# Multivariate Gaussian Definition when Covariance matrix is singular, What's wrong?

Given $$\mathbf{\Sigma} \in \mathbb R^{k \times k}$$ $$\mathbf{u} \in \mathbb R^k$$ The multivariate Gaussian pdf can be determined By definition: $$f(\mathbf{x})=\frac{1}{2\pi^{\frac{-k}{2}}|\Sigma|^{\frac{1}{2}}}e^{\frac{1}{2}(\mathbf{x-u})^T\mathbf{\Sigma}^{-1}(\mathbf{x-u})}$$

The Covariance matrix is only limited to be positive semidefinite.

So it could be singular (Non-invertible)

This will also lead to a zero in the denumerator, and also the$\Sigma^{-1}$ doesn't exist.

What we do in that case to write the joint pdf?