Complex eigenvalues of real matrices mean rotations

The real vector space $R^{2}$ can be seen as $C$ in several ways: it depends on what we set as the $i$ (fixed a complex structure $J$ , $i$ is identified with $J ((1, 0)^{t})$ and reciprocally, $J$ is determined by the choice of the analogous of $i$ ). When you have an $R$ -linear operator $A$ in $R^{2}$ with complex eigenvalues (that must be conjugate of each other) you can look for the appropriate way of see $R^{2}$ as $C$ in such a manner that $A$ becomes like multiplying by a complex number. Which one? The eigenvalue!! (one of them to your choice). Formally:

Lemma
Let $A$ be a $R$ -linear operator in $R^{2}$ . If $A$ has two conjugate complex eigenvalues $λ$ and $\overset{―}{λ}$ , then we can find a complex structure in $R^{2}$ in such a way that $A$ is like multiplication by $λ$ in $C = R^{2}$ .
$◼$
Proof
The complex structure must be a $J$ such that $A = a I + b J$ , where $a + i b$ is an eigenvalue of $A$ with by hypothesis $b \neq 0$ . To see that $J^{2} = - I$ we use the complexification of $R^{2}$ :

J^{C} : (R^{2})^{C} \to (R^{2})^{C}

The key is that the map

A^{C} : (R^{2})^{C} \to (R^{2})^{C} $ $ h a s t h e s a m e e i g e n v a l u e s A N D E I G E N V E C T O R S (w h i l e $ A $ d o e s n^{'} t h a v e t h e l a t t e r o n e s) . M o r e o v e r, w e c a n f i n d a b a s i s o f e i g e n v a l u e s i n $ (R^{2})^{C} $, s a y $ {v_{1}, v_{2}} $ . A n y $ w \in (R^{2})^{C} $ s a t i s f y $ w = x v_{1} + y v_{2} $ . O b s e r v e t h a t

A^{\mathbb{C}}(xv_1+yv_2)=x(a+ib)v_1+y(a-ib)v_2

s o

J^{\mathbb{C}}(w)=\frac{1}{b}A^{\mathbb{C}}-aI^{\mathbb{C}}=\ldots=xiv_1-yiv_2

a n d s i n c e

A^{\mathbb{C}}(xiv_1-yiv_2)=\ldots=xiav_1-xbv_1-yaiv_2-ybv_2

w e g e t

J^{\mathbb{C}}(J^{\mathbb{C}}(w))=J^{\mathbb{C}}(w)(xiv_1-yiv_2)=\ldots=-w=-I^{\mathbb{C}}(w)

A n d s i n c e $ A^{C} = B^{C} $ i m p l i e s $ A = B $, w e o b t a i n

J^2=-I

T h e r e f o r e, $ J $ i s a c o m p l e x s t r u c t u r e a n d $ $ A : R_{J}^{2} \mapsto R_{J}^{2}

is such that

A (v) = (a + i b) v

$◼$

Visually:

Assume the eigenvalues have module one. In the picture above we see in red the real vector space $R^{2}$ , with the usual base $e_{1}, e_{2}$ in red. If we take the complexified $(R^{2})^{C}$ we extend $e_{1}$ and $e_{2}$ not as a line, but each one as a plane $C$ (the solid orange plane and the blue one). The whole system is a four dimensional real space (two dimensional complex space), the complex plane $C^{2}$ . The map $A$ has eigenvectors $v_{1}$ and $v_{2}$ (the black ones in the picture). When we apply $A$ this eigenvectors rotate inside their complex line $C v_{1}$ y $C v_{2}$ (the yellow and the blue transparent planes in the picture). This rotations (one of angle $θ$ and other of $- θ$ since they are conjugates eigenvalues) compensate'' each other in such a way that they produce a rotation in the original real plane $R^{2}$ (the transparent red one).
This rotation of the original plane induces a complex structure in $R^{2}$ which is, in some sense, transversal: the red transparent plane is not a complex line, so it has no natural complex structure...