Inverse of tridiagonal Toeplitz matrix

Question

Consider the following tridiagonal Toeplitz matrix. Let $n$ be even. $${A_{n \times n}} = \left[ {\begin{array}{*{20}{c}} {0}&{1}&{}&{}&{}\\ {1}&{0}&{1}&{}&{}\\ {}&{1}&{\ddots}&{\ddots}&{}\\ {}&{}&{\ddots}&{\ddots}&{1}\\ {}&{}&{}&{1}&{0} \end{array}} \right]$$ What is the inverse $A^{-1}$?

Clearly, $A^{-1}$ is symmetric.

I look for a proof of the following conjecture that $A^{-1}$ is given as follows:

If $A_{i, j}^{-1}$ such that $j$ is odd and $i =1+j + 2 m$ with $m\in \cal N_0$, then $A_{i, j}^{-1} = (-1)^m$. From which follows by symmetry:

If $A_{i, j}^{-1}$ such that $j$ is even and $i =-1+j - 2 m$ with $m\in \cal N_0$, then $A_{i, j}^{-1} = (-1)^m$.

All other $A_{i, j}^{-1} = 0$.

Here is an example, computed with Matlab, for $n=10$ which shows the structure:

$${A_{10 \times 10}^{-1}} = \left[ {\begin{array}{*{20}{r}} 0 & 1 & 0 & -1 & 0 & 1 & 0 & -1 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & -1 & 0 & 1 & 0 & -1 \\ -1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & -1 & 0 & 1 \\ 1 & 0 & -1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & -1 \\ -1 & 0 & 1 & 0 & -1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & -1 & 0 & 1 & 0 & -1 & 0 & 1 & 0 \end{array}} \right]$$

Answer 1

The most direct way to establish the conjecture is to perform the multiplication $E = A^{-1} A$ with the conjectured $A^{-1}$ and show that the result is the unit matrix. Since $A$ is bi-diagonal, each entry of $E$ is the sum of at most two terms, so there is little confusion.

$$E_{ik} = \sum_{j=1}^n A^{-1}_{ij} A_{j k} = A^{-1}_{ij} A_{j, j-1} \delta_{j-1,k} + A^{-1}_{ij} A_{j, j+1} \delta_{j+1,k}\\ = A^{-1}_{i, k+1} A_{k+1, k} + A^{-1}_{i, k-1} A_{k-1, k} \\ = A^{-1}_{i, k+1} + A^{-1}_{i, k-1} $$

Now the diagonal is $$E_{ii} = A^{-1}_{i, i+1} + A^{-1}_{i, i-1} $$ From the structure of $A^{-1}$, if $i$ is odd, then $A^{-1}_{i, i+1} = 1 $ and $ A^{-1}_{i, i-1} = 0 $. If $i$ is even, then $A^{-1}_{i, i+1} = 0 $ and $ A^{-1}_{i, i-1} = 1 $. So in all cases, $E_{ii} = 1$.

For $k \ne i $, i.e the off-diagonal terms, we have $E_{ik} = A^{-1}_{i, k+1} + A^{-1}_{i, k-1} $. From the structure of $A^{-1}$, either both of these summands are zero, or one is $+1$ and the other $-1$, as we have values of $(-1)^m$ for an index difference of $2m$. Hence, always $E_{ik} =0$.

This completes the proof. $\qquad \Box$

Answer 2

I found an algorithm that I hope it help you. Let the inverse be$$A^{-1}=\begin{bmatrix}c_1&c_2&\cdots&c_n\end{bmatrix}$$In other words, $c_i$s are the columns of $A^{-1}$. We build the $c_i$s recursively as following: $$c_{2}=e_1\\c_{2k+2}=-c_{2k}+e_{2k+1}\qquad,\qquad 1\le k\le {n\over 2}-1$$also $$c_{n-1}=e_n\\c_{2k-1}=e_{2k}-c_{2k+1}\qquad,\qquad 1\le k\le {n\over 2}-1$$where $e_i$s are the columns of $I_n$

Answer 3

Firstly Matrix is Toeplitz. This means it represents multiplication by power series expansion. This means matrix inversions corresponds to multiplicative inversion

Therefore, consider $$x+x^{-1}=\frac{x^2+1}x$$ Now it's multiplicative inverse:

$$(x+x^{-1})^{-1}=\frac x{x^2+1}$$

Now you can expand with geometric series / Taylor expansion for $$\frac 1{x^2+1}=\frac 1{1-(-1\cdot x^2)}$$ And substitute with $$\frac{1}{1-t}=1+t+t^2+\cdots, t=-x^2$$ and then finish. You will notice the flipping sign pattern and that the odd exponents disappear when you substitute and do the expansion.