diagonalizability, real and complex vector spaces problem 
Let $A \in M_{n \times n}(\Bbb R)$.
  Let $T_{\Bbb R}:\Bbb R^n \to \Bbb R^n$ and $T_{\Bbb C}:\Bbb C^n \to \Bbb C^n$ be the corresponding linear mapps(defined by $T_{\Bbb R}(x)=Ax$ and $T_{\Bbb C}(v)=Av$ for $v \in \Bbb R^n$ and $v \in \Bbb C^n)$. Assume $T_{\Bbb C}$ is diagonalizable, and let the distinct eigenvalues of $T_{\Bbb C}$ be
  $$\lambda_1,\ldots,\lambda_r,\ \mu_1, \ \bar{\mu_1},\ldots,\mu_s,\ \bar{\mu_s}
$$
  where $\lambda_1,\ldots,\lambda_r\in\Bbb R$ and $\mu_1,\ldots,\mu_s\notin\Bbb R$.
  Prove that
  $$
\Bbb{R}^n=E_{\lambda_1} \oplus \cdots E_{\lambda_r} \oplus Re(E_{\mu_1}) \oplus \cdots \oplus Re(E_{\mu_s})
$$
  Where we define $Re(W)=W \oplus \overline{W} \cap \Bbb{R}^n$.

Let $\beta={v_1,...,v_n}$ be a $\Bbb{C}$ baasis for V, and $\gamma={v_1,...,v_n,iv_1,...,iv_n}$ an $\Bbb{R}$ basis for V. Since $T_{\Bbb{C}}$ is diagonalizable, we can find $\Bbb{R}-bases$ $\beta_j$ for $E_{\lambda_j}$ and $\gamma_j$ for $Re(E_{\mu_j})$ so that the union of both forms a $\Bbb{C}$-basis for $\Bbb{C}^n$. I am not sure how to put it rigorously in a proof. How am I to lead to that they also form an $\Bbb{R}$-basis for $\Bbb{R}^n$?
 A: You are right that $\Bbb C^n$ is the direct sum of the eigenspaces as $A$ is diagonalizable, i.e.: 
$$\Bbb C^n = \hat{E}_{\lambda_1} \oplus \dots \oplus \hat{E}_{\lambda_r} \oplus E_{\mu_1} \oplus E_{\bar{\mu}_1} \oplus \dots E_{\mu_s} \oplus E_{\bar{\mu}_s} \ , $$
where $\hat{E}_{\lambda_i} = \{\pmb{v}\in\Bbb C^n: (A-\lambda_i)\pmb{v} = 0 \}$ is the eigenspaces over complex numbers. 
Then you can find the intersection $\Bbb R^n = \Bbb C^n \cap \Bbb R^n $. Next, find the intersection of each eigenspace as follows,
$$ \hat{E}_{\lambda_1} \cap \Bbb R^n = E_{\lambda_i}, \ i =1,\dots, r \ ,$$
where $E_{\lambda_i} = \{\pmb{v}\in\Bbb R^n: (A-\lambda_i)\pmb{v} = 0 \}$ is the eigenspaces over real numbers. Note that $\dim_{\Bbb C}(\hat{E}_{\lambda_i}) = \dim_{\Bbb R} (E_{\lambda_i})$ because we can always find the set of real eigenvectors for real $\lambda_i$. However, $\hat{E}_{\lambda_i} \neq {E}_{\lambda_i}$ as, for example, $i\pmb{v}$ is also eigenvector for any real eigenvector $\pmb{v}$.
Further, $E_{\bar{\mu}_i} = \bar{E}_{\mu_i}$ as $A$ is real ($\bar{\pmb{v}_i}$ is eigenvector corresponding to $\bar{\mu}_i$ for any eigenvector $\pmb{v}$ corresponding to $\mu_i$) such that
$$ E_{\mu_i} \oplus E_{\bar{\mu}_i} \cap \Bbb R^n = Re(E_{\mu_i}), \ i=1,\dots,s \ . $$
