Square root of compact positive operator

Question

‎Let $T\geq 0$ and $T\in K(H)$, where $K(H)$ denotes the compact operators on $H$.

I want to show the following statement:

There ‎is a‎ ‎compact and unique positive ‎operator‎‎ ‎‎$‎A‎$‎ so that‎ $A‎ =‎ ‎T‎^{2}‎ $‎.‎‎

Answer 1

The spectral decomposition shows that there is an orthonormal sequence $(u_n) $of eigenvectors of $T$ such that

$Tx= \sum \lambda_n<x,u_n>u_n$ for all $x \in H$.

$(Tu_n= \lambda_n u_n)$

Since $T \ge 0$, we have $\lambda_n \ge 0$ for all $n$

Put $Ax:= \sum \sqrt{\lambda_n}<x,u_n>u_n$ and show, that $A$ has the desired properties.