If $v=\sum_{i=1}^n\lambda_{1i}e_i$ and $w=\sum_{i=1}^n\lambda_{2i}e_i$ are $2$ vectors in $\mathbb R^n$, then one can write $v\wedge w$ as a linear combination of $\{e_i\wedge e_j:1\le i<j\le n\}$, what are then coefficients in that case, i.e. $v=\sum\limits_{1\le i<j\le n} a_{ij}e_i\wedge e_j$, how can I express $a_{ij}$ with $\lambda_{1i}$ and $\lambda_{2i}$ ?

Well there are $\binom{n}{2}$ possibilities for $e_i\wedge e_j$ so at least the combination of $\lambda_{1i}$ and $\lambda_{2i}$ cannot be just the product

  • $\begingroup$ Use the skew-symmetry of the wedge product to reduce the number further. $\endgroup$ – Paul May 19 '16 at 13:55
  • $\begingroup$ @Paul but $i<j$ so it is already ordered $\endgroup$ – user257 May 19 '16 at 13:57
  • $\begingroup$ $\binom{n}{2}$ is for $i \leq j$, not $i<j$. $\endgroup$ – Paul May 19 '16 at 14:03
  • $\begingroup$ @Paul yes so $n^2-2n$ possibilites $\endgroup$ – user257 May 19 '16 at 14:05

Why don't we try a little example and see if we can figure this out?

Suppose $v = a e_1 + b e_2$ and $w = c e_1 + d e_2$ (so either $n = 2$ or we decide all the other coefficients are zero). Then \begin{align*} v \wedge w &= (a e_1 + b e_2) \wedge (c e_1 + d e_2) \\ &= ac e_1 \wedge e_1 + ad e_1 \wedge e_2 + bc e_2 \wedge e_1 + db e_2 \wedge e_2 \\ &= (ad - bc) e_1 \wedge e_2 \end{align*} because $e_2 \wedge e_1 = - e_1 \wedge e_2$ (and $e_i \wedge e_i = 0$ for all $i$).

Can you now work out the general case? If not try another example with only three nonzero coefficients. It'll come.

  • $\begingroup$ Using this answer math.stackexchange.com/a/900866/337759 I get $a_{ij}=\lambda_{1i}\lambda_{2j}-\lambda_{1j}\lambda_{2i}$, I think its in accordance with your example, isn't it ? $\endgroup$ – user257 May 19 '16 at 15:00
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    $\begingroup$ It is, but you really should work through the calculations to get a feel for how they work. Once it clicks you'll know it forever. $\endgroup$ – Gunnar Þór Magnússon May 19 '16 at 15:05

You're right that there are $\binom{n}{2}$ possibilities. However, remember that $e_k$ are all basis vectors (This is an assumption I'm making based on notation. If it were not for this, then I doubt there is a neat expression). Therefore, all of the $$a_{ij} = \lambda_{1i}\lambda_{2j}e_i \wedge e_j$$ Will be $0$ when $i \neq j$ since $e_i \wedge e_j = 0$

However, when $i = j$, then $e_i \wedge e_j = 1$


$$ a_{ij} = \lambda_{1i} \lambda_{2j} \ \ \ \text{when} \ \ \ i = j \\ a_{ij} = 0 \ \ \ \text{when} \ \ \ i \neq j $$

However, since your index runs in a way that $i < j$, you can never have $i = j$. Therefore, the entire product collapses to $0$

  • $\begingroup$ but $i$ is never equal to $j$ in this representation as you can see in the middle line of the yellow box $\endgroup$ – user257 May 19 '16 at 14:11
  • $\begingroup$ @yasabur - I added a final note saying why this collapses. Thanks for pointing this out! $\endgroup$ – Siddharth Bhat May 19 '16 at 14:20
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    $\begingroup$ $e_i \wedge e_j = 0$ if $i = j$, but it is nonzero when $i \not= j$. The question is about the wedge product, which you don't seem to understand. $\endgroup$ – Gunnar Þór Magnússon May 19 '16 at 14:51

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