# Eigenvalues of discretized linear integral operator

Suppose I have the following kernel operator:

$Af(x) = \int_{-1}^1 K(x-y)f(y)dy$

which is also positive and compact. Hence, it has a countable set of positive eigenvalues. Suppose those eigenvalues are known and denoted by $\lambda_1, \lambda_2,...$

Now suppose I take a discrete version of the operator, which is represented by a matrix of size $N$ whose entries are given by $K_{mn} = K(x_m-x_n)$. where $x_m, x_n$ are at equispaced points (or any other way if it helps).

What can we say about the eigenvalues of the matrix $K$ with respect to the eigenvalues of the continuous operator $\lambda_1,\lambda_2,...$ ?

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## 1 Answer

While such a compact operator on the closed interval is a Hilbert space so is a subset of limits of all matrices, the actual matrix required is weighted.

e.g. elements K/n

The problem in practice is naive numerical integration and doesn't behave well for general cases. Smoother, i.e. high-order differentiable, K and f converge better but ideally matrix elements would be from a less naive quadrature method.

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