🚧 under construction -> 反對稱矩陣, T(p) = pq -> [T]
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數系 ⟩ 四元數 ⟩ 運算 ⟩ 乘法
設： p=w1+x1i+y1j+z1k\mathbf{p} = w_1+x_1\mathbb{i} +y_1\mathbb{j} +z_1\mathbb{k}p=w1+x1i+y1j+z1k, q=w2+x2i+y2j+z2k\mathbf{q} = w_2+x_2\mathbb{i} +y_2\mathbb{j} +z_2\mathbb{k}q=w2+x2i+y2j+z2k
定義 pq\mathbf{pq}pq 為：
+i(w1w2−x1x2−y1y2−z1z2)+i (w1x2+x1w2+y1z2−z1y2)+j (w1y2+y1w2+z1x2−x1z2)+k (w1z2+z1w2+x1y2−y1x2)\begin{matrix} \phantom{+} & \phantom{\mathbf{i}} (w_1 w_2 - x_1 x_2 - y_1 y_2 -z_1 z_2) \\ + & \mathbf{i} \ (w_1 x_2 + x_1 w_2 + y_1 z_2 - z_1 y_2) \\ + & \mathbf{j} \ (w_1 y_2 + y_1 w_2 + z_1 x_2 - x_1 z_2)\\ + & \mathbf{k} \ (w_1 z_2 + z_1 w_2 + x_1 y_2 - y_1 x_2) \end{matrix}++++i(w1w2−x1x2−y1y2−z1z2)i (w1x2+x1w2+y1z2−z1y2)j (w1y2+y1w2+z1x2−x1z2)k (w1z2+z1w2+x1y2−y1x2)
註：可先借用四元數的分配律推導 👉
若用矩陣乘法則可寫成： [w1−x1−y1−z1x1w1−z1y1y1z1w1−x1z1−y1x1w1][w2x2y2z2]\begin{bmatrix} w_{1} & -x_{1} & -y_{1} & -z_{1}\\ x_{1} & w_{1} & -z_{1} & y_{1}\\ y_{1} & z_{1} & w_{1} & -x_{1}\\ z_{1} & -y_{1} & x_{1} & w_{1} \end{bmatrix}\begin{bmatrix} w_{2}\\ x_{2}\\ y_{2}\\ z_{2} \end{bmatrix}w1x1y1z1−x1w1z1−y1−y1−z1w1x1−z1y1−x1w1w2x2y2z2
⭐️ 注意：前面這個矩陣類似「反對稱矩陣」
(結合律)：(pq)r=p(qr)(\mathbf{p}\mathbf{q})\mathbf{r} = \mathbf{p}(\mathbf{q}\mathbf{r})(pq)r=p(qr)
(左分配律)：p(q+r)=pq+pr\mathbf{p} (\mathbf{q} + \mathbf{r}) = \mathbf{p}\mathbf{q} + \mathbf{p}\mathbf{r}p(q+r)=pq+pr (右分配律也成立)
若：p=s+u\mathbf{p} = s + \mathbf{u}p=s+u, q=t+v\mathbf{q} = t + \mathbf{v}q=t+v，則：
pq=(s+u)(t+v)=st−(u⋅v)⏟scalar part+sv+tu+(u×v)⏟vector part\mathbf{p} \mathbf{q} = ( {\color{orange}{s}} + \mathbf{u})( {\color{orange}{t}} + \mathbf{v}) = \underbrace{ {\color{orange}{st}} - (\mathbf{u} \cdot \mathbf{v}) }_{\text{scalar part}} + \underbrace{ {\color{orange}{s}} \mathbf{v} + {\color{orange}{t}} \mathbf{u} + (\mathbf{u} \times \mathbf{v})} _{\text{vector part}} pq=(s+u)(t+v)=scalar partst−(u⋅v)+vector partsv+tu+(u×v)
p‾q=(s−u)(t+v)=st+(u⋅v)⏟scalar part+sv−tu−(u×v)⏟vector part\mathbf{\overline{p}} \mathbf{q} = ( {\color{orange}{s}} - \mathbf{u})( {\color{orange}{t}} + \mathbf{v}) = \underbrace{ {\color{orange}{st}} + (\mathbf{u} \cdot \mathbf{v}) }_{\text{scalar part}} + \underbrace{ {\color{orange}{s}} \mathbf{v} - {\color{orange}{t}} \mathbf{u} - (\mathbf{u} \times \mathbf{v})} _{\text{vector part}} pq=(s−u)(t+v)=scalar partst+(u⋅v)+vector partsv−tu−(u×v)
(可交換條件)： qp=pq ⟺ u∥v\mathbf{q} \mathbf{p} = \mathbf{p} \mathbf{q} \iff \mathbf{u} \parallel \mathbf{v} qp=pq⟺u∥v
kq=qk{\color{orange}k}\mathbf{q} = \mathbf{q}{\color{orange}k}kq=qk (純數一定可交換)
⬆️ 需要： 平行向量性質
🎖 證明： (5)
uv=(u×v)−(u⋅v)\mathbf{u} \mathbf{v} = (\mathbf{u} \times \mathbf{v})-(\mathbf{u} \cdot \mathbf{v})uv=(u×v)−(u⋅v)
👉 比較：複數乘法性質 、 四元數外積 、四元數旋轉
(1) p‾q=(p⋅q)⏟scalar part−(p×q)⏟vector part(1) \ \mathbf{\overline{p}} \mathbf{q} = \underbrace{ (\mathbf{p} \cdot \mathbf{q}) }_{\text{scalar part}} \underbrace{ - (\mathbf{p} \times \mathbf{q}) }_{\text{vector part}}(1) pq=scalar part(p⋅q)vector part−(p×q) (2) pq=(p‾⋅q)⏟scalar part−(p‾×q)⏟vector part(2) \ \mathbf{p} \mathbf{q} = \underbrace{ (\mathbf{\overline{p}} \cdot \mathbf{q}) }_{\text{scalar part}} \underbrace{ - (\mathbf{\overline{p}} \times \mathbf{q}) }_{\text{vector part}}(2) pq=scalar part(p⋅q)vector part−(p×q)
🎖 證明： 可由 (4) 式與四元數內積、外積定義而來。
(8) 式可視為是 (7) 式的擴充。
注意：四元數乘法沒有「交換律」❗
q1q2≠q2q1\mathbf{q}_1 \mathbf{q}_2 \neq \mathbf{q}_2 \mathbf{q}_1q1q2=q2q1
通常省略四元數的乘法符號，以避免與向量空間的內積符號混淆。
向量係數積 sq{\color{orange}s} \mathbf{q}sq 與四元數乘法 (s,0,0,0)q({\color{orange}s},0,0,0) \mathbf{q}(s,0,0,0)q 兩者結果一樣，因此運算時，我們通常不區分彼此。
四元數乘法可以同時處理兩空間向量的內積與外積。
👉 四元數旋轉
Math for 3D Games, 3.6 Quaternions
Is there a relationship between the cross product and quaternion multiplication?
