Why is $\sqrt{T^*T}$ self-adjoint?

Question

Let $T$ be a bounded linear operator over some Hilbert space $H$.

Since $T^*T$ is a positive operator, it has a square root. Let $R=\sqrt{T^*T}$. Prove that $\forall u\in H, ||Ru||=||Tu||$.

Here's my approach:

$$||Tv||^2=\langle Tv,Tv\rangle=\langle T^*Tv,v \rangle = \langle R^2v,v \rangle= \langle Rv,R^*v \rangle$$

It would be convenient to have $R^*=R$, but I don't see why that should hold. Obviously $T^*T$ is self-adjoint, but why should its square root be self-adjoint as well?

Answer 1

Given a positive operator $A$, $\sqrt{A}$ is defined to be the unique positive square root of $A$.

The uniqueness and existence of such an operator is covered in most functional analysis texts.