Let $X=[X_1,\dots,X_n]$ be an $n\times 1$ vector of square integrable real random variables on a probability space, and let $\mathcal F$ be a sub-$\sigma$-algebra. Consider the following conditional covariance matrix:
$$\Sigma_\mathcal F:=E[XX'|\mathcal F]$$
This matrix is positive semi-definite a.s. (see here). Am wondering if the following result is true:
Proposition. $\Sigma_\mathcal F$ fails to positive definite a.s. if and only if there exists $\mathcal F$-measurable random variables $W_1,\dots,W_n$, not all zeros a.s., such that $W_1X_1+\dots +W_nX_n=0$ a.s..
One direction is not too hard. Suppose $W$ is an $n\times 1$ $\mathcal F$ measurable vector such that $W^\top X=0$ a.s.. Then, from the linearity and pull-out property of conditional expectations,
$$0=E[(W^\top X)^2 |\mathcal F]=E[W^\top XX^\top W |\mathcal F]=W^\top \Sigma_\mathcal F W \quad \quad (1)$$
almost surely. So if $A$ is an event of positive probability on which $W\neq 0$, then $\Sigma_\mathcal F $ fails to be positive definite for almost all $\omega \in A$.
The other direction seems more tricky. Let $A$ be the event that $\Sigma_\mathcal F$ is symmetric positive semi-definite, but not positive definite. Note that $A$ is the intersection of the following three sets $$\bigcap_{i\neq j}\{\Sigma^{ij}_\mathcal F=\Sigma^{ji}_\mathcal F\}, \quad \bigcap_{a \in \mathbf{Q}^n} \{a^\top\Sigma _\mathcal Fa\geq 0\},\quad \{\det(\Sigma _\mathcal F)= 0\} $$ and so is $\mathcal F$-measurable. If $P(A)>0$, then for each $\omega \in A$ we can find a nonzero vector $a(\omega)$ statisfying
$$a(\omega)^\top\Sigma _\mathcal F (\omega) a(\omega)=0$$
and the idea is to define $W(\omega):=a(\omega)$ if $\omega \in A$ and $W(\omega):=0$ otherwise. Assuming $W$ is $\mathcal F$-measurable, then the same calculation as in $(1)$ shows that $W^\top X=0$ a.s..
But how can I ensure $\mathcal F$-measurability of $W$?
Thanks a lot for your help.