Properties of matrix $X=\left[\frac{1}{1-\bar\alpha_i \alpha_j}\right]_{ij}$

Question

Let $\{\alpha_i\}_{i=1}^n$ be complex numbers such that $|\alpha_i|<1$, and consider the following $n\times n$ structured matrix $$ X=\left[\frac{1}{1-\bar\alpha_i \alpha_j}\right]_{ij}. $$ Such matrix arises in the solution of particular Stein matrix equations (e.g., see p. 11 of Bhatia, "Positive definite matrices", Princeton University Press, 2007), and seems to be somehow related to the class of Cauchy matrices. $X$ seems to have a interesting structure (and, I guess, it may feature some "nice" properties), however I couldn't find any additional information on this matrix on the web. So my question:

Does matrix $X$ have a name? Are there any known properties of $X$ (such as formulas for the determinant, inverse, etc.)?

P.S. A property that follows from the connection with Stein equations is that $X$ is always positive semidefinite.

Answer 1

This matrix is also related to the Nevanlinna-Pick Theorem. Namely, if $z_i, \lambda_i \in \mathbb D, 1\leq i\leq n$ then $$\left[\begin{matrix} \frac{1- \overline{z_j}z_i}{1-\overline{\lambda_j}\lambda_i}\end{matrix}\right]_{i,j=1}^n \geq 0$$ if and only if there is a holomorphic function $\varphi : \mathbb D \rightarrow \overline{\mathbb D}$ such that $\varphi(\lambda_i) = z_i, 1\leq i\leq n$.

In this theory $K(\lambda_j,\lambda_i) = (1-\overline{\lambda_j}\lambda_i)^{-1}$ is called the Szego kernel. So one could refer to your matrix as the Szego kernel matrix.

Answer 2

Assuming all the $\alpha_j$ are nonzero, the matrices $X$ are Cauchy-like matrices, since you can rewrite them as $$ X_{ij} = \frac{\alpha_j^{-1}}{\alpha_j^{-1}-\bar{\alpha}_i} $$ so there are analogous formulas for their determinant and inverse. In particular, $XA^{-1}$ is a Cauchy matrix, where $A = diag(\alpha_i)$, so these formulas follow directly from those for Cauchy matrices.

A variant of this equivalence that does not require the invertibility of $A$ and better exploits symmetry/Hermitianity is the following. Let $B = (I+A)(A-I)^{-1}$ (note that $\alpha_i\neq 1$, otherwise there would be a zero denominator in $X_{ii}$, so $A-I$ is invertible). Then, $$ B^*X+XB = -2(A- I)^{-*}E(A- I)^{-1} $$ expands to $$ ( I+A^*)X(A- I) + (A^*- I)X( I + A) = -2E $$ which reduces to $$ 2A^*XA - 2X = -2E, $$ which is the Stein equation that you used to define $X$.

So $X$ solves a Lyapunov equation with a diagonal $B$, and hence one can write the more symmetric formula

$$X_{ij} = \frac{-2(\alpha_i-1)^{-*}(\alpha_j-1)^{-1}}{\bar{\beta}_i + \beta_j},$$

where $\beta_i = \frac{\alpha_i + 1}{\alpha_i-1}$ are the diagonal entries of $B$. Note that $\Re\beta_i < 0$ iff $|\alpha_i| < 1$.

Alternatively, the Hermitian matrix $(A^*-I)^{-1}X(A-I)^{-1}$ is a Cauchy matrix wrt the two sequences $\bar{\beta}_i$ and $-\beta_j$.

This trick to convert between Lyapunov and Stein equations is classical (bilinear transform, or Cayley transform).

Answer 3

These matrices are a special case of a much broader class of positive semidefinite (psd) matrices constructed as follows. Let $(\alpha_{i,j})_{i,j=1}^n$ be any psd matrix satisfying $|\alpha_{i,j}| < 1$ for all $i,j \in \{1,\ldots,n\}$. Then the matrix $((1-\alpha_{i,j})^{-c})_{i,j=1}^n$ is positive semidefinite for all $c \geq 0$. This follows immediately from the geometric series formula since an entrywise power of a psd matrix is psd. The matrix you mention is constructed in this way from the matrix $(\alpha_1,\ldots,\alpha_n)^*(\alpha_1,\ldots,\alpha_n)$ which is transparently psd.