Pick matrix

Remarks

The denominators are not $0$, because $\abs{z_i \overline{z_j}} < 1$. One has $\overline{P_{ji}} = P_{ij}$, so $P$ is a Hermitian matrix. By Nevanlinna–Pick interpolation, with all $w_i \in \mathbb{D}$, there is a holomorphic $f \colon \mathbb{D} \to \mathbb{D}$ with $f(z_i) = w_i$ if and only if $P$ is positive semidefinite.

Example: $n = 2$, $z_1 = 0$, $w_1 = 0$, $z_2 = a$, $w_2 = b$ with $a \ne 0$. Then $P = \begin{pmatrix} 1 & 1 \\ 1 & \frac{1 - \abs{b}^2}{1 - \abs{a}^2} \end{pmatrix}$, and $\det P = \frac{\abs{a}^2 - \abs{b}^2}{1 - \abs{a}^2}$. So $P$ is positive semidefinite exactly when $\abs{b} \le \abs{a}$, which matches the Schwarz lemma.

Non-example: the data must have $z_i \in \mathbb{D}$. For $z_1 = 1$ on the unit circle the entry $P_{11}$ would have denominator $1 - \abs{z_1}^2 = 0$, so no Pick matrix is defined.

The entries are the values $(1 - w_i \overline{w_j}) k(z_i, z_j)$, where $k(z, w) = \frac{1}{1 - z \bar w}$ is the reproducing kernel of the Hardy space $H^2$.