If $\operatorname{var}\left(\sum_{ij}A_{ij}v_{i}v_{j}\right)=0$ for any $v_i$, then $\operatorname{var}(A_{ij})=0$?

Question

Let $A_{ij}$ be a random matrix, satisfying $A_{ii}=0$ and $A_{ij}=A_{ji}$. Suppose we know that $\operatorname{var}\left(\sum_{ij}A_{ij}v_{i}v_{j}\right)=0$ for any vector $v_i$. What can we say about the statistics of the entries $A_{ij}$?

The title of this question is meant only as an example: Does it follow that $\operatorname{var}(A_{ij})=0$ ? If this is not true, I am still interested what kind of constraints can be derived on the distribution of $A_{ij}$ from this assumption.

We know that

$$\operatorname*{var}\left( \sum_{ij}v_{i}v_{j}A_{ij} \right) = \sum_{ij}\sum_{i'j'} v_{i}v_{j} v_{i'}v_{j'} \operatorname*{cov}(A_{ij},A_{i'j'}) = 0$$

for any $v_i$. This must impose some constraints on the distribution of the entries $A_{ij}$. However I'm not sure this equation is sufficient to conclude that $\operatorname*{cov}(A_{ij},A_{i'j'}) = 0$, for example.

Answer 1

Consider the vector $v$ such that all the entries are zero except for two elements, label them $v_m$ and $v_n$, which equal one for arbitrary $m,n$. The sum

$$\sum_{ij}A_{ij}v_{i}v_{j}$$

equals $A_{mn}$, and, since the sum's variance equals zero, the variance of $A_{mn} = 0$ as well. Since $m$ and $n$ are arbitrary, it must be that the variances of all the entries in $A$ are individually equal to zero.