Sum of positive operators. Each must be proportional to right hand side

In these notes, on page 13, there is a claim:

we have $$\sum_{\mu, a} N_\mu M_a |\psi \rangle \langle \psi|M_a^\dagger N_\mu^\dagger = |\psi \rangle \langle \psi|.$$ Since the left-hand side is a sum of positive operators, the equation can only hold if each of these terms is proportional to $|\psi\rangle \langle \psi|$, hence $$N_\mu M_a = \lambda_{\mu a} I.$$

I do not understand how it follows from positivity (and the form of the terms) that they are all proportional to the RHS.

• Hint: What happens if you apply each side to a state orthogonal to $|\psi\rangle$ Jan 31, 2018 at 14:52
• @BySymmetry you get 0 on RHS and on left-hand side sum of $| \langle v|N_\mu M_a |\psi \rangle |^2$ terms which are all $\geq 0$ so that they each must be zero. Therefore, $|v \rangle$ is orthogonal to $N_\mu M_a |\psi \rangle$ for each $\mu, a$ and since $|v \rangle$ was any orthogonal state it implies that $N_\mu M_a |\psi \rangle$ is proportional to $|\psi \rangle$. Is this correct? Also, I think that positivity is implicitly used due to the form of the LHS terms but that it itself is not enough. Is this true? Thank you for the hint. Jan 31, 2018 at 15:47