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

You can also do it in a more algebraic way. Since after rotation ($$(x, y)$$ is rotated to $$(x', y')$$), the length of the vector doesn't change, which means $$n = \sqrt{x'^2 + y'^2} = \sqrt{x^2 + y^2}$$ (see in figure attached).

Therefore we can get the following equation:

\begin{aligned} y' & = n \cdot \sin(\theta + \alpha) & (1)\\ y & = n \cdot \sin \alpha & (2) \end{aligned}

\begin{aligned} x' & = n \cdot \cos(\theta + \alpha) & (3) \\ x & = n \cdot \cos \alpha & (4) \end{aligned}

Then use the trigonometric identities to expand (1) and (3): \begin{aligned} y' & = n \cdot (\sin \theta \cos \alpha + \cos \theta \sin \alpha) & (5) \\ x' & = n \cdot (\cos \theta \cos \alpha - \sin \theta \sin \alpha) & (6) \end{aligned}

By substituting (2) and (4) into (5) and (6), we can get: \begin{aligned} y' & = x \cdot \sin \theta + y \cdot \cos \theta \\ x' & = x \cdot \cos \theta - y \cdot \sin \theta \end{aligned}

From here we can easily see:

$$\begin{bmatrix}x'\\y'\end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}$$