= Commuting positive operators with equal squares are equal
{title2=$ST=TS,\ S^2=T^2\Longrightarrow S=T$}
For bounded <positive operators> satisfying the hypotheses, $(S-T)(S+T)=0$, so $S-T$ vanishes on the closure of the range of $S+T$. The <kernel of a sum of positive operators> identity shows that it also vanishes on $\ker(S+T)$. Self-adjointness and the <Hilbert projection theorem> decompose the whole <Hilbert space> as that closed range plus that kernel. Thus $S-T=0$. This proves the commuting case of uniqueness of a positive square root without using a spectral representation.
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