What is the right hand side in this definition of $n$-dimensional cross product

Question

Let ${\bf e_1}, \dots, {\bf e_n}$ be the standard basis for $\mathbb{R}^n$ and let ${\bf w_1} = (w_{11},\dots,w_{1n}), \dots, {\bf w_{n-1}}=(w_{n-1\;1},\dots,w_{n-1\;n}) \in \mathbb{R}^n$. Then one can define

$${\bf w_1} \times {\bf w_2} \times \cdots \times {\bf w_{n-1}} = \begin{vmatrix} {\bf e_1} & {\bf e_2} & \cdots & {\bf e_n} \cr w_{11} & w_{12} & \cdots & w_{1n} \cr \vdots & \vdots & & \vdots \cr w_{n-1\;1} & w_{n-1\;2} & \cdots & w_{n-1\;n} \end{vmatrix}$$

see e.g. Wikiepdia.

The problem I have is: if the right hand side denotes determinant then this is calculating the determinant of a non square matrix and I don't know how this is defined. (Note the dimensions are $(n+(n-1))\times n = (2n-1) \times n$

How is the determinant for non square matrices defined?

Answer 1

Symbolically, it is an $n\times n$ matrix. Don't expand the $\vec{e}_i$ into coordinates. Just take the determinant according to however you normally do so, and whenever multiplication involves the scalars from below, multiply accordingly, and when it involves a scalar times one of these vectors from the top row, multiply the scalar times the vector accordingly.

---

Also, I don't think it's helpful to think that $\vec{e}_i=\begin{bmatrix}1\\0\end{bmatrix}$. Sometimes that is what is meant, but consider this: just let the collection of $\vec{e}_i$ be abstract vectors. Then, for example if there are only $2$ of these, when you see $\begin{bmatrix}1\\0\end{bmatrix}$, that is just code for $1\vec{e}_1+0\vec{e}_2$. So $\begin{bmatrix}1\\0\end{bmatrix}$ isn't exactly the same as $\vec{e}_1$; rather, $\begin{bmatrix}1\\0\end{bmatrix}$ is the coordinate vector for $\vec{e}_1$ with respect to the basis $\{\vec{e}_1,\vec{e}_2\}$.

I find this more helpful than oversimplifying to $\vec{e}_1=\begin{bmatrix}1\\0\end{bmatrix}$. It may feel strange, but that is because in earlier math classes you worked with coordinates form the start instead of pure vectors. If that point that is one unit to the right of the origin had always been just called $\vec{e}_1$ instead of $(1,0)$, what I am describing would feel more natural.

Answer 2

An equivalent definition is the unique element ${w_1} \times {w_2} \times \cdots \times {w_{n-1}}$ such that $\langle x, {w_1} \times {w_2} \times \cdots \times {w_{n-1}} \rangle = \det\begin{bmatrix} {x_1} & {x_2} & \cdots & {x_n} \cr w_{11} & w_{12} & \cdots & w_{1n} \cr \vdots & \vdots & & \vdots \cr w_{n-1\;1} & w_{n-1\;2} & \cdots & w_{n-1\;n} \end{bmatrix}$ for all $x$.

This avoids taking determinants with vectors as elements.

To define the quantity in the question, we let $\det\begin{bmatrix} {e_1} & {e_2} & \cdots & {e_n} \cr w_{11} & w_{12} & \cdots & w_{1n} \cr \vdots & \vdots & & \vdots \cr w_{n-1\;1} & w_{n-1\;2} & \cdots & w_{n-1\;n} \end{bmatrix} = \sum_k \langle e_k, {w_1} \times {w_2} \times \cdots \times {w_{n-1}} \rangle e_k$.