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](https://zhenghezhang.wordpress.com/category/schrodinger-cocycles/) 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 $x\_0 \le x\_{\text{middle}} \le x\_1.$ 



Discrete dynamical system and bound on norm
This is an extension and correction of my comment, made more explicit.




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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,
$$ \|x\_i\| \leqslant C(z) \|x\_0\|^{1 - i/N} \|x\_N\|^{i/N} . $$




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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 $C(z) = 2 / \sqrt{1 - |p|}$.


