= Solution
A normalized <quantum state> satisfies $\langle\psi,\psi\rangle=1$. Its <expectation value> is $\boxed{\mathbb E_\psi(A)=\langle\psi,A\psi\rangle}$. The conjugate symmetry of the <inner product> and <self-adjointness> give
$$
\overline{\langle\psi,A\psi\rangle}=\langle A\psi,\psi\rangle=\langle\psi,A\psi\rangle,
$$
so the <expectation value> is real. If $A\phi=\lambda\phi$ with $\phi\ne0$, normalize $\phi$ by dividing by its <norm>. Then
$$
\lambda=\langle\phi/\|\phi\|,A\phi/\|\phi\|\rangle\in\mathbb R.
$$
\b[Every eigenvalue of a self-adjoint operator is real.]
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