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