内积空间通过内积赋予向量空间几何结构，定义长度、正交性，统一代数与几何，推广 R n ~\mathbb{R^n}~ R n
1. 内积空间的背景和定义 纯向量空间（如 R n ~\mathbb{R^n}~ R n P 2 ~\mathbb{P}_2~ P 2 R 2 ~\mathbb{R^2}~ R 2 u = [ 1 2 ] T ~\mathbf{u} = \begin{bmatrix}1 & 2\end{bmatrix}^T~ u = [ 1 2 ] T v = [ 3 4 ] T ~\mathbf{v} = \begin{bmatrix}3 & 4\end{bmatrix}^T~ v = [ 3 4 ] T 内积空间 ( inner product space ) (\textbf{inner product space}) ( inner product space ) ⟨ u , v ⟩ ~\langle \mathbf{u}, \mathbf{v} \rangle~ ⟨ u , v ⟩
定义
内积空间
内积空间是添加了内积运算的向量空间 V ~V~ V V ~V~ V u ~\mathbf{u}~ u v ~\mathbf{v}~ v ⟨ u , v ⟩ ~\langle \mathbf{u} , \mathbf{v} \rangle~ ⟨ u , v ⟩ u , v , w ∈ V ~\mathbf{u},\mathbf{v},\mathbf{w} \in V~ u , v , w ∈ V c ~c~ c
对称性 ： ⟨ u , v ⟩ = ⟨ v , u ⟩ \langle \mathbf{u}, \mathbf{v} \rangle = \langle \mathbf{v}, \mathbf{u} \rangle ⟨ u , v ⟩ = ⟨ v , u ⟩ 加法线性 ： ⟨ u + v , w ⟩ = ⟨ u , w ⟩ + ⟨ v , w ⟩ \langle \mathbf{u} + \mathbf{v}, \mathbf{w} \rangle = \langle \mathbf{u}, \mathbf{w} \rangle + \langle \mathbf{v}, \mathbf{w} \rangle ⟨ u + v , w ⟩ = ⟨ u , w ⟩ + ⟨ v , w ⟩ 标量线性 ： ⟨ c u , v ⟩ = c ⟨ u , v ⟩ \langle c\mathbf{u}, \mathbf{v} \rangle = c \langle \mathbf{u}, \mathbf{v} \rangle ⟨ c u , v ⟩ = c ⟨ u , v ⟩ 非负性与正定性 ： ⟨ u , u ⟩ ≥ 0 , 且 ⟨ u , u ⟩ = 0 ⇔ u = 0 \langle \mathbf{u}, \mathbf{u} \rangle \geq 0, \quad \small{且} \quad \langle \mathbf{u}, \mathbf{u} \rangle = 0 \Leftrightarrow \mathbf{u} = \mathbf{0} ⟨ u , u ⟩ ≥ 0 , 且 ⟨ u , u ⟩ = 0 ⇔ u = 0 内积公理通过其对称性、线性性和正定性特性，为向量空间提供了定义长度 ∥ u ∥ = ⟨ u , u ⟩ \|\mathbf{u}\| = \sqrt{\langle \mathbf{u},\mathbf{u} \rangle} ∥ u ∥ = ⟨ u , u ⟩ ∥ u − v ∥ ~\|\mathbf{u} - \mathbf{v}\|~ ∥ u − v ∥ ⟨ u , v ⟩ = 0 ~\langle \mathbf{u},\mathbf{v} \rangle = 0~ ⟨ u , v ⟩ = 0
2. 内积定义的验证与示例 接下来，我们通过具体示例验证这些公理，确保内积空间的定义在不同空间中具有普适性和有效性，从而解决几何结构缺失的问题。
2.1 R 2 \mathbb{R^2} R 2 对于 R 2 \mathbb{R^2} R 2 u = [ u 1 u 2 ] T , v = [ v 1 v 2 ] T \mathbf{u} = \begin{bmatrix} u_1 & u_2 \end{bmatrix}^T,~\mathbf{v} = \begin{bmatrix} v_1 & v_2 \end{bmatrix}^T u = [ u 1 u 2 ] T , v = [ v 1 v 2 ] T
⟨ u , v ⟩ = 4 u 1 v 1 + 5 u 2 v 2 (1) \langle \mathbf{u}, \mathbf{v}\rangle = 4u_1v_1 + 5u_2v_2\tag{1} ⟨ u , v ⟩ = 4 u 1 v 1 + 5 u 2 v 2 ( 1 ) 验证 ( 1 ) ~(1)~ ( 1 )
满足 ⟨ u , v ⟩ = 4 u 1 v 1 + 5 u 2 v 2 = 4 v 1 u 1 + 5 v 2 u 2 = ⟨ v , u ⟩ ~\langle \mathbf{u}, \mathbf{v} \rangle = 4u_1 v_1 + 5u_2 v_2 = 4v_1 u_1 + 5v_2 u_2 = \langle \mathbf{v}, \mathbf{u} \rangle~ ⟨ u , v ⟩ = 4 u 1 v 1 + 5 u 2 v 2 = 4 v 1 u 1 + 5 v 2 u 2 = ⟨ v , u ⟩
满足：
⟨ u + v , w ⟩ = 4 ( u 1 + v 1 ) w 1 + 5 ( u 2 + v 2 ) w 2 = 4 u 1 w 1 + 5 u 2 w 2 + 4 v 1 w 1 + 5 v 2 w 2 = ⟨ u , w ⟩ + ⟨ v , w ⟩ \begin{aligned}
\langle \mathbf{u} + \mathbf{v}, \mathbf{w} \rangle &= 4(u_1 + v_1)w_1 + 5(u_2 + v_2)w_2 \\[1ex]
&= 4u_1 w_1 + 5u_2 w_2 + 4v_1 w_1 + 5v_2 w_2 \\[1ex]
&= \langle \mathbf{u}, \mathbf{w} \rangle + \langle \mathbf{v}, \mathbf{w} \rangle
\end{aligned} ⟨ u + v , w ⟩ = 4 ( u 1 + v 1 ) w 1 + 5 ( u 2 + v 2 ) w 2 = 4 u 1 w 1 + 5 u 2 w 2 + 4 v 1 w 1 + 5 v 2 w 2 = ⟨ u , w ⟩ + ⟨ v , w ⟩ 满足：
⟨ c u , v ⟩ = 4 ( c u 1 ) v 1 + 5 ( c u 2 ) v 2 = c ( 4 u 1 v 1 + 5 u 2 v 2 ) = c ⟨ u , v ⟩ \langle c\mathbf{u}, \mathbf{v} \rangle = 4(cu_1)v_1 + 5(cu_2)v_2 = c(4u_1v_1 + 5u_2v_2) = c\langle \mathbf{u}, \mathbf{v} \rangle ⟨ c u , v ⟩ = 4 ( c u 1 ) v 1 + 5 ( c u 2 ) v 2 = c ( 4 u 1 v 1 + 5 u 2 v 2 ) = c ⟨ u , v ⟩
⟨ u , u ⟩ = 4 u 1 2 + 5 u 2 2 ≥ 0 \langle \mathbf{u}, \mathbf{u} \rangle = 4u_1^2 + 5u_2^2 \geq 0 ⟨ u , u ⟩ = 4 u 1 2 + 5 u 2 2 ≥ 0 u 1 = u 2 = 0 ~u_1 = u_2 = 0~ u 1 = u 2 = 0 u = 0 \mathbf{u} = \mathbf{0} u = 0 0 ~0~ 0
这个示例说明 R n ~\mathbb{R^n}~ R n 标准内积 ( standard inner product ) ⟨ u , v ⟩ = u 1 v 1 + u 2 v 2 ~(\textbf{standard inner product}) ~~ \langle \mathbf{u}, \mathbf{v}\rangle = u_1v_1 + u_2v_2~ ( standard inner product ) ⟨ u , v ⟩ = u 1 v 1 + u 2 v 2 加权内积 ( weighted inner product ) ⟨ u , v ⟩ = 4 u 1 v 1 + 5 u 2 v 2 ~(\textbf{weighted inner product})~~ \langle \mathbf{u}, \mathbf{v}\rangle = 4u_1v_1 + 5u_2v_2~ ( weighted inner product ) ⟨ u , v ⟩ = 4 u 1 v 1 + 5 u 2 v 2 R 2 ~\mathbb{R^2}~ R 2 R 2 ~\mathbb{R^2}~ R 2 新的几何结构（如椭圆单位球）
