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Rank of a 2-vector (exterior algebra)

  1. Oct 28, 2009 #1
    I understand that there is a way to find a basis [tex]\{e_1,...,e_n\}[/tex] of a vector space [tex] V[/tex] such that a 2-vector [tex] A [/tex] can be expressed as

    [tex] A = e_1\wedge e_2 + e_3\wedge e_4 + ...+e_{2r-1}\wedge e_{2r}[/tex]

    where 2r is denoted as the rank of [tex]A[/tex]. However the way that I know to prove this seems sort of inelegant. I'm wondering what other proofs people have.
     
  2. jcsd
  3. Oct 28, 2009 #2
    I'm especially curious if there is a geometric interpretation of the result.
     
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