 Solution

Commutativity and equality of squares giveThus $S-T$ vanishes on $\mathrm{Ran}(S+T)$, and by continuity on its closure. The preceding kernel of a sum of positive operators identity shows that it also vanishes on $\ker (S+T)$. Since $S+T$ is self-adjoint, part (i) and the Hilbert projection theorem give the orthogonal direct sumTherefore $S-T$ vanishes on all of $X$, provingThis proves commuting positive operators with equal squares are equal without assuming a square-root uniqueness theorem.