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

  • 2
    $\begingroup$ You're proving the forward direction here, no? $\endgroup$ – Alex Provost Jul 30 '16 at 17:15

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.

  • $\begingroup$ I can't believe I missed that stupid "dependent", not "independent"... Sooo much time wasted for nothing... But then it's exactly like I thought, yes? (and that's what the lemma says also), the vectors of dependent wedge products span the same space, right? $\endgroup$ – JuliusL33t Jul 30 '16 at 17:24
  • $\begingroup$ @JuliusL33t It might be instructive to think about the geometry of bivectors in $\bigwedge^2 \mathbb{R}^3$, for example. Such a bivector is characterised by its "position" in space (i.e., what plane it lies in) and its magnitude (oriented area of the parallelogram spanned by the two vectors). Then if two such bivectors lie in the same plane, you may rescale one of them by a scalar to get both magnitudes to coincide. $\endgroup$ – Alex Provost Jul 30 '16 at 17:33

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