正交矩阵的指数表示和雅可比
对于任意 n × n n\times n n×n 斜对称(反对称)矩阵 B B B,存在正交阵 P P P和块对角阵 E E E,使以下等式成立 (see Matrix analysis, Horn and Johnson , Corollary 2.5.14)
B = P ′ E P B=P'EP B=P′EP
其中
E = ( E 1 ⋯ ⋮ ⋱ ⋮ ⋯ E m ⋯ O n − 2 m ) E=\begin{pmatrix} E_1&\cdots&\\ \vdots & \ddots & \vdots\\ &\cdots &E_m &\\ \cdots&&&O_{n-2m} \end{pmatrix} E=⎝⎜⎜⎜⎛E1⋮⋯⋯⋱⋯⋮EmOn−2m⎠⎟⎟⎟⎞
这里前面的 m m m 个块 E i , ( i = 1 , ⋯ , m ) E_i, \quad (i=1,\cdots, m) Ei,(i=1,⋯,m), 为
E i = ( 0 − θ i θ i 0 ) , 0 < θ i ≤ π E_i=\begin{pmatrix} 0& - \theta_i\\ \theta_i &0 \end{pmatrix}, \quad 0<\theta_i \le\pi Ei=(0θi−θi0),0<θi≤π对于任意旋转矩阵 R ∈ S O ( n ) R\in SO(n) R∈SO(n),存在正交阵 P P P和块对角阵 D D D,使以下等式成立 (see Matrix analysis, Horn and Johnson , Corollary 2.5.14)
X = P ′ D P X=P'DP X=P′DP
其中
D = ( D 1 ⋯ ⋮ ⋱ ⋮ ⋯ D m ⋯ I n − 2 m ) D=\begin{pmatrix} D_1&\cdots&\\ \vdots & \ddots & \vdots\\ &\cdots &D_m &\\ \cdots&&&I_{n-2m} \end{pmatrix} D=⎝⎜⎜⎜⎛D1⋮⋯⋯⋱⋯⋮DmIn−2m⎠⎟⎟⎟⎞
这里前面的 m m m 个块 D i , ( i = 1 , ⋯ , m ) D_i, \quad (i=1,\cdots, m) Di,(i=1,⋯,m), 为
D i = ( cos θ i − sin θ i sin θ i cos θ i ) , 0 < θ i ≤ π D_i=\begin{pmatrix} \cos \theta_i&-\sin \theta_i\\ \sin \theta_i & \cos \theta_i \end{pmatrix}, \quad 0<\theta_i \le\pi Di=(cosθisinθi−sinθicosθi),0<θi≤π
3 旋转矩阵 R R R 可以表示成斜对称矩阵 B B B 的指数形式:
R = exp ( B ) (1) R=\exp(B) \tag 1 R=exp(B)(1)
证明: exp ( B ) = exp ( P ′ E P ) = I + P ′ E P + ( P ′ E P ) 2 2 ! + ⋯ = P ′ ( I + E + E 2 2 ! + ⋯ ) P \begin{aligned}\exp(B)=\exp(P'EP)&=I+P'EP+{(P'EP)^2\over 2!}+\cdots\\ &=P'\left(I+E+{E^2\over 2!}+\cdots\right )P \end{aligned} exp(B)=exp(P′EP)=I+P′EP+2!(P′EP)2+⋯=P′(I+E+2!E2+⋯)P
注意到
I + ( 0 − θ θ 0 ) + 1 2 ! ( 0 − θ θ 0 ) 2 + ⋯ = ( cos θ − sin θ sin θ cos θ ) \begin{aligned} I+\begin{pmatrix}0& - \theta\\ \theta &0\end{pmatrix}+{1\over 2!}\begin{pmatrix}0& - \theta\\ \theta &0\end{pmatrix}^2+\cdots =\begin{pmatrix}\cos \theta&-\sin \theta\\ \sin \theta & \cos \theta\end{pmatrix} \end{aligned} I+(0θ−θ0)+2!1(0θ−θ0)2+⋯=(cosθsinθ−sinθcosθ)
等式 (1) 成立。
- 设斜对称矩阵 S = H E H ′ S=HEH' S=HEH′, H H H 和 E E E 分别是正交阵和块对角阵,下面求 ( d S ) (dS) (dS) (see Anderson, G. A. 1965)
d S = d H E H ′ + H d E H ′ + H E d H ′ dS=dHEH'+HdEH'+HEdH' dS=dHEH′+HdEH′+HEdH′
等式两边同时左乘 H ′ H' H′右乘 H H H,得到
