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Determinant of (A+B) in GA: where is the mistake?

  1. Nov 17, 2009 #1
    I am quite new to Geometric Algebra, this is the reason for the silly question.
    In Geometric Algebra, the following implicit definition of determinant is given:


    where f is a linear function extended as an outermorphism, and [tex]\mathbf{I_n}[/tex] is the unit n-blade for [tex]\wedge\mathcal{R}^n[/tex], for example [tex]e_1\wedge\ldots\wedge e_n[/tex]. It is also shown that f can be represented as a square matrix. We also know that in general: [tex]det(A+B)\neq det(A) + det(B)[/tex].

    However if we introduce the function [tex]h(X) = f(X)+g(X)[/tex] we have that:

    [tex]h(\mathbf{I_n})= f(\mathbf{I_n}) + g(\mathbf{I_n}) = det(h)\mathbf{I_n} = det(f)\mathbf{I_n} + det(g)\mathbf{I_n}[/tex]

    We have essentially proved that det(F+G)=det(F)+det(G). There must be a trivial mistake in this: where is it???
  2. jcsd
  3. Nov 19, 2009 #2
    After discussing about this issue I think I see where the mistake is.
    The mistake was in introducing a function for general multivectors

    [tex]h(X) = f(X)+g(X)[/tex]

    In this case f and g are not necessarily linear functions extended as outermorphisms, so we cannot treat them in principle in GA, and they do not have matrix representations.
    One must define f(x) as a mapping vector->vector and then extend it to outermorphism. Then one would have:

    [tex]h(a\wedge b) = h(a)\wedge h(b) = (f(a)+f(b))\wedge (g(a)+g(b))[/tex]

    That clearly implies that [tex]h(\mathbf{I})\neq f(\mathbf{I})+g(\mathbf{I})[/tex]
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