<s>[INST] Inverse of tridiagonal Toeplitz matrix 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: 
$${A_{10 \times 10}^{-1}} = \left[ {\begin{array}{*{20}{r}}
0 &     1 &     0 &    -1 &      0 &     1 &     0 &    -1 &      0 &     1 \\
     1 &     0 &     0 &     0 &     0 &     0 &     0 &     0 &     0 &     0 \\
     0 &     0 &     0 &     1 &     0 &    -1 &     0 &     1 &     0 &    -1  \\
    -1 &      0 &     1 &     0 &     0 &     0 &     0 &     0 &     0 &     0 \\
     0 &     0 &     0 &     0 &     0 &     1 &     0 &    -1 &      0 &     1 \\
     1 &     0 &    -1 &     0 &     1 &     0 &     0 &     0 &     0 &     0 \\
     0 &     0 &     0 &     0 &     0 &     0 &     0 &     1 &     0 &    -1  \\
    -1 &     0 &     1 &     0 &    -1 &     0 &     1 &     0 &     0 &     0 \\
     0 &     0 &     0 &     0 &     0 &     0 &     0 &     0 &     0 &     1 \\
     1 &     0 &    -1 &     0 &     1 &     0 &    -1 &     0 &     1 &     0 
\end{array}} \right]$$
 [/INST] 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. $\qquad \Box$
</s>