實內積、誘導範數與所需不等式
實向量空間上的內積是取值於實數的映射 ⟨ u , v ⟩ \langle u,v\rangle ⟨ u , v ⟩ v ≠ 0 v\ne0 v = 0 ⟨ v , v ⟩ > 0 \langle v,v\rangle>0 ⟨ v , v ⟩ > 0 ⟨ 0 , v ⟩ = 0 \langle0,v\rangle=0 ⟨ 0 , v ⟩ = 0 ∥ v ∥ = ⟨ v , v ⟩ \|v\|=\sqrt{\langle v,v\rangle} ∥ v ∥ = ⟨ v , v ⟩ ∥ c v ∥ = ∣ c ∣ ∥ v ∥ \|cv\|=|c|\|v\| ∥ c v ∥ = ∣ c ∣∥ v ∥ R m \mathbb R^m R m ⟨ u , v ⟩ = ∑ i = 1 m u i v i \langle u,v\rangle=\sum_{i=1}^mu_iv_i ⟨ u , v ⟩ = ∑ i = 1 m u i v i
若 v ≠ 0 v\ne0 v = 0 t = ⟨ u , v ⟩ / ∥ v ∥ 2 t=\langle u,v\rangle/\|v\|^2 t = ⟨ u , v ⟩ /∥ v ∥ 2 ∥ u − t v ∥ 2 \|u-tv\|^2 ∥ u − t v ∥ 2 ∥ u ∥ 2 − ⟨ u , v ⟩ 2 / ∥ v ∥ 2 ≥ 0 \|u\|^2-\langle u,v\rangle^2/\|v\|^2\ge0 ∥ u ∥ 2 − ⟨ u , v ⟩ 2 /∥ v ∥ 2 ≥ 0 ∣ ⟨ u , v ⟩ ∣ ≤ ∥ u ∥ ∥ v ∥ |\langle u,v\rangle|\le\|u\|\|v\| ∣ ⟨ u , v ⟩ ∣ ≤ ∥ u ∥∥ v ∥ v = 0 v=0 v = 0 v ≠ 0 v\ne0 v = 0 ∥ u − t v ∥ = 0 \|u-tv\|=0 ∥ u − t v ∥ = 0 u = t v u=tv u = t v ∥ u + v ∥ 2 \|u+v\|^2 ∥ u + v ∥ 2 ∥ u + v ∥ 2 ≤ ( ∥ u ∥ + ∥ v ∥ ) 2 \|u+v\|^2\le(\|u\|+\|v\|)^2 ∥ u + v ∥ 2 ≤ ( ∥ u ∥ + ∥ v ∥ ) 2
不預設計數結論的有限基存在與擴充
設實向量空間 V V V t 1 , … , t N t_1,\ldots,t_N t 1 , … , t N u 1 , … , u k u_1,\ldots,u_k u 1 , … , u k N N N j j j u 1 , … , u j u_1,\ldots,u_j u 1 , … , u j N − j N-j N − j u j + 1 u_{j+1} u j + 1 u i u_i u i u j + 1 u_{j+1} u j + 1 k > N k>N k > N N N N N N N V V V u N + 1 u_{N+1} u N + 1 N = 0 N=0 N = 0
每個子空間 W W W W W W W W W N N N W W W W = { 0 } W=\{0\} W = { 0 } W W W W W W p , q p,q p , q p ≤ q p\le q p ≤ q q ≤ p q\le p q ≤ p
正交規範基，以及保留指定向量的擴充
再設 V V V v 1 , … , v s v_1,\ldots,v_s v 1 , … , v s
w j = v j − ∑ i < j ⟨ v j , q i ⟩ q i , q j = w j / ∥ w j ∥ . w_j=v_j-\sum_{i<j}\langle v_j,q_i\rangle q_i,\qquad q_j=w_j/\|w_j\|. w j = v j − i < j ∑ ⟨ v j , q i ⟩ q i , q j = w j /∥ w j ∥. 歸納假設先前 q i q_i q i l < j l<j l < j q l q_l q l ⟨ w j , q l ⟩ = ⟨ v j , q l ⟩ − ⟨ v j , q l ⟩ = 0 \langle w_j,q_l\rangle=\langle v_j,q_l\rangle-\langle v_j,q_l\rangle=0 ⟨ w j , q l ⟩ = ⟨ v j , q l ⟩ − ⟨ v j , q l ⟩ = 0 w j = 0 w_j=0 w j = 0 v j v_j v j j j j
正交基令內積和坐標計算更容易，但題目給出的基底通常只是一組任意的線性無關向量。
Gram-Schmidt 按次序逐步扣除已由較早方向解釋的成分，將它變成正交向量組。輸入次序
很重要：次序改變可能令輸出不同，但輸出仍然張成同一個子空間。
投影想法
先考慮一個非零向量 v v v w w w w w w v v v v v v c c c
