Plücker matrix - Rank 2 proof

Question

How can I proof that the Plücker matrix of the form

$$ \begin{pmatrix} 0 & -L_{01} & -L_{02} & -L_{03}\\ L_{01} & 0 & -L_{12} & -L_{13}\\ L_{02} & L_{12} & 0 & -L_{23}\\ L_{03} & L_{13} & L_{23} & 0 \end{pmatrix} $$

has only rank 2?

Answer 1

It's not that a generic 4x4 skew-symmetric matrix would have rank 2, but that a generic 4x4 skew-symmetric matrix in the form of $uv^T−vu^T$ has rank 2. To prove this, suppose $u,v$ be linearly independent. Can you show that $u,v$ are in the image of $uv^T−vu^T$? So, what is the image of $uv^T−vu^T$?

Answer 2

The elmenets of L are not arbitrary. They can be written as $L_{ij} = A_iB_j - B_iA_j.$ with any two distinct points A, B the line contains. The answer lies in the Grassman-Plücker relations. The simple proof is to compute $L_{01} L_{23} -L_{02} L_{13}+L_{03} L_{12}=0$ (just expand by hand)

An geometric, although less obvious way to see this is that the moment and direction vectors must be orthogonal (which is exactly the statement above) see Wkipedia https://en.wikipedia.org/wiki/Pl%C3%BCcker_matrix#Geometric_interpretation Section "Geometric interpretation"

P.S: What's called "displacement" in the article is really a vector pointing in the direction from B to A...