CH3-8 證明旋轉(zhuǎn)后的四元數(shù)虛部為零，實(shí)部為羅德里格斯公式結(jié)果

前面已經(jīng)推導(dǎo)過(guò)

$$v^{\prime }=pv{p}^{?}=pv{p}^{?1}$$

其中，$v=[0,\vec{v}]$，$p=[\cos \frac{\theta }{2},\sin \frac{\theta }{2}\vec{u}]$，代入上式

$$\begin{array}{rl}{v}^{\prime }& =pv{p}^{?}\\ & =[\cos \frac{\theta }{2},\sin \frac{\theta }{2}\vec{u}][0,\vec{v}][\cos \frac{\theta }{2},?\sin \frac{\theta }{2}\vec{u}]\\ & =[0?\sin \frac{\theta }{2}\vec{u}?\vec{v},\cos \frac{\theta }{2}\vec{v}+0+\sin \frac{\theta }{2}\vec{u}\times \vec{v}][\cos \frac{\theta }{2},?\sin \frac{\theta }{2}\vec{u}]\\ & =[?\sin \frac{\theta }{2}\vec{u}?\vec{v},\cos \frac{\theta }{2}\vec{v}+\sin \frac{\theta }{2}\vec{u}\times \vec{v}][\cos \frac{\theta }{2},?\sin \frac{\theta }{2}\vec{u}]\end{array}\text{\hspace{1em}(3-8-1)}$$

分別計(jì)算實(shí)部和虛部

$$\begin{array}{rl}\mathrm{Re}& =?\sin \frac{\theta }{2}\cos \frac{\theta }{2}\vec{u}?\vec{v}+(\cos \frac{\theta }{2}\vec{v}+\sin \frac{\theta }{2}\vec{u}\times \vec{v})?\sin \frac{\theta }{2}\vec{u}\\ & =?\sin \frac{\theta }{2}\cos \frac{\theta }{2}\vec{u}?\vec{v}+\cos \frac{\theta }{2}\sin \frac{\theta }{2}\vec{u}?\vec{v}+\sin \frac{\theta }{2}(\vec{u}\times \vec{v})?\vec{u}\\ & =0+0\\ & =0\end{array}\text{\hspace{1em}(3-8-2)}$$

$$\begin{array}{rl}\mathrm{Im}& =(?\sin \frac{\theta }{2}\vec{u}?\vec{v})?(?\sin \frac{\theta }{2}\vec{u})+(\cos \frac{\theta }{2}\vec{v}+\sin \frac{\theta }{2}\vec{u}\times \vec{v})\cos \frac{\theta }{2}\\ & +(\cos \frac{\theta }{2}\vec{v}+\sin \frac{\theta }{2}\vec{u}\times \vec{v})\times (?\sin \frac{\theta }{2}\vec{u})\end{array}\text{\hspace{1em}(3-8-3)}$$

我們希望將其寫(xiě)成矩陣乘法形式。

先證明公式：$\vec{a}\times (\vec{b}\times \vec{c})=(\vec{a}?\vec{c})?\vec{b}?(\vec{a}?\vec{b})?\vec{c}$。

證明：

$$\begin{array}{rl}\vec{a}\times \vec{b}& ={\mathbf{a}}^{\wedge }\mathbf{b}\\ & =\left[\begin{array}{ccc}0& ?a_3& a_2\\ a_3& 0& ?a_1\\ ?a_2& a_1& 0\end{array}\right]\left[\begin{array}{c}{b}_1\\ b_2\\ b_3\end{array}\right]\stackrel{△}{=}{\mathbf{T}}_a\mathbf{b}\end{array}\text{\hspace{1em}(3-8-4)}$$

那么（矩陣乘法滿(mǎn)足結(jié)合律）

$$\vec{a}\times (\vec{b}\times \vec{c})=({\mathbf{T}}_a{\mathbf{T}}_b)\mathbf{c}={\mathbf{T}}_a{\mathbf{T}}_b\mathbf{c}$$