2.2 P 2 \mathbb{P}_2 P 2 对于 p , q ∈ P n ~p,q \in \mathbb{P}_n~ p , q ∈ P n
⟨ p , q ⟩ = p ( t 0 ) q ( t 0 ) + p ( t 1 ) q ( t 1 ) + ⋯ + p ( t n ) q ( t n ) (2) ~\langle p, q \rangle = p(t_0) q(t_0) + p(t_1) q(t_1) + \cdots + p(t_n) q(t_n)~\tag{2} ⟨ p , q ⟩ = p ( t 0 ) q ( t 0 ) + p ( t 1 ) q ( t 1 ) + ⋯ + p ( t n ) q ( t n ) ( 2 ) 其中 t 0 , ⋯ t n ~t_0, \cdots t_n~ t 0 , ⋯ t n n + 1 ~n+1~ n + 1
对于 p , q ∈ P n ~p,q\in \mathbb{P}_n~ p , q ∈ P n
⟨ p , q ⟩ = p ( t 0 ) q ( t 0 ) + p ( t 1 ) q ( t 1 ) + ⋯ + p ( t n ) q ( t n ) ⟨ q , p ⟩ = q ( t 0 ) p ( t 0 ) + q ( t 1 ) p ( t 1 ) + ⋯ + q ( t n ) p ( t n ) \begin{align*}\langle p, q \rangle &= p(t_0) q(t_0) + p(t_1) q(t_1) + \cdots + p(t_n) q(t_n)\\[1ex]
\langle q, p \rangle &= q(t_0) p(t_0) + q(t_1) p(t_1) + \cdots + q(t_n) p(t_n)\end{align*} ⟨ p , q ⟩ ⟨ q , p ⟩ = p ( t 0 ) q ( t 0 ) + p ( t 1 ) q ( t 1 ) + ⋯ + p ( t n ) q ( t n ) = q ( t 0 ) p ( t 0 ) + q ( t 1 ) p ( t 1 ) + ⋯ + q ( t n ) p ( t n ) 显然 ⟨ p , q ⟩ = ⟨ q , p ⟩ ~\langle p, q \rangle = \langle q, p \rangle~ ⟨ p , q ⟩ = ⟨ q , p ⟩
设 p , q , r ∈ P n ~p,q,r \in \mathbb{P}_n~ p , q , r ∈ P n
⟨ p + q , r ⟩ = ( p ( t 0 ) + q ( t 0 ) ) r ( t 0 ) + ( p ( t 1 ) + q ( t 1 ) ) r ( t 1 ) + ⋯ + ( p ( t n ) + q ( t n ) ) r ( t n ) = [ p ( t 0 ) r ( t 0 ) + q ( t 0 ) r ( t 0 ) ] + [ p ( t 1 ) r ( t 1 ) + q ( t 1 ) r ( t 1 ) ] + ⋯ + [ p ( t n ) r ( t n ) + q ( t n ) r ( t n ) ] = [ p ( t 0 ) r ( t 0 ) + ⋯ + p ( t n ) r ( t n ) ] + [ q ( t 0 ) r ( t 0 ) + ⋯ + q ( t n ) r ( t n ) ] = ⟨ p , r ⟩ + ⟨ q , r ⟩ \begin{align*}
\langle p + q, r \rangle &= (p(t_0) + q(t_0))r(t_0) + (p(t_1) + q(t_1))r(t_1) + \cdots + (p(t_n) + q(t_n))r(t_n)\\[1.5ex]
&=[p(t_0)r(t_0) + q(t_0)r(t_0)] + [p(t_1)r(t_1) + q(t_1)r(t_1)] + \cdots + [p(t_n)r(t_n) + q(t_n)r(t_n)]\\[1.5ex]
&= [p(t_0)r(t_0) + \cdots + p(t_n)r(t_n)] + [q(t_0)r(t_0) + \cdots + q(t_n)r(t_n)] \\[1.5ex]
&= \langle p, r \rangle + \langle q, r \rangle
\end{align*} ⟨ p + q , r ⟩ = ( p ( t 0 ) + q ( t 0 )) r ( t 0 ) + ( p ( t 1 ) + q ( t 1 )) r ( t 1 ) + ⋯ + ( p ( t n ) + q ( t n )) r ( t n ) = [ p ( t 0 ) r ( t 0 ) + q ( t 0 ) r ( t 0 )] + [ p ( t 1 ) r ( t 1 ) + q ( t 1 ) r ( t 1 )] + ⋯ + [ p ( t n ) r ( t n ) + q ( t n ) r ( t n )] = [ p ( t 0 ) r ( t 0 ) + ⋯ + p ( t n ) r ( t n )] + [ q ( t 0 ) r ( t 0 ) + ⋯ + q ( t n ) r ( t n )] = ⟨ p , r ⟩ + ⟨ q , r ⟩ 加法线性成立。
设 c ~c~ c
⟨ c p , q ⟩ = ( c p ( t 0 ) ) q ( t 0 ) + ( c p ( t 1 ) ) q ( t 1 ) + ⋯ + ( c p ( t n ) ) q ( t n ) = c p ( t 0 ) q ( t 0 ) + c p ( t 1 ) q ( t 1 ) + ⋯ + c p ( t n ) q ( t n ) = c [ p ( t 0 ) q ( t 0 ) + ⋯ + p ( t n ) q ( t n ) ] = c ⟨ p , q ⟩ \begin{align*}\langle c p, q \rangle &= (c p(t_0)) q(t_0) + (c p(t_1)) q(t_1) + \cdots + (c p(t_n)) q(t_n)\\[1.5ex]
&= c p(t_0) q(t_0) + c p(t_1) q(t_1) + \cdots + c p(t_n) q(t_n) \\[1.5ex]
&= c [p(t_0) q(t_0) + \cdots + p(t_n) q(t_n)] \\[1.5ex]
&= c \langle p, q \rangle
\end{align*} ⟨ c p , q ⟩ = ( c p ( t 0 )) q ( t 0 ) + ( c p ( t 1 )) q ( t 1 ) + ⋯ + ( c p ( t n )) q ( t n ) = c p ( t 0 ) q ( t 0 ) + c p ( t 1 ) q ( t 1 ) + ⋯ + c p ( t n ) q ( t n ) = c [ p ( t 0 ) q ( t 0 ) + ⋯ + p ( t n ) q ( t n )] = c ⟨ p , q ⟩
验证非负性 ： ⟨ p , p ⟩ = [ p ( t 0 ) ] 2 + [ p ( t 1 ) ] 2 + ⋯ + [ p ( t n ) ] 2 ≥ 0 \langle p, p \rangle = [p(t_0)]^2 + [p(t_1)]^2 + \cdots + [p(t_n)]^2 \geq 0 ⟨ p , p ⟩ = [ p ( t 0 ) ] 2 + [ p ( t 1 ) ] 2 + ⋯ + [ p ( t n ) ] 2 ≥ 0
验证正定性 ：
如果 p = 0 ~p = \mathbf{0}~ p = 0 0 ~0~ 0 t i ~t_i~ t i p ( t i ) = 0 ~p(t_i) = 0~ p ( t i ) = 0
⟨ 0 , 0 ⟩ = 0 2 + 0 2 + ⋯ + 0 2 = 0 \langle 0, 0 \rangle = 0^2 + 0^2 + \cdots + 0^2 = 0 ⟨ 0 , 0 ⟩ = 0 2 + 0 2 + ⋯ + 0 2 = 0 反过来，如果 ⟨ p , q ⟩ = 0 ~\langle p, q \rangle = 0~ ⟨ p , q ⟩ = 0
