# Counterclockwise rotation matrix

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

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...

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 $a=\cos\theta,c=\sin\theta$.

Similarly, you can find $b,d$.

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