1
$\begingroup$

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 :)

$\endgroup$
  • $\begingroup$ $\Sigma$ is diagonal? Any more information? $\endgroup$ – Robin Goodfellow Sep 17 '14 at 15:42
  • $\begingroup$ I forgot: all its elements are greater than zero. $\endgroup$ – bayer Sep 17 '14 at 16:19
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.