Solution of the quadratically hyponormal completion problem.

For $m\ge 1$, let $\alpha : \alpha _{0}<\cdots <\alpha _{m}$ be a collection of ($m+1$) positive weights. The \emph{Quadratically Hyponormal Completion Problem} seeks necessary and sufficient conditions on $\alpha $ to guarantee the existence of a quadratically hyponormal unilateral weighted shift $W$ with $\alpha $ as the initial segment of weights. We prove that $\alpha $ admits a quadratically hyponormal completion if and only if the self-adjoint $m\times m$ matrix \begin{equation*}D_{m-1}(s):= \begin{pmatrix}q_{0}&\bar r_{0}&0&\hdots &0&0 r_{0}&q_{1}&\bar r_{1}&\hdots &0&0 0&r_{1}&q_{2}&\hdots &0&0 \vdots &\vdots &\vdots &\ddots &\vdots &\vdots 0&0&0&\hdots &q_{m-2}&\bar r_{m-2} 0&0&0&\hdots &r_{m-2}&q_{m-1}\end{pmatrix} \end{equation*} is positive and invertible, where $q_{k}:=u_{k}+|s|^{2} v_{k}$, $r_{k}:=s\sqrt {w_{k}}$, $u_{k}:=\alpha _{k}^{2}-\alpha _{k-1}^{2}$, $v_{k}:=\alpha _{k}^{2}\alpha _{k+1}^{2}-\alpha _{k-1}^{2}\alpha _{k-2}^{2}$, $w_{k}:=\alpha _{k}^{2}(\alpha _{k+1}^{2}-\alpha _{k-1}^{2})^{2}$, and, for notational convenience, $\alpha _{-2}=\alpha _{-1}=0$. As a particular case, this result shows that a collection of {\em four} positive numbers $\alpha _{0}<\alpha _{1}<\alpha _{2}<\alpha _{3}$ {\em always} admits a quadratically hyponormal completion. This provides a new qualitative criterion to distinguish quadratic hyponormality from 2-hyponormality. [ABSTRACT FROM AUTHOR]