*To show that the variance-covariance matrix has zero eigenvalues if and only if the variables are linearly dependent, it only remains to be shown that "if the matrix has zero eigenvalues then the variables are not linearly independent".*

If you have a zero eigenvalue for $C_{ij} = \text{Cov}(X_i,X_j)$ then there is some linear combination (defined by the eigenvector $v$)

$$Y = \sum_{j=1}^n v_j (X_i) $$

such that for every $X_i$

$$\text{Cov}(X_i,Y)  = \sum_{j=1}^n v_j \text{Cov}(X_i,X_j) = \sum_{j=1}^n v_j C_{ij} = 0$$

and 

$$\text{Cov}(Y,Y) = \sum_{i=1}^n v_i\text{Cov}(X_i,Y) =\sum_{i=1}^n v_i \cdot 0= 0$$

thus $Y$ is a constant and the variables $X_i$ are either constants or not independent.