H ′ d S H = H ′ d H E + d E + E d H ′ H H'dSH=H'dHE+dE+EdH'H H′dSH=H′dHE+dE+EdH′H
考虑到 ( H ′ d S H ) = ( d S ) (H'dSH)=(dS) (H′dSH)=(dS), d H ′ H = − H ′ d H dH'H=-H'dH dH′H=−H′dH,(因为 det H = 1 \det H=1 detH=1, H ′ H = I H'H=I H′H=I)
( d S ) = ( H ′ d H E − E H ′ d H + d E ) (dS)=(H'dHE-EH'dH+dE) (dS)=(H′dHE−EH′dH+dE)
H ′ d H = ( 0 − h 2 ′ d h 1 − h 3 ′ d h 1 − h 4 ′ d h 1 − h 5 ′ d h 1 ⋯ − h n ′ d h 1 h 2 ′ d h 1 0 − h 3 ′ d h 2 − h 4 ′ d h 2 − h 5 ′ d h 2 ⋯ − h n ′ d h 2 h 3 ′ d h 1 h 3 ′ d h 2 0 − h 4 ′ d h 3 − h 5 ′ d h 3 ⋯ − h n ′ d h 3 h 4 ′ d h 1 h 4 ′ d h 2 h 4 ′ d h 3 0 − h 5 ′ d h 4 ⋯ − h n ′ d h 4 h 5 ′ d h 1 h 5 ′ d h 2 h 5 ′ d h 3 h 5 ′ d h 4 0 ⋯ − h n ′ d h 5 ⋮ ⋱ ⋮ h n ′ d h 1 h n ′ d h 2 h n ′ d h 3 h n ′ d h 4 h n ′ d h 5 ⋯ − h n ′ d h n ) H'dH=\begin{pmatrix}0& -h_2'dh_1& -h_3'dh_1& -h_4'dh_1& -h_5'dh_1&\cdots & -h_n'dh_1\\ h_2'dh_1& 0 & -h_3'dh_2& -h_4'dh_2&-h_5'dh_2&\cdots & -h_n'dh_2 \\ h_3'dh_1 & h_3'dh_2& 0& -h_4'dh_3& -h_5'dh_3&\cdots & -h_n'dh_3 \\ h_4'dh_1& h_4'dh_2& h_4'dh_3& 0& -h_5'dh_4& \cdots & -h_n'dh_4\\ h_5'dh_1& h_5'dh_2& h_5'dh_3& h_5'dh_4&0& \cdots & -h_n'dh_5\\ \vdots & & & & &\ddots&\vdots\\ h_n'dh_1& h_n'dh_2& h_n'dh_3& h_n'dh_4&h_n'dh_5& \cdots & -h_n'dh_n \end{pmatrix} H′dH=⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎛0h2′dh1h3′dh1h4′dh1h5′dh1⋮hn′dh1−h2′dh10h3′dh2h4′dh2h5′dh2hn′dh2−h3′dh1−h3′dh20h4′dh3h5′dh3hn′dh3−h4′dh1−h4′dh2−h4′dh30h5′dh4hn′dh4−h5′dh1−h5′dh2−h5′dh3−h5′dh40hn′dh5⋯⋯⋯⋯⋯⋱⋯−hn′dh1−hn′dh2−hn′dh3−hn′dh4−hn′dh5⋮−hn′dhn⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎞
H ′ d H E = ( − h 2 ′ d h 1 θ 1 0 − h 4 ′ d h 1 θ 2 h 3 ′ d h 1 θ 2 ⋯ 0 − h 2 ′ d h 1 θ 1 − h 4 ′ d h 2 θ 2 h 3 ′ d h 2 θ 2 ⋯ h 3 ′ d h 2 θ 1 − h 3 ′ d h 1 θ 1 − h 4 ′ d h 3 θ 2 0 ⋯ h 4 ′ d h 2 θ 1 − h 4 ′ d h 1 θ 1 0 − h 4 ′ d h 3 θ 2 ⋯ h 5 ′ d h 2 θ 1 − h 5 ′ d h 1 θ 1 h 5 ′ d h 4 θ 2 − h 5 ′ d h 3 θ 2 ⋯ ⋮ ⋱ h n ′ d h 2 θ 1 − h n ′ d h 1 θ 1 h n ′ d h 4 θ 2 − h n ′ d h 3 θ 2 ⋯ ) H'dHE=\begin{pmatrix} -h_2'dh_1\theta_1& 0& -h_4'dh_1\theta_2& h_3'dh_1\theta_2&\cdots \\ 0 &-h_2'dh_1\theta_1& -h_4'dh_2\theta_2& h_3'dh_2\theta_2&\cdots \\ h_3'dh_2\theta_1&-h_3'dh_1 \theta_1 & -h_4'dh_3\theta_2&0&\cdots \\ h_4'dh_2\theta_1&-h_4'dh_1\theta_1& 