det\left(\begin{array}& 0&1&1\\0&0&0\\1&0&0\end{array}\right) \cdot \det\left(\begin{array}& 0&0&1\\1&0&0\\1&0&0\end{array}\right) = \det\left(\begin{array}& 1&0&0\\0&1&1\\0&1&1\end{array}\right) $$

The equality follows from the fact that $\det$ is multiplicative, that means $\det (A) \cdot \det (B) = \det (A\cdot B) = \det (B\cdot A)$ for $A,B$ square matrices.

If you multiply the given matrices you will see that the matrix on the RHS is nothing else than the product of the matrices on the LHS.