let say we have $A$ to be symmetric positive-definite (SPD), moreover block tridiagonal Toeplitz matrix and $D$ is block diagonal SPD (both with full rank). Let say we know everything about $A$ and $D$, e.g. theirs (eigenvalues, singular, QR, LDLT) decompositions, inverse, etc.. Is there any more efficient way to solve the given linear system then to decompose the sum $(A+D)$?
I was looking around and it seems that if the matrices have full rank, there is nothing to do with that. On the other hand, this case seems to be so special for me that I would be surprised if any particular method doesn't exist.
In some cases it could be useful to compute approximate solution by the following way:
\begin{align} (A+D)x &= b\\ x &= (D(D^{-1}A+I))^{-1}b=\\ &= (I+X)^{-1}D^{-1}b,\\ \end{align}
where $X = D^{-1}A$ and since we know that
$\|X\|<1:\quad(I+X)^{-1} = I-X+X^2-X^3\ldots$
we can write our solution as
$\|D^{-1}A\|<1:\quad x = (D^{-1}-D^{-1}AD^{-1}+D^{-1}AD^{-1}AD^{-1}-D^{-1}AD^{-1}AD^{-1}AD^{-1}\ldots)b$ or similarly $\|A^{-1}D\|<1:\quad x = (A^{-1}-A^{-1}DA^{-1}+A^{-1}DA^{-1}DA^{-1}-A^{-1}DA^{-1}DA^{-1}DA^{-1}\ldots)b$
and then approximate it by only finite many terms. Noticed that the smaller eigenvalues $\|D^{-1}A\|$ has in the first case (or $\|A^{-1}D\|$ in the second one) the faster series converge to zero and thus less terms may be used to reach the same accuracy.