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I have a question concerning the definition of the square root of bounded linear operators. To introduce some notation: tr denotes the trace of linear operators and $\mathcal{L}(H)$ denotes the set of bounded linear operators, from H to H, where H symbolizes a Hilbert space. L' stands for the adjoint operator of L. We introduced the the space of trace class operators in the following way: $\{L \in \mathcal{L}(H) \ : \ tr((LL')^{\frac{1}{2}} < \infty \}$. The problem is, that I don't know how how square root of the above mentioned operator is defined.

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It is a well-known fact that every bounded positive operator on a Hilbert space (which are exactly those of the form $LL'$ for some bounded $L$) has a unique positive square root. This is a consequence of the continuous functional calculus; you can find about in any Functional Analysis book that talks about C$^*$-algebras.

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Notice that $LL'$ is positive definite. Square root is well-defined for bounded, positive definite operators (by functional calculus theorems).

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  • $\begingroup$ I only read, what you have stated, but I can't actually imagine, what a quare root in the context of Hilbert spaces means, e.g. how to 'calculate' $Q^{\frac{1}{2}} h$ for an $h \in H$. $\endgroup$
    – David
    Commented Jan 11, 2013 at 13:37
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    $\begingroup$ @David: for $\lvert x\rvert\leq 1$ we have $\sqrt{1-x}=1+\sum_{n=1}^\infty \binom {1/2} n (-x)^n$ (and the convergence is uniform). Using this series expansion you can define square root for positive operators $Q$ of norm not greater than $1$, and for others you can put $\sqrt Q= \sqrt {Q/\lVert Q\rVert }\cdot \sqrt {\lVert Q\rVert}$. For more detailed discussion, I suggest you check some functional analysis textbook. It should be explained in detail in proximity of polar decomposition of bounded operators. $\endgroup$
    – tomasz
    Commented Jan 11, 2013 at 13:57

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