Proof that $T^nx$ converges. 
Let $B$ be a real two times two matrix with eigenvalues $\lambda\in (1,\infty)$ and $\mu\in (0,1)$ and we define $$T:S^1\rightarrow S^1;~~x\mapsto \frac{Bx}{\|Bx\|}$$
I want to show that for all $x\in S^1$, the sequence $T^nx$ converges to a fixed point of $T$.

I know that the only fix points of $T$ are $$O:=\left\{\frac{x}{\|x\|},-\frac{x}{\|x\|},\frac{y}{\|y\|},-\frac{y}{\|y\|}\right\}$$ where $x,y$ are the eigenvectors for $\lambda$ respectively $\mu$. Additionally I know that $T^nx=\frac{B^nx}{\|B^nx\|}$. But now if I take $x\in S^1$ how can I show that $T^nx\rightarrow z$ where $z\in O$.
 A: Let $x$ and $y$ be unit eigenvectors for $T$ corresponding to eigenvalues $\lambda$ and $\mu$ respectively.
Then for any scalars $a, b$ we have
$$B^n(ax+by) = aB^n x + bB^n y = (a\lambda^n) x + (b\mu^n) y.$$
Furthermore,
$$\|B^n(ax+by)\|^2 = \|(a\lambda^n) x + (b\mu^n) y\|^2
= a^2 \lambda^{2n} \|x\|^2 + b^2 \mu^{2n} \|y\|^2 + 2ab\lambda^n \mu^n \langle x, y \rangle.$$
Thus,
$$T^n(ax + by)= \frac{B^n(ax+by)}{\|B^n(ax+by)\|} = \frac{1}{\sqrt{a^2 \lambda^{2n} + b^2 \mu^{2n} + 2ab\lambda^n \mu^n \langle x, y \rangle}}((a\lambda^n)x + (b\mu^n) y).$$
If $a \ne 0$, then the coefficient of $x$ is
$$\frac{a\lambda^n}{\sqrt{a^2 \lambda^{2n} + b^2 \mu^{2n} + 2ab\lambda^n \mu^n \langle x, y \rangle}}
= \frac{\text{sign}(a)}{\sqrt{1 + (b/a)^2 (\mu/\lambda)^{2n} + 2 (b/a) (\mu/\lambda)^n \langle x, y \rangle}} \to \text{sign}(a)$$
which implies $T^n(ax+by)$ converges to $x$ or to $-x$.
On the other hand, if $a=0$ and $b \ne 0$, then we have $T^n (by) = y$ for all $n$.
The remaining case $a=b=0$ can be ignored since $0 \notin S^1$.