0& -h_4'dh_3\theta_2&\cdots \\ h_5'dh_2\theta_1&-h_5'dh_1\theta_1& h_5'dh_4\theta_2& -h_5'dh_3\theta_2&\cdots \\ \vdots & & & &\ddots\\ h_n'dh_2\theta_1& -h_n'dh_1\theta_1&h_n'dh_4\theta_2& -h_n'dh_3\theta_2& \cdots \end{pmatrix} H′dHE=⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎛−h2′dh1θ10h3′dh2θ1h4′dh2θ1h5′dh2θ1⋮hn′dh2θ10−h2′dh1θ1−h3′dh1θ1−h4′dh1θ1−h5′dh1θ1−hn′dh1θ1−h4′dh1θ2−h4′dh2θ2−h4′dh3θ20h5′dh4θ2hn′dh4θ2h3′dh1θ2h3′dh2θ20−h4′dh3θ2−h5′dh3θ2−hn′dh3θ2⋯⋯⋯⋯⋯⋱⋯⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎞
− E H ′ d H = ( h 2 ′ d h 1 θ 1 0 − h 3 ′ d h 2 θ 1 − h 4 ′ d h 2 θ 1 − h 5 ′ d h 2 θ 1 ⋯ − h n ′ d h 2 θ 1 0 h 2 ′ d h 1 θ 1 h 3 ′ d h 1 θ 1 h 4 ′ d h 1 θ 1 h 5 ′ d h 1 θ 1 ⋯ h n ′ d h 1 θ 1 h 4 ′ d h 1 θ 2 h 4 ′ d h 2 θ 2 h 4 ′ d h 3 θ 2 0 − h 5 ′ d h 4 θ 2 ⋯ − h n ′ d h 4 θ 2 − h 3 ′ d h 1 θ 2 − h 3 ′ d h 2 θ 2 0 h 4 ′ d h 3 θ 2 h 5 ′ d h 3 θ 2 ⋯ h n ′ d h 3 θ 2 ⋮ ⋱ ⋮ ) -EH'dH=\begin{pmatrix} h_2'dh_1\theta_1& 0 & -h_3'dh_2\theta_1& -h_4'dh_2\theta_1&-h_5'dh_2\theta_1&\cdots & -h_n'dh_2 \theta_1\\ 0& h_2'dh_1\theta_1& h_3'dh_1\theta_1& h_4'dh_1\theta_1& h_5'dh_1\theta_1&\cdots & h_n'dh_1\theta_1\\ h_4'dh_1\theta_2& h_4'dh_2\theta_2& h_4'dh_3\theta_2& 0& -h_5'dh_4\theta_2& \cdots & -h_n'dh_4\theta_2\\ -h_3'dh_1\theta_2 & -h_3'dh_2\theta_2& 0& h_4'dh_3\theta_2& h_5'dh_3\theta_2&\cdots & h_n'dh_3\theta_2 \\ \vdots & & & & &\ddots&\vdots \end{pmatrix} −EH′dH=⎝⎜⎜⎜⎜⎜⎛h2′dh1θ10h4′dh1θ2−h3′dh1θ2⋮0h2′dh1θ1h4′dh2θ2−h3′dh2θ2−h3′dh2θ1h3′dh1θ1h4′dh3θ20−h4′dh2θ1h4′dh1θ10h4′dh3θ2−h5′dh2θ1h5′dh1θ1−h5′dh4θ2h5′dh3θ2⋯⋯⋯⋯⋱−hn′dh2θ1hn′dh1θ1−hn′dh4θ2hn′dh3θ2⋮⎠⎟⎟⎟⎟⎟⎞
d E = ( 0 − d θ 1 ⋯ d θ 1 0 ⋯ ⋯ 0 − d θ 2 ⋯ d θ 2 0 ⋮ ⋱ ⋮ ) dE=\begin{pmatrix} 0&-d\theta_1&\cdots \\ d\theta_1&0&\cdots \\ \cdots&&0&-d\theta_2& \\ \cdots&&d\theta_2 & 0 \\ \vdots & & & &\ddots&\vdots \end{pmatrix} dE=⎝⎜⎜⎜⎜⎜⎛0dθ1⋯⋯⋮−dθ10⋯⋯0dθ2−dθ20⋱⋮⎠⎟⎟⎟⎟⎟⎞
如果 n = 2 k n=2k n=2k,
