四元数
2023-04-09 22:05:35 0 举报AI智能生成
四元数
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四元数
定义与性质
四元数的表示
代数形式<font face="KaTeX_Main, Times New Roman, serif"><span style="font-size: 18.15px; white-space: nowrap;"> </span></font><span class="equation-text" contenteditable="false" data-index="0" data-equation="\begin{aligned}&q=a+b i+c j+d k, \quad(a, b, c, d \in \mathbb{R})\\&i^2=j^2=k^2=i j k=-1\end{aligned}"><span></span><span></span></span>
向量表示 <span class="equation-text" contenteditable="false" data-index="0" data-equation="q=\left[\begin{array}{l}a \\b \\c \\d\end{array}\right]"><span></span><span></span></span>
标量和向量的有序对形式 <span class="equation-text" contenteditable="false" data-index="0" data-equation="q=[s, \mathbf{v}] . \quad\left(\mathbf{v}=\left[\begin{array}{l}x \\y \\z\end{array}\right], s, x, y, z \in \mathbb{R}\right)"><span></span><span></span></span>
模长<br><span class="equation-text" contenteditable="false" data-index="0" data-equation="\|q\|=\sqrt{a^{2}+b^{2}+c^{2}+d^{2}}"><span></span><span></span></span><br> <span class="equation-text" data-index="1" data-equation="\begin{aligned}\|q\| & =\sqrt{s^{2}+\|\mathbf{v}\|^{2}} \\& =\sqrt{s^{2}+\mathbf{v} \cdot \mathbf{v}} \quad\left(\mathbf{v} \cdot \mathbf{v}=\|\mathbf{v}\|^{2}\right)\end{aligned}" contenteditable="false"><span></span><span></span></span>
加法和减法
标量乘法
四元数乘法
不遵守交换律
代数形式<br>
矩阵形式
<span class="equation-text" data-index="0" data-equation="q_1 q_{2}=\left[\begin{array}{cccc}a & -b & -c & -d \\b & a & -d & c \\c & d & a & -b \\d & -c & b & a\end{array}\right]\left[\begin{array}{l}e \\f \\g \\h\end{array}\right] ." contenteditable="false"><span></span><span></span></span><br><span class="equation-text" contenteditable="false" data-index="1" data-equation="q_{2} q_{1}=\left[\begin{array}{cccc}a & -b & -c & -d \\b & a & d & -c \\c & -d & a & b \\d & c & -b & a\end{array}\right]\left[\begin{array}{l}e \\f \\g \\h\end{array}\right]"><span></span><span></span></span><br>
Graßmann 积
纯四元数
如果有两个纯四元数<span class="equation-text" data-index="0" data-equation="v=[0,\mathbf{v}],u=[0,\mathbf{u}]" contenteditable="false"><span></span><span></span></span>，那么<br><span class="equation-text" contenteditable="false" data-index="1" data-equation="\begin{aligned}vu&=[0-\mathbf{v}\cdot\mathbf{u},0+\mathbf{v}\times\mathbf{u}]\\ &=[-\mathbf{v\cdot}\mathbf{u},\mathbf{v\times u}].\end{aligned}"><span></span><span></span></span><br>
逆和共轭
将乘法的逆运算定义为 𝑝𝑞-1 或者 𝑞-1𝑝，注意它们的结果一般是不同的.<br><span class="equation-text" contenteditable="false" data-index="0" data-equation="q q^{-1}=q^{-1} q=1 \quad(q \neq 0)"><span></span><span></span></span><br>
<span class="equation-text" data-index="0" data-equation="𝑞=𝑎+𝑏𝑖+𝑐𝑗+𝑑𝑘" contenteditable="false"><span></span><span></span></span> 的共轭为<br><span class="equation-text" contenteditable="false" data-index="1" data-equation="𝑞^∗ =𝑎−𝑏𝑖−𝑐𝑗−𝑑𝑘"><span></span><span></span></span>
<span class="equation-text" data-index="0" data-equation="q=\left[s,\mathbf{v}\right] " contenteditable="false"><span></span><span></span></span>的共轭为<br><span class="equation-text" contenteditable="false" data-index="1" data-equation="q^{*}=[s,-\mathbf{v}]"><span></span><span></span></span><br>
<span class="equation-text" data-index="0" data-equation="\begin{aligned}qq^*&=[s^2+\mathbf{v}\cdot\mathbf{v},\mathbf{0}]\\ &=s^2+x^2+y^2+z^2\\ &=\|q\|^2.\end{aligned}" contenteditable="false"><span></span><span></span></span><br><b>结果是一个实数，而它正是四元数模长的平方</b><br>
