Let $V$ be a complex vector space. A statement in Complex Geometry of Huybrechts p.25 is, that one could define an almost complex structure on the underlying real vector space of $V$ by $v \mapsto iv$.
If I get this right the underlaying real vector space of $V$, which I am going to call $W$, is the vector space one gets simply by taking all the vectors in $V$ and restricting the scalars to be in $\mathbb{R}$. Also if I get Huybrechts right, the idea of the definition of the almost complex structure is that every vector $v \in W$ is mapped to the vector one would get if multiplied with $i$ in the complex space $V$.
Now however I can't proof that this in fact is a linear mapping, which it should be in order to be an Endomorphism on $V$. My problem is that I'm not quite sure what a linear mapping between a complex vector space and a real vector space should be. Therefore my idea to understand this as a linear mapping
$$W \rightarrow V \overset{i}{\rightarrow} V \rightarrow W$$ fails in the last mapping $V \rightarrow W$.
I would be thankful for Your advice on what I am missing.
I think calling the real vector space "$W$" is a bit confusing here. After all, it's still the same set! We just forget that there is a multiplication by $\mathbb C$, and only remember multiplication by $\mathbb R$. So let's keep the name $V$. Now we have to show that the map $$V \to V, v \mapsto iv$$ is $\mathbb R$-linear. But as it is even $\mathbb C$-linear, and $\mathbb R \subset \mathbb C$, it is also $\mathbb R$-linear. Spelled out $\mathbb R$-linearity means $$i(v+w) = iv + iw \quad \text{for} \quad v,w \in V$$ and $$i(\lambda v) = \lambda(iv) \quad \text{for} \quad v \in V, \lambda \in \mathbb R,$$ and both properties follow from the axioms of vector spaces and the commutativity of $\mathbb C$.