[ p ( t 0 ) ] 2 + [ p ( t 1 ) ] 2 + ⋯ + [ p ( t n ) ] 2 = 0 [p(t_0)]^2 + [p(t_1)]^2 + \cdots + [p(t_n)]^2 = 0 [ p ( t 0 ) ] 2 + [ p ( t 1 ) ] 2 + ⋯ + [ p ( t n ) ] 2 = 0 由于每一项 [ p ( t i ) ] 2 ≥ 0 [p(t_i)]^2 \geq 0 [ p ( t i ) ] 2 ≥ 0 0 ~0~ 0
p ( t 0 ) = p ( t 1 ) = ⋯ = p ( t n ) = 0 p(t_0) = p(t_1) = \cdots = p(t_n) = 0 p ( t 0 ) = p ( t 1 ) = ⋯ = p ( t n ) = 0 这表明 p ~p~ p n + 1 ~n+1~ n + 1 t 0 , t 1 , ⋯ , t n ~t_0,t_1,\cdots,t_n~ t 0 , t 1 , ⋯ , t n 0 ~0~ 0
( 2 ) (2)~ ( 2 ) 将 P n ~\mathbb{P}_n~ P n R n + 1 ~\mathbb{R^{n+1}}~ R n + 1 p ∈ P n ~p \in \mathbb{P}_n~ p ∈ P n
p ↦ [ p ( t 0 ) p ( t 1 ) ⋯ p ( t n ) ] T ∈ R n + 1 (3) p \mapsto \begin{bmatrix}p(t_0) & p(t_1) & \cdots & p(t_n)\end{bmatrix}^T \in \mathbb{R}^{n+1}\tag{3} p ↦ [ p ( t 0 ) p ( t 1 ) ⋯ p ( t n ) ] T ∈ R n + 1 ( 3 ) 通过此定义赋予了多项式空间几何结构。这种映射使多项式之间的内积 ⟨ p , q ⟩ ~\langle p,q \rangle~ ⟨ p , q ⟩ R n + 1 ~\mathbb{R^{n+1}}~ R n + 1 ∥ p ∥ = ⟨ p , p ⟩ \|p\| = \sqrt{\langle p,p \rangle} ∥ p ∥ = ⟨ p , p ⟩ ∥ p − q ∥ \|p - q\| ∥ p − q ∥ ⟨ p , q ⟩ = 0 \langle p, q \rangle = 0 ⟨ p , q ⟩ = 0 P n ~\mathbb{P}_n~ P n
2.3 P 2 \mathbb{P}_2 P 2 设 V ∈ P 2 V \in \mathbb{P}_2 V ∈ P 2 ( 2 ) ~(2)~ ( 2 ) t 0 = 0 , t 1 = 1 / 2 , t 2 = 1 ~t_0 = 0,~ t_1 = 1/2,~t_2 = 1~ t 0 = 0 , t 1 = 1/2 , t 2 = 1 p ( t ) = 12 t 2 ~p(t) = 12t^2~ p ( t ) = 12 t 2 q ( t ) = 2 t − 1 ~q(t) = 2t - 1~ q ( t ) = 2 t − 1
1
计算 ⟨ p , q ⟩ ~\langle p,q \rangle ⟨ p , q ⟩
⟨ p , q ⟩ = p ( 0 ) q ( 0 ) + p ( 1 2 ) q ( 1 2 ) + p ( 1 ) q ( 1 ) = 0 ⋅ ( − 1 ) + 3 ⋅ 0 + 12 ⋅ 1 = 12 \begin{align*}\langle p, q \rangle &= p(0)q(0) + p\left(\frac{1}{2}\right)q\left(\frac{1}{2}\right) + p(1)q(1) \\[3ex]
&= 0 \cdot (-1) + 3 \cdot 0 + 12 \cdot 1 \\[2ex]
&= 12
\end{align*} ⟨ p , q ⟩ = p ( 0 ) q ( 0 ) + p ( 2 1 ) q ( 2 1 ) + p ( 1 ) q ( 1 ) = 0 ⋅ ( − 1 ) + 3 ⋅ 0 + 12 ⋅ 1 = 12 2
计算 ⟨ q , q ⟩ ~\langle q,q \rangle ⟨ q , q ⟩
⟨ q , q ⟩ = [ q ( 0 ) ] 2 + [ q ( 1 2 ) ] 2 + [ q ( 1 ) ] 2 = ( − 1 ) 2 + 0 2 + 1 2 = 2 \begin{align*}\langle q, q \rangle &= [q(0)]^2 + \left[q\left(\frac{1}{2}\right)\right]^2 + [q(1)]^2 \\[3ex]
&= (-1)^2 + 0^2 + 1^2 \\[2ex]
&= 2 \end{align*} ⟨ q , q ⟩ = [ q ( 0 ) ] 2 + [ q ( 2 1 ) ] 2 + [ q ( 1 ) ] 2 = ( − 1 ) 2 + 0 2 + 1 2 = 2 此例验证 ( 2 ) ~(2)~ ( 2 ) ⟨ p , q ⟩ = 12 , ⟨ q , q ⟩ = 2 ~\langle p,q \rangle = 12,~\langle q,q \rangle = 2~ ⟨ p , q ⟩ = 12 , ⟨ q , q ⟩ = 2
3. 长度、距离与正交性 基于内积空间的定义，我们可以将向量间的长度（范数）、距离和正交性这些几何概念扩展到更加广泛的向量空间，如多项式空间 P n ~\mathbb{P}_n~ P n C [ a , b ] ~C[a,b]~ C [ a , b ] R n ~\mathbb{R^n}~ R n Gram-Schmidt \text{Gram-Schmidt} Gram-Schmidt
3.1 长度的定义 一个向量 v ~\mathbf{v}~ v
∥ v ∥ = ⟨ v , v ⟩ \| \mathbf{v} \| = \sqrt{\langle \mathbf{v},\mathbf{v} \rangle} ∥ v ∥ = ⟨ v , v ⟩ 这一定义基于内积的第四个公理（ ⟨ v , v ⟩ ≥ 0 \langle \mathbf{v} , \mathbf{v} \rangle \geq 0 ⟨ v , v ⟩ ≥ 0 ∥ v ∥ \|\mathbf{v}\| ∥ v ∥ R n ~\mathbb{R^n}~ R n ∥ v ∥ = v 1 2 + ⋯ + v n 2 \|\mathbf{v}\| = \sqrt{v_1^2 + \cdots + v_n^2} ∥ v ∥ = v 1 2 + ⋯ + v n 2 任意 内积空间。例如，在多项式空间 P 2 ~\mathbb{P}_2~ P 2 p ( t ) = 12 t 2 ~p(t) = 12t^2~ p ( t ) = 12 t 2 ( 2 ) ~(2)~ ( 2 ) t 0 = 0 , t 1 = 1 / 2 , t 2 = 1 ~t_0 = 0, ~ t_1 = 1/2, ~ t_2 = 1~ t 0 = 0 , t 1 = 1/2 , t 2 = 1
⟨ p , p ⟩ = [ p ( 0 ) ] 2 + [ p ( 1 2 ) ] 2 + [ p ( 1 ) ] 2 = 0 2 + 3 2 + 12 2 = 153 \langle p, p \rangle = [p(0)]^2 + \left[p\left(\frac{1}{2}\right)\right]^2 + [p(1)]^2 = 0^2 + 3^2 + 12^2 = 153 ⟨ p , p ⟩ = [ p ( 0 ) ] 2 + [ p ( 2 1 ) ] 2 + [ p ( 1 ) ] 2 = 0 2 + 3 2 + 1 2 2 = 153 因此，长度 ∥ p ∥ = 153 ≈ 12.37 ~\|p\| = \sqrt{153} \approx 12.37~ ∥ p ∥ = 153 ≈ 12.37 R n ~\mathbb{R^n}~ R n p ( t ) ~p(t)~ p ( t )
3.2 距离的定义 一个长度为 1 ~1~ 1 u ~\mathbf{u}~ u v ~\mathbf{v}~ v