H ′ d H E − E H ′ d H + d E = ( 0 − d θ 1 ∗ ⋯ d θ 1 0 ∗ ⋯ h 3 ′ d h 2 θ 1 + h 4 ′ d h 1 θ 2 − h 3 ′ d h 1 θ 1 + h 4 ′ d h 2 θ 2 0 − d θ 2 ⋯ h 4 ′ d h 2 θ 1 − h 3 ′ d h 1 θ 2 − h 4 ′ d h 1 θ 1 − h 3 ′ d h 2 θ 2 d θ 2 0 ⋯ ⋮ ⋱ h n − 1 ′ d h 2 θ 1 + h n ′ d h 1 θ k − h n − 1 ′ d h 1 θ 1 + h n ′ d h 2 θ k ⋯ h n ′ d h 2 θ 1 − h n − 1 ′ d h 1 θ k − h n ′ d h 1 θ 1 − h n − 1 ′ d h 2 θ k ⋯ ) \begin{aligned}&H'dHE-EH'dH+dE=\\&\begin{pmatrix}0& -d\theta_1&*&\cdots \\ d\theta_1 &0& *& &\cdots \\ h_3'dh_2\theta_1+h_4'dh_1\theta_2&-h_3'dh_1 \theta_1+h_4'dh_2\theta_2 & 0&-d\theta_2&\cdots \\ h_4'dh_2\theta_1-h_3'dh_1\theta_2&-h_4'dh_1\theta_1-h_3'dh_2\theta_2& d\theta_2& 0&\cdots \\ \vdots & & & &\ddots \\ h_{n-1}'dh_2\theta_1+h_n'dh_1\theta_k&-h_{n-1}'dh_1 \theta_1+h_n'dh_2\theta_k & \cdots \\ h_n'dh_2\theta_1-h_{n-1}'dh_1\theta_k&-h_n'dh_1\theta_1-h_{n-1}'dh_2\theta_k&\cdots \\ \end{pmatrix} \end{aligned} H′dHE−EH′dH+dE=⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎛0dθ1h3′dh2θ1+h4′dh1θ2h4′dh2θ1−h3′dh1θ2⋮hn−1′dh2θ1+hn′dh1θkhn′dh2θ1−hn−1′dh1θk−dθ10−h3′dh1θ1+h4′dh2θ2−h4′dh1θ1−h3′dh2θ2−hn−1′dh1θ1+hn′dh2θk−hn′dh1θ1−hn−1′dh2θk∗∗0dθ2⋯⋯⋯−dθ20⋯⋯⋯⋱⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎞
如果 n = 2 k + 1 n=2k+1 n=2k+1,
H ′ d H E − E H ′ d H + d E = ( 0 − d θ 1 ∗ ⋯ d θ 1 0 ∗ ⋯ h 3 ′ d h 2 θ 1 + h 4 ′ d h 1 θ 2 − h 3 ′ d h 1 θ 1 + h 4 ′ d h 2 θ 2 0 − d θ 2 ⋯ h 4 ′ d h 2 θ 1 − h 3 ′ d h 1 θ 2 − h 4 ′ d h 1 θ 1 − h 3 ′ d h 2 θ 2 d θ 2 0 ⋯ ⋮ ⋱ h n − 1 ′ d h 2 θ 1 − h n − 2 ′ d h 1 θ k − h n − 1 ′ d h 1 θ 1 − h n − 2 ′ d h 2 θ k ⋯ h n ′ d h 2 θ 1 − h n ′ d h 1 θ 1 h n ′ d h 4 θ 2 − h n ′ d h 3 θ 2 ⋯ ) \begin{aligned}&H'dHE-EH'dH+dE=\\&\begin{pmatrix}0& -d\theta_1&*&\cdots \\ d\theta_1 &0& *& &\cdots \\ h_3'dh_2\theta_1+h_4'dh_1\theta_2&-h_3'dh_1 \theta_1+h_4'dh_2\theta_2 & 0&-d\theta_2&\cdots \\ h_4'dh_2\theta_1-h_3'dh_1\theta_2&-h_4'dh_1\theta_1-h_3'dh_2\theta_2& d\theta_2& 0&\cdots \\ \vdots & & & &\ddots \\ h_{n-1}'dh_2\theta_1-h_{n-2}'dh_1\theta_k&-h_{n-1}'dh_1 \theta_1-h_{n-2}'dh_2\theta_k & \cdots \\ h_n'dh_2\theta_1& -h_n'dh_1\theta_1&h_n'dh_4\theta_2& -h_n'dh_3\theta_2& \cdots \end{pmatrix} \end{aligned} H′dHE−EH′dH+dE=⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎛0dθ1h3′dh2θ1+h4′dh1θ2h4′dh2θ1−h3′dh1θ2⋮hn−1′dh2θ1−hn−2′dh1θkhn′dh2θ1−dθ10−h3′dh1θ1+h4′dh2θ2−h4′dh1θ1−h3′dh2θ2−hn−1′dh1θ1−hn−2′dh2θk−hn′dh1θ1∗∗0dθ2⋯hn′dh4θ2⋯−dθ20−hn′dh3θ2⋯⋯⋯⋱⋯⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎞
其中对角线上部 * 表示该项已在对角线下出现过,对角线上部的此项对 ( d S ) (dS) (dS) 无贡献。
