<s>[INST] How to find the eigenvalues of tridiagonal Toeplitz matrix? Assume the tridiagonal matrix $T$ is in this form:
$$
T = \begin{bmatrix}
a       & c        &              &              &   &\\
b       & a        & c            &              &   &\\
        & b        & a            & c            &   &\\
        &          &              &\ddots        &   &\\
        &          &              & b            & a & c\\
        &          &              &              & b & a\\
\end{bmatrix}
$$
we must show that its eigenvalues are of the form 
$$a + 2 \sqrt{bc} \, \cos \left( \frac{k \pi}{n+1} \right)$$
where $$a=qh^2−1, ~~ b=1- \frac{ph}{2}, ~~ c =1+\frac{ph}{2} , ~~q \leq 0.$$
 [/INST] Once you get a recursion formula there are at least three different ways to
solve the problem. Using the Z transform, using Chebyshev polynomials, and using the traditional (which I include here) method to solve difference equations.
This document illustrate this. From it, I will
rewrite here one of the methods.
The eigenvalues of $U$ are roots of the characteristic polynomial That is
\begin{eqnarray*}
  p(\lambda) = \det( U - \lambda I) = 0.
\end{eqnarray*}
That is,
\begin{eqnarray*}
  p_n(\lambda) = \det \begin{pmatrix}
    a-\lambda & c &  \\
    b & a-\lambda   & c \\
    &  b & a-\lambda  & c \\
    & &    \ddots &\ddots &\ddots  \\
    & &    & \ddots &\ddots &\ddots  \\
    & &    & &  b & a-\lambda  & c \\
    & &    & &  &  b & a-\lambda
  \end{pmatrix} = 0
\end{eqnarray*}
where $n$ is the number of rows (or columns) of $U$. At the moment
we assume $n \ge 2$. We expand the determinant through the first row to find
\begin{eqnarray*}
  p(\lambda) = (a- \lambda) p_{n-1} - c \det B
\end{eqnarray*}
where $B$ is the matrix
\begin{eqnarray*}
  \begin{pmatrix}
    b & c &  \\
    & a-\lambda   & c \\
    &  b & a-\lambda  & c \\
    & &    \ddots &\ddots &\ddots  \\
    & &    & \ddots &\ddots &\ddots  \\
    & &    & &  b & a-\lambda  & c \\
    & &    & &  &  b & a-\lambda
  \end{pmatrix} 
\end{eqnarray*}
Then we expand the determinant of $B$ through the first column to find
\begin{eqnarray*}
  p_{n}(\lambda) = (a- \lambda) p_{n-1} - bc p_{n-2} ,
\end{eqnarray*}
and
\begin{eqnarray}
  p_{n}(\lambda) - (a- \lambda) p_{n-1} - bc p_{n-2}  = 0.
  (1)
\end{eqnarray}
This is a  difference equation.
We want to find an analytic solution for the function $p_n(\lambda)$ which is
valid for all $n$. This equation needs two initial conditions due to the fact
that $p_n$ depends on two previous instances $p_{n-1}$ and $p_{n-2}$.
That is, for $n=1$, $p_1(\lambda)=a-\lambda$ and  for $n=2$
\begin{eqnarray}
  p_2(\lambda)= \det
  \begin{pmatrix}
    a - \lambda & c \\
    b & a-\lambda
  \end{pmatrix}
  = (a-\lambda)^2 - bc,
 (2)
\end{eqnarray}
This could be the initial conditions. All other values of $p_n(\lambda)$
could be found starting with these two initial conditions on the
recursive equation (1). However, and to simplify operations
we can state simpler initial conditions. Although there are no matrices
with $n=0$ rows we could define $p_0(\lambda)=1$, and keep the initial
