Why did I get $J \bar{v} = \sqrt{-1}\bar{v}$ ?? Q. I cannot deduce the equation $J \bar{v}= -\sqrt{-1}\bar{v}$ for $v$ satisfying $Jv= \sqrt{-1}v$.
Details
Let $V$ be a vector space on $\mathbb{R}$ equipped with a complex structure $J$. Using $J$, the action of $\mathbb{C}$ to $V$ is defined by $(a+b\sqrt{-1})v := av + bJv$ for $v \in V$ and $a,b \in \mathbb{R}$.
Let $V_{\mathbb{C}}:= V \otimes_ \mathbb{R} \mathbb{C}$.
For  $v= w\otimes_ \mathbb{R} \alpha \in V_{\mathbb{C}}$, its complex conjugate is defined by $\bar{v}:= w\otimes_ \mathbb{R} \bar{\alpha} \in V_{\mathbb{C}}$.
I cannot understand the statement that  $Jv= \sqrt{-1}v$ implies $J \bar{v}= -\sqrt{-1}\bar{v}$, where $v \in V_{\mathbb{C}}$.
In my view,
$J \bar{v} = J( w\otimes_ \mathbb{R} \bar{\alpha} )= (J w) \otimes_ \mathbb{R} \bar{\alpha} =  (\sqrt{-1} w) \otimes_ \mathbb{R} \bar{\alpha}$
I get stuck, because $\sqrt{-1}$ cannot pass the $\otimes_ \mathbb{R}$. So,.. but maybe
$J \bar{v} = J( w\otimes_ \mathbb{R} \bar{\alpha} )= (J w) \otimes_ \mathbb{R} \bar{\alpha} =  (\sqrt{-1} w) \otimes_ \mathbb{R} \bar{\alpha}:=  \sqrt{-1}( w \otimes_ \mathbb{R} \bar{\alpha})=  \sqrt{-1} \bar{v}$ ??
Where do I misunderstand??
 A: I suggest you forget about the $\mathbb C$-action that you defined on $V$. It has nothing to do with what you want to prove, and it seems to be the root of the confusion.
Low tech argument: In general, let $A : \mathbb R^n \to \mathbb R^n$ be a real matrix, then by tensoring $\mathbb C$, you have the same real matrix, but now it's acting on $\mathbb C^n$:
$$A: \mathbb C^n \to \mathbb C^n.$$
(As an operator I should have used another notation, say $A_{\mathbb C}$, but as a matrix they are exactly the same matrix). If $\lambda$ is an eigenvalue of $A$ with eigenvector $v$, then
$$Av = \lambda v.$$
Taking complex conjugate, since $\bar A = A$,
$$\overline {Av} = \overline{\lambda v} \Rightarrow A \bar v = \bar \lambda \bar v.$$
thus $\bar\lambda$ is an eigenvalue with eigenvector $\bar v$. That's exactly what you want.
Using tensor notations There are several mistake in your argument,

*

*first, you are assuming that the eigenvector $v\in V_{\mathbb C}$ is of the form $v = w\otimes \alpha$, but in general, elements in $V_\mathbb C$ are of the form
$$ w_1\otimes \alpha_1 + \cdots + w_k \otimes \alpha_k.$$


*Second, the equality $(Jw ) \otimes \alpha = \sqrt{-1} w \otimes \alpha$ is also wrong. Indeed $w \in V$ is real and $J$ acts on $V$, there is no way we have
$$ Jw = \sqrt {-1} w,$$
since the RHS is complex and the LHS is real.
The correct argument is: let
$$ v = w_1\otimes \alpha_1 + \cdots + w_k \otimes \alpha_k$$
be an eigenvector corresponding to $\sqrt{-1}$, then
\begin{align} 
J \bar v &= J (w_1 \otimes \bar\alpha_1 + \cdots + w_k \otimes \bar \alpha_k) \\
&= Jw_1\otimes \bar \alpha_1 + \cdots + Jw_1 \otimes \bar\alpha_k \\
&= \overline {Jw_1 \otimes \alpha_1 + \cdots + Jw_k \otimes \alpha_k}
\end{align}
the last equality can be argued by written $\alpha_j = a_j + \sqrt {-1} b_j$. Then
\begin{align}
J \bar v &= \overline {J(w_1 \otimes \alpha_1 + \cdots w_k\otimes \alpha_k)}\\
&= \overline {Jv} = \overline{\sqrt{-1} v} = -\sqrt{-1} \bar v. 
\end{align}
