*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".*

###Linear independence is not just sufficient but *also* a neccesary condition

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 about derivation

 - 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 $$\scriptsize \sum_{j=1}^nv_jC_{ij} = 0$$


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### Non-linear constraints

So, since linear constraints are a *necessary* condition (not just sufficient),  non-linear constraints will only be relevant when they indirectly imply a (necessary) linear constraint. 

This is only possible when the constraints lead to variables that are constants (the trivial case), or when the surface related to the non-linear constraints has zero curvature (intersections of surfaces with non-zero curvature can not have zero curvature thus can not be a hyper-plane which has zero curvature). 

Your example in the comments is an example where the non-linear constraints lead to degenerate variables. With the matrix $$M=\begin{bmatrix} X_1 & X_2 \\ X_3 & X_4\end{bmatrix}$$ the constraint $M \cdot M^T$ can be translated to the variables parameterized as: 

$$\begin{array}{rcl} X_1 &=& cos(\alpha) \\
X_2 &=& sin(\alpha) \\
X_3 &=& cos(\alpha +\frac{1+2k}{2} \pi) \\
X_4 &=& sin(\alpha +\frac{1+2k}{2} \pi) \\\end{array}$$

a fourth constraint (although $\det M=0$ won't do it) may restrict this surface to one or a few points.