当 n = 2 k + 1 n=2k+1 n=2k+1,
( d S ) = ∏ i = 1 k θ i 2 g 2 ( Θ ) ∏ i = 1 k d θ i ( ( H ′ d H ) ) (dS)=\prod_{i=1}^k\theta_i^2 g_2(\Theta) \prod_{i=1}^k d\theta_i((H'dH)) (dS)=i=1∏kθi2g2(Θ)i=1∏kdθi((H′dH))
当 n = 2 k n=2k n=2k,
( d S ) = g 2 ( Θ ) ∏ i = 1 k d θ i ( ( H ′ d H ) ) (dS)= g_2(\Theta) \prod_{i=1}^k d\theta_i((H'dH)) (dS)=g2(Θ)i=1∏kdθi((H′dH))
这里
g 2 ( Θ ) = ∏ i < j k ( θ i + θ j ) 2 ( θ i − θ j ) 2 ( ( H ′ d H ) ) = ∏ i < j , i ≠ j − 1 , j even n h i ′ d h j \begin{aligned}&g_2(\Theta)= \prod_{i<j}^k (\theta_i+\theta_j)^2(\theta_i-\theta_j)^2\\&((H'dH))=\prod_{i<j,i\neq j-1,j \hspace{0.2em} \textrm{even}}^n h_i'dh_j\end{aligned} g2(Θ)=i<j∏k(θi+θj)2(θi−θj)2((H′dH))=i<j,i=j−1,jeven∏nhi′dhj
此处所定义 ( ( H ′ d H ) ) ((H'dH)) ((H′dH)) 不同于 ( H ′ d H ) (H'dH) (H′dH)。
- 设正交矩阵 R = H D H ′ R=HDH' R=HDH′, H H H 和 D D D 分别是正交阵和块对角阵,下面求 ( R ′ d R ) (R'dR) (R′dR) (see Anderson, G. A. 1965)
R ′ d R = H D ′ H ′ ( d H D H ′ + H d D H ′ + H D d H ′ ) R'dR=HD'H'(dHDH'+HdDH'+HDdH') R′dR=HD′H′(dHDH′+HdDH′+HDdH′)
( R ′ d R ) = ( H ′ R ′ d R H ) = ( D ′ H ′ d H D + D ′ d D + D ′ D d H ′ H ) (R'dR)=(H'R'dRH)=(D'H'dHD+D'dD+D'DdH'H) (R′dR)=(H′R′dRH)=(D′H′dHD+D′dD+D′DdH′H)
D ′ D d H ′ H = d H ′ H = − H ′ d H D'DdH'H=dH'H=-H'dH D′DdH′H=dH′H=−H′dH
D i ′ d D i = ( 0 − d θ i d θ i 0 ) D_i'dD_i=\begin{pmatrix}0 & -d\theta_i\\ d\theta_i&0 \end{pmatrix} Di′dDi=(0dθi−dθi0)
D ′ H ′ d H D = ( 0 ∗ ∗ ⋯ h 2 ′ d h 1 0 ∗ ⋯ A 1 A 2 0 ∗ ⋯ A 3 A 4 h 4 ′ d h 3 0 ⋯ ⋮ ⋱ A 5 A 6 ⋯ ) \begin{aligned}&D'H'dHD=\\&\begin{pmatrix}0& *&*&\cdots \\ h_2'dh_1 &0& *& &\cdots \\ A1& A2& 0& *&\cdots \\ A3& A4 & h_4'dh_3&0&\cdots \\ \vdots & & & &\ddots \\ A5&A6&\cdots \\ \end{pmatrix} \end{aligned} D′H′dHD=⎝⎜⎜⎜⎜⎜⎜⎜⎛0h2′dh1A1A3⋮A5∗0A2A4A6∗∗0h4′dh3⋯⋯∗0⋯⋯⋯⋱⎠⎟⎟⎟⎟⎟⎟⎟⎞
这里
A 1 = ( h 3 ′ d h 1 cos θ 2 + h 4 ′ d h 1 sin θ 2 ) cos θ 1 + ( h 3 ′ d h 2 cos θ 2 + h 4 ′ d h 2 sin θ 2 ) sin θ 1 A 2 = − ( h 3 ′ d h 1 cos θ 2 + h 4 ′ d h 1 sin θ 2 ) sin θ 1 + ( h 3 ′ d h 2 cos θ 2 + h 4 ′ d h 2 sin θ 2 ) cos θ 1 A 3 = ( − h 3 ′ d h 1 sin θ 2 + h 4 ′ d h 1 cos θ 2 ) cos θ 1 + ( − h 3 ′ d h 2 sin θ 2 + h 4 ′ d h 2 cos θ 2 ) sin θ 1 A 4 = ( h 3 ′ d h 1 sin θ 2 − h 4 ′ d h 1 cos θ 2 ) sin θ 1 + ( − h 3 ′ d h 2 sin