Does every positive semidefinite hankel matrix obeys one Vandermonde decomposition?

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I'm reading the paper(I can't find one arXiv version of this paper...) and suspect the correctness of one theorem inside. A hankel matrix $H$ is a square matrix in which each ascending skew-diagonal from left to right is constant. Vandemonde decomposition is to decompose $H$ as $H=V^TDV$ with $D$ diagonal matrix and $V$ the Vandemonde matrix.

Theorem I.1 of the paper states that

Theorem I.1. For any positive semidefinite Hankel matrix $H \in \mathbb{R}^{n \times n}$ with rank $r, 1 \leq r<n$, there exists a Vandermonde matrix $V \in \mathbb{R}^{n \times r}$ and a diagonal matrix $D \in \mathbb{R}^{r \times r}$ with positive diagonal entries such that $H=$ $V^T D V$.

My question is, is this statement really true? For example, if we set $n=3,r=1$ and choose $H=\left( \begin{matrix} 0& 0& 0\\ 0& 0& 0\\ 0& 0& 1\\ \end{matrix} \right) $. Then $H$ is hankel and does not have a Vandemonde decomposition if we require elements of $D$ to be positive.