Diagonalizablity and complex conjugates linear algebra proof problem

Question

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 $T_{\Bbb R}(v)=Av$ for $x \in \Bbb R^n$ and $v \in \Bbb R^n)$. 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 $\mu_1,\ldots,\mu_s\notin\Bbb R$. For each $j=1,\ldots,s$, prove that $E_{\mu_j} \cap \Bbb R^n=0$ and $\overline{E_{\mu_j}}=E_{\bar{\mu_j}}$.

Definition: Let $T$ be a linear operator on a vector space $V$, and let $\lambda\in\sigma(T)$. $E_{\lambda}:=\{x \in V:T(x)= \lambda x\}=\ker(T-\lambda I_V)$. The set $E_{\lambda}$ is called the eigenspace of $T$ corresponding to the eigenvalue $\lambda$.

From the context, since $T_C$ is diagonalizable, it is clear that there exists distinct eigenspaces corresponding to distinct eigenvalues. Since they are of complex conjugates, it is clear that $E_{\mu j} \cap \Bbb R^n=0$. I am not sure what does $\overline{E_{\mu_j}}=E_{\bar{\mu_j}}$ want me to do.

Answer 1

Note that $\mu_i \notin \mathbb R$ means $\mu_i = a + bi$ with $b \neq 0$. Now if $c \neq 0$ is real then $c\mu_i = ac + bci$ and $bc \neq 0$ so $c\mu_i \notin \mathbb R$.

Let $x \in E_{\mu_i}$, so $Ax = \mu_ix$. If $x \in \mathbb R^n$ then $A$ having real entries means $Ax \in \mathbb R^n$. But that means $\mu_ix \in \mathbb R^n$ so the coordinates of $x$, when multiplied by $\mu_i$ are real. By what we've said above this means the coordinates of $x$ must be $0$.

Now assume again that $x \in E_{\mu_i}$, so $Ax = \mu_ix$. Conjugating fixes $A$, so conjugating $Ax = \mu_ix$ gives $A\overline x = \overline{\mu_i}\overline x$, i.e. $x \in E_{\overline{\mu_i}}$. So $Ax = \mu_ix$ if and only if $A\overline x = \overline{\mu_i}\overline x$. This proves $\overline{E_{\mu_i}} = E_{\overline{\mu_i}}$.

Finally, note that $E_{\mu_i} \cap \mathbb R^n = 0$ actually follows from $\overline{E_{\mu_i}} = E_{\overline{\mu_i}}$. This is because $E_{\mu_i} \cap \mathbb R^n$ lies in both $E_{\mu_i}$ and $\overline{E_{\mu_i}}$. But eigenspaces for distinct eigenvalues have trivial intersection.

Answer 2

I guess $\overline{E_{\mu_j}}=\{\bar x:x\in E_{\mu_j}\}$, where $\bar x=\overline{(x_1,\dots,x_n)}= (\overline{x_1},\dots, \overline{x_n})$ for each $x=(x_1,\dots,x_n)\in\Bbb C^n$. Let $x=(x_1,\dots,x_n)\in\Bbb C^n$ be any vector. Using that all entries of $A$ are real, it is easy to check that $T\bar x=\overline{Tx}$. Thus for each $1\le j\le s$ we have $$x\in E_{\overline{\mu_j}}\Leftrightarrow Ax=\overline{\mu_j}x \Leftrightarrow A\bar x=\overline{Ax}=\overline{\overline{\mu_j}x}=\mu_j\bar{x} \Leftrightarrow \bar{x}\in E_{\mu_j} \Leftrightarrow x\in \overline{ E_{\mu_j}}.$$