θ 2 + h 4 ′ d h 2 cos θ 2 ) cos θ 1 \begin{aligned} A1&=(h_3'dh_1\cos\theta_2+h_4'dh_1\sin\theta_2)\cos\theta_1+(h_3'dh_2\cos\theta_2+h_4'dh_2\sin\theta_2)\sin\theta_1\\ A2&=-(h_3'dh_1\cos\theta_2+h_4'dh_1\sin\theta_2)\sin\theta_1+(h_3'dh_2\cos\theta_2+h_4'dh_2\sin\theta_2)\cos\theta_1\\ A3&=(-h_3'dh_1\sin\theta_2+h_4'dh_1\cos\theta_2)\cos\theta_1+(-h_3'dh_2\sin\theta_2+h_4'dh_2\cos\theta_2)\sin\theta_1\\ A4&=(h_3'dh_1\sin\theta_2-h_4'dh_1\cos\theta_2)\sin\theta_1+(-h_3'dh_2\sin\theta_2+h_4'dh_2\cos\theta_2)\cos\theta_1 \end{aligned} A1A2A3A4=(h3′dh1cosθ2+h4′dh1sinθ2)cosθ1+(h3′dh2cosθ2+h4′dh2sinθ2)sinθ1=−(h3′dh1cosθ2+h4′dh1sinθ2)sinθ1+(h3′dh2cosθ2+h4′dh2sinθ2)cosθ1=(−h3′dh1sinθ2+h4′dh1cosθ2)cosθ1+(−h3′dh2sinθ2+h4′dh2cosθ2)sinθ1=(h3′dh1sinθ2−h4′dh1cosθ2)sinθ1+(−h3′dh2sinθ2+h4′dh2cosθ2)cosθ1
如果 n = 2 k n=2k n=2k,
A 5 = ( − h n − 1 ′ d h 1 sin θ k + h n ′ d h 1 cos θ k ) cos θ 1 + ( h n − 1 ′ d h 2 sin θ k + h n ′ d h 2 cos θ k ) sin θ 1 A 6 = ( h n − 1 ′ d h 1 sin θ k − h n ′ d h 1 cos θ k ) sin θ 1 + ( − h n − 1 ′ d h 2 sin θ k + h n ′ d h 2 cos θ k ) cos θ 1 \begin{aligned} A5&=(-h_{n-1}'dh_1\sin\theta_k+h_n'dh_1\cos\theta_k)\cos\theta_1+(h_{n-1}'dh_2\sin\theta_k+h_n'dh_2\cos\theta_k)\sin\theta_1\\ A6&=(h_{n-1}'dh_1\sin\theta_k-h_n'dh_1\cos\theta_k)\sin\theta_1+(-h_{n-1}'dh_2\sin\theta_k+h_n'dh_2\cos\theta_k)\cos\theta_1 \end{aligned} A5A6=(−hn−1′dh1sinθk+hn′dh1cosθk)cosθ1+(hn−1′dh2sinθk+hn′dh2cosθk)sinθ1=(hn−1′dh1sinθk−hn′dh1cosθk)sinθ1+(−hn−1′dh2sinθk+hn′dh2cosθk)cosθ1
如果 如果 n = 2 k + 1 n=2k+1 n=2k+1,
A 5 = h n ′ d h 1 cos θ 1 + h n ′ d h 2 sin θ 1 A 6 = h n ′ d h 2 cos θ 1 − h n ′ d h 1 sin θ 1 \begin{aligned} A5&= h_n'dh_1\cos\theta_1+h_n'dh_2\sin\theta_1 \\ A6&=h_n'dh_2\cos\theta_1-h_n'dh_1\sin\theta_1 \end{aligned} A5A6=hn′dh1cosθ1+hn′dh2sinθ1=hn′dh2cosθ1−hn′dh1sinθ1
D ′ H ′ d H D + D ′ d D + D ′ D d H ′ H = ( 0 ∗ ∗ ⋯ d θ 1 0 ∗ ⋯ A 1 − h 3 ′ d h 1 A 2 − h 3 ′ d h 2 0 ∗ ⋯ A 3 − h 4 ′ d h 1 A 4 − h 4 ′ d h 2 d θ 2 0 ⋯ ⋮ ⋱ A 5 − h n ′ d h 1 A 6 − h n ′ d h 2 ⋯ ) \begin{aligned}&D'H'dHD+D'dD+D'DdH'H=\\&\begin{pmatrix}0& *&*&\cdots \\ d\theta_1 &0& *& &\cdots \\ A1-h_3'dh1& A2-h_3'dh2& 0& *&\cdots \\ A3-h_4'dh1& A4-h_4'dh2 & d\theta_2&0&\cdots \\ \vdots & & & &\ddots \\ A5-h_n'dh1&A6-h_n'dh2&\cdots \\ \end{pmatrix} \end{aligned} D′H′dHD+D′dD+D′DdH′H=⎝⎜⎜⎜⎜⎜⎜⎜⎛0dθ1A1−h3′dh1A3−h4′dh1⋮A5−hn′dh1∗0A2−h3′dh2A4−h4′dh2A6−hn′dh2∗∗0dθ2⋯⋯∗0⋯⋯⋯⋱⎠⎟⎟⎟⎟⎟⎟⎟⎞
