= Solution
For a normalized <pure state>, expand $|\psi\rangle=\sum_m c_m|m\rangle$ in an <orthonormal basis> of <eigenvectors> of the measured <Hermitian operator>. The <Born rule> assigns outcome $m$ the <probability> $p_m=|c_m|^2$. Under the ideal <projective measurement> update, a nondegenerate outcome leaves $|m\rangle$, up to a <global phase>. The sum of the outcome probabilities is one, and the <expectation value> is $\sum_m m|c_m|^2$; a single trial reports one eigenvalue, not that average.
If $M|m\rangle=m|m\rangle$, then
$$
m\langle m|m\rangle=\langle m|M|m\rangle
=\overline{\langle m|M|m\rangle}
=\overline m\langle m|m\rangle,
$$
since $M=M^\dagger$. A nonzero <eigenvector> has positive squared <norm>, so \b[$m$ is real]. For two eigenvectors,
$$
\langle n|M|m\rangle=m\langle n|m\rangle
=\langle Mn|m\rangle=n\langle n|m\rangle.
$$
Thus $(m-n)\langle n|m\rangle=0$: \b[distinct eigenvalues have orthogonal eigenvectors].
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