# Fredholm determinant

In the study of measure transformation in Gaussian space, there is a fundamental issue that I want to write down here, which is Fredholm determinant. Let me try to use half an hour to explain the intuition of this determinant in matrix form with finite dimension.

So suppose a matrix $A=(a_{ij})\in \mathbb{R}_{n\times n}$, our goal is to find out what definition $det(I+A)$ is that makes sense for trace class operator(i.e.$Tr(A)<\infty$). We know that its trace $Tr(A)=\sum a_{ii}$, and assume the eigenvalues of $A$ are $\lambda_1,\lambda_2, \dots, \lambda_n$, we have

$AX=\begin{pmatrix} \lambda_1 & & &\\ & \lambda_2 & 0 \\ & & \ddots &\\ & & & \lambda_n \end{pmatrix} X$, so

$Tr(A)=Tr(X^{-1}AX)=\sum \lambda_i$.

And because $det(I+A)=\prod (1+\lambda_i)=\sum_{k=0}^{n}(\sum_{i_1,\dots,i_k}\lambda_{i_1}\dots\lambda_{i_k})$

Iet’s take a look at these terms in the summation,

when $k=0$, it’s 1,

when $k=1$, it’s $\lambda_1+\lambda_2+\dots+\lambda_n$,

when $k=2$, it’s $\sum_{i\neq j}\lambda_i\lambda_j$,

which is the trace of operator $\Lambda^2(A)$ on a linear space with basis $e_i\wedge e_j, i where is $\wedge$ is the wedge product form.

Similarly, we have the expression given in widipedia about “Fredholm determinant” that for a general trace-class operator $A$

$det(I+A)=\sum_{k=0}^{\infty}Tr(\Lambda^k(A))$ and this new operator $\Lambda^k(A)$ is a linear operator on space formed by the basis $\{e_{i_1}\wedge e_{i_2}\wedge\dots\wedge e_{i_k}| i_1

Finally, I want to mention that when $Tr(A)=\sum |\lambda_i|<\infty$, easy to check that $det(A)<\infty$ indeed.