Given a random vector $X = \begin{bmatrix} X_1 \\ \vdots \\ X_n\end{bmatrix}$
Suppose that $R = \mathbb{E}[XX^T]$ is the correlation matrix of the random vector $X$
We claim that if $R$ is singular, then the components of $X$ must not be linearly independent (i.e. they are linearly dependent).
Does anyone see how to prove this claim? I've tried some thought experiments (e.g. when $R$ is diagonal) but I couldn't see how you would go about proving this.
If $R$ is singular, then there exists a non-zero vector $v$ so that $Rv = 0$; multiplying on the left by $v^T$ then gives $v^T Rv = 0$. Substituting the definition of $R$ in gives $$0 = v^T R v = v^T\mathbb{E}[XX^T]v = \mathbb{E}[v^TX X^Tv] = \mathbb{E}[| X \cdot v |^2].$$ Since $\mathbb{E}[|X \cdot v|^2] = 0,$ this implies that $X \cdot v = 0$ almost surely, i.e. $X$ is linearly dependent.