I've come across this problem asking to do a double wedge product, and my professor didn't cover it.
It asks me to find the coordinates of the wedge product of a∧b∧c
$a = [8,-6,7,8], b = [9,-3,8,4], c = [4,6,5,-5]$
I would've thought, on intuition, that the solution would be $(a∧b)∧(b∧c)$. That gives something along the lines of $15$ coordinates, and I know, from the input interface, that the answer has $4$ coordinates.
The answer having $4$ coordinates confused me, considering $a∧b$ alone gives a $6$-coordinate output.
Hope you can help!
The exterior product of $n-1$ vectors of dimension $n$ is a generalization of the cross product in dimension three (if you identify $\Lambda^{n-1} \Bbb R^n$ to $\Bbb R^n$ by identifying $e_2 \wedge e_3 \wedge ... \wedge e_n \to e_1$, $e_3 \wedge e_3 \wedge ... \wedge e_1 \to e_2$ and so on, that's called Hodge duality)
You can compute it by generalizing the way you'd compute a cross product.
For instance you can compute the determinant of $$ \begin{pmatrix} e_1&e_2&e_3&e_4\\ 8&−6&7&8\\ 9&−3&8&4\\ 4&6&5&−5 \end{pmatrix} $$