How do I find explicit formula? The sequence $(x_n,y_n)$, $n\in \mathbb{N}_0$ is recursively defined:
$\begin{pmatrix}x_{n+1}\\ y_{n+1}\end{pmatrix}=\left (\begin{matrix}0 & 1\\ 1/4 &0 \end{matrix} \right )\begin{pmatrix}x_n\\ y_n\end{pmatrix}$
So for $(x_0,y_0)=(4,-2)$. I have found some follow members: $(x_1,y_1)=(-2,1),(x_2,y_2)=(1,-1/2),(x_3,y_3)=(-1/2,1/4),(x_4,y_4)=(1/4,-1/8)$
But still I don’t know how to find explicit formula and show convergency to (0,0).
 A: $x_{n+1} = y_n, y_{n+1} = \dfrac{x_n}{4}\implies y_n = \dfrac{x_{n-1}}{4}\implies x_{n+1} = \dfrac{x_{n-1}}{4}\implies 4x_{n+1} + 0x_n + (-1)x_{n-1} = 0$. The characteristic equation is: $4x^2 - 1 = 0\implies x = \pm \dfrac{1}{2}\implies x_n = A\left(\dfrac{1}{2}\right)^2+B\left(\dfrac{-1}{2}\right)^n$. Can you take it from here?
A: Notice, by the definition, that $x_{n+1}=y_n, y_{n+1}=x_n/4$. Hence $x_{n+2}=x_n/4$, so in general $x_{2n}=\frac{x_0}{4^n}=\frac{1}{4^{n-1}},x_{2n+1}=\frac{x_1}{4^n}=\frac{-2}{4^n}, y_{2n}=\frac{-2}{4^{n}}, y_{2n+1}=\frac{1}{4^{n}}$
A: If $A=\left (\begin{matrix}0 & 1\\ 1/4 &0 \end{matrix} \right )$ then $A^n=(\frac 1 4)^{n-1}I_2$ if n even and $A^n=(\frac 1 4)^{n-1}A$ for n odd. 
Then use $\begin{pmatrix}x_{n+1}\\ y_{n+1}\end{pmatrix}=\left (\begin{matrix}0 & 1\\ 1/4 &0 \end{matrix} \right )^{n+1}\begin{pmatrix}x_0\\ y_0\end{pmatrix}$
A: Hint: calculate $A^n$ Where $A$ is your matrix
A: You have
$$\begin{pmatrix}x_{n+1}\\ y_{n+1}\end{pmatrix}=\left (\begin{matrix}0 & 1\\ 1/4 &0 \end{matrix} \right )\begin{pmatrix}x_n\\ y_n\end{pmatrix} = A \begin{pmatrix}x_n\\ y_n\end{pmatrix},$$ so that you can write $\begin{pmatrix}x_{n}\\ y_{n}\end{pmatrix}= A^n \begin{pmatrix}x_0\\ y_0\end{pmatrix}$ and $A^2=\left (\begin{matrix}1/4 & 0\\ 0 &1/4 \end{matrix} \right ), A^3 = ....$
