Proving that self-adjoint operators have only real eigenvalues

Question

Does anyone know how to prove that if $T$ is a self-adjoint operator, then all of the eigenvalues of $T$ are real?

Answer 1

$Tx=\lambda x,\ x\not=0\implies \langle Tx,x\rangle=\langle \lambda x,x\rangle =\lambda\langle x,x\rangle$ so $\overline{\langle Tx,x\rangle}=\overline{\langle \lambda x,x\rangle}=\overline{\lambda\langle x,x\rangle}=\overline{\lambda}\langle x,x\rangle$. But by conjugate symmetry of the inner product followed by the fact that $T$ is self-adjoint, $\overline{\langle Tx,x\rangle}=\langle x,Tx\rangle=\langle Tx,x\rangle$ which, from the work above, implies $\lambda=\overline{\lambda}$, i.e. $\lambda\in\mathbb R$.