定义: 给定 m × p m\times p m×p大小的矩阵 A A A, p × n p\times n p×n大小的矩阵 B B B,则矩阵 A A A, B B B的乘积为 m × n m\times n m×n大小的矩阵 C C C,计算过程如下,
C = A ⋅ B C=A\cdot B C=A⋅B
其中, C i , j = ∑ k = 1 p A i , k ⋅ B k , j C_{i,j}=\sum_{k=1}^{p}{A_{i,k}\cdot B_{k,j}} Ci,j=∑k=1pAi,k⋅Bk,j, 1 ≤ i ≤ m , 1 ≤ j ≤ n 1\leq i\leq m,1\leq j\leq n 1≤i≤m,1≤j≤n
举例: 给定 2 × 3 2\times 3 2×3大小的矩阵 A = [ 0 1 2 3 4 5 ] A=\begin{bmatrix}0 & 1 & 2\\ 3 & 4 & 5\end{bmatrix} A=[031425], 3 × 2 3\times 2 3×2大小的矩阵 B = [ 0 3 1 4 2 5 ] B=\begin{bmatrix}0 & 3\\ 1 & 4\\ 2 & 5\end{bmatrix} B=⎣⎡012345⎦⎤,则矩阵 A A A, B B B的乘积为 2 × 2 2×2 2×2大小的矩阵 C C C,计算过程如下,
C = A ⋅ B C=A\cdot B C=A⋅B
= [ 0 1 2 3 4 5 ] ⋅ [ 0 3 1 4 2 5 ] =\begin{bmatrix} 0 & 1 & 2\\ 3 & 4 & 5\end{bmatrix}\cdot \begin{bmatrix}0 & 3\\ 1 & 4\\ 2 & 5\end{bmatrix} =[031425]⋅⎣⎡012345⎦⎤
= [ 0 × 0 + 1 × 1 + 2 × 2 0 × 3 + 1 × 4 + 2 × 5 3 × 0 + 4 × 1 + 5 × 2 3 × 3 + 4 × 4 + 5 × 5 ] =\begin{bmatrix}0\times0+1\times1+2\times2 & 0\times3+1\times4+2\times5\\ 3\times0+4\times1+5\times2 & 3\times3+4\times4+5\times5\end{bmatrix} =[0×0+1×1+2×23×0+4×1+5×20×3+1×4+2×53×3+4×4+5×5]
= [ 5 14 14 50 ] =\begin{bmatrix}5 & 14\\ 14 & 50\end{bmatrix} =[5141450]
定义: 给定 m × n m\times n m×n大小的矩阵 A A A, m × n m\times n m×n大小的矩阵 B B B,则矩阵 A A A, B B B的Hadamard product为 m × n m\times n m×n大小的矩阵 C C C,计算过程如下,
C = A ⨀ B C=A\bigodot B C=A⨀B
其中, C i , j = A i , j ⋅ B i , j C_{i,j}=A_{i,j}\cdot B_{i,j} Ci,j=Ai,j⋅Bi,j, 1 ≤ i ≤ m , 1 ≤ j ≤ n 1\leq i\leq m,1\leq j\leq n 1≤i≤m,1≤j≤n
举例: 给定 2 × 3 2\times 3 2×3大小的矩阵 A = [ 0 1 2 3 4 5 ] A=\begin{bmatrix}0 & 1 & 2\\ 3 & 4 & 5\end{bmatrix} A=[031425], 3 × 2 3\times 2 3×2大小的矩阵 B = [ 3 4 5 0 1 2 ] B=\begin{bmatrix}3 & 4 & 5\\ 0 & 1 & 2\end{bmatrix} B=[304152],则矩阵 A A A, B B B的乘积为 3 × 2 3×2 3×2大小的矩阵 C C C,计算过程如下,
C = A ⨀ B C=A\bigodot B C=A⨀B
= [ 0 1 2 3 4 5 ] ⨀ [ 3 4 5 0 1 2 ] =\begin{bmatrix}0 & 1 & 2\\ 3 & 4 & 5\end{bmatrix}\bigodot \begin{bmatrix}3 & 4 & 5\\ 0 & 1 & 2\end{bmatrix} =[031425]⨀[304152]
= [ 0 × 3 1 × 4 2 × 5 3 × 0 4 × 1 5 × 2 ] =\begin{bmatrix}0 \times 3 & 1\times 4 & 2 \times 5\\ 3\times 0 & 4\times 1 & 5 \times 2\end{bmatrix} =[0×33×01×44×12×55×2]
= [ 0 4 10 0 4 10 ] =\begin{bmatrix}0 & 4 & 10\\ 0 & 4 & 10\end{bmatrix} =[00441010]
定义: 给定 1 × n 1\times n 1×n大小的向量 a ⃗ \vec{a} a, 1 × n 1\times n 1×n大小的向量 b ⃗ \vec{b} b, 则向量 a ⃗ \vec{a} a, b ⃗ \vec{b} b的点积为 1 × 1 1\times 1 1×1大小的数值 c c c, 计算过程如下,
