*To show that the variance-covariance matrix has zero eigenvalues if and only if the variables are not linearly independent, 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 

$$\begin{array}{rcl}
\text{Cov}(Y,Y) &=& \sum_{i=1}^n \sum_{j=1}^n v_i v_j \text{Cov}(X_i,X_j) &\quad^{note\, (1)}\\ 
&=&\sum_{i=1}^n v_i\sum_{j=1}^n v_j C_{ij} &\quad^{note\, (2)} \\
&= &\sum_{i=1}^n v_i \cdot 0 &\quad^{note\,(2)}\\ 
&=& 0 \end{array}$$

which means that $Y$ needs to be a constant and thus the variables $X_i$ add up to a constant and are either constants themselves (the trivial case) or not linearly independent.

<sup> ${(1)}:$ property of covariance $\scriptsize\text{Cov}(aU+bV,cW+dX) = ac\,\text{Cov}(U,W) + bc\,\text{Cov}(V,W) +ad\, \text{Cov}(U,X) + bd \,\text{Cov}(V,X) $</sup>

<sup> ${(2)}:$ property of zero eigenvalue $\scriptsize \sum_{j=1}^nv_jC_{ij} = 0$</sup>