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I am interested in inverting a symmetric banded matrix with the following structure:

\begin{equation}\mathbf A(\epsilon)= \left(\begin{array}{*{20}c} 2+\epsilon &0 & -1&0 &0&0& -1&0\\ 0 &2+\epsilon & 0 &-1 &0&0&0&-1\\ -1 &0 & 2+\epsilon&0 & -1&0&0&0\\ 0 &-1 & 0&2+\epsilon &0& -1&0&0\\ 0 &0 & -1&0 &2+\epsilon&0& -1&0\\ 0 &0 & 0& -1 &0&2+\epsilon&0& -1\\ -1 &0 & 0&0 & -1&0&2+\epsilon&0\\ 0 &-1 & 0&0 &0& -1&0&2+\epsilon\\ \end{array}\right) \end{equation} for an arbitrary separation between the bands. We consider here $\epsilon>0$. The matrix would become singular for $\epsilon\to 0$ (as the sum of the elements of each row becomes zero). Is it possible to determine analytically $\mathbf A^{-1}(\epsilon)$? If so, how?

Here is what I have been trying so far. I decomposed $\mathbf A$ in two (upper $\mathbf A_U$ and lower $\mathbf A_L$) triangular matrices which are both invertible if one splits the main diagonal in two symmetric contribitions with positive entries. I then tried to use some results to deal with the inverse of $(\mathbf A_U+\mathbf A_L)^{-1}$. Moreover, I tried to reduce the inverse of this sum in the form $(\mathbf B+\mathbb I)^{-1}$ for a matrix $||\mathbf B||\ll1$ in order to expand the inverse of the last sum in series and at least obtaining a perurbative expression of the inverse.

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  • $\begingroup$ I tried to decompose $\mathbf A$ in two (upper $\mathbf A_U$ and lower $\mathbf A_L$) triangular matrices which are both invertible if one splits the main diagonal in two symmetric contribitions with positive entries. I then tried to use some results to deal with the inverse of $(\mathbf A_U+\mathbf A_U)^-1$. Moreover, I tried to reduce the inverse of the sum in the form $(\mathbf B+\mathbb I)^{-1}$ for a matrix $||\mathbf B||\ll1$ in order to expand the inverse of the last sum in series and at least obtaining a perurbative expression of the inverse $\endgroup$ – Graz Feb 7 at 11:28
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    $\begingroup$ Your matrices are circulant, not singular. This work ee.stanford.edu/~gray/toeplitz.pdf can be useful, in particular Theorem 3.1(3) ... it is related to the answer by Florian. $\endgroup$ – user376343 Feb 7 at 12:05
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Wolfram Alpha has one for you.

To find out how to get there, this may be helpful:

  • It seems the eigenvalues are $\epsilon$, $\epsilon+2$, and $\epsilon+4$.
  • The corresponding eigenvectors are structured nicely: $(0,1,0,1,0,1,0,1)$ and $(1,0,1,0,1,0,1,0)$ belong to eigenvalue $\epsilon$, $(-1,0,0,0,1,0,0,0)$ and its cyclically shifted copies give eigenvalue $\epsilon+2$ and $(0,-1,0,1,0,-1,0,1)$ as well as $(1,0,-1,0,1,0,-1,0)$ give eigenvalue $\epsilon+4$.

Based on this you can find a full diagonalization into eigenvectors and eigenvalues and once you have that, it should be easy to invert it.

Can you take it from here?

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    $\begingroup$ I can continue from here thanks. It's a good point to diagonalize it, especially given that one can get an analytic expression for the eigenvalues of circulant matrices as mentioned in the work suggested by user376343 $\endgroup$ – Graz Feb 7 at 12:25
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    $\begingroup$ The Woodbury matrix identity should be helpful for the rest. $\endgroup$ – Bertrand Feb 7 at 12:56

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