c = a ⃗ ∙ b ⃗ c=\vec{a}\bullet \vec{b} c=a∙b
= ∑ i = 1 n a i b i =\sum_{i=1}^{n}{a_ib_i} =∑i=1naibi
举例: 给定 1 × 3 1\times 3 1×3大小的向量 a ⃗ = ( 0 1 2 ) \vec{a}=\begin{pmatrix}0 & 1 & 2\end{pmatrix} a=(012), 1 × 3 1\times 3 1×3大小的向量 b ⃗ = ( 3 4 5 ) \vec{b}=\begin{pmatrix}3 & 4 & 5\end{pmatrix} b=(345),则向量 a ⃗ \vec{a} a, b ⃗ \vec{b} b的点积为数值 c c c,计算过程如下,
c = a ⃗ ∙ b ⃗ c=\vec{a}\bullet \vec{b} c=a∙b
= 0 × 3 + 1 × 4 + 2 × 5 =0\times 3 + 1 \times 4 + 2 \times 5 =0×3+1×4+2×5
= 4 + 10 = 4 + 10 =4+10
= 14 =14 =14
定义: 叉积是依赖于欧几里德三维空间的一个度量单位。给定三维空间中 1 × 3 1\times 3 1×3大小的向量 a ⃗ \vec{a} a和 1 × 3 1\times 3 1×3大小的向量 b ⃗ \vec{b} b,则向量 a ⃗ \vec{a} a, b ⃗ \vec{b} b的叉积为 1 × 3 1\times 3 1×3大小的向量 c ⃗ \vec{c} c,计算过程如下,
c ⃗ = a ⃗ × b ⃗ \vec{c}=\vec{a}\times \vec{b} c=a×b
= ∥ a ∥ ∥ b ∥ s i n ( θ ) ζ = \begin{Vmatrix}a \end{Vmatrix}\begin{Vmatrix}b \end{Vmatrix}sin(\theta) \zeta =∥∥a∥∥∥∥b∥∥sin(θ)ζ
其中, θ \theta θ为向量 a a a和向量 b b b之间的夹角, ζ \zeta ζ为叉积 c ⃗ \vec{c} c的方向,即右手定则
下,向量 a ⃗ \vec{a} a和向量 b ⃗ \vec{b} b之间法线的方向。
衍生: 根据叉积的定义,给定三维坐标, y y y, z z z的基坐标分别为 i ⃗ \vec{i} i, j ⃗ \vec{j} j , k ⃗ \vec{k} k,根据右手定则有,
i ⃗ × j ⃗ = k ⃗ \vec{i}\times \vec{j}=\vec{k} i×j=k
j ⃗ × k ⃗ = i ⃗ \vec{j}\times \vec{k}=\vec{i} j×k=i
k ⃗ × i ⃗ = j ⃗ \vec{k}\times \vec{i}=\vec{j} k×i=j
j ⃗ × i ⃗ = − k ⃗ \vec{j}\times \vec{i}=-\vec{k} j×i=−k
k ⃗ × j ⃗ = − i ⃗ \vec{k}\times \vec{j}=-\vec{i} k×j=−i
i ⃗ × k ⃗ = − j ⃗ \vec{i}\times \vec{k}=-\vec{j} i×k=−j
其中,当 θ = 0 \theta =0 θ=0时候,有,
i ⃗ × i ⃗ = 0 ⃗ \vec{i}\times \vec{i}=\vec{0} i×i=0
j ⃗ × j ⃗ = 0 ⃗ \vec{j}\times \vec{j}= \vec{0} j×j=0
k ⃗ × k ⃗ = 0 ⃗ \vec{k}\times \vec{k}= \vec{0} k×k=0
举例一 (向量形式): 给定三维空间中的两个向量 u ⃗ \vec{u} u, v ⃗ \vec{v} v,令 u ⃗ = u 1 i ⃗ + u 2 j ⃗ + u 3 k ⃗ \vec{u} = u_1\vec{i} + u_2\vec{j} + u_3\vec{k} u=u1i+u2j+u3k, v ⃗ = v 1 i ⃗ + v 2 j ⃗ + v 3 k ⃗ \vec{v} = v_1\vec{i} + v_2\vec{j} + v_3\vec{k} v=v1i+v2j+v3k,根据定义,有,
u ⃗ × v ⃗ = ( u 1 i ⃗ + u 2 j ⃗ + u 3 k ⃗ ) × ( v 1 i ⃗ + v 2 j ⃗ + v 3 k ⃗ ) \vec{u} \times \vec{v} = \left ( u_1\vec{i} + u_2\vec{j} + u_3\vec{k} \right ) \times \left ( v_1\vec{i} + v_2\vec{j} + v_3\vec{k} \right ) u×v=(u1i+u2j+u3k)×(v1i+v2j+v3k)
