Picard iterations of a matrix I need help with this problem. I think i got the first three questions of the exercise, but i'm stuck at the fourth one.
We consider the map $T:{\mathbb{R}^2}\longrightarrow{\mathbb{R}^2}$ defined by:
$T(a,b):= \left( \begin{array}{cc}
cos(\alpha) & sin(\alpha)  \\
-sin(\alpha) & cos(\alpha)  \end{array} \right)$$\left( \begin{array}{c}
a \\
b  \end{array} \right)$
Prove that:
1-T has only one fixed point,$(0,0)$.
2-T maps $B[(0,0),1]$ to itself.
3-Picard iterates $(T^n(a,b))$ don't converge to the fixed point, unless $(a,b)=(0,0)$
4-If we consider $U=\frac{1}{2}(I+T)$ (where I is the identity matrix), prove that $U$ has the same fixed point, that still maps  $B[(0,0),1]$ to itself, but in this case, the iterates $(U^n(a,b))$ converge to $(0,0)$.
 A: Show that, for $v\in B[(0,0),1]$, $||U(v)||< k||v||$ for some $k\in (0,1)\subset\mathbb{R}$. 
This will mean that $U$ is a contration; therefore its image does not abandon $B[(0,0),1]$.
If you don't know what a contraction is, try showing that
$$
||U^n(v)||< k^n||v||
$$
and since $k<1$, when $n\rightarrow\infty$, $||U^n(v)||\rightarrow 0$. Now, since $U(0,0)=(0,0)$ in $B[(0,0),1]$ and $U$ is linear, the last limit, through continuity, implies that $U^n(x,y)\rightarrow (0,0)$.
If you're using the contraction argument, since $(0,0)$ it's the unique fixed point of $U$ and is a contraction, $U^n(v)$ converges to $(0,0)$ as $n\rightarrow\infty$ for every $v\in B[(0,0),1]$.
NOTE: To compute $||v||$ just use the Euclidean metric, that is, if $v=(x_1, x_2,...,x_n)\in\mathbb{R}^n$, then
$$
||v||=\sqrt{x_1^2+x_2^2+...+x_n^2}
$$
To compute $||U(v)||$, write $U(v)$ as a vector and apply the above expression. You will get a lot of numbers but since you're dealing with sines and cosines, remember that $cos(x)^2+sin(x)^2=1$.
A: It might be helpful to get a geometric understanding of this transformation. This makes 1,2,3 obvious.

$T$ is the rotation of angle $-\alpha$. 

For 4, use the observation above to translate this in $\mathbb{C}$.

$$T(z)=e^{-i\alpha}z\qquad\Rightarrow\qquad U(z)=\frac{1+e^{-i\alpha}}{2}z.$$

You will see that for $\alpha\not\in 2\pi\mathbb{Z}$, $U$ is a strict contraction ($M$ Lipschitz for some $0<M<1$). Of course, 4 is false when $\alpha\in 2\pi\mathbb{Z}$.
If you don't feel comfortable using the isometry between $\mathbb{R}^2$ Euclidean and $\mathbb{C}$, just perform the more tedious trigonometric reductions that show $U$ is a contraction, using the Euclidean norm.
