**Theorem:** Hermitian matrices have real eigenvalues. **Proof:** If $Av = \lambda v$ with $v \neq 0$, then $v^*Av = \lambda v^*v$ and also $v^*Av = v^*A^*v = (Av)^*v = \overline{\lambda}v^*v$. So $\lambda = \overline{\lambda}$, meaning $\lambda$ is real.