---
tag: 3JVT83K8A1
kind: concept
name: Trace of an integral operator
msc: [45C05, 47G10, 47B10]
sources:
  - url: https://en.wikipedia.org/wiki/Trace_class
---

## Summary

The trace of an integral operator with a continuous kernel $K(s, t)$ on $[a, b]$ is the integral of the kernel along the diagonal, $\int_a^b K(s, s) \, ds$.
It is the continuous analogue of the sum of the diagonal entries of a matrix.
In Fredholm theory, minus the trace is the coefficient of $\lambda$ in the Fredholm determinant $D(\lambda) = 1 - \lambda \int_a^b K(s, s) \, ds + \cdots$.
For a [[trace class operator|Z2CP530WDE]] with continuous kernel, it equals the operator trace and the sum of the [[eigenvalues|0BQAA2BK3Q]].

## Explanations

### L1 human

Think of the kernel $K(s, t)$ as a matrix with a continuum of rows $s$ and columns $t$.
The diagonal of this matrix is the set of points with $s = t$.
Summing the diagonal entries becomes integrating $K(s, s)$ over $s$.
For example, the kernel $K(s, t) = st$ on $[0, 1]$ has trace $\int_0^1 s^2 \, ds = 1/3$.
This operator sends $x$ to $s \mapsto s \int_0^1 t x(t) \, dt$, and its only nonzero eigenvalue is also $1/3$, with eigenfunction $s$.