Infinite dimensional vector space has almost complex structure if and only if it is 'even-dimensional'? I started studying the book of Daniel Huybrechts, Complex Geometry An Introduction. I tried studying backwards as much as possible, but I have been stuck on the concepts of almost complex structures and complexification. I have studied several books and articles on the matter including ones by Keith Conrad, Jordan Bell, Gregory W. Moore, Steven Roman, Suetin, Kostrikin and Mainin, Gauthier
I have several questions on the concepts of almost complex structures and complexification. Here is one:
I understand for a finite dimensional $\mathbb R-$vector space $V=(V,\text{Add}_V: V^2 \to V,s_V: \mathbb R \times V \to V)$, the following are equivalent


*

*$\dim V$ even

*$V$ has an almost complex structure $J: V \to V$

*$V$ has a complex structure $s_V^{\#}: \mathbb C \times V \to V$ that agrees with its real structure: $s_V^{\#} (r,v)=s_V(r,v)$, for any $r \in \mathbb R$ and $v \in V$

*if and only if $V \cong \mathbb R^{2n} \cong (\mathbb R^{n})^2$ for some positive integer $n$ (that turns out to be half of $\dim V$) if and only if $V \cong$ (maybe even $=$) $W^2=W \bigoplus W$ for some $\mathbb R-$vector space $W$.


The last condition makes me think that the property 'even-dimensional' for finite-dimensional $V$ is generalised by the property '$V \cong W^2$ for some $\mathbb R-$vector space $W$' for finite or infinite dimensional $V$.
Question: For $V$ finite or infinite dimensional $\mathbb R-$vector space, are the following equivalent?


*$V$ has an almost complex structure $J: V \to V$

*Externally, $V \cong$ (maybe even $=$) $W^2=W \bigoplus W$ for some $\mathbb R-$ vector space $W$

*Internally, $V=S \bigoplus U$ for some $\mathbb R-$ vector subspaces $S$ and $U$ of $V$ with $S \cong U$ (and $S \cap U = \{0_V\}$)
 A: Yes, they are. Note that 6. and 7. are clearly equivalent (if we have 6. take for $S$ and $U$ the images of $W\times \{0\}$ and $\{0\}\times W$ under an isomorphism $W^2\overset{\sim}{\to} V$. If we have 7., then $V\simeq S\times U\simeq S\times S$, so take $W=S$.)
Assume that we have $7.$ Since $S$ and $U$ are isomorphic , their bases have same cardinality (countable or not). Pick $(s_i)_{i\in I}$ a basis of $S$, and $(u_i)_{i\in I}$ a basis of $U$ (we can index the two bases by the same set, thanks to the previous remark). 
Setting $J(e_i)=u_i$ and $J(u_i)=-e_i$ for all $i\in I$ yields an endomorphism $J$ satisyfing $J^2=-Id_V$.
Conversely, assume that we have an endomorphism $J$ of $V$ satisfying $J^2=-Id_V$.
The map $\mathbb{C}\times V\to {V}$ sending $(a+bi,v)$ to $av+ bJ(v)$ endows $V$ with the structure of a complex vector space which agrees on $\mathbb{R}\times V$ to its real structure.
Now pick a complex basis $(s_i)_{i\in I}$ of $V$, and set $u_i=i\cdot s_i=J(s_i)$.
Then, gluing $(s_i)_{i\in I}$ and $(u_i)_{i\in I}$, we obtain a real basis of $V$. The real subspaces $S=Span_\mathbb{R}(s_i)$ and $U=Span_\mathbb{R}(u_i)$ then satisfy the conditions of 7. 
A: GreginGre's solution is, of course, perfectly lovely, but if we're just killing this with choice, I guess you can also prove it as follows:
Let $V$ be infinite dimensional and, using Zorn's Lemma, let $\{e_i\}_{i\in I}$ be a basis for $V$. Using choice again, there exists $I_1$ and $I_2$ such that both $I_1\cap I_2=\emptyset,$ $I_1\cup I_2=I$ and there exists a bijection $\varphi: I_1\to I_2$. Thus, let $S=\textrm{span}\{e_i\}_{i\in I_1}$ and $U=\textrm{span}\{e_i\}_{i\in I_2}$. Then, $V=S\oplus U$ and $A:S\to U$ given by $e_i\mapsto e_{\varphi(i)}$ is a linear isomorphism of the two. This just proves that any infinite dimensional vector space admits such a decomposition, so there is only something to prove in the finite dimensional case.
A: As a supplement to the other answers, I'm going to prove (6 or) 7 implies 5 without axiom of choice. This is based on Joppy's answer and WoolierThanThou's comment:
Given an isomorphism $\theta: S \to U$, define $J: V \to V$ on the direct sum $V = S \bigoplus U$ by setting $J(s \oplus u) := - \theta^{-1}(u) \oplus \theta(s)$.
