- #1
Peeter
- 303
- 3
Anybody know a good approach to factor a wedge product (blade) into a set of vectors? Loosely, I'd describe the prodlem as finding a basis for the hypervolume that the wedge product "spans".
Example to illustrate the question, taking a grade 2 blade, suppose one had something like:
<br /> A = (e_1 + e_2 + e_3) \wedge ( e_1 + 2 e_4 ) = 2 e_1 \wedge e_4 + e_2 \wedge e_1 + 2 e_2 \wedge e_4 + e_3 \wedge e_1 + 2 e_3 \wedge e_4<br />
The problem is to work backwards to find some pair of vectors that would generate the same wedge product.
One approach that I believe would work, but doesn't seem too efficient, is to utilize the wedge product as a projector onto the space:
<br /> Proj_A(x) = \frac{1}{A} (A \cdot x) = A^\dagger \frac{1}{A^\dagger A}(A \cdot x)<br />
(observe the geometric algebra dot product above).
If you applied this to the basis vectors for the space, you have essentially computed the coordinate vectors for the matrix of the projection operator, so these columns form a basis of the space, and are thus factors of the original wedge product. A column reduction to reduce the rank would provide the factorization desired (disregarding the scaling factor).
As an aside, I found it pretty interesting how similar the projection product is to the projection matrix from traditional matrix algebra, where provided the matrix A is of full column rank, a projection onto it's columns, can be written:
<br /> Proj_A(x) = \left( A \frac{1}{A^\text{T} A} A^\text{T} \right) x<br />
A further comparison. In the above if the columns are othonormal you have:
<br /> Proj_A(x) = \left( A A^\text{T} \right) x<br />
Similarily, for a unit magnitude blade, the inverse term drops out leaving:
<br /> Proj_A(x) = A \left(A^\dagger \cdot x\right)<br />
Example to illustrate the question, taking a grade 2 blade, suppose one had something like:
<br /> A = (e_1 + e_2 + e_3) \wedge ( e_1 + 2 e_4 ) = 2 e_1 \wedge e_4 + e_2 \wedge e_1 + 2 e_2 \wedge e_4 + e_3 \wedge e_1 + 2 e_3 \wedge e_4<br />
The problem is to work backwards to find some pair of vectors that would generate the same wedge product.
One approach that I believe would work, but doesn't seem too efficient, is to utilize the wedge product as a projector onto the space:
<br /> Proj_A(x) = \frac{1}{A} (A \cdot x) = A^\dagger \frac{1}{A^\dagger A}(A \cdot x)<br />
(observe the geometric algebra dot product above).
If you applied this to the basis vectors for the space, you have essentially computed the coordinate vectors for the matrix of the projection operator, so these columns form a basis of the space, and are thus factors of the original wedge product. A column reduction to reduce the rank would provide the factorization desired (disregarding the scaling factor).
As an aside, I found it pretty interesting how similar the projection product is to the projection matrix from traditional matrix algebra, where provided the matrix A is of full column rank, a projection onto it's columns, can be written:
<br /> Proj_A(x) = \left( A \frac{1}{A^\text{T} A} A^\text{T} \right) x<br />
A further comparison. In the above if the columns are othonormal you have:
<br /> Proj_A(x) = \left( A A^\text{T} \right) x<br />
Similarily, for a unit magnitude blade, the inverse term drops out leaving:
<br /> Proj_A(x) = A \left(A^\dagger \cdot x\right)<br />