Conformal Geometric Algebra in Stochastic Optimization Problems ...

42 CHAPTER 2. GEOMETRIC ALGEBRA
Definition 2.5 (Blade ):
The outer product of a set of 0 ≤ k ≤ n linearly independent vectors {a 1...k } ⊂ � p,q
is called a blade of grade k or simply k-blade. It is denoted by A 〈k〉 .
A 〈k〉 = a1 ∧ a2 ∧ ... ∧ ak
A scalar is considered as a blade of grade 0.
Subsequently, every appearance of blades A 〈k〉 , B 〈l〉 , etc will implicitly stand for
� k
i=1 ai, � l
i=1 bi, etc if no deviant declaration is given.
As a matter of course, every blade is a certain κ-vector, which is why every computation
rule for κ-vectors is applicable to blades as well. Likewise, any κ-vector
A [k] is a k-blade A 〈k〉 iff it has an outer product representation a1 ∧ a2 ∧ ...∧ak.
Example 2.4 ( κ-vector vs. blade ):
(e1 + e2) ∧ (e2 + e3) ⇐⇒ e1e2 + e1e3 + e2e3 blade
?? ⇐= e1e2 + e1e3 + e2e4 κ-vector.
On page 23 it is stated that the outer product a ∧ b, n > 2, represents a linear 2Dsubspace,
i.e. a plane. In addition, it is shown on page 32 that the outer product
of vectors is zero if the vectors are linearly dependent. Hence the outer product of
a vector x with a blade A 〈k〉 = �k i=1 ai is zero if the vectors {x,a1,a2,...,ak} are
linearly dependent, or rather if x lies in the k-dimensional subspace spanned by the
{a1...k }. The spanning vectors are of course termed the basis or the frame of the
subspace.
Corollary 2.7
A blade A 〈k〉 = � k
i=1 ai represents a linear k-dimensional subspace in that
x ∧ A 〈k〉 = 0 ⇐⇒ x lies in the span of the {a 1...k }
for every vector x ∈ � p,q
Proof: An outline for a direct proof of the ‘only-if’ direction is given on page 32.
The ‘if’ direction can be deduced from equation (2.31): let x ∈ � n and A ∈ � n×k be
the matrices, the columns of which hold the coefficients of x ∈ � p,q and A 〈k〉 ∈ �p,q,
respectively. Then x ∧ A 〈k〉 may be expressed as
x ∧ A 〈k〉 = �
v∈I k+1/n
det([x, A]|v) ev1ev2 ...evk+1 ,
where [x,A] ∈ � n×k+1 symbolizes the horizontal concatenation of the matrices x
and A. If at least one k+1-minor det([x, A]|v) was non-zero, the matrix [x, A] would
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