Show that if $\lambda$ is an eigenvalue of $T$ such that $\mathcal{M}(T) \in \mathcal{M}(\mathbb{R})$...

Question

Let $V$ be a finite-dimensional complex vector space. Let $T \in \mathcal{L}(V)$ be an operator such that the matrix of $T$ with respect to some basis of $V$ contains only real entries. Show that if $\lambda$ is an eigenvalue of $T$ then so is $\overline{\lambda}$. Suggestions?

Answer 1

Let the matrix of $T$ denoted by $A$. If $Ax= \lambda x$, then $A \overline{x}=\overline{Ax}=\overline{\lambda x}=\overline{\lambda} \overline{x}$

Answer 2

The characteristic polynomial has real coefficients if $P(c)=0, \overline P(\overline c)=P(\overline c)=0$.