而

$$\begin{array}{rl}{\mathbf{T}}_a{\mathbf{T}}_b& =\left[\begin{array}{ccc}0& ?a_3& a_2\\ a_3& 0& ?a_1\\ ?a_2& a_1& 0\end{array}\right]\left[\begin{array}{ccc}0& ?b_3& b_2\\ b_3& 0& ?b_1\\ ?b_2& b_1& 0\end{array}\right]\\ & =\left[\begin{array}{ccc}?a_3{b}_3?a_2{b}_2& a_2{b}_1& a_3{b}_1\\ a_1{b}_2& ?a_3{b}_3?a_1{b}_1& a_3{b}_2\\ a_1{b}_3& a_2{b}_3& ?a_2{b}_2?a_1{b}_1\end{array}\right]\\ & =?(\vec{a}?\vec{b})\mathbf{I}+\mathbf{b}{\mathbf{a}}^{\mathrm{T}}\end{array}$$

則（用到了矩陣結(jié)合律）

$$\begin{array}{rl}\vec{a}\times (\vec{b}\times \vec{c})={\mathbf{T}}_a{\mathbf{T}}_b\mathbf{c}& =(?(\vec{a}?\vec{b})\mathbf{I}+\mathbf{b}{\mathbf{a}}^{\mathrm{T}})\mathbf{c}\\ & =?(\vec{a}?\vec{b})\mathbf{c}+\mathbf{b}({\mathbf{a}}^{\mathrm{T}}\mathbf{c})\\ & =?(\vec{a}?\vec{b})\mathbf{c}+(\vec{a}?\vec{c})\mathbf{b}\end{array}\text{\hspace{1em}(3-8-5)}$$

同理可證

$$(\vec{a}\times \vec{b})\times \vec{c}=(\vec{a}?\vec{c})\mathbf{b}?(\vec{b}?\vec{c})\mathbf{a}\text{\hspace{1em}(3-8-6)}$$

證畢。

下面繼續(xù)推導(dǎo)式（3-8-3）

$$\begin{array}{rl}\mathrm{Im}& ={\sin }^2\frac{\theta }{2}\vec{u}?\vec{v}?\vec{u}+{\cos }^2\frac{\theta }{2}\vec{v}+\sin \frac{\theta }{2}\cos \frac{\theta }{2}\vec{u}\times \vec{v}?\sin \frac{\theta }{2}\cos \frac{\theta }{2}\vec{v}\times \vec{u}?{\sin }^2\frac{\theta }{2}\vec{u}\times \vec{v}\times \vec{u}\\ & ={\sin }^2\frac{\theta }{2}(\vec{u}?\vec{v})?\vec{u}+{\cos }^2\frac{\theta }{2}\vec{v}+\sin \theta \vec{u}\times \vec{v}?{\sin }^2\frac{\theta }{2}[(\vec{u}?\vec{u})\mathbf{v}?(\vec{v}?\vec{u})\mathbf{u}]\\ & =\underline{{\sin }^2\frac{\theta }{2}(\vec{u}?\vec{v})?\mathbf{u}}+{\cos }^2\frac{\theta }{2}\mathbf{v}+\sin \theta \vec{u}\times \vec{v}?{\sin }^2\frac{\theta }{2}\mathbf{v}+\underline{{\sin }^2\frac{\theta }{2}(\vec{v}?\vec{u})\mathbf{u}}\\ & =\cos \theta \mathbf{v}+2{\sin }^2\frac{\theta }{2}(\vec{v}?\vec{u})\mathbf{u}+\sin \theta \vec{u}\times \vec{v}\\ & =\cos \theta \mathbf{v}+(1?\cos \theta )(\vec{v}?\vec{u})\mathbf{u}+\sin \theta \vec{u}\times \vec{v}\end{array}\text{\hspace{1em}(3-8-7)}$$

注意：$\vec{u}$ 是單位向量，故 $\vec{u}?\vec{u}=1$。

也就是拉格朗日公式結(jié)果。證畢。

CH4 李群與李代數(shù)

CH4-1 SO(3) 上的指數(shù)映射

將指數(shù)函數(shù) $e^x$ 在 $x=0$ 處泰勒展開(kāi)，即

$$\begin{array}{rl}{e}^x& =1+x+\frac{1}{2!}{x}^2+\frac{1}{3!}{x}^3+...+\frac{1}{n!}{x}^n\\ & =\sum _{n=0}^{\infty }\frac{x^n}{n!}\end{array}\text{\hspace{1em}(4-1-1)}$$

