旋转变换矩阵求逆

旋转变换矩阵求逆

背景

坐标系之间相互转换涉及到变换矩阵的求逆，求逆是一个野蛮的过程，世界坐标系到观察

坐标系之间的坐标转换，实际上就是坐标系的平移加旋转，而旋转与平移变换都要以简单

得到其逆变换，从而绕过了对矩阵求逆的过程。下面求旋转变换的逆变换。

绕各坐标轴旋转的矩阵的逆等于其的转置

以绕z旋转为例

$$x' = \rho cos(\alpha+\theta) = \rho (cos\alpha \cdot cos\theta - sin\alpha \cdot sin\theta) = x \cdot cos\theta - y \cdot sin\theta$$

$$y' = \rho sin(\alpha+\theta) = \rho (cos\alpha \cdot sin\theta + sin\alpha \cdot cos\theta) = x \cdot sin\theta + y \cdot cos\theta$$

$$z' = z$$

$$\begin{bmatrix} x' \\ y' \\ z' \\ \end{bmatrix} = \left[\begin{array}{cc} cos\theta & -sin\theta & 0 \\ sin\theta & cos\theta & 0 \\ 0 & 0 & 1 \\ \end{array}\right] \left[\begin{array}{cc} x \\ y \\ z \\ \end{array}\right] = R_z(\theta) \left[\begin{array}{cc} x \\ y \\ z \\ \end{array}\right]$$

相应的还原矩阵：

$$R_z(-\theta) = \left[\begin{array}{cc} cos(-\theta) & -sin(-\theta) & 0 \\ sin(-\theta) & cos(-\theta) & 0 \\ 0 & 0 & 1 \\ \end{array}\right] = \left[\begin{array}{cc} cos\theta & sin\theta & 0 \\ -sin\theta & cos\theta & 0 \\ 0 & 0 & 1 \\ \end{array}\right] = R_z(\theta)^T$$

类似地，如果绕x, y轴旋转，有类似变换矩阵：

$$\begin{bmatrix} x' \\ y' \\ z' \\ \end{bmatrix} = \left[\begin{array}{cc} 1 & 0 & 0 \\ 0 & cos\theta & -sin\theta \\ 0 & sin\theta & cos\theta \\ \end{array}\right] \left[\begin{array}{cc} x \\ y \\ z \\ \end{array}\right]$$

$$R_x(-\theta) = R_x(\theta)^T$$

$$\begin{bmatrix} x' \\ y' \\ z' \\ \end{bmatrix} = \left[\begin{array}{cc} cos\theta & 0 & -sin\theta \\ 0 & 1 & 0 \\ sin\theta & 0 & cos\theta \\ \end{array}\right] \left[\begin{array}{cc} x \\ y \\ z \\ \end{array}\right]$$

$$R_y(-\theta) = R_y(\theta)^T$$

要还原某个旋转：

$$R(\theta) = R_x(\theta) \cdot R_y(\theta) \cdot R_z(\theta)$$

应该反先后顺序，反角度旋转：

$$R(-\theta) = R_z(-\theta) \cdot R_y(-\theta) \cdot R_x(-\theta)$$

使用转置替换之后：

$$R(-\theta) = R_z(\theta)^T \cdot R_y(\theta)^T \cdot R_x(\theta)^T$$

$$= ( R_y(\theta) \cdot R_z(\theta) )^T \cdot R_x(\theta)^T$$

$$= (R_x(\theta) \cdot R_y(\theta) \cdot R_z(\theta) )^T$$

$$= R(\theta)^T$$

$$R(-\theta) = R(\theta)^T$$