condition $p_1(\lambda)=a - \lambda$. Using the recursion (1)
we find that
\begin{eqnarray*}
  p_2(\lambda) = (a - \lambda)^2- bc 
\end{eqnarray*}
which is exactly equation (2). The solution of $p_n(\lambda)$ for higher $n$
values will not change.
We then  have the initial conditions
\begin{eqnarray}
  p_0(\lambda)=1 \quad , \quad p_1(\lambda)=a -\lambda.
  (3)
\end{eqnarray}
We apply the theory about difference equations to obtain the solution of our problem.
Returning to equation (1)  and replacing $p_i(\lambda)$ by $t^i$
we find the characteristic equation
\begin{eqnarray*}
  t^n - (a- \lambda) t^{n-1} - bc t^{n-2}  = t^{n-2}[t^2 - (a - \lambda) t - bc] = 0.
\end{eqnarray*}
That is, we need to solve the quadratic equation
\begin{eqnarray}
  t^2 - (a - \lambda) t - bc = 0.
  (4)
\end{eqnarray}
The two solutions are
\begin{eqnarray}
  t_{\pm} = \frac{(a - \lambda) \pm \sqrt{(a-\lambda)^2 - 4 bc}}{2}.
  (5)
\end{eqnarray}
The roots could be equal or different. We consider two cases;
1.
No repeated roots:
The general solution  $p_n(\lambda)$ can be written as
\begin{eqnarray}
  p_n(\lambda)= c_1 t^n_+ + c_2 t^n_-.
  \label{pnl}
\end{eqnarray}
From the initial conditions (3) we have
\begin{eqnarray}
  1 &=& c_1 + c_2  \nonumber \\
  \quad \quad \quad (5a) \\
  a - \lambda  &=& c_1 t_+ + c_2 t_-  . \nonumber
\end{eqnarray}
This system has solution if its determinant does not vanish. That is, if
\begin{eqnarray*}
  \Delta = \det
  \begin{pmatrix}
    1 & 1 \\
    t_+ & t_- \\
  \end{pmatrix}
  =  t_- - t_+ = -\sqrt{(a-\lambda)^2 - 4 bc} \ne 0.
\end{eqnarray*}
Please observe that this is consistent with the fact that  $t_+ \ne t_-$.
We have then that  $(a-\lambda)^2 - 4 bc \ne 0$.   To solve system (5a)
we observe that $c_2=1-c_1$  and
\begin{eqnarray*}
  a-\lambda = c_1 t_+ + (1-c_1) t_- = c_1(t_+ - t_-) + t_-,
\end{eqnarray*}
That is,
\begin{eqnarray}
  c_1 = \frac{(a-\lambda) - t_-}{t_+ - t_-}
  \quad , \quad
  c_2 = 1-c_1 = \frac{t_+ - (a - \lambda)}{t_+-t_-},
  \label{ec1}
\end{eqnarray}
so that the solution for $p_n(\lambda$) is given by
\begin{eqnarray}
  p_n(\lambda) &=& t^n_+  \frac{(a-\lambda) - t_-}{t_+ - t_-}  + 
  t^n_-  \frac{t_+ - (a - \lambda)}{t_+-t_-}.
\end{eqnarray}
Now, since the sum of the roots of equation (5)
is given by  $t_+ + t_-=(a-\lambda)$ we write
\begin{eqnarray*}
  p_n(\lambda) = t_+^n \frac{t_+ + t_- - t_-}{t_+ - t_-}
  + t_-^n \frac{t_+ - t_+ - t_-}{t_+ - t_-}
  &=& \frac{t_+^{n+1} - t_-^{n+1}}{t_+ - t_-}.
\end{eqnarray*}
This is the analytic solution of (1) with initial
conditions (3) .  Since we need to know $\lambda$ such
that $p_n(\lambda=0)$ we have that the eigenvalues $\lambda$ satisfy
\begin{eqnarray}
  t_+^{n+1} = t_-^{n+1}.
  (6)
\end{eqnarray}
Note that equation $t_+^{n+1} - t_-^{n+1}=0$ is of degree $n+1$ in $\lambda$ since
$t_+^{n+1} - t_-^{n+1} = (t_+ - t_-) p_n(\lambda)$.  We introduced a new root
to the original problem. This root satisfy the equation  $t_+ - t_-=0$.
We are assuming  $t_+ - t_- \ne 0$, $\lambda$ can not be a root
of $\Delta = 0$. Please observe that $\Delta=0$ implies
$(a-\lambda)^2 - 4 bc =0$. That is, the roots