將矩陣 $\mathbf{A}$ 代入上式， 則

$$e^{\mathbf{A}}=\sum _{n=0}^{\infty }\frac{{\mathbf{A}}^n}{n!}$$

同樣的，也有

$$e^{{?}^{\wedge }}=\sum _{n=0}^{\infty }\frac{({?}^{\wedge }{)}^n}{n!}\text{\hspace{1em}(4-1-2)}$$

令 $?=\theta \mathbf{a}$，$\theta$為模長(zhǎng)，$\mathbf{a}$ 為單位方向向量。則上式可寫(xiě)為

$$e^{(\theta \mathbf{a}{)}^{\wedge }}=\sum _{n=0}^{\infty }\frac{(\theta {\mathbf{a}}^{\wedge }{)}^n}{n!}$$

我們知道

$${\mathbf{a}}^{\wedge }=\left[\begin{array}{ccc}0& ?a_3& a_2\\ a_3& 0& ?a_1\\ ?a_2& a_1& 0\end{array}\right]$$

則

$$\begin{array}{rl}{\mathbf{a}}^{\wedge }{\mathbf{a}}^{\wedge }& =\left[\begin{array}{ccc}0& ?a_3& a_2\\ a_3& 0& ?a_1\\ ?a_2& a_1& 0\end{array}\right]\left[\begin{array}{ccc}0& ?a_3& a_2\\ a_3& 0& ?a_1\\ ?a_2& a_1& 0\end{array}\right]\\ & =\left[\begin{array}{ccc}?a_3^2?a_2^2& a_1{a}_2& a_1{a}_3\\ a_1{a}_2& ?a_3^2?a_1^2& a_2{a}_3\\ a_1{a}_3& a_2{a}_3& ?a_2^2?a_1^2\end{array}\right]\end{array}\text{\hspace{1em}(4-1-3)}$$

因?yàn)?$\mathbf{a}$ 是單位向量，則有 $a_1^2+a_2^2+a_3^2=1$，可得

$$\begin{array}{rl}\mathbf{a}{\mathbf{a}}^{\mathrm{T}}?\mathbf{I}=\left[\begin{array}{c}{a}_1\\ a_2\\ a_3\end{array}\right]\left[\begin{array}{ccc}{a}_1& a_2& a_3\end{array}\right]& =\left[\begin{array}{ccc}{a}_1^2& a_1{a}_2& a_1{a}_3\\ a_2{a}_1& a_2^2& a_2{a}_3\\ a_1{a}_3& a_2{a}_3& a_3^2\end{array}\right]?\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ & =\left[\begin{array}{ccc}?a_3^2?a_2^2& a_1{a}_2& a_1{a}_3\\ a_1{a}_2& ?a_3^2?a_1^2& a_2{a}_3\\ a_1{a}_3& a_2{a}_3& ?a_2^2?a_1^2\end{array}\right]\\ & ={\mathbf{a}}^{\wedge }{\mathbf{a}}^{\wedge }\end{array}\text{\hspace{1em}(4-1-4)}$$

$$\begin{array}{rl}{\mathbf{a}}^{\wedge }{\mathbf{a}}^{\wedge }{\mathbf{a}}^{\wedge }& =\left[\begin{array}{ccc}?a_3^2?a_2^2& a_1{a}_2& a_1{a}_3\\ a_1{a}_2& ?a_3^2?a_1^2& a_2{a}_3\\ a_1{a}_3& a_2{a}_3& ?a_2^2?a_1^2\end{array}\right]\left[\begin{array}{ccc}0& ?a_3& a_2\\ a_3& 0& ?a_1\\ ?a_2& a_1& 0\end{array}\right]\\ & =\left[\begin{array}{ccc}0& a_2^2{a}_3+a_3^3+a_1^2{a}_3& ?a_2^3?a_2{a}_3^2?a_1{a}_2^2\\ ?a_1^2{a}_3?a_3^3?a_1^2{a}_3& 0& a_1{a}_2^2+a_1^3+a_1{a}_3^2\\ a_2{a}_3^2+a_1^2{a}_2+a_2^3& ?a_1{a}_3^2?a_1^3?a_1{a}_2^2& 0\end{array}\right]\end{array}$$

