How do I find explicit formula for $(x_n,y_n)$ and show that the sequence tends to converge to $(0,0)$? How do I find explicit formula for  $(x_n,y_n)$ and show that the sequence tends to converge to $(0,0)$?
In $\mathbb{R}^2$, 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:$(x_1,y_1)=(2,1),(x_2,y_2)=(1,1/2),(x_3,y_3)=(1/2,1/4),(x_4,y_4)=(1/4,1/8)...$
At this point I see that $x_{n+1}=x_n/2$.
But how do I write it correctly? How to show that it converges in $(0,0)$?
 A: We have $x_1=4$ and $x_{n+1}=x_n \div 2$. This means to get to the next term, we divide by $2$. Therefore, if we want to go three terms forward, we divide by $2$ three times, or divide by $2^3=8$. If we want to go five terms forward, we divide by $2$ five times, or divide by $2^5=32$. We have $x_0$ and want to go to $x_n$, so we want to go $n$ terms forward. This means we need to divide by $2$ for $n$ times, or divide by $2^n$, so we have:
$$x_n=\frac{x_0}{2^n}=\frac{4}{2^n}$$
Similarly, with $y_n$, we have:
$$y_n=\frac{y_0}{2^n}=\frac{2}{2^n}$$
Hopefully, you can see how both of these sequences converge to $0$ as $n \to \infty$ (hint: look at denominator), meaning we have:
$$\lim_{n \to \infty} (x_n, y_n)=(0, 0)$$
A: $
\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}
=
\left (\begin{matrix}0 & 1\\ 1/4 &0 \end{matrix}  \right )\left (\begin{matrix}0 & 1\\ 1/4 &0 \end{matrix}  \right )\begin{pmatrix}x_{n-1}\\ y_{n-1}\end{pmatrix}
=
\dots
=
\left (\begin{matrix}0 & 1\\ 1/4 &0 \end{matrix}  \right )^{n+1}\begin{pmatrix}x_0\\ y_0\end{pmatrix}
$
and you need to power the matrix.
Compute few powers ($n=1,2,3,4$), you'll notice that for even exponent, it is a multiplicity of $\left (\begin{matrix}1&0\\ 0&1 \end{matrix}\right)$.
