 Solution

For a normalized pure state, expand $|\psi ⟩={\sum }_m{c}_m|m⟩$ 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 $|m⟩$, 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 $M|m⟩=m|m⟩$, thensince $M=M^{†}$. A nonzero eigenvector has positive squared norm, so $m$ is real. For two eigenvectors,Thus $(m-n)⟨n|m⟩=0$: distinct eigenvalues have orthogonal eigenvectors.