旋转矩阵
01 旋转矩阵
- [ cosθ−sinθ sinθcosθ]\displaystyle{ \displaystyle{\left[\begin{array}{lll} \ \cos\theta & -\sin\theta\ \\ \ \sin\theta & \cos\theta\end{array}\right]} }[ cosθ sinθ−sinθ cosθ] .
02 如何记忆?
(1) 特殊情形
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点 DDD 旋转 θ\thetaθ 角度到点 EEE .
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设 E (u,v) , D (x,y)E\ (u,v)\ ,\ D\ (x,y)E (u,v) , D (x,y) ,则 u/cosθ=x , v/sinθ=xu/\cos\theta = x\ ,\ v/\sin\theta = xu/cosθ=x , v/sinθ=x .
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[ u v ]=[ cosθ⋅x sinθ⋅x ]=[ cosθ? sinθ?][ x 0 ]=[ cosθ−sinθ sinθcosθ][ x y ]\displaystyle{ \displaystyle{\left[\begin{array}{lll} \ u\ \\ \ v \ \end{array}\right]=\left[\begin{array}{lll} \ \cos\theta\cdot x\ \\ \ \sin\theta\cdot x \ \end{array}\right]=\left[\begin{array}{lll} \ \cos\theta & ?\ \\ \ \sin\theta & ?\end{array}\right]}\left[\begin{array}{lll} \ x\ \\ \ 0 \ \end{array}\right]=\left[\begin{array}{lll} \ \cos\theta & -\sin\theta\ \\ \ \sin\theta & \cos\theta\end{array}\right]\left[\begin{array}{lll} \ x\ \\ \ y \ \end{array}\right] }[ u v ]=[ cosθ⋅x sinθ⋅x ]=[ cosθ sinθ? ?][ x 0 ]=[ cosθ sinθ−sinθ cosθ][ x y ] .
(2) 矩阵分析
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[ cosθ? sinθ?]\displaystyle{ \displaystyle{\left[\begin{array}{lll} \ \cos\theta & ?\ \\ \ \sin\theta & ?\end{array}\right]} }[ cosθ sinθ? ?] ,怎么记忆?
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旋转矩阵的行列式 = 1=\ 1= 1,那么 cos\coscos 的对角线就是 cos\coscos ,
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sin\sinsin 的对角线就是 sin\sinsin ,而且副对角线需要有一个带负号。
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只有这样,行列式才会得到 cos2+sin2\cos^2+\sin^2cos2+sin2 的形式。
(3) 极坐标记忆
- 极坐标变换中,有 x=rcosθx=r\cos\thetax=rcosθ,y=rsinθy=r\sin\thetay=rsinθ,
- 那么 x,yx,yx,y 对 r,θr,\thetar,θ 的雅可比矩阵为:
- [ ∂x∂r∂x∂θ ∂y∂r∂y∂θ]=[ cosθ−rsinθ sinθrcosθ]\displaystyle{ \displaystyle{\left[\begin{array}{lll} \ \displaystyle{\frac{\partial x}{\partial r}} & \displaystyle{\frac{\partial x}{\partial \theta}}\ \\ \ \displaystyle{\frac{\partial y}{\partial r}} & \displaystyle{\frac{\partial y}{\partial \theta}}\end{array}\right]}=\displaystyle{\left[\begin{array}{lll} \ \cos\theta & -r\sin\theta\ \\ \ \sin\theta & r\cos\theta\end{array}\right]} } ∂r∂x ∂r∂y∂θ∂x ∂θ∂y=[ cosθ sinθ−rsinθ rcosθ] ,
- 当 r=1r=1r=1 时,即为旋转矩阵:[ cosθ−sinθ sinθcosθ]\displaystyle{ \displaystyle{\left[\begin{array}{lll} \ \cos\theta & -\sin\theta\ \\ \ \sin\theta & \cos\theta\end{array}\right]} }[ cosθ sinθ−sinθ cosθ] .