 Solution

For a bounded self-adjoint $T$, nonnegative spectrum is equivalent to $⟨Tx,x⟩\ge 0$. Here is a useful elementary justification of the nontrivial direction. Let $a={\inf }_{\Vert x\Vert =1}⟨Tx,x⟩$ and $S=T-aI$, so $S$ has nonnegative quadratic form. Cauchy-Schwarz for that positive form gives $\Vert Sx{\Vert }^2\le \Vert S\Vert ⟨Sx,x⟩$. Taking unit vectors approaching the infimum makes $S{x}_j\to 0$, hence $a\in \sigma (T)$. Thus nonnegative spectrum forces $a\ge 0$. The reverse direction follows from the lower bound and adjoint argument for $T-\lambda$ when $\lambda <0$, together with reality of the self-adjoint spectrum.

If $T=0$ its positive square root is zero. Otherwise set $M=\Vert T\Vert$ and $C=I-T/M$, so $0\le C\le I$ and $\Vert C\Vert \le 1$. Write the scalar binomial seriesThe last equality follows by taking $z\uparrow 1$ in this positive-coefficient series. Consequentlyconverges absolutely in operator norm. Each partial sum is positive, because $0\le C^n\le I$ and the sum of its coefficients is at most one. The limit is positive and self-adjoint. Multiplying the absolutely convergent series gives $B^2=M(I-C)=T$.

For uniqueness let $R$ be another positive square root. It commutes with $T=R^2$, hence with the polynomial-norm-limit $B$. Thus $(R-B)(R+B)=0$. The kernel of $R+B$ is $\ker R\cap \ker B$: zero quadratic form for a positive operator implies it annihilates the vector, by positive-form Cauchy-Schwarz. The difference $R-B$ vanishes there and on $\mathrm{ran}(R+B)$; the latter is dense in the kernel complement because $R+B$ is self-adjoint. Therefore $R=B$. The positive square root of an operator exists and is unique.