Let be $(E,\mathbb{K}),(F,\mathbb{K})$ Banach space and $U\subset E$ open set. If $f_1,...,f_n\in\Omega_1(U;F)$ differential 1-forms are linearly independents, where
$$\Omega_1(U;F)=\{f:U\rightarrow \mathcal{A}_1(E;F)=\mathcal{L}(E;F)\},$$
then $f_1\wedge...\wedge f_n$ is different of the null application.
So, I have some queston about this.
First, $\Omega_1(U;F)$ is really a vector space over $\mathbb{K}$, where $\mathbb{K}$ is a field? (Since I have a vectorial space structure, I can to talk of linear independence and dependence of vectors) Cause, I know just that $$\bigcup_{p\geq 0}\Omega_p(U;F)=\Omega(U;F)$$ is a vector space.
I do that when $E$ has finite dimension, because I can to use the dual basis, but I don't know to do to infinite dimension.
The back sense of the affirmation is true and easy.
Note that, given $x_1,...,x_n\in E$, then $f_1\wedge...\wedge f_n(x_1,...,x_n)=\det\{f_i(x_j)\}.$