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where $B$ is a $k$-vector in $n$ dimensional space and $ \mathbf I $ is the pseudoscalar.

Muphrid
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user997712
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  • What have you tried to do so far on this? It's standard policy for homework problems to show a little preliminary work or thoughts. – Muphrid Jun 24 '13 at 01:16
  • Will keep that in mind for future questions; i can imagine how that will make for a better discussion. It isn't a homework question strictly speaking though. – user997712 Jun 24 '13 at 19:19

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I don't quite recall how Macdonald defines a blade, but Hestenes has a quite useful definition: a $k$-blade is composed of $k$ orthogonal vectors in a geometric product. So $B = b_1 b_2 \ldots b_k$ where $b_i b_j = -b_j b_i$ for any $i \neq j \in 1, 2, \ldots, k$.

Similarly, the psedoscalar $I$ should be, within some constant factor, $I = \alpha B c_{k+1} c_{k+2} \ldots c_n$ for some constant scalar $\alpha$ and some anticommuting vectors $c_{k+1}, c_{k+2}, \ldots$.

From here, the number of times you need to commute anticommuting vectors with each other should be clear. Remember, though, that each vector that forms $B$ does appear in $I$ exactly once. As such, there will be one swap for each vector that is a commuting swap, not incurring a sign change.

Edit: I see Macdonald was talking about a $k$-vector, not necessarily a blade. The same logic basically applies, but the $k$-vector can be broken down into some sum of linearly independent $k$-blades.

Muphrid
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  • Thanks Murphid, I was stuck at the last bit of logic there: one of the swaps will be a commuting swap. I was getting $ (-1)^{kn} $ because of neglecting that. – user997712 Jun 24 '13 at 19:17