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I was wondering if there are closed formulas for the singularvalues of a partial Hankel matrix (by partial I mean $\ell<n$)

\begin{align*} H= \begin{bmatrix} c_1 & c_2 & \ldots & c_\ell \\ c_2 & c_3 & \ldots & c_{\ell+1} \\ c_3 & c_4 & \ldots & c_{\ell+2}\\ \vdots & \vdots & \ddots & \vdots\\ c_n & c_{n+1} &\ldots & c_{n+\ell-1} \end{bmatrix} \end{align*} What does it mean for this matrix to be low-rank?

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Typically, one does "Prony method": considers an infinite (or just long enough) sequence $c=(c_1,c_2,\dots)$ and a system of equations of the form $V(x)Z=c$, with $Z$ a vector of non-0 unknowns, and $V(x)$ the Vandermonde matrix $$ V(x)=\begin{pmatrix} 1&1&\dots&1\\ x_1& x_2&\dots &x_k\\ x_1^2& x_2^2&\dots &x_k^2\\ &\dots&\dots&\dots\\ \end{pmatrix} $$ By multiplying both sides of $V(x)Z=c$ on the left by the row vectors $a_\ell=(\underbrace{0,\dots,0}_\ell,a_0,a_1,\dots)$, for $\ell\geq 0$, one relates the polynomials $x^\ell a(x)=x^\ell\sum_{i\geq 0} a_i x^i$ with zeros $\{x_1,\dots,x_k\}$ and the kernel of a submatrix of the the infinite Hankel matrix corresponding to the sequence $c$.

By controlling the degree of $a(x)=k$, you should be able to get something interesting...

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  • $\begingroup$ Although, my comment is not related directly to this question, the answer of @ Dima Pasechnik motivated me to ask here. Consider the matrix $H$ be semi-Hadamard matrix that means the first row of $H$ be the vector $(c_1,c_2,\cdots,c_n)$ and other rows of $H$ be different permutations of vecor $(c_1,c_2,\cdots,c_n)$. My question: Is there a way to write $H$ matrix by the Vandermonde matrices. I appreciate for any suggestions. $\endgroup$
    – Amin235
    Commented Mar 19, 2017 at 12:05
  • $\begingroup$ it seems to depend a lot on the kind of permutations you allow; in general this appears to be impossible to say much. $\endgroup$
    – Dima Pasechnik
    Commented Mar 20, 2017 at 15:44
  • $\begingroup$ What about circular matrix as a special case of permutation matrices. I mean the first row be the vector $(c_1,c_2,\cdots, c_n)$ and the second row be $(c_2,c_3,\cdots,c_1)$ and finally the last row be $(c_n,c_1,\cdots, c_{n-1})$. $\endgroup$
    – Amin235
    Commented Mar 20, 2017 at 19:52
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    $\begingroup$ in this case my answer directly applies. $\endgroup$
    – Dima Pasechnik
    Commented Mar 20, 2017 at 21:30
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    $\begingroup$ The sequence $c$ I talk about in the answer will be periodic: $c=(c_1,c_2,..,c_n,c_1,c_2,...,c_n,c_1,c_2,...)$ $\endgroup$
    – Dima Pasechnik
    Commented Mar 20, 2017 at 22:58

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