Show that the square root of a non-negative operator is unique Let $H$ be a Hilbert space, and $A\in B(H\to H)$ be a bounded non-negative operator (i.e. $\langle Ax,x\rangle \geq 0$ for all $x\in H$). The square root of $A$ is a  bounded non-negative operator $B\geq $ such that $B^2=A$.
First, We can assume without loss of generality that $0\leq A\leq I$. Note that $B^2=A$ if and only if 
$$I-B=\frac{1}{2}((I-A)+(I-B)^2).$$
Hence, we definite inductively a sequence $C_n$ of operators as follows: $C_0:=0$, and $C_{n+1}:=1/2((I-A)+C_n^2))$. Then it is easy to see that $C_n$ converges to a bounded non-negetive operator $B$ in the strong operator topology and we also have $B^2=A$, thus the square root exists, but I don't know how to show that it is unique. 
 A: Suppose $A$ is a bounded nonnegative operator on a Hilbert space. Let $(p_n)$ be a sequence of polynomials such that
$$
p_n(x) \to \sqrt{x}
$$
uniformly for $x$ in the interval $[0, \|A\|]$ (the Weierstass Approximation Theorem implies the existence of such a sequence of polynomials).
Now suppose $B$ is a nonnegative square root of $A$. Let $\mathcal{B}$ denote the norm closed algebra generated by $B$. Then $\mathcal{B}$ is a commutative $C^*$-algebra that contains $B$ and $A$ (because $A = B^2$). Thus there is a compact Hausdorff space $K$ such that $\mathcal{B}$ is isomorphic as a $C^*$-algebra to $C(K)$. This isomorphism preserves all $C^*$ properties. Thus $A$ corresponds to some nonnegative function $f \in C(K)$ taking values in $[0, \|A\|]$ and $B$ must correspond to the function $\sqrt{f}$ (which is the only nonnegative square root of $f$ in $C(K)$).
Because $p_n \circ f$ converges uniformly to $\sqrt{f}$ uniformly on $K$, we conclude that $p_n(A)$ converges in operator norm to $B$. But the polynomials $(p_n)$ were chosen independently of $B$. Thus $B$ is uniquely determined as a nonnegative square root of $A$.