又 $a_1^2+a_2^2+a_3^2=1$，上式寫(xiě)為

$${\mathbf{a}}^{\wedge }{\mathbf{a}}^{\wedge }{\mathbf{a}}^{\wedge }=\left[\begin{array}{ccc}0& a_3& ?a_2\\ ?a_3& 0& a_1\\ a_2& ?a_1& 0\end{array}\right]=?{\mathbf{a}}^{\wedge }\text{\hspace{1em}(4-1-5)}$$

對(duì)式（4-1-2）

$$\begin{array}{rl}{e}^{{?}^{\wedge }}=e^{(\theta \mathbf{a}{)}^{\wedge }}& =\sum _{n=0}^{\infty }\frac{(\theta {\mathbf{a}}^{\wedge }{)}^n}{n!}\\ & =\mathbf{I}+\theta {\mathbf{a}}^{\wedge }+\frac{1}{2!}{\theta }^2{\mathbf{a}}^{\wedge }{\mathbf{a}}^{\wedge }+\frac{1}{3!}{\theta }^3{\mathbf{a}}^{\wedge }{\mathbf{a}}^{\wedge }{\mathbf{a}}^{\wedge }+\frac{1}{4!}{\theta }^4{\mathbf{a}}^{\wedge }{\mathbf{a}}^{\wedge }{\mathbf{a}}^{\wedge }{\mathbf{a}}^{\wedge }+...\\ & =(\mathbf{a}{\mathbf{a}}^{\mathrm{T}}?{\mathbf{a}}^{\wedge }{\mathbf{a}}^{\wedge })+\theta {\mathbf{a}}^{\wedge }+\frac{1}{2!}{\theta }^2{\mathbf{a}}^{\wedge }{\mathbf{a}}^{\wedge }?\frac{1}{3!}{\theta }^3{\mathbf{a}}^{\wedge }?\frac{1}{4!}{\theta }^4{\mathbf{a}}^{\wedge }{\mathbf{a}}^{\wedge }+...\\ & =\mathbf{a}{\mathbf{a}}^{\mathrm{T}}+(\theta ?\frac{1}{3!}{\theta }^3+\frac{1}{5!}{\theta }^5+...){\mathbf{a}}^{\wedge }+(?1+\frac{1}{2!}{\theta }^2?\frac{1}{4!}{\theta }^4+...){\mathbf{a}}^{\wedge }{\mathbf{a}}^{\wedge }\\ & =({\mathbf{a}}^{\wedge }{\mathbf{a}}^{\wedge }+\mathbf{I})+\sin \theta {\mathbf{a}}^{\wedge }?\cos \theta ({\mathbf{a}}^{\wedge }{\mathbf{a}}^{\wedge })\\ & =(1?\cos \theta ){\mathbf{a}}^{\wedge }{\mathbf{a}}^{\wedge }+\mathbf{I}+\sin \theta {\mathbf{a}}^{\wedge }\\ & =(1?\cos \theta )(\mathbf{a}{\mathbf{a}}^{\mathrm{T}}?\mathbf{I})+\mathbf{I}+\sin \theta {\mathbf{a}}^{\wedge }\\ & =\mathbf{a}{\mathbf{a}}^{\mathrm{T}}?\mathbf{I}?\cos \theta \mathbf{a}{\mathbf{a}}^{\mathrm{T}}+\cos \theta \mathbf{I}+\mathbf{I}+\sin \theta {\mathbf{a}}^{\wedge }\\ & =\cos \theta \mathbf{I}+(1?\cos \theta )\mathbf{a}{\mathbf{a}}^{\mathrm{T}}+\sin \theta {\mathbf{a}}^{\wedge }\end{array}$$

