# Proof: If a matrix is semi-definite and symmetric positive then it is a covariance matrix

Anyone have the following proof? If a matrix is semi-definite positive and symmetric then it is a covariance matrix.

• a covariance matrix is a matrix that is a positive semi definite matrix. (any) definiteness can be a property of any matrix. whether a covariance matrix obtains a specific type of definiteness depends on the nature of the data the covariances are being computed for Aug 31, 2020 at 15:16
• it may not be symmetric Aug 31, 2020 at 15:19
• One proof is that (assuming it is symmetric) it is the covariance matrix of the (multivariate) Normal distribution it determines, qed.
– whuber
Aug 31, 2020 at 15:21

Let $$\mathbf{\Sigma}$$ be an arbitrary $$n \times n$$ real symmetric positive semi-definite matrix. Consider the normal random vector $$\mathbf{X} \sim \text{N}(\mathbf{0}, \mathbf{\Sigma})$$ with density function:
$$p(\mathbf{x}) = (2 \pi)^{-n/2} \det(\mathbf{\Sigma})^{1/2} \exp \Big( -\frac{1}{2} \mathbf{x}^\text{T} \mathbf{\Sigma} \mathbf{x} \Big) \quad \quad \quad \text{for all } \mathbf{x} \in \mathbb{R}^n$$
$$\mathbb{V}(\mathbf{X}) = \mathbf{\Sigma}.$$