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If I take the basis $(\vec{e_x},\vec{e_y})$ and make a rotation counterclockwise of angle $\theta$, I end up with two new vectors $(\vec{u},\vec{v})$ such that :

$\vec{u} = \cos\theta \vec{e_x} + \sin\theta \vec{e_y}$

$\vec{v} = \cos\theta \vec{e_x} - \sin\theta \vec{e_y}$

so \begin{equation} \left( \begin{array}{ccc} \vec{u} \\ \vec{v}\end{array} \right) = \left( \begin{array}{ccc} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{array} \right) \left( \begin{array}{ccc} \vec{e_x} \\ \vec{e_y}\end{array} \right) \end{equation}

I don't understand why the counterclockwise rotation is defined as : \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}

EDIT:

When I look at my picture, it looks like a counterclockwise rotation... enter image description here

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Suppose the rotation matrix is

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

Since it rotate every vector by angle $\theta$, we will look at what it does to the basis $\begin{bmatrix}1\\0\end{bmatrix}$, $\begin{bmatrix}0\\1\end{bmatrix}$.

$$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}a\\c\end{bmatrix}$$

By the following picture, we could see that $a=\cos\theta,c=\sin\theta$.

enter image description here

Similarly, you can find $b,d$.

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  • $\begingroup$ But what is wrong with what I did ? $\endgroup$ – user1234161 May 11 '15 at 9:33
  • $\begingroup$ @user1234161: Your matrix gives you clockwise rotation. You can use the same geometric method to see that. $\endgroup$ – KittyL May 11 '15 at 9:36

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