Results of invertibility of a matrix involving the Szego kernel

Question

In the context of reproducing kernel Hilbert spaces, the Szego kernel is the function $k(z_i,z_j)=\frac{1}{1-z_j\overline{z_j}}$. Given $2n$ points $\{z_1,\ldots,z_n\},\{w_1,\ldots,w_n\}\in\mathbb{D}\cap\mathbb{R}$, (with all $z_i$'s different and all $w_i$'s different, but the sets may not be disjoint), are there any known result of the invertibility of the matrix: $$\big[k(z_i,w_j)]_{i,j=1}^n$$

For all sets of $4$ points I have tested, this matrix came out invertible. I wasn't able, however, to determine if it is always the case. Moreover, I didn't find any information on whether there are conditions on the sets of points for which this matrix is indeed invertible.

Any help would be greatly appreciated.

Answer 1

Expand the determinant $D=\det k(x_j,y_k)$ along the first row. This shows that as a function of $x=x_1$, it is of the form $$ D(x) = \sum_{j=1}^n \frac{c_j}{1-y_j x} , $$ with $c_j$ independent of $x=x_1$ (and of $y_j$, but that doesn't matter here).

This rational function has $n$ poles at $1/y_1,\ldots, 1/y_n$. On the other hand, clearly $D(x_2)=\ldots =D(x_n)=D(\infty)=0$, and since this is a total of $n$ zeros, we have found all of them. Thus $D\not= 0$ under your assumptions.

Small details added later: If $y_j=0$ for some $j$, then we need to slightly modify the argument: now we have $n-1$ poles, and also $n-1$ zeros since now $D(\infty)=c_j\not= 0$. This last fact we could also have obtained directly from an inductive argument since the $c_j$ are determinants of $(n-1)\times (n-1)$ submatrices of the same type. Strictly speaking, we actually need something along these lines no matter what since the argument obviously breaks down if we could have $c_1=\ldots =c_n=0$.

Answer 2

Sorry for my previous answer. This is a partial answer for the $2\times 2$ case. Notice first that we can assume without loss of generality that $z_1=0$. Otherwise we can apply the Moebius transformation $\varphi(z)=\frac{z-z_1}{1-\overline{z_1}z}$ to the points and the new matrix is going to be \begin{equation*} \Big[ \frac{(1-z_1\overline{z_i})(1-\overline{z_1}w_j)}{1-|z_1|^2} k(z_i,w_j) \Big]_{i,j=1}^n \end{equation*} which is invertible if and only if the original one is. But when $z_1=0$ we have \begin{equation*} det \begin{bmatrix} 1 & 1 \\ \frac{1}{1-z_2\overline{w_1}} & \frac{1}{1-z_2\overline{w_2}} \end{bmatrix} = \frac{z_1(\overline{w_1-w_2})}{(1-z_2\overline{w_1})(1-z_2\overline{w_2})} \end{equation*} which is zero if and only if $z_2=0=z_1$ or $w_1=w_2$.