Family of explicitly diagonalizable weighted Hankel matrices generalizing the Hilbert matrix

Abstract

A three-parameter family $B=B(a,b,c)$ of weighted Hankel matrices is introduced with the entries \[ B_{j,k}=\frac{\Gamma(j+k+a)}{\Gamma(j+k+b+c)}\,\sqrt{\frac{\Gamma(j+b)\Gamma(j+c)\Gamma(k+b)\Gamma(k+c)}{\Gamma(j+a)\, j!\,\Gamma(k+a)\, k!}}\,, \] $j,k\in\mathbb{Z}_{+}$, supposing $a$, $b$, $c$ are positive and $a<b+c$, $b<a+c$, $c\leq a+b$. The famous Hilbert matrix is included as a particular case. The direct sum $B(a,b,c)\oplus B(a+1,b+1,c)$ is shown to commute with a discrete analog of the dilatation operator. It follows that there exists a three-parameter family of real symmetric Jacobi matrices, $T(a,b,c)$, commuting with $B(a,b,c)$. The orthogonal polynomials associated with $T(a,b,c)$ turn out to be the continuous dual Hahn polynomials. Consequently, a unitary mapping $U$ diagonalizing $T(a,b,c)$ can be constructed explicitly. At the same time, $U$ diagonalizes $B(a,b,c)$ and the spectrum of this matrix operator is shown to be purely absolutely continuous and filling the interval $[0,M(a,b,c)]$ where $M(a,b,c)$ is known explicitly. If the assumption $c\leq a+b$ is relaxed while the remaining inequalities on $a$, $b$, $c$ are all supposed to be valid, the spectrum contains also a finite discrete part lying above the threshold $M(a,b,c)$. Again, all eigenvalues and eigenvectors are described explicitly.