I'm trying to prove, or disprove, that the following matrix is positive (semi-)definite and symmetric (so a covariance matrix); $$ M = C_M^{-1} + G^T C_D^{-1} G. $$ Here, $C_M$ is a covariance matrix (which is positive semi-definite and symmetric, sometimes positive definite) of dimensions $n \times n$, which is not necessarily diagonal. The matrix $C_D$, of dimensions $m \times m$, is also a covariance matrix which in my application is always diagonal, but could be extended to any arbitrary covariance matrix (so always positive semi-definite, symmetric).
G is an arbitrary matrix of dimensions $m \times n$. It does not necessarily have full rank. I think this already excludes positive definiteness of the second term, since $x^T G^T G x = (G x)^T(G x) > 0$ only if G is full rank.
Can we extend this to $G^T C_D^{-1} G$ to prove that M is positive semi-definite symmetric? I'd suggest to use the property $x^T A x \geq 0$;
$$ \begin{eqnarray} x^T G^T C_D^{-1} G x &=& (G x)^T C_D^{-1} (G x) \nonumber \\ & \overset{?}{\geq} & 0 \end{eqnarray} $$
I guess this boils down to proving that the inverse of $C_D$ (and $C_M$) is positive semi-definite, because afterwards we can again use the same definition on $(G x)^T C_D^{-1} (G x) $. And the sum of PSD matrices is also PSD (that proof is rather easy, see here).
And then a small follow up; if the second term is PSD, but the first term is PD, the resulting matrix M is also PD right?
For a little context, from the first equation matrix $M$ is used as the mass matrix in a Hamiltonian Monte Carlo sampler, and I need to Cholesky decompose it to propose momentum samples. Cholesky decomposition can (I think) only be done on a positive definite matrix. If a matrix is positive semi-definite, it is still possible to Cholesky decompose it if the diagonal entries of $L$ are allowed to be zero, but I don't know if my algorithm breaks down in this case.