∥ u − v ∥ \|\mathbf{u} - \mathbf{v}\| ∥ u − v ∥ 这个定义通过内积自然导出，因为距离是基于范数 ∥ w ∥ = ⟨ w , w \|\mathbf{w}\| = \sqrt{\langle \mathbf{w}, \mathbf{w}} ∥ w ∥ = ⟨ w , w w = u − v \mathbf{w} = \mathbf{u} - \mathbf{v} w = u − v u ~\mathbf{u}~ u v ~\mathbf{v}~ v R n ~\mathbb{R^n}~ R n ( u 1 − v 1 ) 2 + ⋯ + ( u n − v n ) 2 ~\sqrt{(u_1 - v_1)^2 + \cdots + (u_n - v_n)^2}~ ( u 1 − v 1 ) 2 + ⋯ + ( u n − v n ) 2
3.3 正交性的定义 两个向量 u ~\mathbf{u}~ u v ~\mathbf{v}~ v
⟨ u , v ⟩ = 0 \langle \mathbf{u}, \mathbf{v} \rangle = 0 ⟨ u , v ⟩ = 0 则称 u ~\mathbf{u}~ u v ~\mathbf{v}~ v R n ~\mathbb{R^n}~ R n P n ~\mathbb{P}_n~ P n C [ a , b ] ~C[a,b]~ C [ a , b ] u ~\mathbf{u}~ u v ~\mathbf{v}~ v Gram-Schmidt ~\text{Gram-Schmidt}~ Gram-Schmidt
4. Gram-Schmidt \text{Gram-Schmidt} Gram-Schmidt Gram-Schmidt \text{Gram-Schmidt} Gram-Schmidt R n ~\mathbb{R^n}~ R n Gram-Schmidt \text{Gram-Schmidt} Gram-Schmidt 在 R n \mathbb{R}^n R n 是一致的。为了更具体地展示 Gram-Schmidt ~\text{Gram-Schmidt}~ Gram-Schmidt R n ~\mathbb{R^n}~ R n P 4 ~\mathbb{P}_4~ P 4 2 , − 1 , 0 , 1 , 2 ~2,-1,0,1,2~ 2 , − 1 , 0 , 1 , 2
⟨ p , q ⟩ = p ( − 2 ) q ( − 2 ) + p ( − 1 ) q ( − 1 ) + p ( 0 ) q ( 0 ) + p ( 1 ) q ( 1 ) + p ( 2 ) q ( 2 ) (4) \langle p, q \rangle = p(-2)q(-2) + p(-1)q(-1) + p(0)q(0) + p(1)q(1) + p(2)q(2)\tag{4} ⟨ p , q ⟩ = p ( − 2 ) q ( − 2 ) + p ( − 1 ) q ( − 1 ) + p ( 0 ) q ( 0 ) + p ( 1 ) q ( 1 ) + p ( 2 ) q ( 2 ) ( 4 ) 我们将 P 2 ~\mathbb{P}_2~ P 2 P 4 ~\mathbb{P}_4~ P 4 P 2 ~\mathbb{P}_2~ P 2 E = { 1 , t 2 , t 2 } ~\mathcal{E} = \{1,~t^2,~t^2\}~ E = { 1 , t 2 , t 2 } Gram-Schmidt \text{Gram-Schmidt} Gram-Schmidt P 4 ~\mathbb{P}_4~ P 4 − 2 , − 1 , 0 , 1 , 2 ~-2,-1,0,1,2~ − 2 , − 1 , 0 , 1 , 2 R 5 ~\mathbb{R^5}~ R 5 P 2 \mathbb{P}_2~ P 2 E = { p 0 ( t ) = 1 , p 1 ( t ) = t , p 2 ( t ) = t 2 } ~\mathcal{E} = \{p_0(t) = 1,p_1(t) = t, p_2(t) = t^2\}~ E = { p 0 ( t ) = 1 , p 1 ( t ) = t , p 2 ( t ) = t 2 } 可以分别映射为以下向量
p 0 ( t ) ↦ [ 1 1 1 1 1 ] , p 1 ( t ) ↦ [ − 2 − 1 0 1 2 ] , p 2 ↦ [ 4 1 0 1 4 ] p_0(t) \mapsto \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}, \quad p_1(t) \mapsto \begin{bmatrix} -2 \\ -1 \\ 0 \\ 1 \\ 2\end{bmatrix}, \quad p_2 \mapsto \begin{bmatrix} 4 \\ 1 \\ 0 \\ 1 \\ 4\end{bmatrix} p 0 ( t ) ↦ 1 1 1 1 1 , p 1 ( t ) ↦ − 2 − 1 0 1 2 , p 2 ↦ 4 1 0 1 4 设这三个向量分别为 p 0 , p 1 , p 2 ~\mathbf{p}_0, ~ \mathbf{p}_1, ~ \mathbf{p}_2~ p 0 , p 1 , p 2
1
选择第一个向量 v 0 ~\mathbf{v}_0~ v 0
直接选择第一个向量： v 0 = p 0 = [ 1 1 1 1 1 ] T \mathbf{v}_0 = \mathbf{p}_0 = \begin{bmatrix} 1 & 1 & 1 & 1 & 1\end{bmatrix}^T v 0 = p 0 = [ 1 1 1 1 1 ] T
2
构造第二个正交向量 v 1 ~\mathbf{v}_1~ v 1
注意到
⟨ p 0 ( t ) , p 1 ( t ) ⟩ = ⟨ 1 , t ⟩ = ( 1 ) ( − 2 ) + ( 1 ) ( − 1 ) + ( 1 ) ( 0 ) + ( 1 ) ( 1 ) + ( 1 ) ( 2 ) = 0 \langle p_0(t), p_1(t) \rangle = \langle 1, t \rangle = (1)(-2) + (1)(-1) + (1)(0) + (1)(1) + (1)(2) = 0 ⟨ p 0 ( t ) , p 1 ( t )⟩ = ⟨ 1 , t ⟩ = ( 1 ) ( − 2 ) + ( 1 ) ( − 1 ) + ( 1 ) ( 0 ) + ( 1 ) ( 1 ) + ( 1 ) ( 2 ) = 0 因为内积为 0 ~0~ 0 p 0 ( t ) ~p_0(t)~ p 0 ( t ) p 1 ( t ) ~p_1(t)~ p 1 ( t ) v 1 = p 1 = [ − 2 − 1 0 1 2 ] \mathbf{v}_1 = \mathbf{p}_1 = \begin{bmatrix} -2 & -1 & 0 & 1 & 2\end{bmatrix} v 1 = p 1 = [ − 2 − 1 0 1 2 ]
3
构造第三个正交向量 v 2 ~\mathbf{v}_2~ v 2
我们需要将 p 2 ( t ) = t 2 ~p_2(t) = t^2~ p 2 ( t ) = t 2 p 0 ( t ) ~p_0(t)~ p 0 ( t ) p 1 ( t ) ~p_1(t)~ p 1 ( t ) Gram-Schmidt ~\text{Gram-Schmidt}~ Gram-Schmidt
p 2 ( t ) ^ = t 2 − Proj Span { p 0 , p 1 } ( t 2 ) \hat{p_2(t)} = t^2 - \text{Proj}_{\text{Span}\{p_0, p_1\}}(t^2) p 2 ( t ) ^ = t 2 − Proj Span { p 0 , p 1 } ( t 2 ) 投影公式为：
Proj Span { p 0 , p 1 } ( t 2 ) = ⟨ t 2 , p 0 ⟩ ⟨ p 0 , p 0 ⟩ p 0 + ⟨ t 2 , p 1 ⟩ ⟨ p 1 , p 1 ⟩ p 1 = ⟨ t 2 , 1 ⟩ ⟨ 1 , 1 ⟩ 1 + ⟨ t 2 , t ⟩ ⟨ t , t ⟩ t \begin{align*} \text{Proj}_{\text{Span}\{p_0, p_1\}}(t^2) &=
\frac{\langle t^2, p_0 \rangle}{\langle p_0, p_0 \rangle} p_0
+ \frac{\langle t^2, p_1 \rangle}{\langle p_1, p_1 \rangle} p_1 \\[3ex]
&=
\frac{\langle t^2, 1 \rangle}{\langle 1, 1 \rangle} 1
+ \frac{\langle t^2, t \rangle}{\langle t, t \rangle} t
