Is it possible to compute $C = AB^{-1}A^T$ without computing the inverse of $B$ explicitly? A is $n \times m$ matrix. B is $m \times m$ matrix ($m \gg n$). Thus $C$ is much smaller than $B$.
In the case where A is also $m \times m$ matrix and invertible, we can compute $C^{-1}$ with $A^{-T}BA^{-1}$ and then invert $C^{-1}$. However, when A is $n \times m$ ($m \gg n$), even though similar relationship still exists ( answered in Inverse of product of matrices), I did not find a way to compute it without computing inverse of $B$ explicitly.
The motivation for avoiding computation of $B^{-1}$ is that in my problem $B$ is a large sparse matrix, while its inverse is generally not sparse.
Finding the inverse of $B$ is equivalent to finding, for each column of the identity matrix, the vector $v$ such that $Bv$ is equal to that column. To find $B^{-1}A^T$, you just need to find, for each column of $A^T$, a $v$ such that $Bv$ is equal to that column. Depending on your algorithms and your $B$, however, this might not be any faster.