于是得到李代數(shù) $?$ 和旋轉(zhuǎn)矩陣 $\mathbf{R}$ 之間的映射關(guān)系，即

$$\mathbf{R}=e^{{?}^{\wedge }}=\cos \theta \mathbf{I}+(1?\cos \theta )\mathbf{a}{\mathbf{a}}^{\mathrm{T}}+\sin \theta {\mathbf{a}}^{\wedge }\text{\hspace{1em}(4-1-6)}$$

也就是 羅德里格斯公式。

CH4-2 SE(3) 上的指數(shù)映射

已知李代數(shù) $\mathbf{\xi }=[\rho ?{]}^{\mathrm{T}}\in {\mathbb{R}}^6$，它的反對(duì)稱(chēng)矩陣為

$${\mathbf{\xi }}^{\wedge }=\left[\begin{array}{cc}{?}^{\wedge }& \rho \\ 0^{\mathrm{T}}& 0\end{array}\right]$$

則李群為

$$\begin{array}{rl}\mathbf{T}=\exp ({\mathbf{\xi }}^{\wedge })& =\left[\begin{array}{cc}{\sum }_{n=0}^{\infty }\frac{({?}^{\wedge }{)}^n}{n!}& {\sum }_{n=0}^{\infty }\frac{({?}^{\wedge }{)}^n}{(n+1)!}\rho \\ 0^{\mathrm{T}}& 0\end{array}\right]\\ & ?\left[\begin{array}{cc}\mathbf{R}& \mathbf{J}\rho \\ 0^{\mathrm{T}}& 1\end{array}\right]\end{array}\text{\hspace{1em}(4-2-1)}$$

下面開(kāi)始證明

同樣，假設(shè) $?=\theta \mathbf{a}$，$\theta$為模長(zhǎng)，$\mathbf{a}$為單位方向向量。將 $\exp ({\mathbf{\xi }}^{\wedge })$ 泰勒展開(kāi)

$$\exp ({\mathbf{\xi }}^{\wedge })=\frac{1}{n!}\sum _{n=0}^{\infty }{\left[\begin{array}{cc}{?}^{\wedge }& \rho \\ 0^{\mathrm{T}}& 0\end{array}\right]}^n=\frac{1}{n!}\sum _{n=0}^{\infty }{\left[\begin{array}{cc}\theta {\mathbf{a}}^{\wedge }& \rho \\ 0^{\mathrm{T}}& 0\end{array}\right]}^n\text{\hspace{1em}(4-2-2)}$$

當(dāng) $n=0$ 時(shí)，

$$\frac{1}{0!}{\left[\begin{array}{cc}{?}^{\wedge }& \rho \\ 0^{\mathrm{T}}& 0\end{array}\right]}^0=\mathbf{I}$$

當(dāng) $n=1$ 時(shí)，

$$\frac{1}{1!}{\left[\begin{array}{cc}\theta {\mathbf{a}}^{\wedge }& \rho \\ 0^{\mathrm{T}}& 0\end{array}\right]}^1=\left[\begin{array}{cc}\theta {\mathbf{a}}^{\wedge }& \rho \\ 0^{\mathrm{T}}& 0\end{array}\right]$$

當(dāng) $n=2$ 時(shí)，

$$\frac{1}{2!}\left[\begin{array}{cc}\theta {\mathbf{a}}^{\wedge }& \rho \\ 0^{\mathrm{T}}& 0\end{array}\right]\left[\begin{array}{cc}\theta {\mathbf{a}}^{\wedge }& \rho \\ 0^{\mathrm{T}}& 0\end{array}\right]=\left[\begin{array}{cc}(\theta {\mathbf{a}}^{\wedge }{)}^2& \theta {\mathbf{a}}^{\wedge }\rho \\ 0^{\mathrm{T}}& 0\end{array}\right]$$

當(dāng) $n=3$ 時(shí)，

$$\frac{1}{3!}\left[\begin{array}{cc}\theta {\mathbf{a}}^{\wedge }& \rho \\ 0^{\mathrm{T}}& 0\end{array}\right]\left[\begin{array}{cc}\theta {\mathbf{a}}^{\wedge }& \rho \\ 0^{\mathrm{T}}& 0\end{array}\right]\left[\begin{array}{cc}\theta {\mathbf{a}}^{\wedge }& \rho \\ 0^{\mathrm{T}}& 0\end{array}\right]=\frac{1}{3!}\left[\begin{array}{cc}(\theta {\mathbf{a}}^{\wedge }{)}^3& (\theta {\mathbf{a}}^{\wedge }{)}^2\rho \\ 0^{\mathrm{T}}& 0\end{array}\right]$$

