Inverse of the matrix product $\boldsymbol{A} \cdot\boldsymbol{S} \cdot \boldsymbol{A}^{T}$ If I have an $n\times n$ symmetric matrix $\boldsymbol{S}$ and a $m\times n$ matrix $\boldsymbol{A}$ is there any relation between  $(\boldsymbol{A} \cdot\boldsymbol{S} \cdot \boldsymbol{A}^{T})^{-1}$ and $\boldsymbol{S}^{-1}$?

We begin by noting that $m \leq n$, and $A$ has rank $m$.
Because $A$ has full rank, there exists an invertible $n \times n$ matrix $B$ such that 
$$
AB^{-1} = \pmatrix{I_m&0} : = J
$$
(we can find this matrix by row-reducing $A^T$).  We then have
$$
(ASA^T)^{-1} = (JB^SB^{T}J^T)^{-1} = ((B SB^{T})[\{1,\dots,m\}])^{-1}
$$
Where $M[S]:= [m_{ij}]_{i,j \in S}$. 
That is, we are finding the inverse of a submatrix of $BSB^T$.
Because there is no nice general relationship between the inverse of a submatrix and the inverse of the entire matrix, 
there is no nice relationship between $((B SB^{T})[\{1,\dots,m\}])^{-1}$ and $(BSB^T)^{-1} = B^{-T}S^{-1}B^{-1}$,
so there is no nice relationship between $(ASA^T)^{-1}$ and $S^{-1}$.
