$4 \times 4$ wave matrices

A similar result applies to the $4 \times 4$ wave matrix of uniform second neighbor interactions,

$$\begin{eqnarray*} \left[ \begin{array}{l} x_{i+2}\\ x_{i+1}\\ x_i\\ x_{i-1}\end... ...array}{l}x_{i+1}\\ x_i\\ x_{i-1}\\ x_{i-2} \end{array} \right]. \end{eqnarray*}$$

The sequence of indices is reversed by the self-inverse

$$\begin{eqnarray*} R & = & \left[ \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array} \right], \end{eqnarray*}$$

so

$$\begin{eqnarray*} \left[ \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \... ... 1 & 0 \\ 0 & 0 & 0 & 1 \\ d & c & b & a \end{array} \right], \end{eqnarray*}$$

while

$$\begin{eqnarray*} \left[ \begin{array}{cccc} a & b & c & d \\ 1 & 0 & 0 & 0 \... ...& 0 \\ 0 & 0 & 0 & 1 \\ 1 & -a & -b & -c \end{array} \right]. \end{eqnarray*}$$

This time it is $d$ which should be $-1$, although $d = 1$ is also a possibility. Then it is also required that $a = c$, so the recursion relation

$$\begin{eqnarray*} x_{i+2} + a x_{i+1} + b x_i + c x_{i-1} + d x_{i-2} = \lambda x_i \end{eqnarray*}$$

has to be symmetric with respect to reversing the order of the indices.

The characteristic equation of a $4 \times 4$ wave matrix reads

$$\begin{eqnarray*} \left\vert \begin{array}{cccc} a - \mu & b & c & d \\ 1 & -... ... -\mu & 0 \\ 0 & 0 & 1 & -\mu \end{array} \right\vert & = & 0 \end{eqnarray*}$$

Laplace expansion by the first row gives

$$\begin{eqnarray*} (a - \mu) (-\mu)^3 - b (-\mu)^2 + c (-\mu) - d & = & 0 \end{eqnarray*}$$

Thus $d$ is the determinant, the product of all the roots, and should be $+1$. Putting the other symmetry condition, $a = c$,

$$\begin{eqnarray*} (-\mu)^4 + a (-\mu)^3 - b(-\mu)^2 + a (-\mu) - d & = & 0 \\ ... ... + \frac{1}{\mu})^2 - a (\mu + \frac{1}{\mu}) - (b + 2) & = & 0 \end{eqnarray*}$$

As before, the roots occur in reciprocal pairs, as they should when a matrix is equivalent to its inverse.