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Assume the semi-infinite Hankel matrix $H$ with entries $$H_{i,j}=\alpha_{i+j-1}, \quad i,j\geq1,$$ where $\alpha_{n}\in\mathbb{R}$, is given. It turns out that in some special situations a semi-infinite Jacobi matrix $T$ commuting (formally) with $H$ can be determined explicitly. For example, if $H$ is the Hilbert matrix, i.e. $\alpha_{n}=1/n$, then one can set $$T_{n,n}=2n(n-1), \quad T_{n,n+1}=T_{n+1,n}=-n^{2} \quad (T_{m,n}=0 \mbox{ otherwise})$$ and $HT=TH$, indeed.

My question is twofold. Is there any systematic way how to determine the diagonal and off-diagonal sequence of the Jacobi matrix $T$ which commutes with a given Hankel matrix $H$ (perhaps a computer based method determining several first entries of the sequences)?

More precisely, the question is: how to find sequences $b_{n}=T_{n,n}$ and $a_{n}=T_{n,n+1}$, such that $$(a_{j-1}-a_{i-1})\alpha_{i+j-1}+(b_{j}-b_{i})\alpha_{i+j}+(a_{j}-a_{i})\alpha_{i+j+1}=0, \quad \forall i,j\geq1,$$ where one sets $a_{0}=0$ and $\alpha_{n}\in\mathbb{R}$ is given. Of course, only non-trivial solution is of interest.

Second, is any other example of commuting Hankel and Jacobi matrix known? For instance, interesting Hankel matrices correspond to the choice $\alpha_{n}=1/n^{2}$ (or more general powers of $n$) or $\alpha_{n}=1/n!$.

Any information related to the post would be useful. Thanks.

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1 Answer 1

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This looks similar to the well-known(?) connection between Hankel determinants and orthogonal polynomials, see e.g. Sect 2.7 of Advanced Determinantal Calculus.

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    $\begingroup$ In fact, this is not the case. Jacobi matrix associated with a family of orthogonal polynomials (via coefficients from the three-term recurrence) need not commute with the corresponding Hankel matrix of moments of the measure of orthogonality for polynomials in question. For example, the Hilbert matrix is the Hankel matrix of moments of Lebesgue measure on $(0,1)$ but the Jacobi matrix $T$ corresponds to a special case of Continuous dual Hahn polynomials whose measure of orthogonality in not the Lebesgue one on $(0,1)$. $\endgroup$
    – Twi
    Oct 9, 2014 at 10:48

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