Diagonal expression for complexified $\mathbb{R}$-representation

Question

This is Example from section 3.6 (p.38) in Ernest Vinberg, "Linear representations of groups"

The question is about finding equivalent(diagonal) expression for complexification of standard real 2-dimensonal rotation representation of $\mathbb{R}$.

Example: Let us have the representation of $\mathbb{R}$ implemented as rotations of $\mathbb{R}^2$. If we complexify the representation, we can change the base s.t. the representation can be expressed as $$t \mapsto \begin{pmatrix} e^{it} & 0 \\ 0 & e^{-it} \end{pmatrix} $$
This expression is obtaines by changing the base of initial representation $(e_1, e_2)$ to $(e_1-ie_2, e_1+ie_2)$.

Why does the new basis have this form?

Answer 1

I don't know which Vinberg you are talking about. But I assume the "representation of $\mathbb{R}$ through rotations" he is mentioning is the representation of the additive group $\left( \mathbb{R},+\right) $ on $\mathbb{R}^{2}$ which sends each $t\in\mathbb{R}$ to the matrix $\begin{pmatrix} \cos t & -\sin t\\ \sin t & \cos t \end{pmatrix} $. This is the matrix that represents a rotation (around the origin) by the angle $t$ with respect to the standard basis $\left( e_{1} ,e_{2}\right) $ of $\mathbb{R}^{2}$. Let me denote this rotation by $R_{t}$. It is an endomorphism of the $\mathbb{R}$-vector space $\mathbb{R}^2$.

Now, extend scalars to $\mathbb{C}$. Thus, the endomorphism $R_{t}$ of $\mathbb{R}^{2}$ induces an endomorphism $C_{t}$ of $\mathbb{C}^{2}$, which is given by the same matrix with respect to the basis $\left( e_{1} ,e_{2}\right) $.

Now, consider the basis $\left( e_{1}-ie_{2},e_{1}+ie_{2}\right) $ of $\mathbb{C}^{2}$. This basis is connected to the basis $\left( e_{1} ,e_{2}\right) $ through the change-of-basis matrix $\begin{pmatrix} 1 & 1\\ -i & i \end{pmatrix} $. Thus, the matrix representing the endomorphism $C_{t}$ with respect to this new basis is \begin{align*} & \begin{pmatrix} 1 & 1\\ -i & i \end{pmatrix} ^{-1}\begin{pmatrix} \cos t & -\sin t\\ \sin t & \cos t \end{pmatrix} \begin{pmatrix} 1 & 1\\ -i & i \end{pmatrix} \\ & =\begin{pmatrix} \cos t+i\sin t & 0\\ 0 & \cos t-i\sin t \end{pmatrix} =\begin{pmatrix} e^{it} & 0\\ 0 & e^{-it} \end{pmatrix} \end{align*} (since $\cos t+i\sin t=e^{it}$ and $\cos t-i\sin t=e^{-it}$).

Answer 2

The example is the combination of two facts:

1. A complex representation has equivalent matrix expressions and all the expressions are inside the orbit of conjugation by any element of $GL(n,\mathbb{C})$. This conjugation is the change-of-base. All the expressions implement the same representation, which is a set of operators that don't know anything about coordinates.
2. Now we find the equivalent representation for rotation. Forget about representation theory for a moment. Look at Vinberg, "Algebra course", Chapter 6, paragraph 2. The paragraph addresses complexification of operators and addresses your question completely. Particularly Proposition 1 and Example 3. (I used Russian edition, hope, that the edition you use has same numeration)

I will type Proposition 1 here, just in case.

$x+iy (x,y \in V)$ is an eigenvector of linear operator $A_\mathbb{C}$ with eigenvalue $\lambda+i\mu (\lambda, \mu \in \mathbb{R}, \mu \neq 0)$ iff $U = span(x,y) \subset V$ is an invariant space of $A$. Furthermore $$Ax = \lambda x - \mu y $$ $$ Ay = \mu x + \lambda y$$

So, now you have the situation from the proposition. By the Euler $\exp$ formula you get the thing.

Can you make out the full answer from here?