In Grassmann algebra a la Browne, why are vectors dependent if their wedge product vanishes? I'm reading John Browne's Grassmann Algebra, Vol 1 : Foundations. Early on, he asserts without proof that if $x$ and $y$ are any two vectors in the underlying (real) vector space such that $x \wedge y = 0$, then $x$ and $y$ are linearly dependent. Take the vector space to be $R^3$, say. The result is equivalent to proving that if $e_i, e_j$ are two of the standard basis vectors, then $e_i \wedge e_j \neq 0$.
In the framework of axioms and or constructions that Browne provides, how does one prove that simple fact?
 A: The point is that the determinant gives you a non-zero linear function from $\wedge^n V$ to the ground field (which can be arbitrary) when $n=\mathrm{dim}(V)$. So this space is non-zero. Now if you have a basis $e_1,\dots,e_n$ of $V$, then multilinearity and skew-commutativity together imply $e_1 \wedge \cdots \wedge e_n$ spans the top exterior power, and must therefore be non-zero. The result you want follows.
On the other hand, as you point out in the comments below and I confirm, it seems that Browne's axioms are not enough to imply this fact about determinant, and indeed the result you want might fail: any quotient of the Grassmann algebra also satisfies his axioms.
A: Mathematicians often use the term "general position" such as in "Assume three points in the plane in general position." That is, the three points are not in a line. Sometimes general position is assumed without explicitly stating it. In any case, in Browne's application any set of distinct symbols representing vectors are taken to be "in general position (up to the dimension of the underlying space)". It's independence unless specified otherwise.
This is stated clearly in the Help page for the ExteriorProduct. "The GrassmannAlgebra and GrassmannCalculus Applications considers any set of distinct 1-element symbols, without further definitions or values, up to the linear dimension of the space, to be independent."
