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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 :)

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  • $\begingroup$ $\Sigma$ is diagonal? Any more information? $\endgroup$
    – Robin Goodfellow
    Commented Sep 17, 2014 at 15:42
  • $\begingroup$ I forgot: all its elements are greater than zero. $\endgroup$
    – bayer
    Commented Sep 17, 2014 at 16:19

1 Answer 1

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

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