How to show $\nu=dx_1\wedge\ldots \wedge dx_n$?

Question

Let $\nu$ be the $n$-form in $\mathbb R^n$ satisfying $\nu(e_1, \ldots, e_n)=1$ where $\{e_1, \ldots, e_n\}$ is the canonical base of $\mathbb R^n$. Let $\displaystyle v_i=\sum_{j=1}^n a_{ij}e_i$. How to show $\nu=dx_1\wedge\ldots \wedge dx_n$?

Answer 1

The space of $n$-form in $\mathbb R^n$ is a vector space of dimension $1$ and since $\omega=dx_1\wedge \cdots \wedge dx_n$ is a non-zero $n$-form, any $n$-form is proportional to $\omega$.

So you get $\nu = \lambda \omega$ with $\lambda \in \mathbb R$.

But $\omega(e_1,\cdots,e_n)=1=\nu(e_1,\cdots,e_n)$ so $\lambda$ is equal to $1$ and consequently $\nu=\omega$.