Let $\mathbf{A}$ be a real-entries full rank $p \times q $ matrix with $p \neq q $ and $\mathbf{B}$ be a non singular $ p \times p $ real symmetric matrix and consider the product:
$$ \mathbf{K}= \mathbf{A}^t \mathbf{B} \mathbf{A} $$
Since everything has full rank, $\mathbf{K}$ is invertible. I'd like to characterize the inverse of $\mathbf{K}$, i.e, if possible, write $\mathbf{K}^{-1}$ as a function of $\mathbf{A},\mathbf{A}^t,\mathbf{B},\mathbf{B}^{-1}$. To this end, I tried to use the left and right inverses of $\mathbf{A}$, for example my first idea was to consider:
$$ \mathbf{J}=\mathbf{A}^{-1}_{r} \mathbf{B}^{-1} \mathbf{A}^{-t}_{r} \implies \mathbf{K} \mathbf{J} = \mathbf{I} $$
where the subscript "$r$" is to indicate the right inverse of the matrix. The problem is that if $\mathbf{A}$ has more rows than columns, then $\mathbf{A}^t$ has more columns than rows so that $\mathbf{A}^{-t}_{r}$ and $\mathbf{A}^{-1}_{r}$ never simultaneously exist. Furthermore, since $\mathbf{B}$ is symmetric, I tried with diagonalization, this yields:
$$ \mathbf{K}=(\mathbf{Q}^t \mathbf{A})^t \mathbf{D} (\mathbf{Q}^t \mathbf{A})=\mathbf{Y}^t \mathbf{D} \mathbf{Y}$$
where $\mathbf{Y}$ has full rank and $\mathbf{D}$ is diagonal. Maybe it's easier to work from here, but I couldn't see how.
My final thought was the following: suppose we can write $\mathbf{B}=\mathbf{H} \mathbf{H}^t $, where $\mathbf{H}$ is a $p \times q $ matrix, then we would have:
$$ \mathbf{K}= (\mathbf{A}^t \mathbf{H}) ( \mathbf{H}^t \mathbf{A}) \implies \mathbf{K}^{-1}=(\mathbf{A}^t \mathbf{H )}^{-t} (\mathbf{A}^t \mathbf{H )}^{-1}$$
Is such decomposition even possible? If so, I'm not even sure this is easier to handle compared to the previous ideas. Every comment or suggestion is appreciated, thanks in advance.