如果 n = 2 k + 1 n=2k+1 n=2k+1,
( R ′ d R ) = ∏ i = 1 k 4 sin ( θ i / 2 ) g 1 ( Θ ) ∏ i = 1 k d θ i ( ( H ′ d H ) ) (R'dR)=\prod_{i=1}^k4\sin(\theta_i/2)g_1(\Theta)\prod_{i=1}^kd\theta_i((H'dH)) (R′dR)=i=1∏k4sin(θi/2)g1(Θ)i=1∏kdθi((H′dH))
如果 n = 2 k n=2k n=2k,
( R ′ d R ) = g 1 ( Θ ) ∏ i = 1 k d θ i ( ( H ′ d H ) ) (R'dR)=g_1(\Theta)\prod_{i=1}^kd\theta_i((H'dH)) (R′dR)=g1(Θ)i=1∏kdθi((H′dH))
这里
g 1 ( Θ ) = ∏ i < j k { 16 sin 2 [ ( θ i + θ j ) / 2 ] sin 2 [ ( θ i − θ j ) / 2 ] } ( ( H ′ d H ) ) = ∏ i < j , i ≠ j − 1 , j even n h i ′ d h j \begin{aligned}&g_1(\Theta)=\prod_{i<j}^k\{16\sin^2[(\theta_i+\theta_j)/2]\sin^2[(\theta_i-\theta_j)/2]\}\\&((H'dH))=\prod_{i<j,i\neq j-1,j \hspace{0.2em} \textrm{even}}^n h_i'dh_j\end{aligned} g1(Θ)=i<j∏k{16sin2[(θi+θj)/2]sin2[(θi−θj)/2]}((H′dH))=i<j,i=j−1,jeven∏nhi′dhj
当 n = 2 k + 1 n=2k+1 n=2k+1,
( R ′ d R ) = ∏ i = 1 k [ sin ( θ i / 2 ) / ( θ i / 2 ) ] 2 f ( Θ ) ∏ i < j k d s i , j (R'dR)=\prod_{i=1}^k\left[\sin(\theta_i/2)/(\theta_i/2)\right]^2f(\Theta)\prod_{i<j}^kds_{i,j} (R′dR)=i=1∏k[sin(θi/2)/(θi/2)]2f(Θ)i<j∏kdsi,j
当 n = 2 k n=2k n=2k,
( R ′ d R ) = f ( Θ ) ∏ i < j k d s i , j (R'dR)=f(\Theta)\prod_{i<j}^kds_{i,j} (R′dR)=f(Θ)i<j∏kdsi,j
这里
f ( Θ ) = ∏ i < j k { [ sin ( ( θ i + θ j ) / 2 ) / ( ( θ i + θ j ) / 2 ] [ sin ( ( θ i − θ j ) / 2 ) / ( ( θ i − θ j ) / 2 ] } 2 f(\Theta)=\prod_{i<j}^k \{[\sin((\theta_i+\theta_j)/2)/((\theta_i+\theta_j)/2][\sin((\theta_i-\theta_j)/2)/((\theta_i-\theta_j)/2]\}^2 f(Θ)=i<j∏k{[sin((θi+θj)/2)/((θi+θj)/2][sin((θi−θj)/2)/((θi−θj)/2]}2
Reference:
[1] J. Gallier and D. Xu. COMPUTING EXPONENTIALS OF SKEW-SYMMETRIC MATRICES AND LOGARITHMS OF ORTHOGONAL MATRICES International Journal of Robotics and Automation, Vol. 17, No. 4, 2002
[2] R.A. Horn & C.R. Johnson, Matrix analysis (Cambridge, UK: Cambridge University Press, 1990).
[3] Anderson, G. A. (1965). An asymptotic expansion for the distribution of the latent roots of the estimated covariance matrix. Ann. Math. Statist., 36, 1153-1173.
