To expand on Zachary's comment, the covariance matrix does not capture the "relation" between two random variables, as "relation" is too broad of a concept. For example, we'd probably want to include the dependence of two variables on each other to be include in any measure of their "relation".
However, we know that $cov(X,Y)=0$ does not imply that they are independent, as for example is the case with two random variables X~U(-1,1) and Y=X^2 (for a short proof, see: https://en.wikipedia.org/wiki/Covariance#Uncorrelatedness_and_independence).
So if we'd think that the covariance includes full information about variable relations, as you ask, zero covariance would suggest no dependence. This is what Zachary means when he says that there can be non-linear dependences that the covariance does not capture.
However, let $X:=(X_{1},...,X_{n})'$ be multivariate normal, X~$N(\mu,\Sigma)$. Then $X_{1},...,X_{n}$ are independent iff $\Sigma$ is a diagonal matrix with all off-diagonal elements = 0 (if all covariances = 0).
To see that this condition is sufficient, observe that the joint density factors,
\begin{equation} f(x_{1},...,x_{n}) =
\dfrac{1}{ \sqrt{(2 \pi)^{n} | \Sigma |}} exp(- \dfrac{1}{2} (x - \mu)' \Sigma^{-1} (x - \mu))= \Pi^{n}_{i=1} \dfrac{1}{\sqrt{2 \pi \sigma_{ii}}} exp(- \dfrac{(x_{i}-\mu_{i})^{2}}{2 \sigma_{ii}})=f_{1}(x_{1})...f_{n}(x_{n})\end{equation}.
To see that the condition is necessary, recall the bivariate case. If $X_{1}$ and $X_{2}$ are independent, then $X_{1}$ and $X_{1}|X_{2} = x_{2}$ must have the same variance, so
\begin{equation} \sigma_{11}=\sigma_{11|2}=\sigma_{11}-\sigma^{2}_{12} \sigma^{-1}_{22} \end{equation}
which implies $\sigma_{12}=0$. By the same argument, all off-diagonal elements of $\Sigma$ must be zero.
(source: prof. Geert Dhaene's Advanced Econometrics slides)
