I have been doggedly searching for a direct proof of the following theorem:

Theorem 1: Let $H$ be a complex nonsingular $n\times n$ Hankel matrix. Then $H$ can be factorized $H = V^\top DV$ where $V$ is a complex $n\times n$ Vandermonde matrix and $D$ is a complex $n\times n$ diagonal matrix.

There exist numerous references stating that this result is well known either without proof or with a citation to one of the references below which I have found insufficient for one reason or another. Here is an abbreviated summary of my research:

  • References (e.g., here, here, and here) that are unavailable to me or in Russian
  • References (e.g., here, here, and here) for a confluent Vandermonde decomposition of a rank-deficient finite and/or finite-rank infinite Hankel matrices: These do no prove the theorem I am after since the Vandermonde decomposition involves confluent Vandermonde matrices.
  • References for a Vandermonde decomposition of positive semidefinite or positive definite Hankel matrix: These don't address the case of a general square Hankel matrix.
  • References (here, here, and here) where the theorem is proven as a corollary of a rather length theory: these results do prove the theorem that I want. However, the result follows as a corollary of several pages of a lengthy theory drawing connections to other structured matrices, polynomials, and rational functions. It seems like this basic and essential fact should have a more direct proof.

I am looking for as simple and direct a proof of Theorem 1 as possible, a reference to one in the literature, or some insight as to why a simple proof does not exist. Below the fold, I have outlined my attempt to provide such a proof.

My Attempt

Consider the entries of the Hankel matrix

$$ H = \begin{bmatrix} h_0 & h_1 & \cdots & h_{n-1} \\ h_1 & h_2 & \cdots & h_{n+1} \\ \vdots & \vdots &\ddots & \vdots \\ h_{n-1} & h_n & \cdots & h_{2n-2} \end{bmatrix}. $$

The Vandermonde decomposition in Theorem 1 is equivalent to finding a representation of the sequence $(h_k)_{0\le k\le 2n-2}$ of the following form:

$$ h_k = \sum_{j=1}^n \alpha_j z_j^k, \quad k = 0,1,2,\ldots,2n-2. \tag{$\star$} $$

Observe that the sequence $(h_k)_{0\le k\le 2n-2}$ has fewer parameters than the parametrization ($\star$). To rectify this, add $h_{2n-1}$, value to be set later, to the end of the sequence and require it to obey the parametrization ($\star$) for $k=2n-1$.

We seek to identify the poles $z_1,\ldots,z_n$ by Prony's method. The key observation is a sequence of the form ($\star$) is a solution to an order-$n$ difference equation:

$$ h_{n+k} = c_{n-1} h_{k+(n-1)} + \cdots + c_0 h_k. $$

We find the difference equation coefficients $c_0,\ldots,c_{n-1}$ by solving the following linear system of equations

$$ \begin{bmatrix} h_n \\ h_{n+1} \\ \vdots \\ h_{2n-1} \end{bmatrix} = \begin{bmatrix} h_0 & h_1 & \cdots & h_{n-1} \\ h_1 & h_2 & \cdots & h_{n+1} \\ \vdots & \vdots &\ddots & \vdots \\ h_{n-1} & h_n & \cdots & h_{2n-2} \end{bmatrix} \begin{bmatrix} c_0 \\ c_1 \\ \vdots \\ c_{n-1}\end{bmatrix}. $$

The matrix in this equation is precisely the matrix $H$ which we know to be nonzero. Thus, this equation has a unique solution $c_0,\ldots,c_{n-1}$. Provided the characteristic equation

$$ z^n - c_{n-1}z^{n-1} - \cdots - c_0z^0 = 0 \quad \text{has distinct roots} \quad z_1,\ldots,z_n, \tag{$\dagger$} $$

there exist coefficients $\alpha_1,\ldots,\alpha_n$ such that ($\star$) holds. The Vandermonde decomposition is proven under the assumption ($\dagger$).

The Vandermonde decomposition thus rests on choosing $h_{2n-1}$ such that ($\dagger$) holds. Since $h_{2n-1}$ is arbitrary, the polynomial in ($\dagger$) can be chosen to be any polynomial of the form

$$ f(z) = g(z) + \alpha h(z) $$


$$ g(z) = z^n - d_{n-1} z^{n-1} - \cdots - d_0 z^0, \quad h(z) = -e_{n-1} z^{n-1} - \cdots - e_0 z^0 \tag{$\blacktriangle$} $$

with $d_0,\ldots,d_{n-1}$ and $e_0,\ldots,e_{n-1}$ the solutions of

$$ \begin{bmatrix} h_n \\ h_{n+1} \\ \vdots \\ 0 \end{bmatrix} = H \begin{bmatrix} d_0 \\ d_1 \\ \vdots \\ d_{n-1}\end{bmatrix}, \quad \begin{bmatrix} 0 \\ 0 \\ \vdots \\ h_{2n-1} \end{bmatrix} = H \begin{bmatrix} e_0 \\ e_1 \\ \vdots \\ e_{n-1}\end{bmatrix}. $$

To this end, we have a helpful lemma (see, e.g., Lemma 7.7):

Lemma 2. If any polynomials $g$ and $h$ are coprime, then $g - \alpha h$ has simple roots except for finitely many values $\alpha$.

Thus, Theorem 1 would be proven if I could show that $g$ and $h$ as defined in ($\blacktriangle$) are coprime.

  • $\begingroup$ Unless I've missed something, one proof for this result is given here. $\endgroup$ May 13, 2022 at 17:19
  • $\begingroup$ @BenGrossmann I don't believe that Theorem 1 is proven in the cited article. This article focuses on the confluent Vandermonde decomposition of infinite Hankel matrices. They address the square nonsingular case at the bottom of page 7. However, as far as I can tell, they do not prove the existence of a $h_{2n-1}$ such that ($\dagger$) has simple roots. As such, they only prove the existence of a confluent Vandermonde decomposition of a nonsingular Hankel matrix. $\endgroup$
    – eepperly16
    May 13, 2022 at 17:21
  • $\begingroup$ $(\dagger)$ must be $\ldots\;=0$, right? $\endgroup$ May 13, 2022 at 20:43
  • $\begingroup$ @ReinhardMeier Thanks for the correction. Fixed $\endgroup$
    – eepperly16
    May 13, 2022 at 20:45


You must log in to answer this question.