Consider $n,d \in \mathbb{N}_{>0}$, there are many functions $y:\mathbb{N}^{n} \to \mathbb{R}$. Now for simplicity, we denote $y(\alpha)$ to be $y_{\alpha}$. Let $|\alpha| = \sum_{i=1}^{n}\alpha_{i}$, where $\alpha \in \mathbb{N}^{n}$. Let $\mathbb{N}_{d}^{n}:=\{ \alpha \in \mathbb{N}^{n}||\alpha|<d \}$. Since $|\mathbb{N}^{n}_{d}| = {n+d \choose d}$, we can fix an arbitrary bijective function $m:\{ 1,\dots, {n+d \choose d} \} \to \mathbb{N}^{n}_{d}$. So now when describing a matrix, we can use the terminology of the $\alpha$'s row or column to mean $m^{-1}(\alpha)$'s row or column.
So after picking a function $y:\mathbb{N}^{n} \to \mathbb{R}$, we can define the moment matrix $M_{d}[y] = [y_{\alpha+\beta}]_{\alpha\in \mathbb{N}^{n}_{d}, \beta \in\mathbb{N}^{n}_{d}}$. After fixing some arbitrary function $m$, we use the terminology above where the $a's$ row and $\beta's$ column has value $y_{\alpha+\beta}$. The ordering, or how we choose $m$ is irrelavent, we can assume from now, that the ordering put smaller degreed tuples first (i.e. 000 001, 010, 100, 002, 011, ..., d00, in the case of $n=3$)
As an example, for the ternary case $n=3$ and $d=2$, we have $$ M_2[y]=\left[\begin{array}{llllllllll} y_{000} & y_{100} & y_{010} & y_{001} & y_{200} & y_{110} & y_{101} & y_{020} & y_{011} & y_{002} \\ y_{100} & y_{200} & y_{110} & y_{101} & y_{300} & y_{210} & y_{201} & y_{120} & y_{111} & y_{102} \\ y_{010} & y_{110} & y_{020} & y_{011} & y_{210} & y_{120} & y_{111} & y_{030} & y_{021} & y_{012} \\ y_{001} & y_{101} & y_{011} & y_{002} & y_{201} & y_{111} & y_{102} & y_{021} & y_{012} & y_{003} \\ y_{200} & y_{300} & y_{210} & y_{201} & y_{400} & y_{310} & y_{301} & y_{220} & y_{211} & y_{202} \\ y_{110} & y_{210} & y_{120} & y_{111} & y_{310} & y_{220} & y_{211} & y_{130} & y_{121} & y_{112} \\ y_{101} & y_{201} & y_{111} & y_{102} & y_{301} & y_{211} & y_{202} & y_{121} & y_{112} & y_{103} \\ y_{020} & y_{120} & y_{030} & y_{021} & y_{220} & y_{130} & y_{121} & y_{040} & y_{031} & y_{022} \\ y_{011} & y_{111} & y_{021} & y_{012} & y_{211} & y_{121} & y_{112} & y_{031} & y_{022} & y_{013} \\ y_{002} & y_{102} & y_{012} & y_{003} & y_{202} & y_{112} & y_{103} & y_{022} & y_{013} & y_{004} \end{array}\right] $$ We also observe that any Hankel matrix of dimension $s$, is a moment matrix with $n=1$, $d = s-1$, and vice versa.
Now since any positive semidefinite Hankel matrix admits a Vandermonde factorization, that is, for any Hankel matrix $H \in \mathbb{R}^{n\times n}$, we have $H = V \Lambda^{2}V^{\top}$, for a diagonal matrix $\Lambda$, and Vandermonde matrix $V^{\top}$, where $$ V^{\mathrm{T}}=\left[\begin{array}{cccc} 1 & x_1 & \cdots & x_1^{n-1} \\ \cdots & \cdots & \cdots & \cdots \\ 1 & x_n & \cdots & x_n^{n-1} \end{array}\right] $$ The intuition is that we can write positive semidefinite matrix $H$ with rank $r$ as sum of $r$ rank 1 matrix. And we can easily show that any rank 1 Hankel matrix of dimension n that is positive semidefinite is in the form of $x:= [1 \;x \;x^{2}\;\dots\;x^{n-1}], H= c x x^{\top}$, for some $c > 0$. The factorization claim shows that we can write any rank $r$ positive semidefinite Hankel as $r$ sum of rank 1 matrix of this form
So now we conjecture for the same for positive semidefinite moment matrix, we observe the any $M_{d}[y] \succeq 0$ with $n$ variables, and rank 1, we have $M_{d}[y] = cx x^{\top}$, where denoting $y_\alpha \in \mathbb{N}^{n}$ with $\alpha$ having only the ith entry being 0 as $x_{i}$, we obtain the $\alpha's$ entry of $x$ is $\sum_{i=1}^{n}x_{i}^{\alpha_{i}}$.
However, can we still make the same conclusion that any positive semidefinite moment matrix of rank $r$ is the sum of $r$ rank 1 matrix of this form? While I think the ordering of the rows/columns of the moment matrix doesn't matter, I'm not completely sure. I really only care about the positive semidefinite case, but I would also love to hear for any idea/suggestion for the other cases as it might be enlightening.
Here is a proof of positive definite Hankel factorization . Maybe with limiting argument the proof can also apply to positive definite case. I'm think about reproducing the proof
Here is links to all sort of Vandermonde factorization claims, including the positive semidefinite case.
Here is a link to maybe related paper. I only skimmed through but I'm not sure it involves more general moment matrix
Any comment/intuition/idea are welcome. The entire question might not be able to be answered so easily. But I'd love to hear any idea, any resources that might helpful, or any result even for tiny cases.
