四元数

代数形式<font face="KaTeX_Main, Times New Roman, serif"><span style="font-size: 18.15px; white-space: nowrap;">&nbsp;</span></font><span class="equation-text" contenteditable="false" data-index="0" data-equation="\begin{aligned}&amp;q=a+b i+c j+d k, \quad(a, b, c, d \in \mathbb{R})\\&amp;i^2=j^2=k^2=i j k=-1\end{aligned}"><span></span><span></span></span>

向量表示&nbsp;<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>

不遵守交换律

代数形式<br>

矩阵形式

Graßmann&nbsp;积

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&amp;=[0-\mathbf{v}\cdot\mathbf{u},0+\mathbf{v}\times\mathbf{u}]\\&nbsp;&amp;=[-\mathbf{v\cdot}\mathbf{u},\mathbf{v\times&nbsp;u}].\end{aligned}"><span></span><span></span></span><br>

将乘法的逆运算定义为&nbsp;𝑝𝑞-1&nbsp;或者&nbsp;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&nbsp;q^{-1}=q^{-1}&nbsp;q=1&nbsp;\quad(q&nbsp;\neq&nbsp;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>&nbsp;的共轭为<br><span class="equation-text" contenteditable="false" data-index="1" data-equation="𝑞^∗&nbsp;=𝑎−𝑏𝑖−𝑐𝑗−𝑑𝑘"><span></span><span></span></span>

<span class="equation-text" data-index="0" data-equation="q=\left[s,\mathbf{v}\right]&nbsp;" 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>

&nbsp;<span class="equation-text" data-index="0" data-equation="\begin{aligned}qq^*&amp;=[s^2+\mathbf{v}\cdot\mathbf{v},\mathbf{0}]\\&nbsp;&amp;=s^2+x^2+y^2+z^2\\&nbsp;&amp;=\|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&nbsp;q^{-1}&nbsp;&amp;&nbsp;=1&nbsp;&amp;&nbsp;\\q^*&nbsp;q&nbsp;q^{-1}&nbsp;&amp;&nbsp;=q^*&nbsp;&amp;&nbsp;&amp;&nbsp;\\\left(q^*&nbsp;q\right)&nbsp;q^{-1}&nbsp;&amp;&nbsp;=q^*&nbsp;&amp;&nbsp;&amp;&nbsp;\text&nbsp;{&nbsp;等式两边同时左乘以&nbsp;}&nbsp;\left.q^*\right)&nbsp;\\\|q\|^2&nbsp;\cdot&nbsp;q^{-1}&nbsp;&amp;&nbsp;=q^*&nbsp;&amp;&nbsp;&amp;&nbsp;\left(q^*&nbsp;q=\|q\|^2\right)&nbsp;\\q^{-1}&nbsp;&amp;&nbsp;=\frac{q^*}{\|q\|^2}&nbsp;.&nbsp;&amp;\end{array}"><span></span><span></span></span>&nbsp;</b><br>

复数的乘法等价于矩阵与向量相乘&nbsp;&nbsp;<span class="equation-text" contenteditable="false" data-index="0" data-equation="\begin{aligned} z_{1} z_{2}= &amp; a c-b d+ \\ &amp; (b c+a d) i \\ = &amp; {\left[\begin{array}{cc}a &amp; -b \\ b &amp; 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&nbsp;<span class="equation-text" contenteditable="false" data-index="0" data-equation="\left[\begin{array}{cc}a&nbsp;&amp;&nbsp;-b&nbsp;\\b&nbsp;&amp;&nbsp;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&nbsp;z_2&nbsp;&amp;&nbsp;=\left[\begin{array}{cc}a&nbsp;&amp;&nbsp;-b&nbsp;\\b&nbsp;&amp;&nbsp;a\end{array}\right]\left[\begin{array}{cc}c&nbsp;&amp;&nbsp;-d&nbsp;\\d&nbsp;&amp;&nbsp;c\end{array}\right]&nbsp;\\&amp;&nbsp;=\left[\begin{array}{cc}a&nbsp;c-b&nbsp;d&nbsp;&amp;&nbsp;-(b&nbsp;c+a&nbsp;d)&nbsp;\\b&nbsp;c+a&nbsp;d&nbsp;&amp;&nbsp;a&nbsp;c-b&nbsp;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}&amp;&nbsp;1=\left[\begin{array}{ll}1&nbsp;&amp;&nbsp;0&nbsp;\\0&nbsp;&amp;&nbsp;1\end{array}\right]=I&nbsp;\\&amp;&nbsp;i=\left[\begin{array}{cc}0&nbsp;&amp;&nbsp;-1&nbsp;\\1&nbsp;&amp;&nbsp;0\end{array}\right]&nbsp;.\end{aligned}"><span></span><span></span></span><span class="equation-text" contenteditable="false" data-index="1" data-equation="i^2=i&nbsp;\cdot&nbsp;i=\left[\begin{array}{cc}0&nbsp;&amp;&nbsp;-1&nbsp;\\1&nbsp;&amp;&nbsp;0\end{array}\right]\left[\begin{array}{cc}0&nbsp;&amp;&nbsp;-1&nbsp;\\1&nbsp;&amp;&nbsp;0\end{array}\right]=\left[\begin{array}{cc}-1&nbsp;&amp;&nbsp;0&nbsp;\\0&nbsp;&amp;&nbsp;-1\end{array}\right]=-I=-1"><span></span><span></span></span>