\begin{eqnarray}
  \lambda = a \pm 2 \sqrt{bc}
  \label{tworaices}
\end{eqnarray}
cannot be included in the set of solutions to the problem. Now,
\begin{eqnarray}
  \frac{t_+}{t_-}  
  = \frac{t_+^2}{t_+ t_-}  
  = \frac{t_+^2}{bc} 
  = \left ( \frac{t_+}{\sqrt{bc}} \right)^2
  (7)
\end{eqnarray}
where we used the product of the two roots of the quadratic equation (4)
$t_+ t_- =bc$.  We find, using (6)
\begin{eqnarray*}
  \left ( \frac{t_+}{\sqrt{bc}} \right)^{2n+2} = 1.
\end{eqnarray*}
The solution of this equation corresponds with $2n+2$ roots located uniformly
along the unit circle. These roots are represented by

\begin{eqnarray*}
  \frac{(t_+)_k}{\sqrt{bc}} = \mathrm{e}^{\frac{ k \pi \mathrm{i}}{2n+2}}
  \quad , \quad k=0, 1, \cdots , 2n-1.
\end{eqnarray*}
so that $(t_+)_k=\sqrt{bc} \, \mathrm{e}^{k \pi \mathrm{i}/(2n+2)}$.
From $t_+ t_- = bc$ we find $(t_-)_k=\sqrt{bc} \, \mathrm{e}^{-k \pi
  \mathrm{i}/(2n+2)}$ and from $\mathrm{e}^{\theta} + \mathrm{e}^{-\theta}=2 \cos
  \theta$,


  \begin{eqnarray*}
    (t_+)_k + (t_-)_k = 2 \sqrt{bc} \cos \frac{k \pi}{2n +2} .
  \end{eqnarray*}
  Now, since  $(t_+)_k + (t_-)_k=a-\lambda$ we have that


  \begin{eqnarray*}
    a - \lambda_k = 2 \sqrt{bc} \cos \frac{k \pi }{2n +2}
  \end{eqnarray*}
  and


  \begin{eqnarray*}
    \lambda_k = a - 2 \sqrt{bc} \cos \frac{k \pi }{2n+2}
    \quad , \quad k=0,1, \cdots ,  2n+1.
  \end{eqnarray*}
  However $p_n(\lambda)=0$ only has $n$ roots and we found $2n+2$.
  We already eliminated two roots, corresponding
  to $k=0$ and $k=n+1$. We need to exclude $n$ other roots.
  If we observe equation (7) we see that the function
  $t_+$ was squared when substituting $t_-$ for $\sqrt{cd}/t_+$. 
  This introduced $n$ new roots, corresponding to $k=1,3,5, \cdots, 2n+1$, 
  which are solutions of the equation $(t_+/\sqrt{bc})^{2n+2}$  instead of the
  corresponding equation $(t_+/\sqrt{bc})^{n+1}$ where $t_+$ is not squared.
  With this, the solutions $\lambda_k$ are given by the set

  \begin{eqnarray*}
    \lambda_k = a - 2 \sqrt{bc} \cos \frac{k \pi }{2n+2}
    \quad , \quad k=2,4, \cdots ,  2n,
  \end{eqnarray*}
  or

  \begin{eqnarray*}
    \lambda_k = a - 2 \sqrt{bc} \cos \frac{k \pi }{n+1}
    \quad , \quad k=1,2, \cdots ,  n.
  \end{eqnarray*}



*

Repeated roots:
We have $t_+=t_-=(a-\lambda)/2$. That  is, the quadratic equation is a
perfect square. The recursive equation (1) has a solution of the form
  \begin{eqnarray*}
    p_n(\lambda) = c_1 t_+^n + c_2 n t_+^n.
  \end{eqnarray*}
  Again, to find $c_1$ y $c_2$ we need to apply the initial conditions to
  find a two-by-two linear system of equations.
  

  \begin{eqnarray*}
    p_0(\lambda) &=& 1 = c_1 \\
    p_1(\lambda) &=&  a-\lambda = (c_1 + c_2 )t_+ 
  \end{eqnarray*}
   and so


  \begin{eqnarray*}
    c_2 = \frac{a-\lambda}{t_+} -1 = 1.
  \end{eqnarray*}
  Then


  \begin{eqnarray*}
    p_n(\lambda) =  t_+^n +  n t_+^n
    = \left ( \frac{a-\lambda}{2}  \right )^n(1 + n).
  \end{eqnarray*}
  From here we find that all eigenvalues are equal.

  Recall that $\Delta=0$ means $\lambda = a \pm 2 \sqrt{bc}$,
  and given that  $\lambda=a$ we have that $b=0$ or $c=0$. That is, the original matrix 
  is lower triangular($b=0$) or upper triangular($c=0$) where all its eigenvalues are sitting
  along the diagonal.

</s>