I am given $\theta_0 \in [0,2\pi]$ fixed and $$ T(\hat{v})=\left( \begin{array}{ccc} cos(\theta_0) & -sin(\theta_0) \\ sin(\theta_0) & cos(\theta_0) \\ \end{array} \right) $$
I want to prove that the angle between $\hat{v}$ and $T(\hat{v})$ is exactly $\theta_0$
I uses the definition of the dot product $\hat{v}\cdot T(\hat{v})= \|\hat{v}\|\|T(\hat{v})\|cos(\theta)$ where $\theta$ is the angle between $\hat{v}$ and $T(\hat{v})$.
I'm got to $cos(\theta)=cos(\theta_0)$ but Idon't know how to conclude that $\theta=\theta_0$.
Thanks.
If $\cos \theta = \cos \theta_0$ with both of them in $[0, 2 \pi]$ then I assume you know that either $\theta = \theta_0$ or $\theta_0 = 2 \pi - \theta$. But of course saying that the angle between two lines is $2 \pi - \theta$ is the same as saying that the angle between them is $\theta$; it just depends on whether you're measuring the long way or the short way around.