Consider the 1-torus $\mathbb{T}$. Let $k$ be a smooth function on $\mathbb{T}^2$ and $K$ be the integral operator on $L^2(\mathbb{T})$ with kernel $k$. One can show that $K$ is of trace class, hence $|K|^{1/2}$ is a Hilbert Schmidt operator=integral operator. But what is the kernel of $|K|^{1/2}$?
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2$\begingroup$ Can you do the matrix case? Given $n \times n$ complex matrix $K = [k_{ij}]$, what are the entries of the matrix $|K|^{1/2}$? $\endgroup$– Gerald EdgarNov 21, 2010 at 23:22
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$\begingroup$ Diagonalize $K*K=UDU*$ and $|K|^{1/2}=UD^{1/4}U*$ $\endgroup$– m07klNov 22, 2010 at 17:30
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It seems to me that you are looking for a formula for the kernel of $|K|^{1/2}$. But, as Gerald, mentioned, such a formula (in the case where the space $\mathbb{T}$ is replaced by a finite set) would give you a formula for the entries of the square root of an arbitrary positive matrix. And I don't think such a thing exists (or, at least, I don't think it is known).
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2$\begingroup$ I agree, although I have neither a proof nor a reference. $\endgroup$ Nov 22, 2010 at 7:38