I have to following statement, which I want to proof. Unfortunatly I am not used to the dual space and know barely anything about it.
I want to proof what seems trivial. Namely:
Let $e_1,\dotso, e_k$ be the canonical basis of $\mathbb{R}^k$ and let $\delta^1,\dotso,\delta^k$ be the dual basis of $\mathbb{R}^{k\ast}$, then is
$\delta^1\wedge\dotso\wedge\delta^k(e_1,\dotso, e_k)=1$ for every $k\geq 1$.
If I am informed right, then $\delta^i(e_j)=\begin{cases}1\quad,\text{for}\,\,j=i\\0\quad,\text{else}\end{cases}$ just by definition, of the dual basis.
The statement should follow easily by induction. The start for k=1 is trivial by definition. Since the $\wedge$-product is associative the inductive step should go like this:
$\delta^1\wedge\dotso\wedge\delta^{k+1}(e_1,\dotso,e_{k+1})=(\delta^1\wedge\dotso\wedge\delta^k)(e_1,\dotso, e_k)\wedge\delta^{k+1}(e_{k+1})=1\wedge\delta^{k+1}(e_{k+1}) \\=1\wedge 1=1$
Is this correct? Thanks in advance.