以此類(lèi)推

$$\frac{1}{n!}{\left[\begin{array}{cc}\theta {\mathbf{a}}^{\wedge }& \rho \\ 0^{\mathrm{T}}& 0\end{array}\right]}^n=\frac{1}{n!}\left[\begin{array}{cc}(\theta {\mathbf{a}}^{\wedge }{)}^n& (\theta {\mathbf{a}}^{\wedge }{)}^{n?1}\rho \\ 0^{\mathrm{T}}& 0\end{array}\right]\text{\hspace{1em}(4-2-3)}$$

那么，式（4-1-8）可化為

$$\begin{array}{rl}\exp ({\mathbf{\xi }}^{\wedge })& =\frac{1}{n!}\sum _{n=0}^{\infty }{\left[\begin{array}{cc}{?}^{\wedge }& \rho \\ 0^{\mathrm{T}}& 0\end{array}\right]}^n\\ & =\mathbf{I}+\left[\begin{array}{cc}\theta {\mathbf{a}}^{\wedge }& \rho \\ 0^{\mathrm{T}}& 0\end{array}\right]+\left[\begin{array}{cc}(\theta {\mathbf{a}}^{\wedge }{)}^2& \theta {\mathbf{a}}^{\wedge }\rho \\ 0^{\mathrm{T}}& 0\end{array}\right]+...+\frac{1}{n!}\left[\begin{array}{cc}(\theta {\mathbf{a}}^{\wedge }{)}^n& (\theta {\mathbf{a}}^{\wedge }{)}^{n?1}\rho \\ 0^{\mathrm{T}}& 0\end{array}\right]\\ & =\left[\begin{array}{cc}{\sum }_{n=0}^{\infty }\frac{1}{n!}(\theta {\mathbf{a}}^{\wedge }{)}^n& {\sum }_{n=0}^{\infty }\frac{1}{(n+1)!}(\theta {\mathbf{a}}^{\wedge }{)}^n\rho \\ 0^{\mathrm{T}}& 1\end{array}\right]\end{array}\text{\hspace{1em}(4-2-4)}$$

其中，左上角為 $SO(3)$ 指數(shù)映射，前面已經(jīng)證明。令

$$\begin{array}{rl}\mathbf{J}& =\sum _{n=0}^{\infty }\frac{1}{(n+1)!}(\theta {\mathbf{a}}^{\wedge }{)}^n\\ & =\mathbf{I}+\frac{1}{2!}\theta {\mathbf{a}}^{\wedge }+\frac{1}{3!}(\theta {\mathbf{a}}^{\wedge }{)}^2+\frac{1}{4!}(\theta {\mathbf{a}}^{\wedge }{)}^3+\frac{1}{5!}(\theta {\mathbf{a}}^{\wedge }{)}^4\\ & =\frac{1}{\theta }(\frac{1}{2!}{\theta }^2?\frac{1}{4!}{\theta }^4+...){\mathbf{a}}^{\wedge }+\frac{1}{\theta }(\frac{1}{3!}{\theta }^3?\frac{1}{5!}{\theta }^5+...)({\mathbf{a}}^{\wedge }{)}^2+\mathbf{I}\\ & =\frac{1?\cos \theta }{\theta }{\mathbf{a}}^{\wedge }+\frac{\theta ?\sin \theta }{\theta }({\mathbf{a}}^{\wedge }{)}^2+\mathbf{I}\\ & =\frac{1?\cos \theta }{\theta }{\mathbf{a}}^{\wedge }+(1?\frac{\sin \theta }{\theta })(\mathbf{a}{\mathbf{a}}^{\mathrm{T}}?\mathbf{I})+\mathbf{I}\\ & =\frac{1?\cos \theta }{\theta }{\mathbf{a}}^{\wedge }+(1?\frac{\sin \theta }{\theta })\mathbf{a}{\mathbf{a}}^{\mathrm{T}}?\mathbf{I}+\frac{\sin \theta }{\theta }\mathbf{I}+\mathbf{I}\\ & =\frac{\sin \theta }{\theta }\mathbf{I}+(1?\frac{\sin \theta }{\theta })\mathbf{a}{\mathbf{a}}^{\mathrm{T}}+\frac{1?\cos \theta }{\theta }{\mathbf{a}}^{\wedge }\end{array}\text{\hspace{1em}(4-2-5)}$$

