向量的两种运算

scalar multiplication and addition，分别为数乘和加法。两种运算一起有个好听的名字叫linear combination，也就是线性组合。线性组合是线代的基石之一。比如v和w的线性组合表示为,其中a,b为常数 a v → + b w → \begin{aligned} a\overrightarrow v + b\overrightarrow w \end{aligned} av +bw ​

向量的三种表示方法

[]的形式，如下面这种

[ 1 2 3 ] \begin{aligned}\left[ \begin{array}{l} 1\\ 2\\ 3 \end{array} \right]\end{aligned} ⎣⎡​123​⎦⎤​​

arrow from 0 → \begin{aligned}\overrightarrow {\rm{0}} \end{aligned} 0 ​

point in the vector space ，如下图：

向量的内积与向量的长度

内积：两个向量的内积定义为： v T w {v^T}w vTw

如 v = [ 1 2 3 ] \begin{aligned} v = \left[ \begin{array}{l} 1\\ 2\\ 3 \end{array} \right] \end{aligned} v=⎣⎡​123​⎦⎤​​， w = [ 4 5 6 ] \begin{aligned}w = \left[ \begin{array}{l} 4\\ 5\\ 6 \end{array} \right]\end{aligned} w=⎣⎡​456​⎦⎤​​， v T w = 1 × 4 + 2 × 5 + 3 × 6 = 32 \begin{aligned}{v^{\rm{T}}}w = 1 \times 4 + 2 \times {\rm{5 + 3}} \times {\rm{6 = 32}}\end{aligned} vTw=1×4+2×5+3×6=32​

向量的长度为 ∣ ∣ v ∣ ∣ ||v|| ∣∣v∣∣，且有 ∣ ∣ v ∣ ∣ 2 = v T v \begin{aligned} ||v|{|^2} = {v^T}v \end{aligned} ∣∣v∣∣2=vTv​，以上面为例， ∣ ∣ v ∣ ∣ 2 = 1 2 + 2 2 + 3 3 = 14 \begin{aligned}||v|{|^2} = {1^2} + {2^2} + {3^3} = 14\end{aligned} ∣∣v∣∣2=12+22+33=14​，其长度为 14 \begin{aligned}\sqrt {14} \end{aligned} 14 ​​

单位向量：unit vector， u = v ∣ ∣ v ∣ ∣ \begin{aligned}u{\rm{ = }}\frac{v}{{||v||}}\end{aligned} u=∣∣v∣∣v​​，与v同方向的单位向量为 u = 1 14 [ 1 2 3 ] \begin{aligned}u = \frac{1}{{\sqrt {14} }}\left[ \begin{array}{l} 1\\ 2\\ 3 \end{array} \right]\end{aligned} u=14 ​1​⎣⎡​123​⎦⎤​​

向量的夹角： u T U = cos ⁡ θ \begin{aligned}{u^T}U = \cos \theta \end{aligned} uTU=cosθ​，U也为单位向量。 cos ⁡ θ = v T w ∣ ∣ v ∣ ∣ ∣ ∣ w ∣ ∣ \begin{aligned}\cos \theta = \frac{{{v^{\rm{T}}}w}}{{||v||||w||}}\end{aligned} cosθ=∣∣v∣∣∣∣w∣∣vTw​​

柯西-许瓦兹不等式

根据向量的夹角公式，由于 − 1 ≤ cos ⁡ θ ≤ 1 \begin{aligned} - 1 \le \cos \theta \le 1 \end{aligned} −1≤cosθ≤1​，推出许瓦兹不等式如下： ∣ v T w ∣ ≤ ∣ ∣ v ∣ ∣ . ∣ ∣ w ∣ ∣ \begin{aligned} |{v^T}w| \le ||v||.||w||\end{aligned} ∣vTw∣≤∣∣v∣∣.∣∣w∣∣​ 其代数表达形式为，下面的v和w为n维向量 ( ∑ i = 1 n v i w i ) 2 ≤ ∑ i = 1 n v i 2 ∑ i = 1 n w i 2 \begin{aligned} {\left( {\sum\limits_{i = 1}^n {{v_i}{w_i}} } \right)^2} \le \sum\limits_{i = 1}^n {{v_i}^2} \sum\limits_{i = 1}^n {{w_i}^2} \end{aligned} (i=1∑n​vi​wi​)2≤i=1∑n​vi​2i=1∑n​wi​2​