<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&nbsp;i"><span></span><span></span></span>

<span class="equation-text" contenteditable="false" data-index="0" data-equation="\|z\|=\sqrt{z&nbsp;\bar{z}}"><span></span><span></span></span>

<span class="equation-text" contenteditable="false" data-index="0" data-equation="\left[\begin{array}{cc}a&nbsp;&amp;&nbsp;-b&nbsp;\\b&nbsp;&amp;&nbsp;a\end{array}\right]=\sqrt{a^2+b^2}\left[\begin{array}{cc}\frac{a}{\sqrt{a^2+b^2}}&nbsp;&amp;&nbsp;\frac{-b}{\sqrt{a^2+b^2}}&nbsp;\\\frac{b}{\sqrt{a^2+b^2}}&nbsp;&amp;&nbsp;\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&nbsp;&amp;&nbsp;-b&nbsp;\\b&nbsp;&amp;&nbsp;a\end{array}\right]&nbsp;}&nbsp;&amp;&nbsp;=\sqrt{a^2+b^2}\left[\begin{array}{cc}\cos&nbsp;(\theta)&nbsp;&amp;&nbsp;-\sin&nbsp;(\theta)&nbsp;\\\sin&nbsp;(\theta)&nbsp;&amp;&nbsp;\cos&nbsp;(\theta)\end{array}\right]&nbsp;\\&amp;&nbsp;=\|z\|\left[\begin{array}{cc}\cos&nbsp;(\theta)&nbsp;&amp;&nbsp;-\sin&nbsp;(\theta)&nbsp;\\\sin&nbsp;(\theta)&nbsp;&amp;&nbsp;\cos&nbsp;(\theta)\end{array}\right]&nbsp;\\&amp;&nbsp;=\|z\|&nbsp;\cdot&nbsp;I\left[\begin{array}{cc}\cos&nbsp;(\theta)&nbsp;&amp;&nbsp;-\sin&nbsp;(\theta)&nbsp;\\\sin&nbsp;(\theta)&nbsp;&amp;&nbsp;\cos&nbsp;(\theta)\end{array}\right]&nbsp;\\&amp;&nbsp;=\left[\begin{array}{cc}\|z\|&nbsp;&amp;&nbsp;0&nbsp;\\0&nbsp;&amp;&nbsp;\|z\|\end{array}\right]\left[\begin{array}{cc}\cos&nbsp;(\theta)&nbsp;&amp;&nbsp;-\sin&nbsp;(\theta)&nbsp;\\\sin&nbsp;(\theta)&nbsp;&amp;&nbsp;\cos&nbsp;(\theta)\end{array}\right]\end{aligned}"><span></span><span></span></span>

如果有一个复数&nbsp;<span class="equation-text" data-index="0" data-equation="z=a+b&nbsp;i" contenteditable="false"><span></span><span></span></span>，那么&nbsp;𝑧&nbsp;与任意一个复数&nbsp;𝑐&nbsp;相乘<br>都会将&nbsp;𝑐&nbsp;逆时针旋转<span class="equation-text" data-index="1" data-equation="\theta=\operatorname{atan}&nbsp;2(b,&nbsp;a)" contenteditable="false"><span></span><span></span></span>度，并缩放&nbsp;<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

复数的极坐标表示<br>

极坐标下的旋转&nbsp;<span class="equation-text" contenteditable="false" data-index="0" data-equation="v^{\prime}=e^{i&nbsp;\theta} v"><span></span><span></span></span><br>