# Finding the trace of $(I + \Sigma^{-1} AA^T)^{-1}$

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 :)

• $\Sigma$ is diagonal? Any more information? – Robin Goodfellow Sep 17 '14 at 15:42
• I forgot: all its elements are greater than zero. – bayer Sep 17 '14 at 16:19

## 1 Answer

One idea: note that if the entries of $\Sigma$ are sufficiently large, then setting $M= \Sigma^{-1}AA^T$we can write $$(I + M)^{-1} = I - M + M^2 - M^3 + \cdots$$