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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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    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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Somehow the construction is similar to the matrix case, but usually more involved.

You first have to clarify between what Hilbert spaces your operator maps. In your case $P: L^2([-1,1]) \to L^2([3,5])$ seems appropriate. Then you have to form the adjoint operator, that is the operator $P^*$ such that for all $\phi\in L^2([-1,1])$ and $f\in L^2([3,5])$ it holds that $ \int_3^5 P(\phi)(x)f(x) dx = \int_{-1}^1 \phi(t) P^*(f)(t)dx. $ Just inserting and changing the order of integration gives $ P^*(f)(t) = \int_3^5 \log(x-t)f(x) dx. $

Now you form the operator $P^*P:L^2([-1,1]) \to L^2([-1,1])$ which is selfadjoint and positive definite. Moreover it is also compact. Hence, you can apply the spectral theorem for compact selfadjoint operator and you know that there is an orthonormal basis of eigenvectors of $P^*P$ (with nonnegative eigenvalues $\sigma_k$). This gives you the right singular vectors $v_k$ and the singular values are $\lambda_k = \sqrt\sigma_k$.

This is the general procedure for compact operators. However, there may be simpler approach for intergral operators.

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    Thank you. The process is really very similar to SVD of matrix.2011-09-15