# 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?

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Can you provide that framework / those axioms, at least in a nutshell? –  Berci May 18 '13 at 22:01
I'm going to guess one of them is that for an $n$-dimensional vector space with basis $v_1,\dots,v_n$, we have $v_1\wedge\dots\wedge v_n\ne 0$. :) –  Ted Shifrin May 18 '13 at 22:04
Well, it's usually defined as a quotient of tensor power. –  Berci May 18 '13 at 22:06
Well if you believe that the $e_i\wedge e_j$, $i\neq j$ are part of the standard basis for the algebra, then you already know they're nonzero... –  rschwieb May 18 '13 at 22:41
@TedShifrin: No, unfortunately, what you guess in included in the axioms in Browne's treatment is not in fact there. Unless there's something much more subtle, I think what's missing is one or another version of what's called Axiom 4 in William Schulz's document cefns.nau.edu/~schulz/grassmann.pdf (pages 50-51). –  murray May 19 '13 at 3:42
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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.
@murray, Well, if you take 2.7-2.18 in Browne's book as the axioms, and assume there are no other relations, then what I wrote is a consequence. That is, linear functions on $\wedge^k V$ are identified with $k$-multilinear alternating functions on $V$. –  S123 May 19 '13 at 15:23