My question is about the last part of the proof (in bold), for more context, I'll quote all the proof:
(Linear basis of $\mathcal{G}_{2}$ ) The elements $\langle 1, e_{1}, e_{2}, e_{1} e_{2} \rangle \in \mathcal{G}_{2}$ are linearly independent. In particular, $\operatorname{dim} \mathcal{G}_{2}=4 .$ Moreover, the linear subspaces $\mathcal{G}_{2}^{0}=\langle 1\rangle=\mathbb{R}$ $\mathcal{G}_{2}^{1}=\left\langle e_{1}, e_{2}\right\rangle=E_{2},$ and $\mathcal{G}_{2}^{2}=\left\langle e_{1} e_{2}\right\rangle$ are independent of the orthonormal basis used to construct them.
Proof Assume a linear relation of the form $$ \lambda+\lambda_{1} e_{1}+\lambda_{2} e_{2}+\lambda_{12} e_{1} e_{2}=0 $$ Multiplying by $e_{1}$ from the left and from the right, we get the relation $$ \lambda+\lambda_{1} e_{1}-\lambda_{2} e_{2}-\lambda_{12} e_{1} e_{2}=0 $$ Adding the last two relations, we conclude that $\lambda+\lambda_{1} e_{1}=0$ and hence $\lambda=\lambda_{1}=0 .$ So we are left with the relation $\lambda_{2} e_{2}+\lambda_{12} e_{1} e_{2}=0$ or, multiplying on the right by $e_{2}, \lambda_{2}+\lambda_{12} e_{1}=0,$ which gives $\lambda_{2}=\lambda_{12}=0$ Since the assertions on $\mathcal{G}_{2}^{0}=\mathbb{R}$ and $\mathcal{G}_{2}^{1}=E_{2}$ are clear, what remains is to see that $\mathcal{G}_{2}^{2}$ is independent of the basis. We will do this by giving a description of it that is basis independent.
Consider the map $E_{2}^{2} \rightarrow \mathcal{G}_{2}$ given by $(v, w) \mapsto \frac{1}{2}(v w-w v)$. Since this map is bilinear and skew-symmetric, it gives a linear map $\wedge^{2} E_{2} \rightarrow G_{2}$ (which is basis independent by definition) such that $v \wedge w \mapsto \frac{1}{2}(v w-w v)$. In particular, $$ e_{1} \wedge e_{2} \mapsto \frac{1}{2}\left(e_{1} e_{2}-e_{2} e_{1}\right)=e_{1} e_{2} $$ which shows that the map in question yields a canonical linear isomorphism $$ \wedge^{2} E_{2} \simeq G_{2}^{2} $$ Thus we actually have a canonical linear isomorphism $$ \wedge E_{2}=\wedge^{0} E_{2} \oplus \wedge^{1} E_{2} \oplus \wedge^{2} E_{2} \simeq \mathcal{G}_{2}^{0}+\mathcal{G}_{2}^{1}+\mathcal{G}_{2}^{2}=\mathcal{G}_{2} $$
My question is pretty basic but I think I am missing something here : How does being skew-symmetric and bilinear gives a linear map $\wedge^{2} E_{2} \rightarrow \mathcal{G}_{2}$ ? if I understood correctly: if I have a bilinear and skew-symmeteric map that takes an element from $E_2$ (2-vector) to $\mathcal{G}_{2}$ then I have a linear map from $\wedge^{2} E_{2} $ to $\mathcal{G}_{2}$. To see this, I've just applied this to a simple exapmle to see
($v \wedge w + z \wedge y) \mapsto 1/2 (vw-wv) + 1/2(zy - yz) = 1/2 ((vw + zy)-(wv + yz)) $
I am assuming that if I take two elements, by linearity I should find something like $(v+z) \wedge (w+y)$, I am probably applying incorrectly, but what's linear with respect to what in the explanation of this proof?
and how does the map in question yields a canonical linear isomorphism?
Thank you.