Show that $af∧bg=(ab)f∧g$

Question

Let $v$ be vector space. For $a$ and $b$ are in IR, $f$ is in $A_{k}(V)$ and $g$ is in $A_{l}(V)$

Show that $af∧bf=(ab)f∧g$

Here Will I use the definition of wedge product? Is ti right? How to use? Thank you.

Answer 1

We will use http://en.wikipedia.org/wiki/Exterior_algebra as a reference

Here $$(f \wedge g) (x_1, \dots, x_{k+l}) = \sum_{\sigma \in S_n} sgn(\sigma) f(x_{\sigma (1)}, \dots, x_{\sigma(k)}) g( x_{\sigma(k+1)}, \dots, x_{\sigma(k+l)}) $$

$$(af \wedge bg) (x_1, \dots, x_{k+l}) = \sum_{\sigma \in S_n} sgn(\sigma) a f(x_{\sigma (1)}, \dots, x_{\sigma(k)}) b g( x_{\sigma(k+1)}, \dots, x_{\sigma(k+l)}) = ab (f \wedge g) (x_1, \dots, x_{k+l})$$