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.