正交矩阵的指数表示和雅可比相关推荐
- 雅可比旋转求解对称二维矩阵的特征值和特征向量
问题描述: 给定一个矩阵,如下: A=[a11a21a12a22] A=\begin{bmatrix} a_{11}&a_{12}\\ a_{21}& a_{22} \end{bmat ...
- “用于无监督图像生成解耦的正交雅可比正则化”论文解读
Tikhonov regularization terms https://blog.csdn.net/jiejinquanil/article/details/50411617 本文是对博客http ...
- 【现代机器人学】基于指数积的机械臂逆运动学
0 引言 基于指数积的正运动学 现代机器人学名词概念 有了以上的基础,我们现在利用指数积来对机器人的逆运动学进行求解,有一点需要注意,需要先对机器人进行指数积的正运动学建模,然后才能利 ...
- 在 VSLAM 的后端优化中的重投影误差的雅可比计算详细推导
对于相机位姿的变换可以通过旋转矩阵或者四元数进行表示,对于旋转矩阵的定义满足: R{∣R∣=1RRT=IR \begin{cases} |R| = 1 \\ RR^T = I\\ \end{cases ...
- 线性代数matlab求一个正交矩阵,线性代数求一个正交矩阵P,是P^-1AP= – 手机爱问...
2018-05-12 请问如何用雅克比法求解矩阵特征值和特征向量 雅可比方法的基本思想是通过一系列的由平面旋转矩阵构成的正交变换将实对称矩阵逐步化为对角阵,从而得到 的全部特征值及其相应的特征向量.首 ...
- 3D图形:矩阵的行列式,矩阵的逆、正交矩阵、齐次矩阵
前言 在前面我们说到关于矩阵的一些计算知识,相信大家已经觉得进入了水深火热之中了,那么为了让大家感到更加刺激的视觉体验和感官体验,这一篇博客,我将对矩阵的行列式,矩阵的逆,正交矩阵,齐次矩阵进行探讨研 ...
- 雅可比主对角线(Jacobi diagonalization)化求对称矩阵的特征值(python,数值积分)
第三十篇 雅可比主对角线化求对称矩阵的特征值 对于标准特征值方程 由特征值问题编程基础可知,对于任何非0解矩阵[P],标准方程可以转化为具有相同特征值的方程 其中 这种转换技术的关键核心在于[A *] ...
- 雅可比旋转(Jacobi法)求对称矩阵的特征值和特征向量
雅可比方法 该方法是求解对称矩阵全部特征值和特征向量的一种方法,它基于以下结论: ①任何实对称矩阵A可以通过正交相似变换成对角型,即存在正交矩阵Q,使得 QTAQ=diag(λ1,λ2,-,λn)Q^ ...
- matlab 雅各比符号,雅可比符号计算
x2 ? 286(mod563) 解:不用考虑563是否为素数,直接计算雅可比符号: 563?563?1 143?1 563?1 ? 563 ? ? 286 ? ? 2 ?? 143 ? 8 2 2 ...
最新文章
- Google Android开发精华教程
- java处理excel(java使用Apache POI处理Excel)
- reactjs组件优化:setState的反复render问题
- ivew 的ajax,iView-Upload组件分析
- Android打开谷歌应用,谷歌确认 Android 12 新增剪贴板访问提醒,将在 Beta 2 上线
- 剖析供应链攻击的防范
- python离散化方法_python中字符串离散化的例子
- Java开发实战经典 目录
- python换行输出三个数中最大数_关于Python 3中print函数的换行详解
- MTK:串口学习和代码调试
- 如何使用@PostConstruct初始化敏感词库和hutool过滤敏感词信息
- 【教程】Win10安装SQLServer2005出现服务启动失败的问题解决
- 解决tar.bz2解压报错
- APU工业控制领域应用
- 金蝶EAS,序时簿ListUI只允许选择一行或至少选择一行记录
- 信息化15年规划推动中国信息化步入深水区
- 高德地图中缩放级别(zoom)和比例尺(getScalePerPixel)之间的计算关系
- 风控贷款---年龄因素
- 联想ghost重装系统_如何使用ghost手动安装系统_手动ghost安装系统图文步骤
- php如何运行vbs文件,HTML_用vbs实现在启动 Windows 资源管理器时打开特定文件夹,my-script.vbs c:\scripts 在文件夹 - phpStudy...
热门文章
- 一阶动态电路的响应测试(一)
- tiktok x-tt-params
- 【聆听】泰戈尔诗集(六)
- 七月在线深度学习错题刷
- linux 卷标,linux 卷标设置与管理
- 这份“插件英雄榜Top40”才是Chrome的正确打开方式!(Github7000+ Stars)
- break在python中什么意思_Python中break 关键字用途说明
- ata考证和计算机一级b,ATA
- 最高估值2641亿港元,京东物流能攻下“物流界”半壁江山吗?
- 深度学习——早停法(Early Stopping)