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

