Linear independence regarding Exterior Power .

Question

I have been trying to learn the proof of dimension of exterior power from this text : http://www.thehcmr.org/issue1_2/poincare_lemma.pdf.( Page 16)
I am not able to understand the part of linear independence ( how its been assured) . It would be nice if you could explain . thank you .

Answer 1

To say that vectors $v_1, \ldots, v_n$ (lets focus on $n \geq 2$) are linearly dependent is equivalent to saying that some fixed $v_i$ is a linear combination of $v_1, \ldots, v_{i-1}, v_{i+1}, \ldots, v_n$. If you know beforehand that $v_i \neq 0$ for every $i$, then evidently the coefficients of this linear combination can't all be zero.

In your particular case, it's not hard to see that $e_{i_1} \wedge \dots \wedge e_{i_s} \neq 0$ by using, for instance, the universal property. Now suppose that for some fixed set of indices $\{i_1, \ldots, i_s\}$ you had $$e_{i_1} \wedge \dots \wedge e_{i_s} = \sum_{\{i_1, \ldots, i_s\} \neq \{j_1, \ldots, j_s\}} a_{j_1, \ldots, j_s} e_{j_1} \wedge \dots \wedge e_{j_s}.$$ In this case, there should be some multi-index $\{j_1, \ldots, j_s\}$ such that $a_{j_1, \ldots, j_s} \neq 0$, because the left-handed side is not zero. Applying the corresponding $B_{j_1, \ldots, j_s}$ (which is a linear map) to both sides gives you $$B_{j_1, \ldots, j_s}(e_{i_1} \wedge \dots \wedge e_{i_s}) = a_{j_1, \ldots, j_s}.$$ But since $\{i_1, \ldots, i_s\} \neq \{j_1, \ldots, j_s\}$, the left-handed side must be zero, by construction, and the right-handed side was supposed to be $\neq 0$. Hence the contradiction.