Eigenvalues and eigenspaces of almost complex structures under each other 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 are some:
Let $L$ be $\mathbb C$-vector space. Let $L_{\mathbb R}$ be its realification, and let the $(L_{\mathbb R})^{\mathbb C} = (L_{\mathbb R}^2,J)$ be the complexification of its realification with almost complex structure $J(l,m):=(-m,l)$ on $L_{\mathbb R}^2$. For every almost complex structure $K$ on $L_{\mathbb R}$, $K \oplus K$ is an almost complex structure on $L_{\mathbb R}^2$. Then $K^{\mathbb C} := (K \oplus K)^J$ (see notation and definitions here, in particular the bullet below 'Definition 4') is $\mathbb C$-linear, i.e. $K \oplus K$ and $J$ commute.
Based on this question, it appears we have that for $K=i^{\sharp}$, we have that $(K \oplus K)^J$ has the same eigenvalues as $J^{K \oplus K}$
Question 1. For any almost complex structure $K$ on $L_{\mathbb R}$, does $(K \oplus K)^J$ always have the same eigenvalues as $J^{K \oplus K}$?
Question 2. For any eigenvalues $(K \oplus K)^J$ and $J^{K \oplus K}$ have in common, do the corresponding eigenspaces have the same underlying sets?
I think the answer to both questions is yes and that this need not be only for the case where we have an almost complex structure on $L_{\mathbb R}^2$ that is the realification of a complexification of a map on $L_{\mathbb R}$ (such map must, I think, be an almost complex structure on $L_{\mathbb R}$):
Question 3. For any almost complex structure $H$ on $L_{\mathbb R}^2$ (not necessarily the realification of a complexification of a map on $L_{\mathbb R}$) such that $H$ and $J$ commute, does $H^J$ always have the same eigenvalues as $J^H$?
Question 4. For any eigenvalues $H^J$ and $J^H$ have in common, do the corresponding eigenspaces have the same underlying sets?
Additional questions:
Question 5. For any almost complex structures $K$ and $M$ on $L_{\mathbb R}^2$ that commute, are the eigenvalues of $K^M$ a subset of $\{ \pm i\}$?
Question 6. If yes to Question 5, then is it that $K^K$ has $i$ as its only eigenvalue if $L \ne 0$ and has no eigenvalues if $L=0$? (I assume $L=0$ iff $L_{\mathbb R} = 0$ iff $(L_{\mathbb R})^{\mathbb C} = 0$ iff $L_{\mathbb R}^2 = 0$)
 A: The answer to all your questions is yes and has nothing to do with complexification. Let $V$ be a real vector space and let $J,H$ be two commuting linear complex structures on $V$ ($J^2 = H^2 = -\operatorname{id}_W$ and $JH = HJ$). 
You can consider $V$ as a complex vector space with respect to $H$. Then, since $J$ commutes with $H$, the map $J$ is actually $\mathbb{C}$-linear as a map $J^H \colon (V,H) \rightarrow (V,H)$. As a real map, $J \colon V \rightarrow V$ doesn't have any eigenvalues since if $Jv = \lambda v$ then $J^2v = \lambda^2 v = -v$ which implies that $\lambda^2 = -1$. Considering $J^H$ as a complex linear map, the above calculation shows that the only possible eigenvalues of $J^H$ are $\pm i$. We also have a direct sum decomposition
$$ V = \{ v \in V \, | \, J^H v = iv \ \iff Jv = Hv \} \oplus \{ v \in V \, | \, J^Hv = -iv \iff Jv = -Hv\} $$
where the first factor is the "eigenspace" of $J^H$ corresponding to the eigenvalue $i$ and the second is the "eigenspace" of $J^H$ corresponding to the eigenvalue $-i$. The only caveat is that one of the factors might be trivial so $J^H$ won't necessarily have both $\pm i$ as eigenvalues.
Similarly, you can consider $V$ as a complex vector space with respect to $J$ and then $H^J \colon (V,J) \rightarrow (V,J)$ is $\mathbb{C}$-linear with the only possible eigenvalues being $\pm i$ and you get a direct sum decomposition
$$ V = \{ v \in V \, | \, H^Jv = iv \iff Hv = Jv \} \oplus \{ v \in V \, | \, H^Jv = -iv \iff Hv = -Jv \} $$
where the first factor is the "eigenspace" of $H^J$ corresponding to the eigenvalue $i$ and the second is the "eigenspace" of $H^J$ corresponding to the eigenvalue $-i$. This shows that $J^H$ and $H^J$ have the same eigenvalues and the same eigenspaces.
Finally, the map $J^J \colon (V,J) \rightarrow (V,J)$ is also complex linear and is just given by multiplication by $i$ so it has only $i$ as an eigenvalue (at least as long as $V \neq \{ 0 \}$).