\end{align*} Proj Span { p 0 , p 1 } ( t 2 ) = ⟨ p 0 , p 0 ⟩ ⟨ t 2 , p 0 ⟩ p 0 + ⟨ p 1 , p 1 ⟩ ⟨ t 2 , p 1 ⟩ p 1 = ⟨ 1 , 1 ⟩ ⟨ t 2 , 1 ⟩ 1 + ⟨ t , t ⟩ ⟨ t 2 , t ⟩ t 根据 ( 4 ) ~(4)~ ( 4 )
Proj Span { p 0 , p 1 } ( t 2 ) = 2 \text{Proj}_{\text{Span}\{p_0, p_1\}}(t^2) = 2 Proj Span { p 0 , p 1 } ( t 2 ) = 2 那么 p 2 ( t ) ^ = t 2 − 2 ~\hat{p_2(t)} = t^2 - 2~ p 2 ( t ) ^ = t 2 − 2
v 2 = p 2 − 2 p 0 = [ 2 − 1 − 2 − 1 2 ] \mathbf{v}_2 = \mathbf{p}_2 - 2 \mathbf{p}_0 = \begin{bmatrix}2 \\ -1 \\ -2 \\ -1 \\2 \end{bmatrix} v 2 = p 2 − 2 p 0 = 2 − 1 − 2 − 1 2 计算得到 V ~V~ V P 2 ~\mathbb{P}_2~ P 2
v 0 = [ 1 1 1 1 1 ] , v 1 = [ − 2 − 1 0 1 2 ] , v 2 = [ 2 − 1 − 2 − 1 2 ] \mathbf{v}_0 = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 1 \end{bmatrix}, \quad \mathbf{v}_1 = \begin{bmatrix} -2 \\ -1 \\ 0 \\ 1 \\ 2 \end{bmatrix}, \quad \mathbf{v}_2 = \begin{bmatrix} 2 \\ -1 \\ -2 \\ -1 \\ 2 \end{bmatrix} v 0 = 1 1 1 1 1 , v 1 = − 2 − 1 0 1 2 , v 2 = 2 − 1 − 2 − 1 2 5. 内积空间中的最佳逼近 最佳逼近定理 同样可以推广到更广泛的向量空间。例如，在函数空间内用低次多项式拟合高次多项式。假设向量空间 V = P 4 ~V = \mathbb{P}_4~ V = P 4 W = P 2 ~W = \mathbb{P}_2~ W = P 2 t = − 2 , − 1 , 0 , 1 , 2 ~t=-2,-1,0,1,2~ t = − 2 , − 1 , 0 , 1 , 2 P 2 ~\mathbb{P}_2~ P 2 v 0 , v 1 , v 2 ~\mathbf{v}_0,\mathbf{v}_1, \mathbf{v}_2~ v 0 , v 1 , v 2 P 4 ~\mathbb{P}_4~ P 4 p ( t ) = 5 − 1 2 t 4 ~p(t) = 5 - \frac{1}{2}t^4~ p ( t ) = 5 − 2 1 t 4 P 2 ~\mathbb{P}_2~ P 2
解
P 4 \mathbb{P}_4~ P 4 P 2 ~\mathbb{P}_2~ P 2
条件 ： 已知 P 2 ~\mathbb{P}_2~ P 2 p 0 ( t ) = 1 , p 1 ( t ) = t , p 2 ( t ) = t 2 − 2 ~p_0(t) = 1,~ p_1(t) = t,~ p_2(t) = t^2 - 2~ p 0 ( t ) = 1 , p 1 ( t ) = t , p 2 ( t ) = t 2 − 2 p ( t ) = 5 − 1 2 t 4 p(t) = 5 - \frac{1}{2} t^4 p ( t ) = 5 − 2 1 t 4 − 2 , − 1 , 0 , 1 , 2 ~-2, ~ -1, ~ 0, ~ 1, ~ 2~ − 2 , − 1 , 0 , 1 , 2 − 3 , 9 / 2 , 5 , 9 / 2 , − 3 ~-3, ~ 9/2, ~ 5, ~ 9/2, ~ -3~ − 3 , 9/2 , 5 , 9/2 , − 3 ( 2 ) ~(2)~ ( 2 )
根据投影公式 ：
p ^ ( t ) = proj P 2 ( p ) = ⟨ p , p 0 ⟩ ⟨ p 0 , p 0 ⟩ p 0 ( t ) + ⟨ p , p 1 ⟩ ⟨ p 1 , p 1 ⟩ p 1 ( t ) + ⟨ p , p 2 ⟩ ⟨ p 2 , p 2 ⟩ p 2 ( t ) \hat{p}(t) = \text{proj}_{P_2}(p) = \frac{\langle p, p_0 \rangle}{\langle p_0, p_0 \rangle} p_0(t) + \frac{\langle p, p_1 \rangle}{\langle p_1, p_1 \rangle} p_1(t) + \frac{\langle p, p_2 \rangle}{\langle p_2, p_2 \rangle} p_2(t) p ^ ( t ) = proj P 2 ( p ) = ⟨ p 0 , p 0 ⟩ ⟨ p , p 0 ⟩ p 0 ( t ) + ⟨ p 1 , p 1 ⟩ ⟨ p , p 1 ⟩ p 1 ( t ) + ⟨ p 2 , p 2 ⟩ ⟨ p , p 2 ⟩ p 2 ( t ) 计算 p ~p~ p p 0 , p 1 , p 2 ~p_0,~p_1,~p_2~ p 0 , p 1 , p 2 5 ~5~ 5
⟨ p , p 0 ⟩ = ⟨ 5 − 1 2 t 4 , 1 ⟩ = 8 ⟨ p , p 1 ⟩ = ⟨ 5 − 1 2 t 4 , t ⟩ = 0 ⟨ p , p 2 ⟩ = ⟨ 5 − 1 2 t 4 , t 2 − 2 ⟩ = − 31 \begin{align*}
\langle p, p_0 \rangle &= \langle 5-\frac{1}{2}t^4, 1 \rangle = 8\\[2ex]
\langle p, p_1 \rangle &= \langle 5-\frac{1}{2}t^4, t \rangle = 0\\[2ex]
\langle p, p_2 \rangle &= \langle 5-\frac{1}{2}t^4, t^2 - 2 \rangle = -31
\end{align*} ⟨ p , p 0 ⟩ ⟨ p , p 1 ⟩ ⟨ p , p 2 ⟩ = ⟨ 5 − 2 1 t 4 , 1 ⟩ = 8 = ⟨ 5 − 2 1 t 4 , t ⟩ = 0 = ⟨ 5 − 2 1 t 4 , t 2 − 2 ⟩ = − 31 基向量函数 p 0 , p 1 , p 2 ~p_0,~p_1,~p_2~ p 0 , p 1 , p 2
⟨ p 0 , p 0 ⟩ = ⟨ 1 , 1 ⟩ = 5 ⟨ p 1 , p 1 ⟩ = ⟨ t , t ⟩ = 14 ⟨ p 2 , p 2 ⟩ = ⟨ t 2 − 2 , t 2 − 2 ⟩ = 14 \begin{align*}
\langle p_0, p_0 \rangle &= \langle 1, 1 \rangle = 5\\[2ex]
\langle p_1, p_1 \rangle &= \langle t, t \rangle = 14\\[2ex]
\langle p_2, p_2 \rangle &= \langle t^2 - 2, t^2 - 2 \rangle = 14
\end{align*} ⟨ p 0 , p 0 ⟩ ⟨ p 1 , p 1 ⟩ ⟨ p 2 , p 2 ⟩ = ⟨ 1 , 1 ⟩ = 5 = ⟨ t , t ⟩ = 14 = ⟨ t 2 − 2 , t 2 − 2 ⟩ = 14
求最佳逼近 p ( t ) ^ ~\hat{p(t)}~ p ( t ) ^
p ^ ( t ) = 8 5 p 0 ( t ) + − 31 14 p 2 ( t ) \hat{p}(t) = \frac{8}{5} p_0(t) + \frac{-31}{14} p_2(t) p ^ ( t ) = 5 8 p 0 ( t ) + 14 − 31 p 2 ( t ) 代入 p 0 ( t ) = 1 , p 1 ( t ) = t ~p_0(t) = 1, ~p_1(t) = t~ p 0 ( t ) = 1 , p 1 ( t ) = t
p ( t ) = 8 5 − 31 14 ( t 2 − 2 ) p(t) = \frac{8}{5} - \frac{31}{14} (t^2 - 2) p ( t ) = 5 8 − 14 31 ( t 2 − 2 )
下面这张图展示了 P 4 ~\mathbb{P}_4~ P 4 P 2 ~\mathbb{P}_2~ P 2
