🚧四元數乘法

🚧 under construction -> 反對稱矩陣, T(p) = pq -> [T]

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}, q=w2+x2i+y2j+z2k\mathbf{q} = w_2+x_2\mathbb{i} +y_2\mathbb{j} +z_2\mathbb{k}

定義 pq\mathbf{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}

  • 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}

  • ⭐️ 注意：前面這個矩陣類似「反對稱矩陣」

  1. (結合律)：(pq)r=p(qr)(\mathbf{p}\mathbf{q})\mathbf{r} = \mathbf{p}(\mathbf{q}\mathbf{r})

  2. (左分配律)：p(q+r)=pq+pr\mathbf{p} (\mathbf{q} + \mathbf{r}) = \mathbf{p}\mathbf{q} + \mathbf{p}\mathbf{r} (右分配律也成立)

若：p=s+u\mathbf{p} = s + \mathbf{u}, q=t+v\mathbf{q} = t + \mathbf{v}，則：

  1. 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}}

  2. 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}}

  1. (可交換條件)： qp=pq  ⟺  u∥v\mathbf{q} \mathbf{p} = \mathbf{p} \mathbf{q} \iff \mathbf{u} \parallel \mathbf{v}

  2. kq=qk{\color{orange}k}\mathbf{q} = \mathbf{q}{\color{orange}k} (純數一定可交換)

  1. uv=(u×v)−(u⋅v)\mathbf{u} \mathbf{v} = (\mathbf{u} \times \mathbf{v})-(\mathbf{u} \cdot \mathbf{v})

  1. (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}} (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}}

  • 🎖 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。

Previous四元數加法Next四元數除法

Last updated

Was this helpful?