The Grassmann algebra provides for an alternative of generalizing the quaternion algebra as opposed to the Clifford algebra. Here we define the wedge product $u \wedge v $.

Now there exists a particular useful theorem:

Vectors $u_1,u_2,\ldots,u_r$ are linearly dependent if, and only if, their wedge product vanishes, $$u_1 \wedge u_2 \wedge \ldots \wedge u_r = 0 \ . $$

Curiously though, this does not give a method of finding a linear relation, given that we have linear depenedent vectors. How would we go about using wedge products to find a linear relation between linear dependent vectors?

In an excercise I showed the following vectors $$(e_1+e_2+e_3) ,\ (e_2+e_3+e_4) ,\ (e_3+e_4+e_5) , \ (e_1+e_3+e_5) \\ (\{e_1,e_2,e_3,e_4,e_5\} \text{ a basis of the underlying vector space}) $$to be linear dependent by computing their wedge product.

Since we end up with 2-vectors cancelling each other out, I suppose we can not get a linear relation among these.

  • $\begingroup$ I don't se how you manage to kill $(e_1+e_2+e_3)\wedge e_4\wedge e_5$-terms. Agree with Rene, those three are independent. May be a typo or something? $\endgroup$ – Jyrki Lahtonen Jul 4 '14 at 18:19
  • $\begingroup$ I am redoing it right now. Hold on.. $\endgroup$ – Mussé Redi Jul 4 '14 at 18:21
  • $\begingroup$ Denote $$u := (e_1+e_2+e_3) \wedge (e_2 +e_3 +e_4) \ , v := (e_3+e_4+e_5) \wedge (e_1 +e_3 +e_5)\ . $$ We have $$ u = e_{12}+e_{13}+e_{14}+e_{24}+e_{34} \ , \\ v = e_{13}+e_{41}+e_{43}+e_{45}+e_{51}. $$ This yields $$ u\wedge v = e_{1243}+e_{1245}+e_{1345}+e_{2431}+e_{2451}+e_{3451} = 0 \ . $$ The last identity follows by considering the permutations of the indices. So I do indeed think that the given vectors are linearly dependent, since the wedge product vanishes. $\endgroup$ – Mussé Redi Jul 4 '14 at 18:52
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    $\begingroup$ So you really have four vectors ? $\endgroup$ – Rene Schipperus Jul 4 '14 at 19:01
  • $\begingroup$ Yes, I edited. Sorry for the typo. $\endgroup$ – Mussé Redi Jul 4 '14 at 19:04

You can. Say you have $$\sum a_i v_i=0$$ then wedge with $u_1 \wedge \ldots \hat{v_i}\wedge \ldots \wedge \hat{v_j} \ldots \wedge u_r$

you get $$a_i u_1 \wedge \ldots \wedge \hat{v_j} \ldots \wedge u_r +a_j u_1 \wedge \ldots \hat{v_i}\wedge \ldots \wedge u_r=0$$ So $a_i:a_j$ is determined.

But there is some problem with your example they are not dependent.

  • $\begingroup$ +1: Just feel like remarking that this works only if the $r$ vectors span a space of dimension $r-1$. If the span has a lower dimension, then all those wedges involving $r-1$ vectors vanish. As they should, of course, because in that case there are two linearly independent dependency relations, so you should not be able to determine $a_i:a_j$. This is an abstraction of the method working on $n$ vectors in $\Bbb{R}^n$: the determinant vanishes, iff they are dependent. You get the (ratios of) coefficients in a non-trivial dependency relation, iff one or more of $(n-1)\times(n-1)$ minors $\neq0$. $\endgroup$ – Jyrki Lahtonen Jul 4 '14 at 18:16
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    $\begingroup$ @JyrkiLahtonen True, I am assuming that $u_1 \wedge u_2 \wedge \ldots \wedge u_{r-1} \neq 0$ for if they were zero then there will be a relation between the first $r-1$ vectors and you could apply my answer in a lower dimension. $\endgroup$ – Rene Schipperus Jul 4 '14 at 18:20
  • $\begingroup$ What do you mean by the sum $\sum a_iv_i$ and the hatted vectors? If we wedge the whole sum with an r-vector we get more than two multivectors (each term of the sum is wedged separately). Could you perhaps illustrate your method with my example? $\endgroup$ – Mussé Redi Jul 5 '14 at 15:27
  • $\begingroup$ I can illustrate with a different example, but it is too much work with your example. If you want to work it yourself, take the $a(e_1+e_2+e_3)+b(e_1+\dots =0$ and wedge with $(e_1+e_2+e_3)\wedge (e_2+e_3+e_4)$ and the other combinations to get the ratios of the coefficients $a:b:c:d$ $\endgroup$ – Rene Schipperus Jul 5 '14 at 16:13

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