我们可以利用 6.1 ~6.1~ 6.1 毕达哥拉斯定理 （也就是勾股定理）推导出内积空间中两个重要的不等式：柯西-施瓦茨不等式 （ Cauchy-Schwarz Inequality \textbf{Cauchy-Schwarz Inequality} Cauchy-Schwarz Inequality 三角不等式 （ Triangle Inequality \textbf{Triangle Inequality} Triangle Inequality
定理 16
柯西-施瓦茨不等式
对于 V ~V~ V u ~\mathbf{u}~ u v ~\mathbf{v}~ v
∣ ⟨ u , v ⟩ ∣ ≤ ∥ u ∥ ∥ v ∥ | \langle \mathbf{u}, \mathbf{v} \rangle | \leq \| \mathbf{u} \| \| \mathbf{v} \| ∣ ⟨ u , v ⟩ ∣ ≤ ∥ u ∥∥ v ∥ 其中 ⟨ u , v ⟩ ~\langle\mathbf{u}, \mathbf{v}\rangle~ ⟨ u , v ⟩ u ~\mathbf{u}~ u v ~\mathbf{v}~ v ∥ u ∥ \|\mathbf{u}\| ∥ u ∥ ∥ v ∥ ~\|\mathbf{v}\|~ ∥ v ∥ u ~\mathbf{u}~ u v ~\mathbf{v}~ v
特殊情况 ： u = 0 \mathbf{u} = \mathbf{0} u = 0
如果 u = 0 ~\mathbf{u} = \mathbf{0}~ u = 0 ⟨ u , v ⟩ = 0 ~\langle \mathbf{u}, \mathbf{v} \rangle = 0~ ⟨ u , v ⟩ = 0 ∥ u ∥ = 0 ~\|\mathbf{u}\| = 0~ ∥ u ∥ = 0
因此，两个不等式的两边都是零，所以不等式自然成立。
一般情况 u ≠ 0 \mathbf{u} \neq 0 u = 0
假设 u ≠ 0 ~\mathbf{u} \neq 0~ u = 0 W ~W~ W u ~\mathbf{u}~ u W = Span { u } ~W = \text{Span}\{\mathbf{u}\}~ W = Span { u }
根据内积空间的基本性质，对于任意标量 c ~c~ c ∥ c u ∥ = ∣ c ∣ ∥ u ∥ ~\|c\mathbf{u}\| = |c|\|\mathbf{u}\|~ ∥ c u ∥ = ∣ c ∣∥ u ∥
投影公式 ：
我们通过计算 v ~\mathbf{v}~ v W ~W~ W
∥ proj W v ∥ = ∥ ⟨ v , u ⟩ ⟨ u , u ⟩ u ∥ = ∣ ⟨ v , u ⟩ ∣ ⟨ u , u ⟩ ∥ u ∥ \| \text{proj}_W \mathbf{v} \| = \left\| \frac{\langle \mathbf{v}, \mathbf{u} \rangle}{\langle \mathbf{u}, \mathbf{u} \rangle} \mathbf{u} \right\| = \frac{|\langle \mathbf{v}, \mathbf{u} \rangle|}{\langle \mathbf{u}, \mathbf{u} \rangle} \| \mathbf{u} \| ∥ proj W v ∥ = ⟨ u , u ⟩ ⟨ v , u ⟩ u = ⟨ u , u ⟩ ∣ ⟨ v , u ⟩ ∣ ∥ u ∥
简化后得到：
∥ proj W v ∥ = ∣ ⟨ v , u ⟩ ∣ ∥ u ∥ \| \text{proj}_W \mathbf{v} \| = \frac{|\langle \mathbf{v}, \mathbf{u} \rangle|}{\| \mathbf{u} \|} ∥ proj W v ∥ = ∥ u ∥ ∣ ⟨ v , u ⟩ ∣
利用不等式推导 ：
由于投影的范数不大于原向量 v ~\mathbf{v}~ v Proj w v ≤ ∥ v ∥ \text{Proj}_w\mathbf{v} \leq \|\mathbf{v}\| Proj w v ≤ ∥ v ∥
∣ ⟨ v , u ⟩ ∣ ∥ u ∥ ≤ ∥ v ∥ \frac{|\langle \mathbf{v}, \mathbf{u} \rangle|}{\| \mathbf{u} \|} \leq \| \mathbf{v} \| ∥ u ∥ ∣ ⟨ v , u ⟩ ∣ ≤ ∥ v ∥
整理后可得柯西-施瓦茨不等式：
∣ ⟨ u , v ⟩ ∣ ≤ ∥ u ∥ ∥ v ∥ | \langle \mathbf{u}, \mathbf{v} \rangle | \leq \| \mathbf{u} \| \| \mathbf{v} \| ∣ ⟨ u , v ⟩ ∣ ≤ ∥ u ∥∥ v ∥
利用柯西-施瓦茨不等式可以进一步推导出三角不等式：
定理 17
三角不等式
对于 V ~V~ V u ~\mathbf{u}~ u v ~\mathbf{v}~ v
∥ u + v ∥ ≤ ∥ u ∥ + ∥ v ∥ \| \mathbf{u} + \mathbf{v} \| \leq \| \mathbf{u} \| + \| \mathbf{v} \| ∥ u + v ∥ ≤ ∥ u ∥ + ∥ v ∥
展开 ∥ u + v ∥ 2 \|\mathbf{u} + \mathbf{v}\|^2 ∥ u + v ∥ 2
我们从定义出发，首先将 ∥ u + v ∥ 2 \|\mathbf{u} + \mathbf{v}\|^2 ∥ u + v ∥ 2
∥ u + v ∥ 2 = ⟨ u + v , u + v ⟩ = ⟨ u , u ⟩ + 2 ⟨ u , v ⟩ + ⟨ v , v ⟩ \| \mathbf{u} + \mathbf{v} \|^2 = \langle \mathbf{u} + \mathbf{v}, \mathbf{u} + \mathbf{v} \rangle = \langle \mathbf{u}, \mathbf{u} \rangle + 2 \langle \mathbf{u}, \mathbf{v} \rangle + \langle \mathbf{v}, \mathbf{v} \rangle ∥ u + v ∥ 2 = ⟨ u + v , u + v ⟩ = ⟨ u , u ⟩ + 2 ⟨ u , v ⟩ + ⟨ v , v ⟩ 其中， ⟨ u , u ⟩ = ∥ u ∥ 2 和 ⟨ v , v ⟩ = ∥ v ∥ 2 \langle \mathbf{u}, \mathbf{u} \rangle = \| \mathbf{u} \|^2 \quad \text{和} \quad \langle \mathbf{v}, \mathbf{v} \rangle = \| \mathbf{v} \|^2 ⟨ u , u ⟩ = ∥ u ∥ 2 和 ⟨ v , v ⟩ = ∥ v ∥ 2
∥ u + v ∥ 2 = ∥ u ∥ 2 + 2 ⟨ u , v ⟩ + ∥ v ∥ 2 \| \mathbf{u} + \mathbf{v} \|^2 = \| \mathbf{u} \|^2 + 2 \langle \mathbf{u}, \mathbf{v} \rangle + \| \mathbf{v} \|^2 ∥ u + v ∥ 2 = ∥ u ∥ 2 + 2 ⟨ u , v ⟩ + ∥ v ∥ 2
应用柯西-施瓦茨不等式
根据柯西-施瓦茨不等式，我们有：
∣ ⟨ u , v ⟩ ∣ ≤ ∥ u ∥ ∥ v ∥ | \langle \mathbf{u}, \mathbf{v} \rangle | \leq \| \mathbf{u} \| \| \mathbf{v} \| ∣ ⟨ u , v ⟩ ∣ ≤ ∥ u ∥∥ v ∥ 因此：
2 ⟨ u , v ⟩ ≤ 2 ∥ u ∥ ∥ v ∥ 2 \langle \mathbf{u}, \mathbf{v} \rangle \leq 2 \| \mathbf{u} \| \| \mathbf{v} \| 2 ⟨ u , v ⟩ ≤ 2∥ u ∥∥ v ∥ 代入上式得到：
∥ u + v ∥ 2 ≤ ∥ u ∥ 2 + 2 ∥ u ∥ ∥ v ∥ + ∥ v ∥ 2 \| \mathbf{u} + \mathbf{v} \|^2 \leq \| \mathbf{u} \|^2 + 2 \| \mathbf{u} \| \| \mathbf{v} \| + \| \mathbf{v} \|^2 ∥ u + v ∥ 2 ≤ ∥ u ∥ 2 + 2∥ u ∥∥ v ∥ + ∥ v ∥ 2
