# 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 $$x_0 \le x_{\text{middle}} \le x_1.$$

• Is $\|x_i\|^N \leqslant C(z) \|x_0\|^{N-i} \|x_N\|^i$ something that you are looking for? This follows quite simply: write $u$, $v$ for the eigenvectors and $\alpha_i$, $\beta_i$ for the coefficients of the eigenvector expansion of $x_i$. Then $\alpha_i^N = \alpha_0^{N-i} \alpha_N^i$, $\beta_i^N = \beta_0^{N-i} \beta_N^i$, and $C_1(z)(|\alpha_i|+|\beta_i|) \|x_i\| \leqslant C_2(z)(|\alpha_i|+|\beta_i|)$. – Mateusz Kwaśnicki Mar 5 at 10:34
• sorry, so what is this $C(z)$ precisely in your notation? – J.Doe Mar 5 at 12:20
• A "constant" that depends on $z$. – Mateusz Kwaśnicki Mar 5 at 12:22
• sorry, I mean is it explicit? – J.Doe Mar 5 at 12:23
• That was too long for a comment, I posted some details as an answer. – Mateusz Kwaśnicki Mar 5 at 12:49

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