<s>[INST] Find spectrum for matrix $A$ Let $A = \left[ \begin{array}{*{20}{c}}
0&b&0&0&0&0\\
c&0&b&0&0&0\\
0&c&0&b&0&0\\
0&0&c&0&b&0\\
0&0&0&c&0&b\\
0&0&0&0&c&0
\end{array} \right]_{n \times n}$.
Why does matrix $A$ have spectrum $\sigma (A) = \left\{ 2\sqrt {bc} \cos \left(\frac{\pi k}{n + 1}\right):k = 1,2,\ldots,n \right\}$?
 [/INST] Let $\lambda$ be an eigenvalue of $A$ and $\mathbf{v}$ be the respective eigenvector. If you expand $(A - \lambda I)\mathbf{v} = \mathbf{0}$, you'll notice that the components $v_k$ ($k = 1, \ldots ,n$) of $\mathbf{v}$ satisfy the second order linear difference equation
$$\tag{1} cv_{k-1} - \lambda v_k + v_{k+1}= 0, \; k = 1, \ldots ,n,$$
with $v_0=v_{n+1}=0$.
We can write $(1)$ as
$$\tag{2} v_{k+2} - \frac{\lambda}{b} v_{k+1} + \frac{c}{b} v_{k}= 0, \; k = 0, \ldots ,n-1,$$
again with $v_0=v_{n+1}=0$.
In order to find the general solution of $(2)$, in the same spirit of one of the techniques we can use to find the general solution of a second order ordinary differential equation, we seek solutions of the form $v_k = \xi r^k$, for some constants $\xi$ and $r$. Substituting in $(2)$, we arrive at the characteristic equation
$$\tag{3} r^2 - \frac{\lambda}{b} r + \frac{c}{b}= 0, $$
which, being a quadratic equation, will have two roots, say $r_1$ and $r_2$.
It can be proven that the general solution of $(2)$ is then given by
$$ v_k =
\left\{
\begin{array}{ll}
        \alpha (r_1)^k + \beta (r_2)^k &, \text{ if } r_1 \neq r_2, \\
        \alpha \rho^k + \beta k \rho^k &, \text{ if } r_1 = r_2 = \rho, 
\end{array}
\right.
$$
where $\alpha$ and $\beta$ are arbitrary constants.
(If you want to learn a general method of solving linear difference equations of second or higher order, you can read this answer.)
However, in our particular case, we must have $r_1 \neq r_2$, since otherwise $v_k = \alpha \rho^k + \beta k \rho^k$ and $v_0=v_{n+1}=0$ would imply that $v_k = 0$, for $k = 1, \ldots ,n$, and hence $\mathbf{v} = \mathbf{0}$, so $\mathbf{v}$ wouldn't be an eigenvector.
Thus, we know that $v_k = \alpha (r_1)^k + \beta (r_2)^k$ and $v_0=v_{n+1}=0$, which yields
$$
\left\{
\begin{array}{l}
        \alpha + \beta = 0 \\
        \alpha (r_1)^{n+1} + \beta (r_2)^{n+1} = 0 
\end{array}
\right.,
$$
and this implies
$$\tag{4} \left( \frac{r_1}{r_2}\right)^{n+1} = 1 \Rightarrow \frac{r_1}{r_2} = \exp \left( i \frac{2 \pi j}{n+1} \right) \Rightarrow r_1 = r_2 \exp \left( i \frac{2 \pi j}{n+1} \right) ,$$
for some $1 \leq j \leq n$.
Now, applying the Viète's formulas to $(3)$, we also have
$$\tag{5}
\left\{
\begin{array}{l}
        r_1 r_2 = c/b\\
        r_1 + r_2 = \lambda / b
\end{array}
\right..
$$
Plugging $(4)$ in the first equation of $(5)$ we get
$$\tag{6} (r_2)^2 \exp \left( 2 \frac{i \pi j}{n+1} \right) = c/b \Rightarrow r_2 = \sqrt{c/b} \exp \left( - i \frac{\pi j}{n+1} \right),$$
and
$$\tag{7} r_1 = \sqrt{c/b} \exp \left( i \frac{\pi j}{n+1} \right).$$
Finally, by plugging $(6)$ and $(7)$ in the second equation of $(5)$, we arrive at
$$\lambda = b (r_1 + r_2) = b \sqrt{c/b}  \underbrace{\left\{ \exp \left( i \frac{\pi j}{n+1} \right) + \exp \left( - i \frac{\pi j}{n+1} \right) \right\}}_{= 2 \cos \left( \frac{\pi j}{n+1} \right)} = 2 \sqrt{bc} \cos \left( \frac{\pi j}{n+1} \right).$$
Therefore, the eigenvalues of $A$ are given by
$$\lambda_j = 2 \sqrt{bc} \cos \left( \frac{\pi j}{n+1} \right), \; j=1,\ldots,n.$$
</s>