Eigenvalues of Complex Stucture in a Complexified Vector Space

Question

For context, this question comes from my reading of Kobayashi+Nomizu's differential geometry book, volume 2 (pages 116-117).

Given a real vector space $V$ with $\mathrm{dim}(V)=2n$, a complex structure is a linear endomorphism satisfying $J^2=-1$.

One can also define the complexification of a vector space as $V^{\mathbb{C}} = V \otimes_{\mathbb{R}} \mathbb{C}$.

Now, $J$ extends naturally to a complex endomorphism on the complexified space, and (Kobayashi and Nomizu claim that) it has eigenvalues $\pm i$.

I have some confusions about this:

1. Is it true that the natural extension of $J$ is just defined by $J(V\otimes z) = J(V)\otimes z$?
2. More crucially, the complexified vector space is treated as a tensor product over the reals, so isn't the coefficient field of this new space also the reals? Isn't, then, the characteristic polynomial viewed as a polynomial over $\mathbb{R}$ (and thus cannot have imaginary roots)?
3. My intuition is to treat application of $J$ as multiplication by $i$ in the original space, but the complexified space offers a bona fide multiplication by $i$, and I'm not sure how these two interact. What would be an example of a $v\in V^{\mathbb{C}}$ such that $Jv = -iv$? What does $Jv=iv$ actually mean? I've tried to answer these questions for $V=\mathbb{R}^2$ with the canonical structure but I'm coming up empty handed.

I greatly appreciate your time and help!

Answer 1

After @KCd wrote the helpful comments regarding the broader context of extending vector spaces, I believe I've come up with the answers, so for completeness to this question I'll write them up here:

1. Yes.

2. This seems to be true on a technicality, but because $V^{\mathbb{C}}$ can also be viewed as a $\mathrm{dim}_{\mathbb{R}}(V)$ space over the complex numbers, we can think of the characteristic polynomial over $\mathbb{C}$, so $\lambda= \pm i$ is perfectly fine.

3. Once I stopped tunnel visioning on simple tensors, the breakdown became clear. For the example of $V=\mathbb{R}^2$ with the canonical complex structure, then in $V^{\mathbb{C}}$, $$J\left(\begin{pmatrix} 1 \\\ 0 \end{pmatrix} \otimes 1 - \begin{pmatrix} 0 \\\ 1 \end{pmatrix} \otimes i \right) = \begin{pmatrix} 0 \\\ 1 \end{pmatrix} \otimes 1 + \begin{pmatrix} 1 \\\ 0 \end{pmatrix} \otimes i = i\left ( \begin{pmatrix} 1 \\\ 0 \end{pmatrix} \otimes 1 - \begin{pmatrix} 0 \\\ 1 \end{pmatrix} \otimes i \right )$$ exhibits an element of the positive eigenspace $V^{1,0}$. Of course, the notation simplifies tremendously if you think of the space as $V(\mathbb{C})$ rather than $V(\mathbb{R})\otimes_{\mathbb{R}} \mathbb{C}$, but computing with the explicit tensor product was quite beneficial to my understanding.