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}


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


2 Answers 2


Suppose the rotation matrix is


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


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

enter image description here

Similarly, you can find $b,d$.

  • $\begingroup$ But what is wrong with what I did ? $\endgroup$ May 11, 2015 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, 2015 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).

enter image description here

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:

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


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