Understanding proof concerning wedge products I'm reading Andreas Gathmanns lecture notes on Algebraic geometry, found here, and on page 68, there is a proof of lemma 8.8 where I don't understand a step, could anyone explain?
Lemma 8.8: Let $v_1,...,v_k \in K^n$ and $w_1...w_k \in K^n$ be both linearly independent. Then $v_1 \wedge ... \wedge v_k$ and $w_1 \wedge ... \wedge w_k$ are linearly dependent in $\Lambda^kK^n$ if and only if $\text{Lin}(v_1,...,v_k) = \text{Lin}(w_1,...,w_k)$
(here I take $\text{Lin}(v_1...v_k)$ to mean the subspace spanned by those vectors)
Proof:("$\leftarrow$") Assume that $\text{Lin}(v_1,...,v_k) \not= \text{Lin}(w_1,...,w_k)$, so without loss of generality that $w_1 \not\in \text{Lin}(v_1,...,v_k) $. 
Comment: OK $v_i$ and $w_i$ don't span the same space, so $w_1 \not\in \text{Lin}(v_1,...,v_k) $. Some $w_i$ could be in the span but at least one is not, for else they would span the same subspace.
Proof: Then $w_1, v_1,...v_k$ are linearly independent and thus $w_1 \wedge v_1 \wedge... \wedge v_k \not= 0$.
Comment: This is fine too, since wedges are $0$ iff their vectors are linearly dependent.
Proof: But by assumption, $v_1 \wedge ... \wedge v_k = \lambda w_1 \wedge ... \wedge w_k$ for some $\lambda \in K$...
And here I don't understand, where is this assumption? I thought this only happened exactly when they DO span the same subspace? For example: $e_1, e_2 \in \mathbb{R}^3 $ span the same space as $2e_1,e_2$ and $(2e_1) \wedge e_2 = 2e_1\wedge e_2$, and we assumed exactly the opposite! Is this wrong? If so, under what condition does this happen and what's the intuitive meaning of $v_1 \wedge ... \wedge v_k = \lambda w_1 \wedge ... \wedge w_k$?
Thanks in advance

He's proving the forward direction. Namely, he's assuming that $\bigwedge v_i$ and $\bigwedge w_i$ are linearly dependent, and two vectors are dependent iff one is a scalar multiple of the other.
