Let $A$ be a positive bounded linear operator on a Hilbert space $\mathscr{H}$, that is, $\langle \psi,A\psi\rangle \ge 0$ holds for every $\psi \in \mathscr{H}$. We write this property as $A \ge 0$.
I am reading Reed & Simon's proof of the following result.
Theorem: Let $A$ be positive, bounded and linear. Then there exists a unique bounded linear operator $B$ such that $B \ge 0$ and $B^{2} = A$. Moreover, $B$ commutes with every bounded linear operator which commutes with $A$.
Suppose $\|A\| \le 1$. To define $B$, one considers the power series: $$B = \sqrt{A} = \sqrt{I - (I-A)} = 1+c_{1}(I-A) + c_{2}(I-A)^{2} + \cdots, $$ which is absolutely convergent when $\|I-A\| \le 1$.
My first question is: how to show that $B^{2} = A$? I know that, because the series for $B$ is absolutely convergent, one can square it and write it as: $$B^{2} = \sum_{n=0}^{\infty}a_{n}(I-T)^{n}$$ with: $$a_{n} = \sum_{m=0}^{n}c_{m}c_{n-m} $$ and $c_{0} = 1$. I also know that: $$1-z = \sum_{n=0}^{\infty}a_{n}z^{n}.$$
Question 1: Is this enough to prove that $B^{2} = I - (I-A) = A$?
Question 2: To prove uniqueness, the authors assume $B'$ is another positive bounded linear operator such that $(B')^{2} = A$. Then, they claim that $B - B'$ is self-adjoint. I don't see, however, why this is true. A positive bounded linear operator is only self-adjoint in the case of complex Hilbert spaces, but there is no mention about the underlying Hilbert space to be complex in their theorem. So, why is this true?