Doran/Lasenby example of determinant of linear geometric product operator?


by Peeter
Tags: determinant, doran or lasenby, geometric, linear, operator, product
Peeter
Peeter is offline
#1
Apr2-08, 09:03 PM
P: 294
Geometric Algebra for Physicists has an example of a linear operator:

[tex]
F(a) = a + \alpha(a \cdot f_1) f_2.
[/tex]

This is used to compute the determinant without putting the operator in matrix form. Their first step is to compute:

[tex]
F(a \wedge b) = F(a) \wedge F(b) = a \wedge b + \alpha((a \wedge b) \cdot f_1) f_2
[/tex]

(this is straightforward enough to do). However, they then state that this implies for any grade blade the following is true:

[tex]
F(A) = A + \alpha(A \cdot f_1) f_2
[/tex]

This is true by inspection for a 2-blade from the above, and since [tex](a \cdot f_1) f_2 = {\langle (a \cdot f_1) f_2 \rangle}_1 = (a \cdot f_1) \wedge f_2[/tex] this is true for the initial linear operator on a vector too.

Sure enough if I calculate [tex]F(a) \wedge F(b) \wedge F(c)[/tex] (not as straightforward as the bivector case) I can confirm their statement for a grade 3 blade.

I assume this could be done inductively for any grade. However, I'm wondering if there was some other principle that I didn't notice in my reading that allows them to make this statement for any grade blade without an inductive proof?

Anybody see what that would be?

I'm guessing this would be related to the matrix concept of rank in some way too, but it's not clear to me exactly how.
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