<span class="equation-text" data-index="0" data-equation="q^*q=qq^*" contenteditable="false"><span></span><span></span></span><br><b>这个特殊的乘法是遵守交换律的</b><br>
<span class="equation-text" data-index="0" data-equation="q^{-1}=\frac{q^*}{\|q\|^2}" contenteditable="false"><span></span><span></span></span><br><b>用这种办法寻找一个四元数的逆会非常高效，<br>我们只需要将一个四元数的虚部改变符号，<br>除以它模长的平方就能获得这个四元数的逆了<br><br><span class="equation-text" contenteditable="false" data-index="1" data-equation="\begin{array}{rlrl}q q^{-1} & =1 & \\q^* q q^{-1} & =q^* & & \\\left(q^* q\right) q^{-1} & =q^* & & \text { 等式两边同时左乘以 } \left.q^*\right) \\\|q\|^2 \cdot q^{-1} & =q^* & & \left(q^* q=\|q\|^2\right) \\q^{-1} & =\frac{q^*}{\|q\|^2} . &\end{array}"><span></span><span></span></span> </b><br>
四元数与3d旋转
因为所有的旋转四元数的实部都只是一个角度的余弦值，假设有一个单位四元数 <span class="equation-text" data-index="0" data-equation="q=[a, \mathbf{b}]" contenteditable="false"><span></span><span></span></span>，<br>其对应的旋转角度为，<span class="equation-text" data-index="1" data-equation="\frac{\theta}{2}=\cos ^{-1}(a)" contenteditable="false"><span></span><span></span></span>，旋转轴为 <span class="equation-text" contenteditable="false" data-index="2" data-equation="\mathbf{u}=\frac{\mathbf{b}}{\sin \left(\cos ^{-1}(a)\right)}"><span></span><span></span></span>
<b>虽然 3D 旋转的矩阵形式可能不如四元数形式简单，而且占用更多的空间，<br>但是对于大批量的变换，使用预计算好的矩阵是比四元数乘法更有效率的</b>
复数
复数的坐标表示<br>向量表示<span class="equation-text" contenteditable="false" data-index="0" data-equation="z=\left[\begin{array}{l}a \\b\end{array}\right]"><span></span><span></span></span>
复数的性质<br>
加法<br>
乘法<br>
复数的乘法等价于矩阵与向量相乘 <span class="equation-text" contenteditable="false" data-index="0" data-equation="\begin{aligned} z_{1} z_{2}= & a c-b d+ \\ & (b c+a d) i \\ = & {\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]\left[\begin{array}{l}c \\ d\end{array}\right] }\end{aligned}"><span></span><span></span></span><br>
复数的矩阵形式 <span class="equation-text" contenteditable="false" data-index="0" data-equation="\left[\begin{array}{cc}a & -b \\b & a\end{array}\right]"><span></span><span></span></span><br>
复数的乘法等价于矩阵乘法<span class="equation-text" contenteditable="false" data-index="0" data-equation="\begin{aligned}z_1 z_2 & =\left[\begin{array}{cc}a & -b \\b & a\end{array}\right]\left[\begin{array}{cc}c & -d \\d & c\end{array}\right] \\& =\left[\begin{array}{cc}a c-b d & -(b c+a d) \\b c+a d & a c-b d\end{array}\right]\end{aligned}"><span></span><span></span></span>
注意，复数的相乘是满足交换律的
特殊复数的矩阵形式<span class="equation-text" contenteditable="false" data-index="0" data-equation="\begin{aligned}& 1=\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]=I \\& i=\left[\begin{array}{cc}0 & -1 \\1 & 0\end{array}\right] .\end{aligned}"><span></span><span></span></span><span class="equation-text" contenteditable="false" data-index="1" data-equation="i^2=i \cdot i=\left[\begin{array}{cc}0 & -1 \\1 & 0\end{array}\right]\left[\begin{array}{cc}0 & -1 \\1 & 0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\0 & -1\end{array}\right]=-I=-1"><span></span><span></span></span>
模长与共轭<br>
<span class="equation-text" contenteditable="false" data-index="0" data-equation="\|z\|=\sqrt{a^2+b^2}"><span></span><span></span></span>
<span class="equation-text" contenteditable="false" data-index="0" data-equation="\bar{z}=a-b i"><span></span><span></span></span>
