---
tag: 7T9ZS932SG
kind: definition
name: Fredholm determinant
msc: [45B05, 47B10]
defines: PPX0AK81QV
uses: [ZM5G144PZG, 00KDZRW4C0, 0G4F0XB9X2]
sources:
  - url: https://en.wikipedia.org/wiki/Fredholm_determinant
---

## Definition

Let $a < b$ and let $K \colon [a, b] \times [a, b] \to \C$ be a [[continuous kernel|ZM5G144PZG]].
For $n \ge 1$ and points $x_1, \dots, x_n \in [a, b]$, let $\det[K(x_i, x_j)]_{i,j=1}^{n}$ be the [[determinant|00KDZRW4C0]] of the $n \times n$ matrix with entries $K(x_i, x_j)$.
For $n \ge 1$ let
$$c_n = \frac{1}{n!} \int_a^b \cdots \int_a^b \det[K(x_i, x_j)]_{i,j=1}^{n} \, dx_1 \cdots dx_n,$$
and let $c_0 = 1$.
The *Fredholm determinant* of $K$ is the function $D \colon \C \to \C$ given by
$$D(\lambda) = \sum_{n=0}^{\infty} (-1)^n c_n \lambda^n.$$
The series converges for every $\lambda \in \C$, so $D$ is an [[entire function|0G4F0XB9X2]].

## Remarks

Example: for $K(x, y) = xy$ on $[0, 1]$, $c_1 = \int_0^1 x^2 \, dx = 1/3$.
For $n \ge 2$ the matrix $[x_i x_j]$ has rank at most $1$, so its determinant is $0$ and $c_n = 0$.
Hence $D(\lambda) = 1 - \lambda/3$.

Non-example: the definition does not apply when $K$ is only square-integrable, because the diagonal values $K(x, x)$ enter $c_1 = \int_a^b K(x, x) \, dx$ and are not defined for an element of $L^2$.
For such kernels one uses a modified determinant in which the diagonal terms are removed.

The convergence for all $\lambda$ follows from Hadamard's inequality: if $\abs{K} \le M$, then $\abs{c_n} \le n^{n/2} M^n (b - a)^n / n!$.
The coefficient $c_1$ is the integral $\int_a^b K(x, x) \, dx$, the trace of the kernel, and $D(\lambda) = 1 - c_1 \lambda + \cdots$.
For a degenerate kernel $K(x, y) = \sum_{i=1}^{m} a_i(x) b_i(y)$, one has $D(\lambda) = \det(I_m - \lambda A)$, where $A_{ij} = \int_a^b b_i(y) a_j(y) \, dy$.