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<j 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<i_2<\dots<i_k\}

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


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