Is uncorrelated normal vector still independent even if covariance matrix is singular?

Question

Let $\mathbf{X}=(X_1,\ldots,X_n)$ be a normal random vector with a mean vector $\mu$ and a covariance matrix $\Sigma$. Suppose that $X_1,\ldots,X_n$ are uncorrelated, so that $\Sigma$ is diagonal. If $\Sigma$ is positive definite, then $\mathbf{X}$ has density $$f(\mathbf{x})=\frac{1}{\sqrt{(2\pi)^n\det\Sigma}}\exp\left(-\frac{1}{2}(\mathbf{x}-\mu)^T\Sigma^{-1}(\mathbf{x}-\mu)\right),$$ which splits into densities of $X_1,\ldots,X_n$, showing that $X_1,\ldots,X_n$ are independent. But what if $\Sigma$ is not positive definite? If $\mathbf{X}$ has no density, how can I show that $X_1,\ldots,X_n$ are indepenent? Is it even true?