5.证明:
Rp∧RT=(Rp)∧.
Rp^{\wedge }R^{T} = (Rp)^{\wedge}.
Rp∧RT=(Rp)∧.
证明:
⇒(Rp)∧n=(Rp)×(RR−1n)=R[p×(R−1n)]=Rp∧R⊤n
\Rightarrow (Rp)^{\wedge}n = (Rp) \times (RR^{-1}n) = R [ p \times (R ^{-1} n) ] = Rp ^ {\wedge}R^{\top}n
⇒(Rp)∧n=(Rp)×(RR−1n)=R[p×(R−1n)]=Rp∧R⊤n
⇒Rp∧R⊤=(Rp)∧
\Rightarrow Rp^{\wedge}R^{\top} = (Rp)^{\wedge}
⇒Rp∧R⊤=(Rp)∧
证明过程如下,纯属个人理解,仅供参考交流。
.
6.1 证明
Rexp(p∧)RT=exp((Rp)∧)
Rexp(p^{\wedge})R^{T}=exp((Rp)^{\wedge})
Rexp(p∧)RT=exp((Rp)∧)
该式称为SO(3) 上的伴随性质。
证明:
Rexp(p∧)RT=R∑n=0∞p∧nn!RT=∑n=0∞(Rp∧RT)nn!=exp(Rp∧RT)=exp((Rp)∧) Rexp(p^{\wedge})R^{T}=R\sum_{n=0}^{\infty }\frac{{p}^{\wedge n}}{n!}R^{T}=\sum _{n=0}^{\infty }\frac{(R {p}^{\wedge} R^{T}) ^{n}}{n!}=exp(R{p}^{\wedge}R^{T})=exp({(Rp)^{\wedge}}) Rexp(p∧)RT=Rn=0∑∞n!p∧nRT=n=0∑∞n!(Rp∧RT)n=exp(Rp∧RT)=exp((Rp)∧)
6.2 证明
Texp(ξ∧)T−1=exp((Ad(T)ξ)∧)
\mathbf{T}\exp(\bm{\xi}^\land)\mathbf{T}^{-1} = \exp((Ad(\mathbf{T})\bm{\xi})^\land)
Texp(ξ∧)T−1=exp((Ad(T)ξ)∧)
其中:
Ad(T)=[Rt∧R0R]
Ad(\textbf{T}) = \begin{bmatrix}
\textbf{R} & \bm{t}^\land \textbf{R} \\
0 & \textbf{R}
\end{bmatrix}
Ad(T)=[R0t∧RR]
证明:
设 ξ=[ρϕ]\bm{\xi}=\begin{bmatrix}\bm{\rho} \\ \bm{\phi} \end{bmatrix}ξ=[ρϕ],T=[Rt0⊤1]\textbf{T}=\begin{bmatrix}\textbf{R} & \bm{t} \\ \textbf{0}^\top &1 \end{bmatrix}T=[R0⊤t1], 则:
Texp(ξ∧)T−1=T∑n=0∞1n!(ξ∧)nT−1=∑n=0∞1n!(Tξ∧T−1)n=exp(Tξ∧T−1)=exp([Rϕ∧R⊤−Rϕ∧R⊤t+Rρ0⊤0])=exp([(Rϕ)∧−(Rϕ)∧t+Rρ0⊤0])=exp([−(Rϕ)∧t+RρRϕ]∧)=exp(([Rt∧R0R][ρϕ])∧)=exp((Ad(T)ξ)∧) \begin{aligned} \textbf{T}\exp(\bm{\xi}^\land)\textbf{T}^{-1} &= \textbf{T} \sum_{n=0}^\infty \frac{1}{n!}(\bm{\xi}^\land)^n\textbf{T}^{-1} \\ &= \sum_{n=0}^{\infty}{\frac{1}{n!}(\textbf{T}\bm{\xi}^\land\textbf{T}^{-1})^n} \\ &= \exp{(\textbf{T}\bm{\xi}^\land\textbf{T}^{-1})}\\ &=\exp({\begin{bmatrix} \textbf{R}\bm{\phi}^\land\textbf{R}^\top & -\textbf{R}\bm{\phi}^\land\textbf{R}^\top\bm{t} + \textbf{R}\bm{\rho} \\ \textbf{0}^\top &0 \end{bmatrix}} )\\ &= \exp(\begin{bmatrix} (\textbf{R}\bm{\phi})^\land & -(\textbf{R}\bm{\phi})^\land\bm{t} + \textbf{R}\bm{\rho} \\ \textbf{0}^\top &0 \end{bmatrix}) \\ &= \exp(\begin{bmatrix} -(\textbf{R}\bm{\phi})^\land\bm{t} + \textbf{R} \bm{\rho} \\ \textbf{R}\bm{\phi} \end{bmatrix} ^\land) \\ &=\exp((\begin{bmatrix} \textbf{R} & \bm{t}^\land \textbf{R} \\ 0 & \textbf{R} \end{bmatrix} \begin{bmatrix} \bm{\rho} \\ \bm{\phi} \end{bmatrix})^\land) \\ &= \exp((Ad(\textbf{T})\bm{\xi})^\land) \end{aligned} Texp(ξ∧)T−1=Tn=0∑∞n!1(ξ∧)nT−1=n=0∑∞n!1(Tξ∧T−1)n=exp(Tξ∧T−1)=exp([Rϕ∧R⊤0⊤−Rϕ∧R⊤t+Rρ0])=exp([(Rϕ)∧0⊤−(Rϕ)∧t+Rρ0])=exp([−(Rϕ)∧t+RρRϕ]∧)=exp(([R0t∧RR][ρϕ])∧)=exp((Ad(T)ξ)∧)
参考:
- https://blog.youkuaiyun.com/qq_17032807/article/details/84942548