How to evaluate this limit involving matrices?

Question

We are given $$a_{ij}(n)=\begin{pmatrix}2 & 1\\\ 1 & 0\end{pmatrix}^n$$ And we have to evaluate $$\lim_{n\to\infty}\frac{a_{12}(n)}{a_{22}(n)}$$

Now I tried to evaluate the first few powers and found the relations: $a_{12}(n)=2a_{22}(n)+a_{22}(n-1)$

$a_{11}(n)=2a_{12}(n)+a_{12}(n-1)$

But I can't relate these for finding the limit. Can someone guide me further or perhaps provide an alternate way?

Answer 1

Observe that $A$ can be diagonalized as it has two distinct eigenvalues $1 \pm \sqrt{2}$. $$A=PDP^{-1}=\begin{bmatrix} -\sqrt{2}+1 & \sqrt{2}+1 \\ 1 & 1 \end{bmatrix} \, \, \begin{bmatrix} -\sqrt{2}+1 & 0 \\ 0 & \sqrt{2}+1 \end{bmatrix}\,\, \begin{bmatrix} \frac{-\sqrt{2}}{4} & \frac{\sqrt{2}+2}{4} \\ \frac{\sqrt{2}}{4} & \frac{-\sqrt{2}+2}{4} \end{bmatrix}.$$ Thus $$A^n=PD^nP^{-1}=\begin{bmatrix} -\sqrt{2}+1 & \sqrt{2}+1 \\ 1 & 1 \end{bmatrix} \, \begin{bmatrix} (-\sqrt{2}+1)^n & 0 \\ 0 & (\sqrt{2}+1)^n \end{bmatrix}\, \begin{bmatrix} \frac{-\sqrt{2}}{4} & \frac{\sqrt{2}+2}{4} \\ \frac{\sqrt{2}}{4} & \frac{-\sqrt{2}+2}{4} \end{bmatrix}.$$ Now you can get expressions for $a_{12}(n)$ and $a_{22}(n)$ to compute the given limit.

Answer 2

Hint:

If you diagonalize, you will have $P^{-1}AP=D\implies A^n=PD^nP^{-1}$.