 Positive square root of an operator

Every bounded positive operator $T$ on a Hilbert space has a unique bounded positive operator $T^{1/2}$ with $(T^{1/2}{)}^2=T$. The continuous functional calculus applies the continuous scalar function $\sqrt{t}$ on the compact spectrum of $T$, givingIt also gives $⟨Tx,x⟩=\Vert T^{1/2}x{\Vert }^2$ and $\ker T^{1/2}=\ker T$.

For uniqueness, any positive $S$ with $S^2=T$ commutes with $T$ and hence with uniform polynomial approximations to $T^{1/2}$. Thus $S$ commutes with $T^{1/2}$, and commuting positive operators with equal squares are equal applies. Positivity is essential: $I$ has both $I$ and $-I$ as self-adjoint square roots, but only the former is positive. For $T=\mathrm{diag}(0,4,9)$, the positive square root is $\mathrm{diag}(0,2,3)$.