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I am reading a paper about SVD of a operator. I know a little about SVD of matrix, and eigenvalue decomposition of operator. But I totally don't know what does SVD of a operator mean. In the following text, could anyone tell me the expressions of $u_k$ or $v_k$ and how to get them? Thanks in advance.

Let $$ f(x) = P(\varphi)(x) = \int_{-1}^1 \log(x-t)\varphi(t) dt,\quad x \in [3,5] $$ Using the language of classical potential theory, P is the operator mapping the charge disctribution on the interval $[-1,1]$ to the induced potential created on the interval $[3,5]$

One can construct the SVD of the operator P (defined above), representing it in the form $$ f(x) = P(\varphi)(x) = \sum_{k=1}^\infty \lambda_k(v_k,\varphi)u_k(x) $$ where $u_k:[3,5]\rightarrow \mathbf{R}, v_k:[-1,1]\rightarrow\mathbf{R} $ are the $k$th left and right singular vectors, respectively, and $\lambda_k$ is the $k$th singular value.

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    I know nothing about SVD decomposition of an operator. But I'm trying to guess. Operators may be represented in a matrix form (in finite basis), then SVD decomposition of an operator is probably the same as of the matrix. If the basis is infinite, or even non-discrete, then there's probably a generalization of the same process.2011-09-15
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    valdo, it is somehow the other way round. The SVD of an operator gives a (diagonal) matrix representation...2011-09-15

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