- #1
Peeter
- 305
- 3
For two blades:
[tex]
A \in {\bigwedge}^r
[/tex]
[tex]
B \in {\bigwedge}^{k-r}
[/tex]
and a corresponding pseudovector for the wedge product of the two:
[tex]
I \in {\bigwedge}^k
[/tex]
intuition tells me that the following scalar quotient:
[tex]
\frac{A \wedge B}{I}
[/tex]
can be reduced to
[tex]
A \cdot \left(B \frac{1}{I}\right) = A \cdot \left(B \cdot \frac{1}{I}\right)
[/tex]
Presuming it this is correct, I curious if it is clear to anybody how to prove this.
Motivation for the question comes from computing the coordinates of a multivector using a basis generated from not necessarily orthonormal vectors (ie: calculation of the reciprocal frame vector/bivector/...). My Doran/Lasenby text has an approach that I have trouble following (I understand the lead up and the end result but not the middle), so I was trying to puzzle it out on my own in a more natural fashion. Such a grade reduction is required as an intermediate step.
[tex]
A \in {\bigwedge}^r
[/tex]
[tex]
B \in {\bigwedge}^{k-r}
[/tex]
and a corresponding pseudovector for the wedge product of the two:
[tex]
I \in {\bigwedge}^k
[/tex]
intuition tells me that the following scalar quotient:
[tex]
\frac{A \wedge B}{I}
[/tex]
can be reduced to
[tex]
A \cdot \left(B \frac{1}{I}\right) = A \cdot \left(B \cdot \frac{1}{I}\right)
[/tex]
Presuming it this is correct, I curious if it is clear to anybody how to prove this.
Motivation for the question comes from computing the coordinates of a multivector using a basis generated from not necessarily orthonormal vectors (ie: calculation of the reciprocal frame vector/bivector/...). My Doran/Lasenby text has an approach that I have trouble following (I understand the lead up and the end result but not the middle), so I was trying to puzzle it out on my own in a more natural fashion. Such a grade reduction is required as an intermediate step.