Let $V$ be a vector space over $K$ with basis $\{e_1,\ldots,e_n\}$.
- For a fixed tuple $I = (i_1,\ldots,i_r)$ with $1 \le i_1 < i_2 < \cdots < i_r \le n$, use multilinear extension to conclude that there exists a multilinear map $\varphi_I: V^r \rightarrow K$ such that
\begin{align} \varphi_I(e_{j_1}, \ldots, e_{j_r}) &= \begin{cases} (-1)^{sgn(\sigma)} \quad \text{if there exists a permutation $\sigma$ of $[r]$ with $\sigma(j_k) = i_k$ for all $k \in [r]$}\\ 0 \qquad \text{else}\end{cases} \end{align}
for all $(j_1,\ldots,j_r) \in [n]^r$.
- Show that $\varphi_I$ is alternating and deduce by factorisation that there exists a linear map $f: \wedge^r V \rightarrow K$ satisfying $f \circ \wedge^r = \varphi_I$.
- Show that the set $A := \{e_{j_1} \wedge \ldots \wedge e_{j_r} : 1 \le j_1 < j_2 < \cdots < j_r \le n \}$ is linearly independent. Hint: Consider a linear relation of these elements and apply suitable maps $f$ to conclude all the coefficients are zero.
I got the first two points, but I do not understand how to use the hint for the third. Could you please help me?
Edit: Here is my attempt: We consider the sum $$\sum_{a \in A}c_aa =0$$ ,where the $c_a$ are scalars.Using the $f$ from above on the above equation yields
$$\sum_{a \in A} c_af(a) = 0.$$
However, when I use the definition of $f$ from above here I get
$$\sum_{\sigma \in \mathcal {S}_n} c_{\sigma}(-1)^\sigma = 0.$$
But this does not imply that all $c_\sigma$ are zero, since $\sum_{\sigma \in \mathcal {S}_n} (-1)^\sigma= 0$. Could you please tell me what I am doing wrong?
In the definition of $f$ the $r$-tuple $I$ is fixed. The particular $I$ appears exactly once in your set $A$. So you get $\pm c_I = 0$.