# Results of invertibility of a matrix involving the Szego kernel

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.

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 $$c_1=\ldots =c_n=0$$.
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 $$w_1=w_2$$.