TITLE: A question about the proof of "a symmetric matrix has real eigenvalues".
QUESTION [6 upvotes]: This question is in reference to this proof of the Spectral Theorem. 
Let $A$ be a symmetric matrix. Take $u$ to be an orthonormal vector (such that $u^Tu=1$). Then if $u$ is an eigenvector of $A$, then $Au=\lambda u$. This implies that $\lambda=u^TAu$. 
The article says that this proves that $\lambda$ is real. I don't follow why this is. Both $u$ and $A$ can contain complex entries! Why would $u^TAu$ have to be real?

REPLY [8 votes]: 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.