 Solution

Let $K={\Sigma }^{-1}$ be the precision matrix, let $R=\{1,\dots ,p\}\setminus \{j,k\}$, and condition on $V_R=v_R$. In the Gaussian exponent, all terms depending jointly on $v_j$ and $v_k$ are contained inIf $K_{jk}=0$, this conditional density is a product of one function of $v_j$ and one function of $v_k$, so the conditional independence of $V_j$ and $V_k$ given $V_R$ holds.

Conversely, conditional independence makes this everywhere-positive conditional density factorize. Its mixed second derivative must therefore vanish:Hence the conditional independence holds exactly when $({\Sigma }^{-1}{)}_{jk}=0$.