Singular covariance matrix identity

Question

Let $Z=(Z_1,...,Z_n)^T$ be a random vector with mean 0, and define its covariance $\Sigma=\mathbb{E}[ZZ^T]$. Assume that $\Sigma$ is singular. Then I need to show something (non-related to the question), but I was told to start with the identity, that $\alpha^T \Sigma \alpha=0$ (I guess its supposed to be that there exists an $\alpha$ such that it is true), but I fail to understand how that is true, any help is appreciated.

Answer 1

A covariance matrix $\Sigma$ is always positive semi-definite, that is $\alpha^T \Sigma \alpha \geq 0 $ for any $\alpha$. Singularity occurs when two variables are fully correlated. In other case, a non-zeros $\alpha$ exists such that $\alpha^T \Sigma \alpha=0$. As a small example, assume $n=2$ and $Z_2 = 2 Z_1$, then $$\Sigma = \begin{bmatrix} \sigma_{1}^2 & \sigma_{1,2} \\ \sigma_{1,2} & \sigma_2^2 \end{bmatrix}$$ where $\sigma_1^2 = E(Z_1^2), \sigma_2^2 = E(Z_2^2) = 4\sigma_1^2$ and $\sigma_{12} = E(Z_1Z_2) = 2\sigma_1^2$, then $$\Sigma = \begin{bmatrix} 1 & 2\\ 2 & 4\end{bmatrix}$$ which is singular.

Answer 2

I think it should be $\alpha^T \Sigma \alpha = 0$ (the dimensions for forming $Z \alpha$ are wrong). If that's the case then any $\alpha$ in the null-space of the singular matrix $\Sigma$ would do.