The way I have proved the Eigen values of Hermitian matrices are real like this:
I considered $H$ is a hermitian matrix.
Operator applied in ket space $ \ \ <\psi|H | \psi> \ \ = \lambda$
Operator applied in ket space $\ \ <\psi|H | \psi> \ \ = \lambda^\star \ $
There is no state in the right hand side because of the normalization.
If we subtract the two equations we get $\lambda = \lambda ^*$ Therefore the eigenvalues are real
Have I done in the right way?
I can't post a comment, but the operator must be applied to the eigenkets, where you have the relation $H |\psi\rangle = \lambda |\psi \rangle$. You do not need to substract, just need to see that the two quantities are equal.
Just for the sake of completeness, I'll do the derivation in terms of inner products.
We have, taking a normalized eigenvector $\psi$ of $H$ such that $H\psi=\lambda \psi$:
$\langle \psi, H \psi \rangle=\lambda \langle \psi, \psi \rangle=\lambda$, (taking linearity in the second argument, as customary in physics).
Since $H$ is hermitian (i.e. $H=H^*$) we have
$\langle \psi, H \psi\rangle = \langle H \psi,\psi \rangle= \lambda^* \langle \psi, \psi \rangle=\lambda^*$ (linear conjugate in the first argument)
So we have that $\lambda$ is equal to its complex conjugate, i.e., it takes real values only.