重写结果 ：
上式可以重写为：
∥ u + v ∥ 2 ≤ ( ∥ u ∥ + ∥ v ∥ ) 2 \| \mathbf{u} + \mathbf{v} \|^2 \leq \left( \| \mathbf{u} \| + \| \mathbf{v} \| \right)^2 ∥ u + v ∥ 2 ≤ ( ∥ u ∥ + ∥ v ∥ ) 2
取平方根 ：
对两边取平方根，得到：
∥ u + v ∥ ≤ ∥ u ∥ + ∥ v ∥ \| \mathbf{u} + \mathbf{v} \| \leq \| \mathbf{u} \| + \| \mathbf{v} \| ∥ u + v ∥ ≤ ∥ u ∥ + ∥ v ∥
7. 连续函数空间的内积 内积空间的概念不仅适用于有限维空间（如多项式空间），也同样适用于无限维的连续函数空间 C [ a , b ] ~C[a,b]~ C [ a , b ] C [ a , b ] ~C[a,b]~ C [ a , b ]
7.1 从有限维 P n \mathbb{P}_n P n C [ a , b ] C[a,b] C [ a , b ] 对于连续函数空间 C [ a , b ] ~C[a,b]~ C [ a , b ] P n ~\mathbb{P}_n~ P n P n ~\mathbb{P}_n~ P n n ~n~ n n + 1 ~n+1~ n + 1 P n ~\mathbb{P}_n~ P n n + 1 ~n+1~ n + 1 t 0 , t 1 , ⋯ , t n ~t_0,t_1,\cdots,t_n~ t 0 , t 1 , ⋯ , t n
⟨ p , q ⟩ = ∑ j = 0 n p ( t j ) q ( t j ) \langle p, q \rangle = \sum_{j=0}^{n} p(t_j) q(t_j) ⟨ p , q ⟩ = j = 0 ∑ n p ( t j ) q ( t j ) 对于连续函数空间 C [ a , b ] ~C[a,b]~ C [ a , b ] 两个主要问题 。首先，上式内积求和依赖点的选取，它不能全面反映函数的整体特性（函数在整个区间上的分布、形状或趋势）。其次，当 n ~n~ n n ~n~ n
为了解决这两个问题，我们可以将区间 C [ a , b ] ~C[a,b]~ C [ a , b ] n + 1 ~n+1~ n + 1
Δ t = b − a n + 1 \Delta t = \frac{b-a}{n+1} Δ t = n + 1 b − a 然后，我们在这些子区间的中点 t j ~t_j~ t j 1 n + 1 ~\frac{1}{n+1}~ n + 1 1
⟨ p , q ⟩ = 1 n + 1 ∑ j = 0 n p ( t j ) q ( t j ) Δ t \langle p, q \rangle = \frac{1}{n+1} \sum_{j=0}^{n} p(t_j) q(t_j) \Delta t ⟨ p , q ⟩ = n + 1 1 j = 0 ∑ n p ( t j ) q ( t j ) Δ t 随着 n → ∞ ~n \to \infty~ n → ∞ Δ t ~\Delta t~ Δ t 黎曼和 （ Riemann sum \textbf{Riemann sum} Riemann sum
lim n → ∞ ∑ j = 0 n p ( t j ) q ( t j ) Δ t = ∫ a b p ( t ) q ( t ) d t \lim_{n \to \infty} \sum_{j=0}^{n} p(t_j) q(t_j) \Delta t = \int_a^b p(t) q(t) \, dt n → ∞ lim j = 0 ∑ n p ( t j ) q ( t j ) Δ t = ∫ a b p ( t ) q ( t ) d t 于是，我们得到函数空间中的积分定义的内积：
⟨ p , q ⟩ = 1 b − a ∫ a b p ( t ) q ( t ) d t \langle p, q \rangle = \frac{1}{b - a} \int_a^b p(t) q(t) \, dt ⟨ p , q ⟩ = b − a 1 ∫ a b p ( t ) q ( t ) d t 这个内积不仅适用于多项式空间 P n ~\mathbb{P}_n~ P n C [ a , b ] ~C[a,b]~ C [ a , b ]
7.2 一般函数内积定义与验证 在一般情况下，对于任意连续函数 f , g ∈ C [ a , b ] f, g \in C[a, b] f , g ∈ C [ a , b ]
⟨ f , g ⟩ = ∫ a b f ( t ) g ( t ) d t \langle f, g \rangle = \int_a^b f(t) g(t) \, dt ⟨ f , g ⟩ = ∫ a b f ( t ) g ( t ) d t 下面验证这一定义满足内积空间的四个基本公理：
根据内积的定义：
⟨ f , g ⟩ = ∫ a b f ( t ) g ( t ) d t \langle f, g \rangle = \int_a^b f(t) g(t) \, dt ⟨ f , g ⟩ = ∫ a b f ( t ) g ( t ) d t 因为乘法是可交换的，即 f ( t ) g ( t ) = g ( t ) f ( t ) ~f(t)g(t) = g(t)f(t)~ f ( t ) g ( t ) = g ( t ) f ( t )
⟨ g , f ⟩ = ∫ a b g ( t ) f ( t ) d t = ∫ a b f ( t ) g ( t ) d t = ⟨ f , g ⟩ \langle g, f \rangle = \int_a^b g(t) f(t) \, dt = \int_a^b f(t) g(t) \, dt = \langle f, g \rangle ⟨ g , f ⟩ = ∫ a b g ( t ) f ( t ) d t = ∫ a b f ( t ) g ( t ) d t = ⟨ f , g ⟩ 因此，对称性成立。
根据内积的定义：
⟨ f + h , g ⟩ = ∫ a b ( f ( t ) + h ( t ) ) g ( t ) d t \langle f + h, g \rangle = \int_a^b (f(t) + h(t)) g(t) \, dt ⟨ f + h , g ⟩ = ∫ a b ( f ( t ) + h ( t )) g ( t ) d t 可以将积分拆分：
⟨ f + h , g ⟩ = ∫ a b f ( t ) g ( t ) d t + ∫ a b h ( t ) g ( t ) d t = ⟨ f , g ⟩ + ⟨ h , g ⟩ \langle f + h, g \rangle = \int_a^b f(t) g(t) \, dt + \int_a^b h(t) g(t) \, dt = \langle f, g \rangle + \langle h, g \rangle ⟨ f + h , g ⟩ = ∫ a b f ( t ) g ( t ) d t + ∫ a b h ( t ) g ( t ) d t = ⟨ f , g ⟩ + ⟨ h , g ⟩ 因此，加法线性成立。
根据内积的定义：
⟨ c f , g ⟩ = ∫ a b ( c f ( t ) ) g ( t ) d t \langle c f, g \rangle = \int_a^b (c f(t)) g(t) \, dt ⟨ c f , g ⟩ = ∫ a b ( c f ( t )) g ( t ) d t 我们可以把常数 c 提到积分外面：
⟨ c f , g ⟩ = c ∫ a b f ( t ) g ( t ) d t = c ⟨ f , g ⟩ \langle c f, g \rangle = c \int_a^b f(t) g(t) \, dt = c \langle f, g \rangle ⟨ c f , g ⟩ = c ∫ a b f ( t ) g ( t ) d t = c ⟨ f , g ⟩ 因此，标量线性成立。
根据内积的定义：
⟨ f , f ⟩ = ∫ a b f ( t ) 2 d t \langle f, f \rangle = \int_a^b f(t)^2 \, dt ⟨ f , f ⟩ = ∫ a b f ( t ) 2 d t 由于 f ( t ) 2 ≥ 0 f(t)^2 \geq 0 f ( t ) 2 ≥ 0 t ∈ [ a , b ] ~t \in [a, b]~ t ∈ [ a , b ] f ( t ) 2 = 0 ~f(t)^2 = 0~ f ( t ) 2 = 0 f ( t ) = 0 f(t) = 0 f ( t ) = 0
