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‎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}‎ $‎.‎‎

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

$Tx= \sum \lambda_nu_n$ 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 $Ax:= \sum \sqrt{\lambda_n}u_n$ and show, that $A$ has the desired properties.

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    I think this only works for the field $\mathbb C$, probably this assumption was omited in the question? – 2017-01-24
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    No. It works also for the field of reel numbers. – 2017-01-24
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    Then how do you conclude that $T$ is self-adjoint? Because your statement that $Tx = \sum\lambda_n\langle x,u_n\rangle u_n$ implies that. – 2017-01-24
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    We have $T \ge 0$. – 2017-01-25
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    Ah, so self-adjointness is part of the definition of $T\geq 0$ for you. I did some googling to find that the definitions are not all the same. For me it is simply $\langle Tx, x\rangle\geq 0$ for each $x$. There are some threads on MSE concerning this issue as well. – 2017-01-25
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    I consider $T$ self-adjoint as part of the definition of $T \geq 0$ as-well and I consider this as pretty natural since then the term agrees with the usual definition of positive element in a c*-algebra. – 2017-01-26