# Is every symmetric positive semi-definite matrix a covariance of some multivariate distribution?

One can easily prove that every covariance matrix is positive semi-definite. I come across many claims that the converse is also true; that is,

Every symmetric positive semi-definite matrix is a covariance marix of some multivariate distribution.

Is it true? If it is, how can we prove it?

The answer is affirmative. Every positive semidefinite matrix $C$ can be orthogonally diagonalised as $QD^2Q^T$, where $Q$ is a real orthogonal matrix and $D$ is a nonnegative diagonal matrix. Let $\mathbf{Z}$ be a random vector following the standard multivariate normal distribution $N(0,I_n)$. It is straightforward to verify that $C$ is the covariance matrix of $\mathbf{X}=QD\mathbf{Z}$.
From the identity just above, let $\mathbf{b}$ be a $(p \times 1)$ real-valued vector, then: $$\operatorname{var}(\mathbf{b}^{\rm T}\mathbf{X}) = \mathbf{b}^{\rm T} \operatorname{var}(\mathbf{X}) \mathbf{b},$$ which must always be nonnegative since it is the variance of a real-valued random variable and the symmetry of the covariance matrix's definition it follows that only a positive-semidefinite matrix can be a covariance matrix. The answer to the converse question, whether every symmetric positive semi-definite matrix is a covariance matrix, is "yes". To see this, suppose $\mathbf{M}$ is a $p\times p$ positive-semidefinite matrix. From the finite-dimensional case of the spectral theorem, it follows that $\mathbf{M}$ has a nonnegative symmetric square root, that can be denoted by $\mathbf{M}^{1/2}$. Let $\mathbf{X}$ be any $p\times 1$ column vector-valued random variable whose covariance matrix is the $p\times p$ identity matrix. Then: $$\operatorname{var}(\mathbf{M}^{1/2}\mathbf{X}) = \mathbf{M}^{1/2} (\operatorname{var}(\mathbf{X})) \mathbf{M}^{1/2} = \mathbf{M}.$$