I need to efficiently compute the trace of $$ B = (I + \Sigma^{-1} AA^T)^{-1} $$ where $\Sigma$ is diagonal and all its elements strictly greater than zero. $A$ is $-1$ on the diagonal and $1$ right below it, thus $AA^T$ will be tridiagonal. Some example for $A$: $$ \left[\begin{matrix}-1 & 0 & 0 & 0 & 0\\1 & -1 & 0 & 0 & 0\\0 & 1 & -1 & 0 & 0\\0 & 0 & 1 & -1 & 0\\0 & 0 & 0 & 1 & -1\end{matrix}\right] $$
I have been scratching my head for a while now. This question made me believe that trying to diagonalize $B$. Some attempts to do so algebraically failed.
Anyone can point me in the right direction?
If it can be shown that this is not possible without explicitly writing down $B$, this would also help me. (It would stop me try to find a solution :)
One idea: note that if the entries of $\Sigma$ are sufficiently large, then setting $M= \Sigma^{-1}AA^T$we can write $$ (I + M)^{-1} = I - M + M^2 - M^3 + \cdots $$