Let $V = {}^{ t } \left( v _ { 1 } , \ldots , v _ { n } \right)$ be an eigenvector of $A _ { n }$ associated with a complex eigenvalue $\lambda$, where $A_n$ is the square matrix of size $n$:
$$A _ { n } = \left( \begin{array} { c c c c c c } 2 & - 1 & 0 & \ldots & \ldots & 0 \\ - 1 & 2 & - 1 & \ddots & & \vdots \\ 0 & - 1 & 2 & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & \ddots & 0 \\ \vdots & & \ddots & \ddots & 2 & - 1 \\ 0 & \ldots & \ldots & 0 & - 1 & 2 \end{array} \right)$$
Show that $\lambda$ is necessarily real and that the components $v _ { i }$ of $V$ satisfy the relation:
$$v _ { i + 1 } - ( 2 - \lambda ) v _ { i } + v _ { i - 1 } = 0, \quad 1 \leq i \leq n$$
where we set $v _ { 0 } = v _ { n + 1 } = 0$.
Let $V = {}^{ t } \left( v _ { 1 } , \ldots , v _ { n } \right)$ be an eigenvector of $A _ { n }$ associated with a complex eigenvalue $\lambda$, where $A_n$ is the square matrix of size $n$:

$$A _ { n } = \left( \begin{array} { c c c c c c } 
2 & - 1 & 0 & \ldots & \ldots & 0 \\
- 1 & 2 & - 1 & \ddots & & \vdots \\
0 & - 1 & 2 & \ddots & \ddots & \vdots \\
\vdots & \ddots & \ddots & \ddots & \ddots & 0 \\
\vdots & & \ddots & \ddots & 2 & - 1 \\
0 & \ldots & \ldots & 0 & - 1 & 2
\end{array} \right)$$

Show that $\lambda$ is necessarily real and that the components $v _ { i }$ of $V$ satisfy the relation:

$$v _ { i + 1 } - ( 2 - \lambda ) v _ { i } + v _ { i - 1 } = 0, \quad 1 \leq i \leq n$$

where we set $v _ { 0 } = v _ { n + 1 } = 0$.