In the context of a Gaussian model, I came across a matrix product $R \, A^{-1} \, R^t$ where $R$ is a $m \times n$ rectangular matrix and as implied $A$ is $n \times n$ and invertible.
On which properties of $R$ does the existence of $(R \, A^{-1} \, R^t)^{-1}$ depend?
First, if $m=n$ and $R$ is invertible, then $RA^{-1}R^T$ is invertible as well.
Second, if $A$ is in addition symmetric and positive definite on the range of $R$, then $RA^{-1}R^T$ is invertible if $R$ has rank $m$. Actually, in this case $RA^{-1}R^T$ is symmetric and positive definite as well.
This does not work without definiteness of $A$: $$ R=\begin{pmatrix}0&1\end{pmatrix}, \quad A^{-1}=\begin{pmatrix}1 & 1 \\ 1 &0\end{pmatrix}, \quad RA^{-1}R^T = 0, $$ which provides an example, where $R$ has full rank, but $RA^{-1}R^T$ is not invertible.