Perhaps by simplifying the notation we can bring out the essential ideas. It turns out we don't need involve expectations or complicated formulas, because everything is purely algebraic.
###The algebraic nature of the mathematical objects
The question concerns relationships between (1) the covariance matrix of a finite set of random variables $X_1, \ldots, X_n$ and (2) linear relations among those variables, considered as vectors.
The vector space in question is the set of all finite-variance random variables (on any given probability space $(\Omega,\mathbb P)$) modulo the subspace of almost surely constant variables, denoted $\mathcal{L}^2(\Omega,\mathbb P)/\mathbb R.$ (That is, we consider two random variables $X$ and $Y$ to be the same vector when there is zero chance that $X-Y$ differs from its expectation.) We are dealing only with the finite-dimensional vector space $V$ generated by the $X_i,$ which is what makes this an algebraic problem rather than an analytic one.
###What we need to know about variances
$V$ is more than just a vector space: it is a quadratic module, because it comes equipped with the variance. All we need to know about variances are two things:
The variance is a scalar-valued function $Q$ with the property that $Q(aX)=a^2Q(X)$ for all vectors $X.$
The variance is nondegenerate.
The second needs some explanation. $Q$ determines a "dot product," which is a symmetric bilinear form given by
$$X\cdot Y = \frac{1}{4}\left(Q(X+Y) - Q(X-Y)\right).$$
(This is of course nothing other than the covariance of the variables $X$ and $Y.$) Vectors $X$ and $Y$ are orthogonal when their dot product is $0.$ The orthogonal complement of any set of vectors $\mathcal A \subset V$ consists of all vectors orthogonal to every element of $\mathcal A,$ written
$$\mathcal{A}^0 = \{v\in V\mid a . v = 0\text{ for all }v \in V\}.$$
It is clearly a vector space. When $V^0 = \{0\}$, $Q$ is nondegenerate.
Allow me to prove that the variance is indeed nondegenerate, even though it might seem obvious. Suppose $X$ is a nonzero element of $V^0.$ This means $X\cdot Y = 0$ for all $Y\in V;$ equivalently,
$$Q(X+Y) = Q(X-Y)$$
for all vectors $Y.$ Taking $Y=X$ gives
$$4 Q(X) = Q(2X) = Q(X+X) = Q(X-X) = Q(0) = 0$$
and thus $Q(X)=0.$ However, we know (using Chebyshev's Inequality, perhaps) that the only random variables with zero variance are almost surely constant, which identifies them with the zero vector in $V,$ QED.
###Interpreting the questions
Returning to the questions, in the preceding notation the covariance matrix of the random variables is just a regular array of all their dot products,
$$T = (X_i\cdot X_j).$$
There is a good way to think about $T$: it defines a linear transformation on $\mathbb{R}^n$ in the usual way, by sending any vector $x=(x_1, \ldots, x_n)\in\mathbb{R}^n$ into the vector $T(x)=y=(y_1, \ldots, x_n)$ whose $i^\text{th}$ component is given by the matrix multiplication rule
$$y_i = \sum_{j=1}^n (X_i\cdot X_j)x_j.$$
The kernel of this linear transformation is the subspace it sends to zero:
$$\operatorname{Ker}(T) = \{x\in \mathbb{R}^n\mid T(x)=0\}.$$
The foregoing equation implies that when $x\in \operatorname{Ker}(T),$ for every $i$
$$0 = y_i = \sum_{j=1}^n (X_i\cdot X_j)x_j = X_i \cdot \left(\sum_j x_j X_j\right).$$
Since this is true for every $i,$ it holds for all vectors spanned by the $X_i$: namely, $V$ itself. Consequently, when $x\in\operatorname{Ker}(T),$ the vector given by $\sum_j x_j X_j$ lies in $V^0.$ Because the variance is nondegenerate, this means $\sum_j x_j X_j = 0.$ That is, $x$ describes a linear dependency among the $n$ original random variables.
You can readily check that this chain of reasoning is reversible:
Linear dependencies among the $X_j$ as vectors are in one-to-one correspondence with elements of the kernel of $T.$
(Remember, this statement still considers the $X_j$ as defined up to a constant shift in location--that is, as elements of $\mathcal{L}^2(\Omega,\mathbb P)/\mathbb R$--rather than as just random variables.)
Finally, by definition, an eigenvalue of $T$ is any scalar $\lambda$ for which there exists a nonzero vector $x$ with $T(x) = \lambda x.$ When $\lambda=0$ is an eigenvalue, the space of associated eigenvectors is (obviously) the kernel of $T.$
###Summary
We have arrived at the answer to the questions: the set of linear dependencies of the random variables, qua elements of $\mathcal{L}^2(\Omega,\mathbb P)/\mathbb R,$ corresponds one-to-one with the kernel of their covariance matrix $T.$ This is so because the variance is a nondegenerate quadratic form. The kernel also is the eigenspace associated with the zero eigenvalue (or just the zero subspace when there is no zero eigenvalue).
###Reference
I have largely adopted the notation and some of the language of Chapter IV in
Jean-Pierre Serre, A Course In Arithmetic. Springer-Verlag 1973.