4
$\begingroup$

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.

$\endgroup$

2 Answers 2

5
$\begingroup$

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 $c_1=\ldots =c_n=0$.

$\endgroup$
0
$\begingroup$

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 $w_1=w_2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.