Wedge product equals 0 implies linear dependence of 1-forms?

Question

Why is it the case that, if $v$ and $w$ are differential 1-forms in $\Bbb R^2$, then if $v \wedge w = 0$ then $v = cw$ for some $c \in \Bbb R$? The converse seems straightforward enough, but I couldn't figure out a way to prove this statement. Does it depend on the alternating nature of the wedge product?

Answer 1

Differential 1-forms on $\mathbb{R}^2$ can be written $$v= f_1 dx + g_1 dy, w = f_2 dx + g_2 dy$$ Where $f_i, g_i$ are smooth functions. Therefore, $$v \wedge w = 0 \iff f_1 g_2 \cdot dx \wedge dy + g_1 f_2 \cdot dy \wedge dx = 0\iff f_1 g_2 - f_2 g_1 = 0$$ In particular, the wedge is $0$ iff the two forms are linearly dependent (since this is merely the determinant 0 condition).