Prove an interesting property of pdf moments? I am examining properties of mass moments of probability densities:
$$ m_{i}\equiv\int_{-\infty}^{\infty}x^{i}f\left(x\right)dx  $$
Define a $\,n\times n\,$ covariance of the first $\,n\,$ moments:
$$ M_{ij}\equiv\int_{-\infty}^{\infty}f\left(x\right)\left(x^{i}-m_{i}\right)\left(x^{j}-m_{j}\right)dx=m_{i+j}-m_{i}m_{j} $$
where $1\leq i\leq n$ and $1\leq j\leq n$
I suspect but have failed to prove that:
$$ \left|M_{n\times n}\right|=0\iff\left\{ \begin{array}{c}
f\left(x\right)=\frac{1}{r}\sum_{j=1}^{r}\delta\left(x-x_{j}\right)\\
r\leq n\end{array}\right\} $$
In other words if $M$ is $n\times n$ and its Determinant is $0$, then $f(x)$ is a set of $n$ or fewer spikes.If $f(x)$ has more than $n$ spikes or has any continuous portion then $|M|>0$ for all $n$.
It's easy enough to demonstrate for chosen values of $r$ that $|M|=0$ ; the Mathematica code below does that. But I need a proof that if $\,|M|=0$ , then $\,f(x)\,$ consists of $\,r\,$ spikes, where $\,r\le n$ .
(*Demonstrate that for any n, Det[M]==0 if r==n, Det[M]>0 if r>n *)

n = 4;
M = Table[m[i + j] - m[i]*m[j], {i, 1, n}, {j, 1, n}];
Print[n, "-particle Det = ", Det[M] /. m[q_] -> (Sum[x[i]^q, {i, 1, n}]/n) // Simplify];
Print[n + 1, "-particle Det = ", Det[M] /. m[q_] -> (Sum[x[i]^q, {i, 1, n + 1}]/(n + 1)) // Simplify];

 A: It's false as stated: the determinant should be $0$ for any $f$ that is supported on a set of cardinality $\le n$, not just a uniform distribution.
In fact, let $X$ be a random variable whose distribution is supported on a set $\{r_1, \ldots, r_n\}$ of cardinality $n$.  Let
$P(z) = \prod_{k=1}^n (z - r_k) = \sum_{j=0}^n a_j z^j$.
If ${\bf a} = (a_1, \ldots, a_n)^T$, then $P(X) = 0$ so 
$(M {\bf a})_j = \text{Cov}(X^j, P(X)) = 0$ (the lack of an $a_0$ term does not matter because $\text{Cov}(X^j, 1) = 0$)
and therefore $M$ is singular.
Conversely, if $M$ is singular, take ${\bf b} \ne 0$ in its null space, and
define $P(z) = \sum_{k=1}^n b_j z^j$.  We have 
$\text{Var}(P(X)) = \text{Cov}(P(X),P(X)) = {\bf b}^T M {\bf b} = 0$, 
which implies $P(X) = 0$ a.s., and that means $X$ is almost surely one of the roots of $P$ (of which there are at most $n$).
A: I posted the Question and awarded the bounty to @RobertIsrael for his great Answer above. But I had to think pretty hard about his answer and "translate" into simpler language for myself, so I thought I'd post my simpler version.
First of all, my assertion wasn't quite correct because the spikes in the pdf $f(x)$ can be differently weighted. So the corrected theorem is:
$$ \left|M_{n\times n}\right|=0\iff\left\{ \begin{array}{c}
f\left(x\right)=\sum_{j=1}^{r}w_{j}\delta\left(x-x_{j}\right)\\
r\leq n\end{array}\right\}  $$
Since any weight can be zero, the case where there are fewer than n spikes is covered. The RHS of the above eqn can be restated as:
$$ f\left(x\notin\left\{ x_{1},...,x_{n}\right\} \right)=0 $$
Now consider this n-th order polynomial:
$$ q\left(x\right)\equiv\prod_{j=1}^{n}\left(x-x_{j}\right)=\sum_{k=0}^{n}a_{k}x^{k}$$
It is trivial but not necessary to calculate the $a_k$'s. For any x, either $f(x)$ or $q(x)$ will equal 0. Therefore, the following integrand and integral both equal zero:
$$ 0=\int f\left(x\right)\left(x^{j}-m_{j}\right)q\left(x\right)dx=\int f\left(x\right)\left(x^{j}-m_{j}\right)\sum_{k=0}^{n}a_{n}x^{k}dx=\sum_{k=0}^{n}a_{n}\left(m_{j+k}-m_{j}m_{k}\right)=\left(M_{n\times n}a\right)_{j} $$
This proves that $a$ is a 0-eigenvector of M, which can only occur if M is singular, QED:
$$ f\left(x\notin\left\{ x_{1},...,x_{n}\right\} \right)=0\Rightarrow|M_{n\times n}|=0 $$
Proof of the converse: If $M$ is singular, then there exists some $a\neq0$ such that $Ma=0$, from which we can construct a polynomial $q\left(x\right)\equiv\sum_{k=0}^{n}a_{k}x^{k}$, the covariance of which is:
$$Cov\left(q\left(x\right)\right)=\int f\left(x\right)\left(q\left(x\right)-\left\langle q\left(x\right)\right\rangle \right)^{2}=\int f\left(x\right)\left(\sum_{k=0}^{n}a_{k}\left(x^{k}-m_{k}\right)\right)^{2}=a^{T}Ma=0$$
$Cov(q(x))=0$ implies that $f(x)=0$ whenever $q(x)\neq0$, i.e., $f(x)$ can only be non-zero at the n roots of $q(x)$, QED:
$$|M_{n\times n}|=0\Rightarrow f\left(x\notin\left\{ x_{1},...,x_{n}\right\} \right)=0$$
...which completes the proof.
