Show that for any measurement operator $M_m$ there exists unitary $U_m$ such that $M_m=U_m\sqrt{E_m}$ with $E_m$ POVM

Question

Exercise 2.63 of Nielsen & Chuang asks one to show that if a measurement is described by measurement operators $M_m$, there exists unitary $U_m$ such that $M_m = U_m \sqrt{E_m}$ where $E_m$ are the POVM associated to the measurement (that is, $E_m = M^{\dagger}_m M_m$).

I can see that, if $\sqrt{E_m}$ is invertible, then $U_m = M_m \sqrt{E_m}^{-1}$ is unitary; indeed, we have (dropping the needless subscript for simplicity) $U^{\dagger} U = (\sqrt{E}^{-1})^\dagger M^\dagger M \sqrt{E}^{-1} = (\sqrt{E}^{-1})^\dagger E \sqrt{E}^{-1} = (\sqrt{E}^{-1})^\dagger \sqrt{E} = (\sqrt{E}^{-1})^\dagger \sqrt{E}^\dagger = (\sqrt{E} \sqrt{E}^{-1})^\dagger = I^\dagger = I$ where I used that $\sqrt{E}$ is Hermitian (since it is positive).

But what if it's not invertible? Perhaps some continuity argument would work?

Answer 1

It is just a polar decomposition of $M_m$.
If $M_m = U P$ then $M^{\dagger}_m M_m = P^2$, hence $P = \sqrt{E_m}$.

Limiting argument, similar to this, also can work.