A question about the proof of "a symmetric matrix has real eigenvalues".
Solution 1:
Lemma: A hermitian matrix has real eigenvalues.
Proof: Let $A$ be a complex matrix such that $A^*=A$. Let $\lambda$ be one of $A$'s eigenvalues and $u\neq 0$ a corresponding eigenvector. $Au=\lambda u\implies u^*A^*=\lambda^*u^*$. Hence $$\lambda^* u^*u=u^*A^*u=u^*Au=\lambda u^*u$$ and therefore $\lambda^*=\lambda$. Therefore $\lambda$ is real.
Since real symmetric matrices are hermitian, their eigenvalues are real.