I am a bit confused about what appears to be basic linear algebra in Voisin's book. I suspect I am misunderstanding maybe some of the notations.
My first confusion is from this passage.
$W_{\mathbb{C}}:=\text{Hom}_{\mathbb{R}}(V,\mathbb{C})$ admits the decomposition $W_{\mathbb{C}}=W^{1,0}\oplus W^{0,1}$ into $\mathbb{C}$-linear and $\mathbb{C}$-antilinear forms. Let $W^{1,1}=W^{1,0}\otimes W^{0,1}\subset \bigwedge^2 W_{\mathbb{C}}$.
So, $W^{1,1}$ consists of a form that is linear in the first component and antilinear in the second as $f\in W^{1,1}$ can be written as $f(u,v)=\sum_{j=1}^n g_j(u)\overline{h_j(v)}$ where the $g_j, h_j$ are $\mathbb{C}$-linear. Why is this in $\wedge^2 W_{\mathbb{C}}$? In other words, why should this be alternating? For example, take $f(u,v)=g(u)\overline{h(v)}=u\overline{v}$ where we take $g,h$ to be the identity map. This is in $W^{1,1}$ but it's not alternating.
A little bit down below, we have the following theorem.
Lemma 3.3: There is a natural identification between Hermitian forms on $V\times V$ and the elements of $W_{\mathbb{R}}^{1,1}$ given by $h\mapsto \omega=-\Im h$.
$W^{1,1}_{\mathbb{R}}$ is defined as $W^{1,1}\cap \bigwedge^2 W_{\mathbb{R}}$. After checking the $\omega$ is alternating, the proof proceeds to show that $\omega$ is in $W^{1,1}$.
But by definition, $\omega$ is in $W^{1,1}$ if and only if the natural extension of $\omega$ (by $\mathbb{C}$-bilinearity) to a 2-form on $V_{\mathbb{C}}$ vanishes on the bivectors $(u,v)$, $u,v\in V^{1,0}$ and on the bivectors $(u,v)$, $u,v\in V^{0,1}$ , the second property following from the first by using complex conjugation.
Here $V^{1,0}$ and $V^{0,1}$ are the eigenspaces of the operator $I: V\to V$ corresponding to eigenvalues $i$ and $-i$.
For example, the hermitian form $h(u,v)=u\overline{v}$ yields $\omega(u,v)=-\Im (u\overline{v})=-(u\overline{v}-v\overline{u})$. Sure, $(u,v)\mapsto u\overline{v}$ is in $W^{1,1}$ (linear in first variable, antilinear in second) but $(u,v)\mapsto v\overline{u}$ is not.
