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

[mnb96]

Hello,
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:

$$f(\mathbf{I_n})=det(f)\mathbf{I_n}$$

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

However if we introduce the function $$h(X) = f(X)+g(X)$$ we have that:

$$h(\mathbf{I_n})= f(\mathbf{I_n}) + g(\mathbf{I_n}) = det(h)\mathbf{I_n} = det(f)\mathbf{I_n} + det(g)\mathbf{I_n}$$

We have essentially proved that det(F+G)=det(F)+det(G). There must be a trivial mistake in this: where is it?

 

[mnb96]

After discussing about this issue I think I see where the mistake is.
The mistake was in introducing a function for general multivectors

$$h(X) = f(X)+g(X)$$

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:

$$h(a\wedge b) = h(a)\wedge h(b) = (f(a)+f(b))\wedge (g(a)+g(b))$$

That clearly implies that $$h(\mathbf{I})\neq f(\mathbf{I})+g(\mathbf{I})$$