Suppose $V$ is a complex vector space, $T \in \mathcal{L}\mathbb{(V)}$, and $\lambda \in \mathbb{C}$ is an eigenvalue of $T$. Show that if there is a basis of $V$ such that the matrix of $T$ has only real entries then $\bar{\lambda}$ is also an eigenvalue of $T$.
Since $\lambda$ is an eigenvalue of $T$, there exists a vector $v \in V$ such that $$Tv_j = \lambda v_j \tag 1$$ Considering the basis of $V$ and the real matrix of $T$, expanding (1) yields $$ \begin{bmatrix}a_{11} &\dots a_{n1} \\ \vdots & \vdots \\ a_{1n} & \dots a_{nn}\end{bmatrix} \begin{bmatrix}v_1 \\ \vdots \\ v_n \end{bmatrix} = \lambda \begin{bmatrix}v_1 \\ \vdots \\ v_n \end{bmatrix}\tag 2$$
If $\bar{\lambda}$ is an eigenvalue of $T$, then there will be a corresponding eigenvector $w \in V$ such that $$ Tv = \lambda v = \bar{\lambda}w\tag3$$ Multiplying $\mathcal{M}(T) v_j$ yields $$ v_j = \sum_{i=1}^{n} a_{ji}v_j \tag4$$ Also multiplying $\bar{\mathcal{M}(T)} w_j$ equals $$ w_j = \sum_{i=1}^{n} \bar{a_{ji}}v_j \tag5$$ Since the matrix of $T$ is real, then $a_j = \bar{a_j}$ for $j = 1, \dots , n$. Thus $\lambda v = \bar{\lambda}w$ and $\bar{\lambda}$ is an eigenvalue of $T$
This proof doesn't feel entirely sturdy especially from (3) - (5). I was wondering what some other opinions or perspectives would think too, thanks
I think the proof can be simplified. Let $M_T$ be the matrix representing the operator $T$, where $M_T$ has only real entries.
$M_T v = \lambda v$
Conjugate both sides.
$\bar{M_T} \bar{v} = \bar{\lambda} \bar{v}$
Then substitute $\bar{M_T} = M_T$.
$M_T \bar{v} = \bar{\lambda} \bar{v}$
This shows $\bar{\lambda}$ is an eigenvalue of $T$.