In reference to the explanation of the exterior or wedge product by Prof. Shifrin’s series (Chapter 8 - here, here and here), and connecting it with the more general tensor algebra operations, here is a summary:

The dual vector space \(V^*=(\mathbb R^n)^*\) is a vector space of linear maps \(\mathbb R^n \to \mathbb R\) that can be represented by a \(1\times n\) matrix (i.e. row vector). It is isomorphic to \(\mathbb R^n,\) and forms a vector space, which basis can be expressed as \(\mathrm e^1,\mathrm e^2,\dots,\mathrm e^n,\) or as \(\mathrm dx_1, \mathrm dx_2, \dots ,\mathrm dx_n,\) such that \(\mathrm dx_i(\vec v)=\vec {\mathrm e}_i\cdot \vec v.\)

The wedge product of two such forms in \(\Lambda^1(\mathbb R^3)^*\) will result in an element in \(\Lambda^2(\mathbb R^3)^*.\) For example, taking the last element of \(\Lambda^1(\mathbb R^3)^*\) in the preceding example, i.e. \(\psi =4\mathrm dx_1 +7 \mathrm dx_2,\) and wedging it with \(\varphi =2\mathrm dx_1 +5 \mathrm dx_3,\) also in the same vector space \(\Lambda^1(\mathbb R^3)^*,\) we obtain

\[\psi\wedge\varphi=4\mathrm dx_1 +7 \mathrm dx_2 \wedge 2\mathrm dx_1 +5 \mathrm dx_3=-14\;\mathrm dx_{12}+20\;\mathrm d_{13}+35\; \mathbb d_{23}\tag 1\]

Critically, this is the same as picturing the two forms in \(\Lambda^1(\mathbb R^3)^*,\) i.e. \(4\mathrm dx_1 +7 \mathrm dx_2=\color{red}{\begin{bmatrix}4&7&0\end{bmatrix}}^\top\) and \(2\mathrm dx_1 +5 \mathrm dx_3=\color{blue}{\begin{bmatrix}2&0&5\end{bmatrix}}^\top\) as

\[\begin{bmatrix} \color{red}4&\color{red}7&\color{red}0\\ \color{blue}2& \color{blue} 0 & \color{blue}5 \end{bmatrix}\]

and realizing that \(\mathrm d_{12}\) is simply the determinant of the first two columns (minor of the submatrix):

\[\det{\begin{bmatrix} \color{red}4&\color{red}7\\ \color{blue}2& \color{blue} 0 \end{bmatrix}}=-14\]

and that \(\mathrm d_{13}\)

\[\det{\begin{bmatrix} \color{red}4&\color{red}0\\ \color{blue}2& \color{blue} 5 \end{bmatrix}}=20\]

with \(\mathrm d_{23}\)

\[\det{\begin{bmatrix} \color{red}7&\color{red}0\\ \color{blue}0& \color{blue} 5 \end{bmatrix}}=35\]

We could express this as

\[\psi\wedge\varphi=\sum \text {minors}_{ij}\; \mathrm d_{ij}\]

The wedge product can be carried out between vector spaces \(\Lambda^l(\mathbb R^n)^*\wedge \Lambda^k(\mathbb R^n)^*\in \Lambda^{l+k}(\mathbb R^n)^*.\)

This all comes together when two vectors in \(\mathbb R^3\) are fed into the expression in Eq (1), for example \[\small\vec p =\begin{bmatrix}-1&-\pi&-\sqrt 2\end{bmatrix}^\top\] and \[\small\vec q =\begin{bmatrix}10^1&10^2&10^3\end{bmatrix}^\top:\]

\[\begin{align}\left(\psi \wedge \varphi\right)(p,q) &= -14\,\mathrm dx_{12}(p,q) + 20 \,\mathrm dx_{13}(p,q) + 35\, \mathrm dx_{23}(p,q)\\[2ex] &=-14 \,\begin{vmatrix}-1&10 \\-\pi & 10^2\end{vmatrix}+ 20\, \begin{vmatrix}-1 & 10 \\-\sqrt 2 & 10^3\end{vmatrix} + 35\, \begin{vmatrix}-\pi & 10^2 \\-\sqrt 2 & 10^3\end{vmatrix} \end{align}\]

As for the equivalence to tensor algebra pointed out in the answer

\[\begin{align} \mathrm dx_{12}(p,q)&=\mathrm dx_1 \wedge \mathrm dx_2 (p,q)\\[2ex] &= \left(\mathrm dx_1 \otimes \mathrm dx_2 - \mathrm dx_2\otimes \mathrm dx_1\right)(p,q)\\[2ex] &= (\mathrm dx_1 \otimes \mathrm dx_2 )(p,q)-(\mathrm dx_2\otimes \mathrm dx_1)(p,q)\\[2ex] &=\mathrm dx_1(p)\,\mathrm dx_2(q) - \mathrm dx_2(p) \,\mathrm dx_1(q)\\[2ex] &= (-1)(10^2) - (-\pi)(10)\\[2ex] &=\begin{vmatrix}-1&10 \\-\pi & 10^2\end{vmatrix}. \end{align}\]

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