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

  • $\begingroup$ 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 $\endgroup$
    – develarist
    Aug 31, 2020 at 15:16
  • 1
    $\begingroup$ it may not be symmetric $\endgroup$
    – gunes
    Aug 31, 2020 at 15:19
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    $\begingroup$ One proof is that (assuming it is symmetric) it is the covariance matrix of the (multivariate) Normal distribution it determines, qed. $\endgroup$
    – whuber
    Aug 31, 2020 at 15:21

1 Answer 1


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:

$$p(\mathbf{x}) = (2 \pi)^{-n/2} \det(\mathbf{\Sigma})^{1/2} \exp \Big( -\frac{1}{2} \mathbf{x}^\text{T} \mathbf{\Sigma} \mathbf{x} \Big) \quad \quad \quad \text{for all } \mathbf{x} \in \mathbb{R}^n$$

The variance/covariance matrix for this random vector is:

$$\mathbb{V}(\mathbf{X}) = \mathbf{\Sigma}.$$

(Hat-tip to whuber in the comments for this answer.)


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