Reduction of a blade pseudovector quotient?

[Peeter]

For two blades:

$$A \in {\bigwedge}^r$$
$$B \in {\bigwedge}^{k-r}$$

and a corresponding pseudovector for the wedge product of the two:

$$I \in {\bigwedge}^k$$

intuition tells me that the following scalar quotient:

$$\frac{A \wedge B}{I}$$

can be reduced to

$$A \cdot \left(B \frac{1}{I}\right) = A \cdot \left(B \cdot \frac{1}{I}\right)$$

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.

 

[genneth]

Wow --- it's been a long time since I've tried to look at the GA stuff. From recollection, isn't there something about projecting and rejecting vectors and multivectors from each other? Your desired step looks something like one of the identities there of that comes from that.

 

[Peeter]

There's a vector blade reduction identity for r< s:

$$(a \wedge A_r) \cdot A_s = a \cdot (A_r \cdot A_s)$$

If I try to prove this for myself I need to use:

$$a \wedge A_r = a A_r - a \cdot A_r$$

So that I can first express this "triple" product as a direct product (with an observation that the dot product term doesn't contribute) :

$$(a \wedge A_r) \cdot A_s = {\langle a A_r A_s \rangle}_{\lvert s - (r+1) \rvert}$$

Reduction follows by expanding this by first taking the products of $$A_r$$ and $$A_s$$ and looking at the grades of the various terms.

But without that first wedge = product - dot result I didn't see how to do this for arbitrary grade blades, but after writing this down I think I now see how to approach it (not much different from above).