# 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

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?

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