# How to construct the Singular Value Decomposition(SVD) of an operator P?

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. – valdo Sep 15 '11 at 7:46
valdo, it is somehow the other way round. The SVD of an operator gives a (diagonal) matrix representation... – Dirk Sep 15 '11 at 7:54

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 $\lambda_k = \sqrt\sigma_k$.