A rotation by an angle $\theta $ in two dimensions Let $x_i$ be the coordinates of a two-dimensional vector and let $x^\prime_i$ be the coordinates
of the vector rotated by an angle $\theta$ in the plane. The components of the two vectors are related by a transformation as
$$x^\prime_j = R_{ij} x_i$$
where $R_{ij}$ is a rotation matrix. This is a representation of the rotation group.

Specifically, a rotation by an angle $\theta$ in two dimensions
  matrix, $$ R(\theta) = \begin{pmatrix}   \cos\theta & \sin \theta\\
 - \sin \theta  & \cos\theta  \\  \end{pmatrix} $$

we do we got this matrix, can someone show me the argument or visualization?
 A: According to the picture, for the first coordinates:

$$A_y=Acos(\theta)$$
$$A_z=Asin(\theta)$$
now write $\mathbf A$ in terms of new coordinates:
$$\bar{A_y}=Acos(\bar{\theta})=Acos(\theta - \phi)=A(cos\theta cos\phi +sin\theta sin\phi)=A_ycos\phi+A_zsin\phi$$
$$\bar{A_z}=Asin \bar{\theta}=Asin(\theta-\phi)=A(sin\theta cos\phi-cos\theta sin\phi)=-sin\phi A_y+A_zcos\phi$$
and writing this relations in  matrix form, we will get the given formula.
$$ \begin{pmatrix}   \cos\phi & \sin \phi\\
 - \sin \phi  & \cos\phi  \\  \end{pmatrix} $$
A: I suggest drawing a diagram and working it through.
To simplify matters, you could consider the unit vectors on the $x$ and $y$ axes and what the rotation does to them, to see that the matrix must be as it is (draw diagrams - they are very easy - and work out the components of the transformed unit vectors).
Then think about why, if you rotate the $x$ and $y$ components of a vector by $\theta$, you rotate the vector itself through $\theta$ (if the orientations are all the same).
[Another way of looking at it - rotating a vector clockwise is the same as rotating the axes anticlockwise by the same amount]
I recommend working at it, because when it clicks it can become one of those beautiful things where different insights play off each other (de Moivre's theorem, for example).
