Almost complex structures and multiplication by $i$ Let $V$ be a real vector space (of finite dimension) with an almost complex structure $I$, namely an endomorphism $I:V\to V$ such that $I^2=-\operatorname{id}$. Consider the extension by scalars $V_{\mathbb C}:=V\otimes_{\mathbb R}\mathbb C$ and the extension by linearity $I:V_{\mathbb C}\to V_{\mathbb C}$ (we mantain the same name for $I$ just by comodity). Now the following things hold:


*

*The eigenvalues of $I$ are $\pm i$.

*If $V^{1,0}$ is the eigenspace of $i$ and $V^{0,1}$ is the eigenspace of $-i$, then $V^{1,0}\cong V^{0,1}$. Moreover $V_{\mathbb C}=V^{1,0}\oplus V^{0,1}$



Now consider the following example: $V$ is a real vector space which is also a complex vector space. On his "real structure'' we consider the endomorphism $v\mapsto i\cdot v$. This clearly defines an almost complex structure $I$ on $V$, but the problem is that $1.$ and $2.$ seems to be false. I mean, here we have only one eigenvalue which is $i$ and $V_{\mathbb C}=V^{1,0}$.
Where is my mistake, probably I'm forgetting someting in the extension by scalars $V_{\mathbb C}$.
 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Cpx}{\mathbf{C}}\newcommand{\Basis}{\mathbf{e}}$For example, take $V = \Cpx^{m}$, and let $(\Basis_{j})_{j=1}^{m}$ denote the standard (complex) basis. The corresponding "real scalars" vector space is isomorphic to $\Reals^{2m}$, with (real) basis $(\Basis_{j})_{j=1}^{m} \cup (i\Basis_{j})_{j=1}^{m}$.
The complexification is (isomorphic to)
$$
\Reals^{2m} \otimes_{\Reals} \Cpx = \Cpx^{2m},
$$
with complex basis $(\Basis_{j} \otimes 1)_{j=1}^{m} \cup (i\Basis_{j} \otimes 1)_{j=1}^{m}$. Complex scalar multiplication acts by $\beta(v \otimes\alpha) = v \otimes \alpha\beta$, so the complexification has real basis
$$
(\Basis_{j} \otimes 1)_{j=1}^{m} \cup (i\Basis_{j} \otimes 1)_{j=1}^{m} \cup (\Basis_{j} \otimes i)_{j=1}^{m} \cup (i\Basis_{j} \otimes i)_{j=1}^{m}.
$$
The complex structure $I$ acts via $I(v \otimes \alpha) = iv \otimes \alpha$. The spaces $V^{1, 0}$ and $V^{0, 1}$ have respective complex bases
$$
\Basis_{j}^{1, 0} = \tfrac{1}{2}(\Basis_{j} \otimes 1 - i\Basis_{j} \otimes i),\qquad
\Basis_{j}^{0, 1} = \tfrac{1}{2}(\Basis_{j} \otimes 1 + i\Basis_{j} \otimes i).
$$
A: If you start with a complex vector space $V$  and realify it  you obtain a  real vector space $V_0$ endowed with a complex structure satisfying $I(v_0)=(iv)_0$, where $v_0$ means $v$ seen as a vector in $V_0$.
If you complexify $V$ you obtain- surprise, surprise!-$$u: (V_0)_\mathbb C \stackrel {\cong}{\longrightarrow}V\oplus \overline  V \quad (\bigstar)$$ where $\overline  V $ is the abstract conjugate of $V$, which  means that $\overline V=V$ as abelian groups but $i* v$ in $\overline V$ is equal to $-iv$ in $V$.    
The isomorphism $u: (V_0)_\mathbb C \stackrel {\cong}{\longrightarrow}   V\oplus \overline  V $ is given by the formula $$u(v_0\otimes z)=(z\cdot v, \bar z\cdot v) \quad (\bigstar)$$ and is $ \mathbf C$-linear.
(Beware however that not every vector in $(V_0)_\mathbb C $ can be written $v_0\otimes z$: you have to take the sum $v_0\otimes 1+w_0\otimes i$ of two such)    
We then have (no longer distinguishing notationally $v$ from $v_0$) $$(V_0)_\mathbb C^{1,0}=\{v\otimes 1-iv\otimes i\vert \:v\in V\},\quad (V_0)_\mathbb C^{0,1}=\{v\otimes 1+iv\otimes i\vert \:v\in V\}$$ 
The  decomposition of $(V_0)_\mathbb C$ is $$(V_0)_\mathbb C=(V_0)_\mathbb C^{1,0}\oplus (V_0)_\mathbb C^{0,1}\\v\otimes 1+w\otimes i=
[\frac {v+iw}{2}\otimes1-i\frac {v+iw}{2}\otimes i ]+[\frac {v-iw}{2}\otimes1+i\frac {v-iw}{2}\otimes i       ]$$ and it satisfies $u[(V_0)_\mathbb C^{1,0}]=V\oplus0,\quad u[(V_0)_\mathbb C^{0,1}]=0 \oplus \overline V$
