$n$-form on $\mathbb{R}^n$ - $w(v_1,v_2,\cdots ,v_n)=\det A \cdot w(e_1,e_2,\cdots, e_n)$

Question

Is not hard to see that for dimension $n=2$ we have, for $w$ a two-form:

$w(v_1,v_2)=\det A \cdot w(e_1,e_2)$, where $A$ is s.t. $(v_1,v_2)=(e_1,e_2)\cdot A.$

I'd like to generalize this result for dimension $n$ and respective $n$-form (same dimension), but I am stucked at calculus.

Many thanks for any clue!

Answer 1

Let $V$ be a vector space of finite dimension $n$, and let $f:V \to V$ be a linear map. The induced map $f: \Lambda^n V \to \Lambda^n V$, given explicitly by \begin{align*} f(v_1 \wedge \cdots \wedge v_n) = f(v_1) \wedge \cdots \wedge f(v_n), \end{align*} is just multiplication by $\det V$ on the $1$-dimensional space $\Lambda^n V$ by definition. Translating that into the notation of forms gives the relation you're looking for.