Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 1 / 8 / a

A bounded self-adjoint operator $T$ is a positive operator when $⟨T\xi ,\xi ⟩\ge 0$ for every $\xi$. Equivalently its quadratic form is nonnegative. We construct its positive square root of an operator directly and prove uniqueness.

If $T=0$ its positive square root is zero. Otherwise set $r=\Vert T\Vert$, $B=T/r$, and $C=I-B$. Then $0\le B\le I$ and $C$ is positive. It is also a contraction. Indeed, the Cauchy-Schwarz inequality for the positive form $⟨B\xi ,\eta ⟩$ gives $\Vert B\xi {\Vert }^2\le ⟨B\xi ,\xi ⟩$, and henceThe scalar binomial series has the formThe last equality follows by taking $z\uparrow 1$ in the nonnegative series. Thereforeconverges in operator norm. Every $C^n$ is a positive contraction: its even powers are squares, its odd powers have form $C^{k}C{C}^k$, and their norms are at most one. Each partial sum for $S/\sqrt{r}$ is consequently positive, since it is at least $(1-{\sum }_{n\le N}{c}_n)I$. The limit is positive. Multiplying the absolutely convergent series and using the scalar coefficient identity gives $S^2=r(I-C)=T$. This construction also shows that $S$ is a norm limit of polynomials in $T$ and thus commutes with every operator commuting with $T$.

Now suppose $R$ is another positive square root. It commutes with $T=R^2$, hence with $S$. ThusThe difference vanishes on the range of $R+S$ and on its closure. It also vanishes on its kernel: if $(R+S)\xi =0$, the fact that $R$ and $S$ are positive operators gives $⟨R\xi ,\xi ⟩=⟨S\xi ,\xi ⟩=0$, and the Cauchy-Schwarz inequality for their positive forms implies $R\xi =S\xi =0$. Since $R+S$ is self-adjoint, the orthogonal sum of its kernel and closed range is all of $H$. Therefore $R=S$, proving