⟨ f , f ⟩ ≥ 0 \langle f, f \rangle \geq 0 ⟨ f , f ⟩ ≥ 0 而且只有当 f ( t ) = 0 f(t) = 0 f ( t ) = 0 t ∈ [ a , b ] t \in [a, b] t ∈ [ a , b ] ⟨ f , f ⟩ = 0 \langle f, f \rangle = 0 ⟨ f , f ⟩ = 0
7.3 连续函数空间的正交化过程 接下来，我们通过一个具体的例子来演示如何在连续函数空间中应用 Gram-Schmidt ~\text{Gram-Schmidt}~ Gram-Schmidt V = C [ 0 , 1 ] ~V = C[0, 1]~ V = C [ 0 , 1 ] p 1 ( t ) = 1 , p 2 ( t ) = 2 t − 1 , p 3 ( t ) = 12 t 2 ~p_1(t) = 1,~p_2(t) = 2t - 1,~p_3(t) = 12t^2~ p 1 ( t ) = 1 , p 2 ( t ) = 2 t − 1 , p 3 ( t ) = 12 t 2 W ~W~ W Gram-Schmidt ~\text{Gram-Schmidt}~ Gram-Schmidt W ~W~ W
解
使用 Gram-Schmidt \text{Gram-Schmidt} Gram-Schmidt
选择第一个多项式作为正交基 q ~q_~ q
选择 q 1 = p 1 ~q_1 = p_1~ q 1 = p 1
构造第二个正交基 q 2 ~q_2~ q 2
通过计算 ⟨ p 2 , q 1 ⟩ \langle p_2, q_1 \rangle ⟨ p 2 , q 1 ⟩
⟨ p 2 , q 1 ⟩ = ∫ 0 1 ( 2 t − 1 ) ( 1 ) d t = [ t 2 − t ] 0 1 = 0 \left\langle p_2, q_1 \right\rangle = \int_0^1 (2t - 1)(1) dt = \left[ t^2 - t \right]_0^1 = 0 ⟨ p 2 , q 1 ⟩ = ∫ 0 1 ( 2 t − 1 ) ( 1 ) d t = [ t 2 − t ] 0 1 = 0 由于内积为零，说明 p 2 ~p_2~ p 2 q 1 ~q_1~ q 1 q 2 = p 2 ~q_2 = p_2~ q 2 = p 2
构造第三个正交基 q 3 ~q_3~ q 3
利用如下公式计算 q 3 ~q_3~ q 3
q 3 = p 3 − proj W 2 p 3 q_3 = p_3 - \text{proj}_{W_2} p_3 q 3 = p 3 − proj W 2 p 3
W 2 W_2~ W 2 q 1 ~q_1~ q 1 q 2 q_2 q 2
proj W 2 p 3 = ⟨ p 3 , q 1 ⟩ ⟨ q 1 , q 1 ⟩ q 1 + ⟨ p 3 , q 2 ⟩ ⟨ q 2 , q 2 ⟩ q 2 \text{proj}_{W_2} p_3 = \frac{\langle p_3, q_1 \rangle}{\langle q_1, q_1 \rangle} q_1 + \frac{\langle p_3, q_2 \rangle}{\langle q_2, q_2 \rangle} q_2 proj W 2 p 3 = ⟨ q 1 , q 1 ⟩ ⟨ p 3 , q 1 ⟩ q 1 + ⟨ q 2 , q 2 ⟩ ⟨ p 3 , q 2 ⟩ q 2 对于 p 3 ( t ) = 12 t 2 ~p_3(t) = 12t^2~ p 3 ( t ) = 12 t 2 ⟨ p 3 , q 1 ⟩ ~\langle p_3, q_1\rangle ⟨ p 3 , q 1 ⟩ ⟨ p 3 , q 1 ⟩ \langle p_3, q_1\rangle ⟨ p 3 , q 1 ⟩ ⟨ q 1 , q 1 ⟩ \langle q_1, q_1\rangle ⟨ q 1 , q 1 ⟩ ⟨ q 2 , q 2 ⟩ \langle q_2, q_2 \rangle~ ⟨ q 2 , q 2 ⟩
⟨ p 3 , q 1 ⟩ = ∫ 0 1 12 t 2 ⋅ 1 d t = [ t 3 ] 0 1 = 4 ⟨ q 1 , q 1 ⟩ = ∫ 0 1 1 ⋅ 1 d t = [ t ] 0 1 = 1 ⟨ p 3 , q 2 ⟩ = ∫ 0 1 12 t 2 ( 2 t − 1 ) d t = ∫ 0 1 ( 24 t 3 − 12 t 2 ) d t = 2 ⟨ q 2 , q 2 ⟩ = ∫ 0 1 ( 2 t − 1 ) 2 d t = 1 6 ∫ 0 1 ( 2 t − 1 ) 3 d t = 1 3 \begin{aligned}
\langle p_3, q_1 \rangle &= \int_0^1 12t^2 \cdot 1 \, dt = \left[t^3 \right]_0^1 = 4 \\[2ex]
\langle q_1, q_1 \rangle &= \int_0^1 1 \cdot 1 \, dt = \left[t \right]_0^1 = 1 \\[2ex]
\langle p_3, q_2 \rangle &= \int_0^1 12t^2(2t - 1) \, dt = \int_0^1 (24t^3 - 12t^2) \, dt = 2 \\[2ex]
\langle q_2, q_2 \rangle &= \int_0^1 (2t - 1)^2 \, dt = \frac{1}{6} \int_0^1 (2t - 1)^3 \, dt = \frac{1}{3}
\end{aligned} ⟨ p 3 , q 1 ⟩ ⟨ q 1 , q 1 ⟩ ⟨ p 3 , q 2 ⟩ ⟨ q 2 , q 2 ⟩ = ∫ 0 1 12 t 2 ⋅ 1 d t = [ t 3 ] 0 1 = 4 = ∫ 0 1 1 ⋅ 1 d t = [ t ] 0 1 = 1 = ∫ 0 1 12 t 2 ( 2 t − 1 ) d t = ∫ 0 1 ( 24 t 3 − 12 t 2 ) d t = 2 = ∫ 0 1 ( 2 t − 1 ) 2 d t = 6 1 ∫ 0 1 ( 2 t − 1 ) 3 d t = 3 1 代入投影公式可得，
proj W 2 p 3 = ⟨ p 3 , q 1 ⟩ ⟨ q 1 , q 1 ⟩ q 1 + ⟨ p 3 , q 2 ⟩ ⟨ q 2 , q 2 ⟩ q 2 = 4 1 q 1 + 2 1 3 q 2 = 4 q 1 + 6 q 2 \begin{aligned}
\text{proj}_{W_2} p_3 &= \frac{\langle p_3, q_1 \rangle}{\langle q_1, q_1 \rangle} q_1 + \frac{\langle p_3, q_2 \rangle}{\langle q_2, q_2 \rangle} q_2 \\[3ex]
&= \frac{4}{1} q_1 + \frac{2}{\frac{1}{3}} q_2 \\[3ex]
&= 4 q_1 + 6 q_2
\end{aligned} proj W 2 p 3 = ⟨ q 1 , q 1 ⟩ ⟨ p 3 , q 1 ⟩ q 1 + ⟨ q 2 , q 2 ⟩ ⟨ p 3 , q 2 ⟩ q 2 = 1 4 q 1 + 3 1 2 q 2 = 4 q 1 + 6 q 2 最后结果：
q 3 = p 3 − proj W 2 p 3 = 12 t 2 − 4 − 6 ( 2 t − 1 ) = 12 t 2 − 12 t + 2 \begin{aligned}
q_3 &= p_3 - \text{proj}_{W_2} p_3\\[2ex]
&= 12t^2 - 4 - 6(2t - 1)\\[2ex]
&= 12t^2 - 12t + 2
\end{aligned} q 3 = p 3 − proj W 2 p 3 = 12 t 2 − 4 − 6 ( 2 t − 1 ) = 12 t 2 − 12 t + 2
最终，子空间 W ~W~ W
{ q 1 , q 2 , q 3 } = { 1 , 2 t − 1 , 12 t 2 − 12 t + 2 } \{ q_1, ~ q_2, ~ q_3 \} = \{ 1, ~ 2t - 1, ~ 12t^2 - 12t + 2 \} { q 1 , q 2 , q 3 } = { 1 , 2 t − 1 , 12 t 2 − 12 t + 2 } 
