# Square root of integral operator

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 $|K|^{1/2}$?

• Can you do the matrix case? Given $n \times n$ complex matrix $K = [k_{ij}]$, what are the entries of the matrix $|K|^{1/2}$? – Gerald Edgar Nov 21 '10 at 23:22
• Diagonalize $K*K=UDU*$ and $|K|^{1/2}=UD^{1/4}U*$ – m07kl Nov 22 '10 at 17:30

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).

• I agree, although I have neither a proof nor a reference. – Denis Serre Nov 22 '10 at 7:38