<span class="equation-text" contenteditable="false" data-index="0" data-equation="\|z\|=\sqrt{z \bar{z}}"><span></span><span></span></span>
复数相乘与2D旋转<br>
复数的相乘是旋转与缩放变换的复合<br>
<span class="equation-text" contenteditable="false" data-index="0" data-equation="\left[\begin{array}{cc}a & -b \\b & a\end{array}\right]=\sqrt{a^2+b^2}\left[\begin{array}{cc}\frac{a}{\sqrt{a^2+b^2}} & \frac{-b}{\sqrt{a^2+b^2}} \\\frac{b}{\sqrt{a^2+b^2}} & \frac{a}{\sqrt{a^2+b^2}}\end{array}\right]"><span></span><span></span></span>
<span class="equation-text" contenteditable="false" data-index="0" data-equation="\begin{aligned}{\left[\begin{array}{cc}a & -b \\b & a\end{array}\right] } & =\sqrt{a^2+b^2}\left[\begin{array}{cc}\cos (\theta) & -\sin (\theta) \\\sin (\theta) & \cos (\theta)\end{array}\right] \\& =\|z\|\left[\begin{array}{cc}\cos (\theta) & -\sin (\theta) \\\sin (\theta) & \cos (\theta)\end{array}\right] \\& =\|z\| \cdot I\left[\begin{array}{cc}\cos (\theta) & -\sin (\theta) \\\sin (\theta) & \cos (\theta)\end{array}\right] \\& =\left[\begin{array}{cc}\|z\| & 0 \\0 & \|z\|\end{array}\right]\left[\begin{array}{cc}\cos (\theta) & -\sin (\theta) \\\sin (\theta) & \cos (\theta)\end{array}\right]\end{aligned}"><span></span><span></span></span>
如果有一个复数 <span class="equation-text" data-index="0" data-equation="z=a+b i" contenteditable="false"><span></span><span></span></span>，那么 𝑧 与任意一个复数 𝑐 相乘<br>都会将 𝑐 逆时针旋转<span class="equation-text" data-index="1" data-equation="\theta=\operatorname{atan} 2(b, a)" contenteditable="false"><span></span><span></span></span>度，并缩放 <span class="equation-text" contenteditable="false" data-index="2" data-equation="\|z\|=\sqrt{a^2+b^2}"><span></span><span></span></span>
二维旋转公式
<span class="equation-text" contenteditable="false" data-index="0" data-equation="\mathbf{v}^{\prime}=\left[\begin{array}{cc}\cos (\theta) & -\sin (\theta) \\\sin (\theta) & \cos (\theta)\end{array}\right] \mathbf{v}"><span></span><span></span></span>
<span class="equation-text" contenteditable="false" data-index="0" data-equation="\begin{aligned}v^{\prime} & =z v \\& =(\cos (\theta)+i \sin (\theta)) v\end{aligned}"><span></span><span></span></span>
极坐标<br>
复数的极坐标表示<br>
欧拉公式 <span class="equation-text" contenteditable="false" data-index="0" data-equation="\cos (\theta)+i \sin (\theta)=e^{i \theta}"><span></span><span></span></span>
<span class="equation-text" contenteditable="false" data-index="0" data-equation="\begin{aligned}z & =\|z\|\left[\begin{array}{cc}\cos (\theta) & -\sin (\theta) \\\sin (\theta) & \cos (\theta)\end{array}\right] \\& =\|z\|(\cos (\theta)+i \sin (\theta)) \\& =\|z\| e^{i \theta} .\end{aligned}"><span></span><span></span></span>
极坐标下的旋转 <span class="equation-text" contenteditable="false" data-index="0" data-equation="v^{\prime}=e^{i \theta} v"><span></span><span></span></span><br>
旋转的复合
当对两个 2D 旋转进行复合时，所得到的变换 𝑧 net 仍是一个旋转， 而且与施加的次序无关．<br>这个等效变换的旋转角是 𝑧1 与 𝑧2 旋转角之和．
三维空间中的旋转
旋转的分解
将向量分解为平行于旋转轴和垂直于旋转轴两个分量
分别绕旋转轴进行旋转，然后合并
<br>
v∥ 的旋转
v⊥ 的旋转
v 的旋转
Rodrigues’ Rotation Formula: <br>3D 空间中任意一个 v 沿着单位向量 u 旋转 θ 角度之后的 v′ 为<br> <span class="equation-text" contenteditable="false" data-index="0" data-equation="\mathbf{v}^{\prime}=\cos (\theta) \mathbf{v}+(1-\cos (\theta))(\mathbf{u} \cdot \mathbf{v}) \mathbf{u}+\sin (\theta)(\mathbf{u} \times \mathbf{v})"><span></span><span></span></span>
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