*To show that the variance-covariance matrix has eigenvalues equal to zero if and only if the variables are not linearly independent, it only remains to be shown that "if the matrix has eigenvalues equal to zero 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 $v$)

$$Y = \sum_{i=1}^n v_i (X_i) $$

such that 

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

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

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Notes:

 - the first line in the equation with $\text{Cov}(Y,Y)$ is due to the property of covariance $$\scriptsize\text{Cov}(aU+bV,cW+dX) = ac\,\text{Cov}(U,W) + bc\,\text{Cov}(V,W) +ad\, \text{Cov}(U,X) + bd \,\text{Cov}(V,X) $$

 - the step from the second to the third line  is due to the property of a zero eigenvalue $$\scriptsize \sum_{j=1}^nv_jC_{ij} = 0$$