A-G不等式（arithmetic-geometry inequality)

二维：令 v = [ a b ] , w = [ b a ] \begin{aligned}v{\rm{ = }}\left[ \begin{array}{l} a\\ b \end{array} \right],w = \left[ \begin{array}{l} b\\ a \end{array} \right]\end{aligned} v=[ab​],w=[ba​]​,根据许瓦兹不等式可得 2 a b ≤ a 2 + b 2 a b ≤ a + b 2 \begin{aligned} \begin{array}{l} 2ab \le {a^2} + {b^2}\\\displaystyle \sqrt {ab} \le \frac{{a + b}}{2} \end{array}\end{aligned} 2ab≤a2+b2ab ​≤2a+b​​​ n维的待证

三角不等式（triangle inequality)

∣ ∣ a + b ∣ ∣ ≤ ∣ ∣ a ∣ ∣ + ∣ ∣ b ∣ ∣ \begin{aligned}||a + b|| \le ||a|| + ||b||\end{aligned} ∣∣a+b∣∣≤∣∣a∣∣+∣∣b∣∣​

证明如下 ∣ ∣ a + b ∣ ∣ 2 = ∣ ∣ a ∣ ∣ 2 + 2 a T b + ∣ ∣ b ∣ ∣ 2 \begin{aligned}||a + b|{|^2} = ||a|{|^2} + 2{a^T}b + ||b|{|^2}\end{aligned} ∣∣a+b∣∣2=∣∣a∣∣2+2aTb+∣∣b∣∣2​ 根据许瓦兹不等式易知 ∣ ∣ a ∣ ∣ 2 + 2 a T b + ∣ ∣ b ∣ ∣ 2 ≤ ( ∣ ∣ a ∣ ∣ + ∣ ∣ b ∣ ∣ ) 2 \begin{aligned}||a|{|^2} + 2{a^T}b + ||b|{|^2} \le {\left( {||a|| + ||b||} \right)^2}\end{aligned} ∣∣a∣∣2+2aTb+∣∣b∣∣2≤(∣∣a∣∣+∣∣b∣∣)2​，所以 ∣ ∣ a + b ∣ ∣ 2 ≤ ( ∣ ∣ a ∣ ∣ + ∣ ∣ b ∣ ∣ ) 2 \begin{aligned}||a + b|{|^2} \le {\left( {||a|| + ||b||} \right)^2}\end{aligned} ∣∣a+b∣∣2≤(∣∣a∣∣+∣∣b∣∣)2​，得证。如果a,b为标量同样也成立

余弦定理

∣ ∣ v − w ∣ ∣ 2 = ∣ ∣ v ∣ ∣ 2 − 2 v T w + ∣ ∣ w ∣ ∣ 2 = ∣ ∣ v ∣ ∣ 2 − 2 ∣ ∣ v ∣ ∣ . ∣ ∣ w ∣ ∣ cos ⁡ θ + ∣ ∣ w ∣ ∣ 2 \begin{aligned} ||v - w|{|^2} = &||v|{|^2} - 2{v^T}w + ||w|{|^2}\\\displaystyle =& ||v|{|^2} - 2||v||.||w||\cos \theta + ||w|{|^2} \end{aligned} ∣∣v−w∣∣2==​∣∣v∣∣2−2vTw+∣∣w∣∣2∣∣v∣∣2−2∣∣v∣∣.∣∣w∣∣cosθ+∣∣w∣∣2​

平行四边形对角线与边长的关系

从图中可以看到 ∣ ∣ v − w ∣ ∣ 2 + ∣ ∣ v + w ∣ ∣ 2 = 2 ∣ ∣ v ∣ ∣ 2 + 2 ∣ ∣ w ∣ ∣ 2 \begin{aligned} ||v - w|{|^2} + ||v + w|{|^2} = 2||v|{|^2} + 2||w|{|^2}\end{aligned} ∣∣v−w∣∣2+∣∣v+w∣∣2=2∣∣v∣∣2+2∣∣w∣∣2​ 从向量的角度很直观地知道了平行四边形对角线与边长的关系。