第四讲 李群与李代数 习题证明 第5题
《SLAM十四讲》,主要是针对第四讲 第5题这个证明题,进行展开。
网上其他的各种证明方法,都稍微有点复杂。所以有如下这个方法。
1. 习题5的证明: R p ∧ R T = ( R p ) ∧ Rp^{\wedge}R^T=(Rp)^{\wedge} Rp∧RT=(Rp)∧
我们提前需要知道:
- 向量叉乘 u × v = u ∧ v u\times v=u^\wedge v u×v=u∧v,空间一点 p ∈ R 3 p\in\mathbb{R}^3 p∈R3,
- 旋转矩阵R: R R T = R T R = I RR^T=R^TR=I RRT=RTR=I。
- 此外,用到了一个向量叉乘的知识点,对任意的2个向量 u , v u,v u,v,向量叉乘的旋转变换不变性,即对2个向量先旋转 R u , R v Ru,Rv Ru,Rv,再叉乘 ( R u ) × ( R v ) (Ru)\times(Rv) (Ru)×(Rv),等于2向量先叉乘 u × v u\times v u×v,再旋转 R ( u × v ) R(u\times v) R(u×v)。
R ( u × v ) = ( R u ) × ( R v ) R(u\times v)=(Ru)\times(Rv) R(u×v)=(Ru)×(Rv).
上述的几何性质,来源于叉乘的基本性质。参考如下。
引用:
From the geometrical definition, the cross product is invariant under proper rotations about the axis defined by a × b a\times b a×b. In formulae:
( R a ) × ( R b ) = R ( a × b ) {\displaystyle (R\mathbf {a} )\times (R\mathbf {b} )=R(\mathbf {a} \times \mathbf {b} )} (Ra)×(Rb)=R(a×b),
where R {\displaystyle R} R is a rotation matrix with det ( R ) = 1 {\displaystyle \det(R)=1} det(R)=1.
这个结论,就是我们上面用到的,对向量的叉乘的旋转不变性。
More generally, the cross product obeys the following identity under matrix transformations:
( T a ) × ( T b ) = ( det T ) ( T − 1 ) T ( a × b ) = cof T ( a × b ) {\displaystyle (T\mathbf {a} )\times (T\mathbf {b} )=(\det T)\left(T^{-1}\right)^{\mathrm {T} }(\mathbf {a} \times \mathbf {b} )=\operatorname {cof} T(\mathbf {a} \times \mathbf {b} )} (Ta)×(Tb)=(detT)(T−1)T(a×b)=cofT(a×b),
where T {\displaystyle T} T is a 3-by-3 matrix and ( T − 1 ) T {\displaystyle \left(T^{-1}\right)^{\mathrm {T} }} (T−1)T is the transpose of the inverse and cof {\displaystyle \operatorname {cof} } cof is the cofactor matrix. It can be readily seen how this formula reduces to the former one if T {\displaystyle T} T is a rotation matrix.
更一般的说,任意的 3 × 3 3\times 3 3×3变换矩阵 T T T都满足上式。