坐标变换

平移

$$\left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right]=\left[\begin{array}{c}x\\ y\end{array}\right]+\left[\begin{array}{c}a\\ b\end{array}\right]$$

用矩阵乘法也表示为：

$$\left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\\ 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& a\\ 0& 1& b\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ 1\end{array}\right]$$

旋转

绕原点的旋转，通过几何或复数($e^{i\theta }z$)来推导：

$$\left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\\ 1\end{array}\right]=\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ 1\end{array}\right]$$

一般变换

先

$$\left[\begin{array}{c}{x}^{\prime }\\ y^{\prime }\\ 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& a\\ 0& 1& b\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0\\ \sin \theta & \cos \theta & 0\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ 1\end{array}\right]=\left[\begin{array}{ccc}\cos \theta & -\sin \theta & a\\ \sin \theta & \cos \theta & b\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\\ 1\end{array}\right]$$

求变换矩阵

只需要两个点的坐标就能知道变换矩阵。

设坐标系 $A$ 中两点 $(x_1,y_1),(x_2,y_2)$, $B$ 中两点 $(x_1^{\prime },y_1^{\prime }),(x_2^{\prime },y_2^{\prime })$ ，则

$$\left[\begin{array}{c}{x}_2-x_1\\ y_2-y_1\end{array}\right]=\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right]\left[\begin{array}{c}{x}_2^{\prime }-x_1^{\prime }\\ y_2^{\prime }-y_1^{\prime }\end{array}\right]$$

可解得 $\theta$ 。则变换矩阵为：

$$\left[\begin{array}{ccc}\cos \theta & -\sin \theta & x_1-x_1^{\prime }\\ \sin \theta & \cos \theta & y_1-y_1^{\prime }\\ 0& 0& 1\end{array}\right]$$