0 = ⟨ w − c v , v ⟩ = ⟨ w , v ⟩ − c ∥ v ∥ 2 , c = ⟨ w , v ⟩ ∥ v ∥ 2 . 0=\langle w-cv,v\rangle=\langle w,v\rangle-c\|v\|^2,
\qquad c=\frac{\langle w,v\rangle}{\|v\|^2}. 0 = ⟨ w − c v , v ⟩ = ⟨ w , v ⟩ − c ∥ v ∥ 2 , c = ∥ v ∥ 2 ⟨ w , v ⟩ . 由於 v ≠ 0 v\ne0 v = 0
假設 v 1 , … , v k v_1,\dots,v_k v 1 , … , v k w ∈ R m w\in\mathbb{R}^m w ∈ R m w w w v i v_i v i
⟨ w , v i ⟩ ∥ v i ∥ 2 . \frac{\langle w,v_i\rangle}{\|v_i\|^2}. ∥ v i ∥ 2 ⟨ w , v i ⟩ . 扣除所有舊有正交方向後留下的部分是
r = w − ∑ i = 1 k ⟨ w , v i ⟩ ∥ v i ∥ 2 v i . r=w-\sum_{i=1}^{k}\frac{\langle w,v_i\rangle}{\|v_i\|^2}v_i. r = w − i = 1 ∑ k ∥ v i ∥ 2 ⟨ w , v i ⟩ v i . 分母合法，是因為正交集不含零向量。當 k = 0 k=0 k = 0 r = w r=w r = w
定理
正交餘量定理 設 S = { v 1 , … , v k } S=\{v_1,\dots,v_k\} S = { v 1 , … , v k } R m \mathbb{R}^m R m w ∈ R m w\in\mathbb{R}^m w ∈ R m
r = w − ∑ j = 1 k ⟨ w , v j ⟩ ∥ v j ∥ 2 v j . r=w-\sum_{j=1}^{k}\frac{\langle w,v_j\rangle}{\|v_j\|^2}v_j. r = w − j = 1 ∑ k ∥ v j ∥ 2 ⟨ w , v j ⟩ v j . 則對每個 i i i ⟨ r , v i ⟩ = 0 \langle r,v_i\rangle=0 ⟨ r , v i ⟩ = 0 r r r S S S
固定一個 v i v_i v i
⟨ r , v i ⟩ = ⟨ w , v i ⟩ − ∑ j = 1 k ⟨ w , v j ⟩ ∥ v j ∥ 2 ⟨ v j , v i ⟩ . \langle r,v_i\rangle
=\langle w,v_i\rangle-
\sum_{j=1}^{k}
\frac{\langle w,v_j\rangle}{\|v_j\|^2}\langle v_j,v_i\rangle. ⟨ r , v i ⟩ = ⟨ w , v i ⟩ − j = 1 ∑ k ∥ v j ∥ 2 ⟨ w , v j ⟩ ⟨ v j , v i ⟩ . 當 j ≠ i j\ne i j = i ⟨ w , v i ⟩ − ⟨ w , v i ⟩ = 0 \langle w,v_i\rangle-\langle w,v_i\rangle=0 ⟨ w , v i ⟩ − ⟨ w , v i ⟩ = 0 r = 0 r=0 r = 0 w w w w w w
投影與張成空間判別
對正交列表 S = { v 1 , … , v k } S=\{v_1,\dots,v_k\} S = { v 1 , … , v k }
r ( w ) = w − ∑ i = 1 k ⟨ w , v i ⟩ ∥ v i ∥ 2 v i . r(w)=w-\sum_{i=1}^{k}\frac{\langle w,v_i\rangle}{\|v_i\|^2}v_i. r ( w ) = w − i = 1 ∑ k ∥ v i ∥ 2 ⟨ w , v i ⟩ v i . 正交餘量定理給出 r ( w ) ⊥ span S r(w)\perp\operatorname{span}S r ( w ) ⊥ span S
r ( w ) = 0 ⟺ w ∈ span S . r(w)=0
\quad\Longleftrightarrow\quad
w\in\operatorname{span}S. r ( w ) = 0 ⟺ w ∈ span S . 正向由投影公式立即得到。反過來，若 w = ∑ i α i v i w=\sum_i\alpha_i v_i w = ∑ i α i v i v i v_i v i α i = ⟨ w , v i ⟩ / ∥ v i ∥ 2 \alpha_i=\langle w,v_i\rangle/\|v_i\|^2 α i = ⟨ w , v i ⟩ /∥ v i ∥ 2 r ( w ) = 0 r(w)=0 r ( w ) = 0 w w w
