By Shafarevich I.R.

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Consider what happens to each term when you take the outer product with Ar and multiply by A−1 r . The term RAr (b1 ) ∧ · · · ∧ RAr (bs ) is orthogonal to Ar , so by Theorem 18 the outer product becomes a product, so the Ar and A−1 r cancel out and you’re left with RAr (b1 ) ∧ · · · ∧ RAr (bs ). On the other hand, each of the other terms contains a factor that lies in Ar , so the outer product with Ar vanishes. Thus the second equation is valid too. Therefore PAr (Bs ) = Bs ⌋ Ar Ar−1 and RAr (Bs ) = Bs ∧ Ar A−1 r for any blade Bs .

5. The dual The next operation is called a duality transformation or taking the dual. Let Ar be an invertible r-blade; then the dual of any multivector B by Ar is B ⌋ A−1 r . ) To understand what taking the dual does, let B be a s-blade Bs . 1. If s > r, the dual of Bs vanishes. 2. If s = r, the dual of Bs is a scalar which is zero iff Bs contains a vector orthogonal to Ar (Theorem 16). 3. If s < r, the dual of Bs is either zero or an r − s-blade representing the orthogonal complement of Bs in Ar (Theorem 16 again).

1. Projecting a vector into a subspace Let’s take a moment to consider the general notion of projection into a subspace. Let S be a subspace (S is not a blade this time; it really is the subspace itself) and let a be a vector not in S. Then for any v ∈ S I can write a = v + (a − v), which is the sum of a vector in S and a vector not in S. So which v is the “projection” of a into S? We can’t say without further information. For example, consider two subspaces S1 and S2 that share only the zero vector; then if a vector lies in their direct sum, it can be expressed only one way as a vector from S1 plus a vector from S2 , and thus has a unique projection into either subspace.