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## Main Question or Discussion Point

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.