Do we assume as part of axioms, that Wedge Product of two linearly independent vectors in $V$ will be non-zero?
If not, can we show just using axioms of Wedge Product that wedge product of two linearly independent vectors of $V$ is non-zero? I know, if we can write $\Lambda^2 V$ as isomorphic to quotient space $\dfrac{F(V\times V)}{S(V\times V)}$, with their usual meaning, then we can easily show it, but I was thinking of proving it directly from bi-linearity and self-annihilation property of Wedge Product on $\Lambda^2 V$.