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

1. Nov 17, 2009

### 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???

2. Nov 19, 2009

### mnb96

$$h(X) = f(X)+g(X)$$
$$h(a\wedge b) = h(a)\wedge h(b) = (f(a)+f(b))\wedge (g(a)+g(b))$$
That clearly implies that $$h(\mathbf{I})\neq f(\mathbf{I})+g(\mathbf{I})$$