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?
 A: 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 $\mathbf{X}=QD\mathbf{Z}$.
A: The wikipedia article on covariance matrices answers that (the excerpt below is taken verbatim from that article):

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:
  $$\operatorname{var}(\mathbf{M}^{1/2}\mathbf{X}) = \mathbf{M}^{1/2} (\operatorname{var}(\mathbf{X})) \mathbf{M}^{1/2} = \mathbf{M}.$$

A: Q: Every Cov is a symmetric PSD. Is every symmetric positive definite matrix a Cov?A: Yes.
Let $M$ be symmetric and PSD. Since it is symmetric PSD there is cholesky factorization $LL^T$. Given a random vector $Z$ with covariance $I$ and zero expectation we construct $X = LZ$. Then
$$
V[X] = E[XX^T] - 0 = E[(LZ) (LZ)^T] = E[LZZ^TL^T] = LE[ZZ^T]L^T = LL^T = M.
$$
So $M$ is the covariance matrix of $X$. Thus, every symmetric positive definite matrix is a covariance matrix.

Note that both the symmetric and the PSD property are required for the cholesky decomposition:
If $A$ decomposes to $LL^T$ then
$$
x A x = x LL^T x = (xL, xL) = \lVert xL \rVert^2 \geq 0
$$
so $A$ must be PSD. Further $A^T = (LL^T)^T = LL^T = A$ so $A$ must be symmetric.
This answer assumes real matrices.