注意，這里用到式（4-1-5） $({\mathbf{a}}^{\wedge }{)}^3=?{\mathbf{a}}^{\wedge }$ 和 $\mathbf{a}{\mathbf{a}}^{\mathrm{T}}?\mathbf{I}={\mathbf{a}}^{\wedge }{\mathbf{a}}^{\wedge }$ 以及泰勒展開(kāi)

$$\cos \theta =1?\frac{1}{2!}{\theta }^2+\frac{1}{4!}{\theta }^4+...$$

$$\sin \theta =\theta ?\frac{1}{3!}{\theta }^3+\frac{1}{5!}{\theta }^5+...$$

綜上，證畢。

CH4-3 李代數(shù)求導(dǎo)

一、（1）$\mathrm{SO}(3)$ 直接求導(dǎo)

對(duì)極幾何：本質(zhì)矩陣奇異值分解

矩陣內(nèi)積和跡

矩陣具有 弗羅比尼烏斯內(nèi)積，類(lèi)似向量的內(nèi)積。它被定義為兩個(gè)相同大小的矩陣 $\mathbf{A}$ 和 $\mathbf{B}$ 的對(duì)應(yīng)元素的積的和， 即

$$<\mathbf{A},\mathbf{B}>=\sum _{i=1}^n\sum _{j=1}^n{a}_{ij}{b}_{ij}$$

以 $3\times 3$ 矩陣為例，設(shè)

$$\mathbf{A}=\left[\begin{array}{ccc}{a}_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\end{array}\right]，\mathbf{B}=\left[\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right]$$

則有

$$<\mathbf{A},\mathbf{B}>=\sum _{i=1}^n\sum _{j=1}^n{a}_{ij}{b}_{ij}$$

對(duì)于 ${\mathbf{A}}^{\mathrm{T}}\mathbf{B}$

$${\mathbf{A}}^{\mathrm{T}}\mathbf{B}=\left[\begin{array}{ccc}{a}_{11}& a_{21}& a_{31}\\ a_{12}& a_{22}& a_{32}\\ a_{13}& a_{23}& a_{33}\end{array}\right]\left[\begin{array}{ccc}{b}_{11}& b_{12}& b_{13}\\ b_{21}& b_{22}& b_{23}\\ b_{31}& b_{32}& b_{33}\end{array}\right]=\left[\begin{array}{ccc}{a}_{11}{b}_{11}+a_{21}{b}_{21}+a_{31}{b}_{31}& ?& ?\\ ?& a_{12}{b}_{12}+a_{22}{b}_{22}+a_{32}{b}_{32}& ?\\ ?& ?& a_{13}{b}_{13}+a_{23}{b}_{23}+a_{33}{b}_{33}\end{array}\right]$$

則 $\mathrm{Tr}({\mathbf{A}}^{\mathrm{T}}\mathbf{B})$等于

$$\mathrm{Tr}({\mathbf{A}}^{\mathrm{T}}\mathbf{B})=a_{11}{b}_{11}+a_{21}{b}_{21}+a_{31}{b}_{31}+a_{12}{b}_{12}+a_{22}{b}_{22}+a_{32}{b}_{32}+a_{13}{b}_{13}+a_{23}{b}_{23}+a_{33}{b}_{33}=\sum _{i=1}^n\sum _{j=1}^n{a}_{ij}{b}_{ij}$$

也就是說(shuō) ${\mathbf{A}}^{\mathrm{T}}\mathbf{B}$ 的跡等于兩矩陣對(duì)應(yīng)元素相乘的積的和。