Discrete dynamical system and bound on norm Let $z \in \mathbb R\backslash \left\{2 \right\}$ then I would like to understand the following:
Consider the dynamical system with $x_i \in \mathbb C^2:$
$$ x_{i} = \left(\begin{matrix} z &&-1 \\ 1 && 0 \end{matrix} \right)^ix_0.$$
I would like to understand whether one can obtain a sharp bound on $\Vert x_i \Vert$ for $N \ge i \ge 0$ just in terms of 
$\Vert x_0 \Vert$, $\Vert x_N \Vert$ and $z \neq 2.$
Observations: It seems that this dynamical system is rather simple in the sense that the matrix $A=\left(\begin{matrix} z &&-1 \\ 1 && 0 \end{matrix} \right)$ is diagonalizable for $z \neq 2.$
More precisely, the eigenvalues are $$\tfrac{1}{2} \left(z \pm \sqrt{z^2-4 }\right)$$ with eigenvectors $$\left(\tfrac{1}{2} \left(z \pm \sqrt{z^2-4 }\right), 1\right).$$
It is also worth noticing that this system is invertible since $\operatorname{det}(A)=1$
so it is believable that once the boudary norms for $x_0$ and $x_N$ are known. Everything else should be fixed as well. 
Of course there are trivial bounds like $\Vert x_i \Vert \le \Vert A \Vert^i \Vert x_0 \Vert$ but I am looking for something more refined.
Motivation: This is the discrete analogue of $-y''(x)=zy(x),$ see for details and in the continuous world it is almost trivial to bound $\vert y(x_{\text{middle}}) \vert$ in terms of $\vert y(x_0) \vert$ and $\vert y(x_1)\vert$ where $x_0 \le x_{\text{middle}} \le x_1.$ 
 A: This is an extension and correction of my comment, made more explicit.

Denote the normalised eigenvectors of $A$ by $u$ and $v$, and the corresponding eigenvalues by $\lambda$ and $\mu$. Since all norms on $\mathbb{R}^2$ are equivalent, we have
$$ C_1(z) \max\{|\alpha|, |\beta|\} \leqslant \|x\| \leqslant C_2(z) \max\{|\alpha|, |\beta|\} $$
for some constants $C_1(z)$, $C_2(z)$ that can be estimated explicitly (see below).
Recall that $x_i = A^i x_0$, and write $x_i = \alpha_i u + \beta_i v$, so that $\alpha_i = \lambda^i \alpha_0$ and $\beta_i = \mu^i \beta_0$. It follows that
$$ |\alpha_i| = |\alpha_0|^{1 - i/N} |\alpha_N|^{i/n} , \qquad |\beta_i| = |\beta_0|^{1 - i/N} |\beta_N|^{i/n} . $$
Thus,
$$ \begin{aligned} \|x_i\| & \leqslant C_2(z) \max\{|\alpha_i|, |\beta_i|\} \\
& = C_2(z) \max\{|\alpha_0|^{1 - i/N} |\alpha_N|^{i/n}, |\beta_0|^{1 - i/N} |\beta_N|^{i/n}\} \\
& \leqslant C_2(z) \times (\max\{|\alpha_0|, |\beta_0|\})^{1 - i/N} \times (\max\{|\alpha_N|, |\beta_N|\})^{i/n} \\
& \leqslant \frac{C_2(z)}{C_1(z)} \, \|x_0\|^{1 - i/N} \|x_N\|^{i/n} . \end{aligned} $$
That is,
$$ \|x_i\| \leqslant C(z) \|x_0\|^{1 - i/N} \|x_N\|^{i/N} . $$

Evaluation of optimal values of $C_1(z)$ and $C_2(z)$, especially in higher dimensions, is not completely obvious. Rougher estimates in dimension 2, however, are quite simple: clearly,
$$ \|\alpha u + \beta v\| \leqslant |\alpha| + |\beta| \leqslant 2 \max\{|\alpha|, |\beta|\} ,$$
so that $C_2(z) = 2$ does the job. On the other hand, if we denote the dot product of $u$ and $v$ by $p = u \cdot v$, then
$$ \begin{aligned} \|\alpha u + \beta v\|^2 & = |\alpha|^2 + |\beta|^2 - 2 p \operatorname{Re}(\alpha \bar\beta) \geqslant |\alpha|^2 + |\beta|^2 - 2 |p| |\alpha| |\beta| \\ & \geqslant (1 - |p|) (|\alpha|^2 + |\beta|^2) \ge (1 - |p|) (\max\{|\alpha|, |\beta|\})^2 , \end{aligned} $$
and so $C_1(z) = \sqrt{1 - |p|}$ works well. It follows that we may take $C(z) = 2 / \sqrt{1 - |p|}$.
