Eigenvalue of a complex matrix

Question

Wonder if this is correct:

If $$ A=\begin{pmatrix}a & b\\c & d \end{pmatrix} $$ is a complex matrix that has a real eigenvalue, then the matrix $$ B=\begin{pmatrix}\overline{a} & \overline{b}\\\overline{c} & \overline{d} \end{pmatrix} $$ also has a real eigenvalue. So, if $\exists\lambda_0\in{\Bbb{R}}\,\exists v_0$ so that $$ \begin{pmatrix}a & b\\c & d \end{pmatrix}\cdot v_0=\lambda_0\cdot v_0, $$ then: $$ \begin{pmatrix}\overline{a} & \overline{b}\\ \overline{c} & \overline{d} \end{pmatrix}\cdot\overline{v_0}=\overline{\lambda_0}\cdot\overline{v_0}, $$ given that $$ \lambda_0\in{\Bbb{R}}\implies\overline{\lambda_0}=\lambda_0 \implies \begin{pmatrix}\overline{a} & \overline{b}\\\overline{c} & \overline{d} \end{pmatrix}\cdot\overline{v_0}=\lambda_0\cdot\overline{v_0}. $$

Answer 1

Yes, this is correct. The reason is that for $v_0=\begin{pmatrix}x\\y\end{pmatrix}$ it is true that, after complex conjugation, $\begin{pmatrix}\overline{ax+by}\\\overline{cx+dy}\end{pmatrix}=\begin{pmatrix}\overline{\lambda_0x}\\\overline{\lambda_0y}\end{pmatrix}$, which is equivalent to $Bv_1=\lambda_0v_1$ where $v_1=\begin{pmatrix}\bar{x}\\\bar{y}\end{pmatrix}$.

(Reacall that $\overline{\alpha+\beta\gamma}=\overline{\alpha}+\overline{\beta}\overline{\gamma}$ for any $\alpha, \beta, \gamma \in \mathbb{C}$. In the upper special case, also note that $\overline{\lambda_0}=\lambda_0$.)

Answer 2

This should work out because we have, for $\alpha, \beta\in \Bbb C$, that $\overline{\alpha\beta}=\overline{\alpha}\overline{\beta}$. (And of course $\overline {\alpha+\beta}=\overline{\alpha}+\overline{\beta}$).

Thus for the characteristic polynomials, we have $c_{\bar A}(\bar z)=\overline {c_A(z)}$. Hence $c_A(\lambda)=0\iff c_{\overline A}(\overline{\lambda})=0$.

So we get that $\lambda $ is an eigenvalue of $A$ iff $\overline{\lambda}$ is an eigenvalue of $\overline A$. But $\lambda $ real implies $\lambda =\overline{\lambda}$.