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I am currently learning (or rather trying to) about tensor and wedge products. In general I find that most books treat the subject a bit too abstractly for my liking and I always prefer to have something I can "compute with" (think of the Kroenecker product for tensors).

In my studies I came across something interesting in the wikipedia article on the exterior product:

https://en.wikipedia.org/wiki/Exterior_algebra

which basically says that if $\{v_i\}_{i=1}^k$ are $k \leq n$ vectors in $\mathbb{F}^n$ then there is an expression for $v_1 \wedge ... \wedge v_k$ in terms of the basis elements $e_i$ involving the minors of a matrix.

I would like to show this / compute the expression.

That this should be related to a determinant seems intuitively clear as for instance for $n=k=2$ \begin{align} (v_1, v_2) \wedge (u_1 , u_2) &= (v_1 e_1 +v_2 e_2 )\wedge (u_1 e_1 + u_2 e_2)\\ &=v_1 u_1 e_1\wedge e_1+v_1 u_2 e_1 \wedge e_2+v_2u_1 e_2 \wedge e_1 + u_2v_2 e_2\wedge e_2\\ &=v_1 u_2 e_1 \wedge e_2+v_2u_1 e_2 \wedge e_1=(v_1u_2-v_2u_1) e_1 \wedge e_2\\ &=det(v,u) e_1\wedge e_2. \end{align}

I should probably be able to find a nice expression using similarly easy computations for the general case.

However, I have a rather big problem doing this. Mainly, my book does not even really give a definition of the wedge product other than as a quotient space of the tensor product (the identification becomes hard to compute with for $n>2$). In particular, this has given me NO idea whatsoever as to how to compute any actual wedges. I guess we should be able to express this quotient projection more directly somehow?

I have read the answers to algebraic manipulation of differential form which seems to treat a similar (or the same?) problem phrased differently. But I don't really understand them (I know nothing of differential forms and my book does not treat them). I have also read other answers referring to clifford algebras but really I know nothing of these so I would appreciate an answer which appeals to a more elementary understanding of linear algebra.

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If $v_j=\sum_{i=1}^na_{ij}e_i\quad (j=1\cdots k, \;a_{ij}\in \mathbb F)$, then the requested formula is: $$ v_1 \wedge ... \wedge v_k=\sum _I A_Ie_I$$ where $I$ runs through all strictly increasing sequence of integers $I=(1\leq i_1\lt\cdots\lt i_k\leq n)$ and where $e_I=e_{i_1}\wedge \cdots \wedge e_{ i_k}\in \Lambda ^k (\mathbb F^n)$.
The coefficient $A_I\in \mathbb F$ of $e_I$ is the determinant of the square $k\times k$ matrix obtained by extracting lines $i_1,\cdots , i_k$ from the $n\times k$ matrix $A=(a_{ij})$.

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  • $\begingroup$ Thanks. How do you arrive at this? $\endgroup$ – Winston Nov 29 '16 at 1:20
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    $\begingroup$ Yokonuma Proposition 3;3, page 84. $\endgroup$ – Georges Elencwajg Nov 29 '16 at 1:51
  • $\begingroup$ Although not really what I was looking for as an answer, I acquired the reference and find that I really like its style. Thanks again! $\endgroup$ – Winston Nov 29 '16 at 9:43
  • $\begingroup$ You are welcome, dear ZMI. $\endgroup$ – Georges Elencwajg Nov 29 '16 at 14:12

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