A tridiagonal matrix is a Toeplitz matrix of the form $T(0,\ldots,0,t_{-1},t_0,t_1,0,\ldots,0)$, i.e. a matrix of the form $$A_n(a,b,c) = \left(\begin{array}{cccc} a & b & & (0) \\ c & a & \ddots & \\ & \ddots & \ddots & b \\ (0) & & c & a \end{array}\right)$$ where $(a,b,c)$ are complex numbers. We fix $(a,b,c)$ three complex numbers such that $bc \neq 0$. Let $\lambda \in \mathbb{C}$ be an eigenvalue of $A_n(a,b,c)$ and $X = \left(\begin{array}{c} x_1 \\ \vdots \\ x_n \end{array}\right) \in \mathbb{C}^n$ be an associated eigenvector.
Show that if we set $x_0 = 0$ and $x_{n+1} = 0$, then $(x_1, \ldots, x_n)$ are the terms of rank varying from 1 to $n$ of a sequence $(x_k)_{k \in \mathbb{N}}$ satisfying $x_0 = 0, x_{n+1} = 0$ and $$\forall k \in \mathbb{N}, \quad bx_{k+2} + (a-\lambda)x_{k+1} + cx_k = 0$$

Recall the expression of the general term of the sequence $(x_k)_{k \in \mathbb{N}}$ as a function of the solutions of the equation $$bx^2 + (a-\lambda)x + c = 0 \tag{I.1}$$

Using the conditions imposed on $x_0$ and $x_{n+1}$, show that (I.1) admits two distinct solutions $r_1$ and $r_2$.