I'm trying to prove that if $\vec{z}$ is an eigenvector of a matrix $A$ with complex entries, then $\bar{\vec{z}}$ is an eigenvector of $\bar{A}$.
My approach: $A\vec{z}=\lambda\vec{z} \implies \bar{A}\bar{\vec{z}} = \overline{A\vec{z}} = \overline{\lambda\vec{z}} = \bar{\lambda}\bar{\vec{z}} $. From this, we can infer that $\bar{\lambda}$ is an eigenvalue of $\bar{A}$. However, I'm afraid this does not prove that $\bar{\vec{z}}$ is an eigenvector of $\bar{A}$.
Please share your thoughts.
The way I parse it, what is being said is, basically, that assuming we have $\vec z \ne 0$ with
$A\vec z = \lambda \vec z, \tag{1}$
then we may take complex conjugates and write
$\overline{A \vec z} = \overline{\lambda \vec z}; \tag{2}$
also,
$\overline{A \vec z} = \bar A \bar{\vec z} \tag{3}$
and
$\overline{\lambda \vec z} = \bar \lambda \bar{\vec z}; \tag{4}$
thus, since as Euclid might say, "things equal to equals are equal to each other" (i.e., in modern terms, the transitivity of the relation "$=$"), we indeed have
$\bar A \bar{\vec z} = \overline {A \vec z} = \overline{\lambda \vec z} = \bar \lambda \bar{\vec z}, \tag{5}$
that is,
$\bar A \bar{\vec z} = \bar \lambda \bar{\vec z}; \tag{6}$
since $\bar{\vec z} \ne 0 \Leftrightarrow \vec z \ne 0$, (6) shows that $\bar{\vec z}$ is an eigenvector of $\bar A$ with eigenvalue $\bar \lambda$.
Note that (3) and (4) hold as a simple extension of $\overline {a + b} = \bar a + \bar b$ and $\overline{ab} = \bar a \bar b$ for $a, b \in \Bbb C$.
If this is what our OP sequence intended, then his/her proof is correct.
