# Invert a symmetric banded matrix

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.

• 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 – Graz Feb 7 at 11:28
• 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. – user376343 Feb 7 at 12:05

## 1 Answer

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?

• 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 – Graz Feb 7 at 12:25
• The Woodbury matrix identity should be helpful for the rest. – Bertrand Feb 7 at 12:56