Let $x\_1,\dots,x\_p$ be $p$ points in $\mathbb{R}^n$ ($n\geq 2$) with $x\_1=0$. Consider the symmetric matrix $M(x)=(m\_{ij}(x))\_{1\leq i,j\leq p}$ where $m\_{ij}(x) = \exp(-\frac{1}{2}\Vert x\_i - x\_j\Vert^2)$. The norm is the standard Euclidean one. I would like to prove that $\det(M(x))^{-\frac{1}{2}}$ is locally integrable on $(\mathbb{R}^n)^{p-1}$.


I already know that $M(x)$ is positive definite if the $(x\_i)\_{1\leq i\leq p}$ are pairwise distinct. More precisely, if $x\_1,\dots,x\_p$ take exactly $q$ different values, then $M(x)$ has rank $q$.


I also managed to prove that, if $\lambda\_1(x)$ is the smallest eigenvalue of $M(x)$ then there exists a constant $K\_{n,p}>0$ depending only on $n$ and $p$ such that:
$$ \frac{1}{\lambda\_1(x)} \leq K\_{n,p}\sum\_{i=1}^p \frac{1}{\prod\_{j\neq i}\Vert x\_i-x\_j\Vert^2} \exp(\Vert x\_i\Vert^2) \prod\_{j\neq i} (1+\Vert x\_j\Vert^2).$$
This is enough to prove that $\frac{1}{\sqrt{\lambda\_1}}$ is locally integrable but not enough to get the integrability of $\det(M(x))^{-\frac{1}{2}}$.


Does anyone know how to prove a similar inequality for $\frac{1}{\det(M(x))}$ instead of $\frac{1}{\lambda\_1(x)}$? Alternatively, does anyone know of a suitable lower bound for $\det(M(x))$ in terms of the $(\Vert x\_i-x\_j\Vert)\_{i\neq j}$? I know that this matrix appears in statistics, but I couldn't find anything concerning its determinant in the literature so far.



Determinant of some covariance matrix (Gaussian kernel process)
Welcome to MO, Thomas.


I am afraid what you would like to prove is false. Namely, $det(M(x))^{-\frac 1 2}$ is locally integrable if and only if $n$ is large enough depending on $p$.


More precisely there is a function $p\_0 \colon \mathbf{N} \to \mathbf{N}$ such that $det(M(x))^{-\frac 1 2}$ if and only if $p \leq p\_0(n)$.


I do not know the precise value of $p\_0(n)$, but I can give the inequalities $n-1\leq p\_0(n) < (n+1) n^{n+1}$ (I do not guarantee that these inequalities are correct).


First remark that by multiplying by the diagonal matrix with diagonal entries $\exp({\frac 1 2 \|x\_i\|^2})$, the problem is the same as whether $det(N(x))^{- \frac 1 2}$ if locally integrable, where $N\_{i,j}(x) = \exp(\langle x\_i,x\_j\rangle)$.


We can write $N(x) = \sum\_{k\geq 0} \frac{1}{k!}\widetilde N\_k(x)$ where $\widetilde N\_k(x)$ if the positive matrix with entries $\langle x\_i,x\_j\rangle^k$. If $N\_k(x) \in M\_{p-1}(\mathbf R)$ is the matrix obtained by removing the first (vanishing if $k \geq 1$) lign and column of $\widetilde N\_k(x)$, we get that $det(N(x))=det(\sum\_{k \geq 1} \frac{1}{k!}N\_k(x))$, and we deduce that $det(N(x)) \geq det(N\_1(x)) = det(\langle x\_i,x\_j\rangle)\_{2\leq i,j\leq p}$. A rapid computation on a piece of paper indicates that $det(N\_1(x))^{-\frac 1 2}$ is integrable if $n \geq p-1$, which would give the first inequality.


For the other inequality, remark that $N\_k(x)$ has rank at most $n^k$. This says that $\sum\_{k \geq 1} \frac{1}{k!}N\_k(x)$ can be decomposed as the sum from $1$ to $K-1$, which has rank less than $n+\dots+n^{K-1} < n^{K}$ plus the rest, which is a matrix with entries of order $O(\max\_i \|x\_i\|^{2K})$. If $p \geq n^K$, we therefore get that at least $p-n^K$ eigenvalues of $\sum\_{k \geq 1} \frac{1}{k!}N\_k(x)$ are less than $\max\_i \|x\_i\|^{2K}$, whereas the other are locally bounded. This tells that $det(N(x))^{- \frac 1 2} \geq C (\max\_i \|x\_i\|)^{-K(p-n^K)}$ locally. This last quantity is not locally integrable if $K(p-n^K)\geq n(p-1)$. Taking $K=n+1$, we get that $det(N(x))^{- \frac 1 2}$ is not locally integrable if $p \geq (n+1)n^{n+1}$.


