Let $E$ be a Banach space over the reals and $T:E\to E$ a linear operator. Suppose that $\mu=re^{i\theta}$ is a eigenvalue of $T$ with $r>0$. How can one find two vector $z_1,z_2\in E$ such that if $R_\theta$ is the rotation by $\theta$ in the plane $P\subset E$ spanned by $z_1,z_2$, then $$T_{|P}z=r R_\theta z$$
where $T_{|P}$ is the restriction of $T$ to $P$ and $z\in P$ can be identified with $z_1+iz_2=(z_1,z_2)$.
Remark: If $E$ is finite dimensional, this is the same as real Jordan decomposition, so my problem really is in infinite dimensional spaces. Maybe we need to add some hypothesis, like $T$ being compact or something like that, if so please feel free to add.
Thank you
If you consider the complexification $T_\mathbb{C} \colon E_\mathbb{C} \to E_\mathbb{C}$, since $T_\mathbb{C}$ is a real operator, for an eigenvector $z$ to the eigenvalue $\mu$, you have $T_\mathbb{C}(\overline{z}) = \overline{\mu}\cdot\overline{z}$, i.e. $\overline{z}$ is an eigenvector to the eigenvalue $\overline{\mu}$.
Thus $(\overline{\mu}I - T_\mathbb{C})\circ(\mu I-T_\mathbb{C})$ vanishes on $\operatorname{span} \{z,\, \overline{z}\} = \operatorname{span} \{\operatorname{Re} z,\, \operatorname{Im} z\} \subset E_\mathbb{C}$.
Now, $(\overline{\mu}I - T_\mathbb{C})\circ(\mu I-T_\mathbb{C}) = r^2 I - 2r\cos\theta\cdot T_\mathbb{C} + {T_\mathbb{C}}^2$ is a real operator since $T_\mathbb{C}$ is one, hence $r^2 I - 2r\cos\theta\cdot T + T^2$ vanishes on $\operatorname{span} \{\operatorname{Re} z,\, \operatorname{Im} z\} \subset E$.
So the existence of $z_1,\,z_2$ spanning a plane on which $T$ acts by $r\cdot\begin{pmatrix}\cos\theta & -\sin\theta\\\sin\theta & \cos\theta\end{pmatrix}$ is assured.
In simple cases, that may even give a practical way to find $z_1,\, z_2$.