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

*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.

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

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


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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. 

In fact, there is a direct correspondence between the eigenvectors associated with the zero eigenvalue and the linear constraints. 

$$C \cdot v = 0 \iff Y = \sum_{i=1}^n v_i X_i = \text{const}$$

Thus non-linear constraints leading to a zero eigenvalue must, together combined, generate some linear constraint.

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### How can non-linear constraints lead to linear constraints

Your example in the comments can show this intuitively how non-linear constraints can lead to linear constraints by reversing the derivation. The following non-linear constraints

 $$\begin{array}{lcr}
   a^2+b^2&=&1\\ 
   c^2+d^2&=&1\\
   ac + bd &=& 0 \\
   ad - bc &=& 1
 \end{array}$$

can be reduced to

 $$\begin{array}{lcr}
   a^2+b^2&=&1\\ 
   c^2+d^2&=&1\\
   a-d&=&0 \\
   b+c &=& 0 
 \end{array}$$

You could *inverse* this. Say you have non-linear plus linear constraints, then it is not strange to imagine how we can replace one of the linear constraints with a non-linear constraint, by filling the linear constraints into the non-linear constraints. E.g when we substitute $a=d$ and $b=-c$ in the non-linear form $a^2+b^2=1$ then you can make another relationship $ad-bc=1$. And when you multiply $a=d$ and $c=-b$ then you get $ac=-bd$.