*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. 

For this it is necessary that the intersections of the constraints lead to separate points, or to a linear relation. For this latter case it is necessary that the constraints relate to surfaces that have zero Riemann curvature (intersections of surfaces with non-zero curvature can not have zero curvature thus can not be a hyper-plane which has zero curvature and lead to linear constraints). 

Your example in the comments is an example where the non-linear constraints must lead to points. 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 $\det M=1$ won't do it) may restrict this surface/curve (which hay non-zero curvature) to one or a few points.