= u 1 v 1 i ⃗ × i ⃗ + u 1 v 2 i ⃗ × j ⃗ + u 1 v 3 i ⃗ × k ⃗ + u 2 v 1 j ⃗ × i ⃗ + u 2 v 2 j ⃗ × j ⃗ + u 2 v 3 j ⃗ × k ⃗ + u 3 v 1 k ⃗ × i ⃗ + u 3 v 2 k ⃗ × j ⃗ + u 3 v 3 k ⃗ × k ⃗ = u_1v_1\vec{i}\times \vec{i} + u_1v_2\vec{i}\times \vec{j} + u_1v_3\vec{i}\times \vec{k} + u_2v_1\vec{j}\times \vec{i} + u_2v_2\vec{j}\times \vec{j} + u_2v_3\vec{j}\times \vec{k} + u_3v_1\vec{k}\times \vec{i} + u_3v_2\vec{k}\times \vec{j} + u_3v_3\vec{k}\times \vec{k} =u1v1i×i+u1v2i×j+u1v3i×k+u2v1j×i+u2v2j×j+u2v3j×k+u3v1k×i+u3v2k×j+u3v3k×k
= 0 ⃗ + u 1 v 2 k ⃗ − u 1 v 3 j ⃗ − u 2 v 1 k ⃗ + 0 ⃗ + u 2 v 3 i ⃗ + u 3 v 1 j ⃗ − u 3 v 2 j ⃗ − u 3 v 2 i ⃗ + 0 ⃗ = \vec{0} + u_1v_2\vec{k} - u_1v_3\vec{j} - u_2v_1\vec{k} + \vec{0} + u_2v_3\vec{i} + u_3v_1\vec{j} - u_3v_2\vec{j} - u_3v_2\vec{i} + \vec{0} =0+u1v2k−u1v3j−u2v1k+0+u2v3i+u3v1j−u3v2j−u3v2i+0
= u 1 v 2 k ⃗ − u 1 v 3 j ⃗ − u 2 v 1 k ⃗ + u 2 v 3 i ⃗ + u 3 v 1 j ⃗ − u 3 v 2 i ⃗ = u_1v_2\vec{k} - u_1v_3\vec{j} - u_2v_1\vec{k} + u_2v_3\vec{i} + u_3v_1\vec{j} - u_3v_2\vec{i} =u1v2k−u1v3j−u2v1k+u2v3i+u3v1j−u3v2i
= ( u 2 v 3 − u 3 v 2 ) i ⃗ + ( u 3 v 1 − u 1 v 3 ) j ⃗ + ( u 1 v 2 − u 2 v 1 ) k ⃗ = \left ( u_2v_3 - u_3v_2 \right )\vec{i} + \left ( u_3v_1 - u_1v_3 \right )\vec{j} + \left ( u_1v_2 - u_2v_1 \right )\vec{k} =(u2v3−u3v2)i+(u3v1−u1v3)j+(u1v2−u2v1)k
举例二 (行列式形式): 给定三维空间中的两个向量 u ⃗ \vec{u} u, v ⃗ \vec{v} v,令 u ⃗ = u 1 i ⃗ + u 2 j ⃗ + u 3 k ⃗ \vec{u} = u_1\vec{i} + u_2\vec{j} + u_3\vec{k} u=u1i+u2j+u3k, v ⃗ = v 1 i ⃗ + v 2 j ⃗ + v 3 k ⃗ \vec{v} = v_1\vec{i} + v_2\vec{j} + v_3\vec{k} v=v1i+v2j+v3k,差积也可以通过构造矩阵,计算行列式而得,如下,
u ⃗ × v ⃗ = ∣ i ⃗ j ⃗ k ⃗ u 1 u 2 u 3 v 1 v 2 v 3 ∣ \vec{u} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ u_1 & u_2 & u_3\\ v_1 & v_2 & v_3 \end{vmatrix} u×v=∣∣∣∣∣∣iu1v1ju2v2ku3v3∣∣∣∣∣∣
高阶方阵的行列式可以通过根据拉普拉斯定理降维计算而得,如下,
u ⃗ × v ⃗ = ∣ u 2 u 3 v 2 v 3 ∣ i ⃗ − ∣ u 1 u 3 v 1 v 3 ∣ j ⃗ + ∣ u 1 u 2 v 1 v 2 ∣ k ⃗ \vec{u} \times \vec{v} = \begin{vmatrix}u_2 & u_3\\ v_2 & v_3\end{vmatrix}\vec{i} - \begin{vmatrix}u_1 & u_3\\ v_1 & v_3\end{vmatrix}\vec{j} + \begin{vmatrix}u_1 & u_2\\ v_1 & v_2\end{vmatrix}\vec{k} u×v=∣∣∣∣u2v2u3v3∣∣∣∣i−∣∣∣∣u1v1u3v3∣∣∣∣j+∣∣∣∣u1v1u2v2∣∣∣∣k
= ( u 2 v 3 − u 3 v 2 ) i ⃗ + ( u 3 v 1 − u 1 v 3 ) j ⃗ + ( u 1 v 2 − u 2 v 1 ) k ⃗ = \left ( u_2v_3 - u_3v_2 \right )\vec{i} + \left ( u_3v_1 - u_1v_3 \right )\vec{j} + \left ( u_1v_2 - u_2v_1 \right )\vec{k} =(u2v3−u3v2)i+(u3v1−u1v3)j+(u1v2−u2v1)k