# functional determinant

Let $w(x,y)$ be a real-valued symmetric function over the $[0,1]$x$[0,1]$ interval. We also know that if $n$ is an integer, and you pick $n$ values $x_i$ in the $[0,1]$ interval, then the matrix $\Sigma_N$ whose entry $(i,j)$ is given by $w(x_i,x_j)$ is symmetric positive definite. To me, $\Sigma_N$ is a covariance matrix so I call $w$ a covariance function.

What is the determinant of $w$? How can it be defined, and more importantly calculated for a given function $w$?

This is as far as I've gotten: If the matrix is diagonal, that is if $w(x,y)=0$ if $x\ne y$, then

$$\det \Sigma_N = \prod_{i=1}^N w(x_i,x_i)$$ so that when $N$ tends to infinity and taking $x_i$ points with even spacing and $x_1=0$ and $x_N=1$ $$\lim_{N\to +\infty} \frac{1}{N-1} \log\det\Sigma_N = \int_0^1 \log w(x,x)\text{d} x$$ How can you generalize that to arbitrary covariance functions?

I'm no mathematician, feel free to make this more rigorous.

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I cannot provide you with a solution of the general case (as I guess is difficult). But there is an important class of matrices for which the limit can be taken rather explicitly (for the other matrices it just remains a limit :-)).

Imagine that the matrix $w(x_i,y_j)$ just depends on the difference $i-j$. These matrices are called Toeplitz matrices and they appear in real life more often than you think. Without the discretization, we introduce $f(x-y) = w(x,y)$.

Take a look at the Fourier transform of $w(x,y)$. It is also a symmetric matrix $$\hat{w}_{m,n}= \int_{[0,1]^2}\! dx dy\, e^{2\pi i (xm + yn)} w(x,y) = \int\! dx dy_{[0,1]^2}\, e^{2\pi i (xm + yn)} f(x-y) = \hat f_{m-n},$$ where $\hat f(m) = \int \!dx\,f(x) e^{2\pi i xm}$. Note that the Fourier transform is unitary (up to a factor $N^{-1}$ see below), so we can calculate the determinant of $\hat{w}$ (not very rigorous but in principle you can write everything out as limits and convince yourself).

Now, we can use a result due to Szegő:

let $w_N$ be the $N\times N$ Toeplitz matrix with the elements $$(\hat w_N)_{m,n} = \hat f_{m-n}.$$ Then (for $N \to \infty$) $$\log \det \hat{w}_N \sim N [\log f]_0 + \sum_{n=1}^\infty n [\log f]_n [\log f]_{-n}$$ where $$[\log f]_n = \int_0^1\!dx\,\log f(x) e^{2\pi i m x}.$$

We obtain that the logarithm of the functional determinant of $w(x,y)$ is given by (the $1/N$ comes from the Fourier transform) $$\lim_{N\to\infty} \frac{1}{N} \log \det \hat w_{N} = \int_0^1\!dx\,\log f(x),$$ which gives your result when all the elements on the diagonal are equal.

Note that all of this is not yet rigorous (I just wanted to give you a short summary). In principle you should in the beginning introduce $N$ by only allowing $w$ to have up to $N$ Fourier components. Then use Szegő and in the end let $N\to\infty$.

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wonderful! Is an extension possible to small deviations from the Toeplitz case? Say to a matrix M=T+E where T is toeplitz and E is any matrix but M and T have almost the same norm? – yannick Jan 13 '12 at 19:18