Ok so I'm studying linear algebra and we went trough the Spectral Theorem, including and proving the fact that for every Herimitian matrix, its eigenvalues have $0$ imaginary parts. I was wondering is it true, and if so is there a proof, that if we have non-Hermitian matrix there always exist at least one (covnersly at least $2$ as it's conjugate will also be) eigenvalue with imaginary part different from $0$. In other words the question is:
Is there any matrix $A$ such that $A^{*} \neq A$ and $\lambda = \overline{\lambda}$ for every eigenvalue $\lambda$.
Consider the matrix:
$A=\begin{pmatrix}1 & 1\\ 0 & 1 \end{pmatrix}$,
which is clearly non-Hermitian. Its only eigenvalue is $1$, with multiplicity $2$.