證明
投影與正交餘量 令 U = span S U=\operatorname{span}S U = span S
p ( w ) = ∑ i = 1 k ⟨ w , v i ⟩ ∥ v i ∥ 2 v i , r = w − p ( w ) . p(w)=\sum_{i=1}^{k}\frac{\langle w,v_i\rangle}{\|v_i\|^2}v_i,
\qquad r=w-p(w). p ( w ) = i = 1 ∑ k ∥ v i ∥ 2 ⟨ w , v i ⟩ v i , r = w − p ( w ) . 則 p ( w ) ∈ U p(w)\in U p ( w ) ∈ U r r r v i v_i v i r ⊥ U r\perp U r ⊥ U w = p + r w=p+r w = p + r S S S 0 0 0 U = { 0 } U=\{0\} U = { 0 } w = 0 + w w=0+w w = 0 + w
若還有 w = p ′ + r ′ w=p'+r' w = p ′ + r ′ p ′ ∈ U p'\in U p ′ ∈ U r ′ ⊥ U r'\perp U r ′ ⊥ U z = p − p ′ = r ′ − r z=p-p'=r'-r z = p − p ′ = r ′ − r z ∈ U z\in U z ∈ U z ⊥ U z\perp U z ⊥ U ⟨ z , z ⟩ = 0 \langle z,z\rangle=0 ⟨ z , z ⟩ = 0 z = 0 z=0 z = 0 p = p ′ p=p' p = p ′ r = r ′ r=r' r = r ′ U U U
這個分解精確說明「刪去舊方向」的含義。p ( w ) p(w) p ( w ) r r r w ∈ U w\in U w ∈ U r = 0 r=0 r = 0 w ∉ U w\notin U w ∈ / U r ≠ 0 r\ne0 r = 0 U U U z ∈ U ∩ U ⊥ z\in U\cap U^\perp z ∈ U ∩ U ⊥ ⟨ z , z ⟩ = 0 \langle z,z\rangle=0 ⟨ z , z ⟩ = 0 z = 0 z=0 z = 0
Gram-Schmidt 過程
定理
Gram-Schmidt 正交化過程 設 k ≥ 1 k\ge1 k ≥ 1 w 1 , … , w k w_1,\dots,w_k w 1 , … , w k R m \mathbb{R}^m R m
v 1 = w 1 , v_1=w_1, v 1 = w 1 , 並對 ℓ = 2 , … , k \ell=2,\dots,k ℓ = 2 , … , k
v ℓ = w ℓ − ∑ j = 1 ℓ − 1 ⟨ w ℓ , v j ⟩ ∥ v j ∥ 2 v j . v_\ell=w_\ell-\sum_{j=1}^{\ell-1}
\frac{\langle w_\ell,v_j\rangle}{\|v_j\|^2}v_j. v ℓ = w ℓ − j = 1 ∑ ℓ − 1 ∥ v j ∥ 2 ⟨ w ℓ , v j ⟩ v j . 則每個 v ℓ v_\ell v ℓ v 1 , … , v k v_1,\dots,v_k v 1 , … , v k ℓ \ell ℓ
span { w 1 , … , w ℓ } = span { v 1 , … , v ℓ } . \operatorname{span}\{w_1,\dots,w_\ell\}
=
\operatorname{span}\{v_1,\dots,v_\ell\}. span { w 1 , … , w ℓ } = span { v 1 , … , v ℓ } . 空輸入列表另行處理：它的張成空間是零子空間 { 0 } \{0\} { 0 } v 1 v_1 v 1
每一步必須使用已經構造的正交向量 v j v_j v j w j w_j w j w j w_j w j
證明
歸納中的三個不變量 我們同時證明第 ℓ \ell ℓ v 1 = w 1 ≠ 0 v_1=w_1\ne0 v 1 = w 1 = 0
假設這三件事已對 ℓ − 1 \ell-1 ℓ − 1 v j v_j v j v ℓ v_\ell v ℓ v ℓ ⊥ v j v_\ell\perp v_j v ℓ ⊥ v j j < ℓ j<\ell j < ℓ
定義式說明 v ℓ v_\ell v ℓ w ℓ w_\ell w ℓ v j v_j v j v v v w w w
w ℓ = v ℓ + ∑ j = 1 ℓ − 1 ⟨ w ℓ , v j ⟩ ∥ v j ∥ 2 v j , w_\ell=v_\ell+
\sum_{j=1}^{\ell-1}
\frac{\langle w_\ell,v_j\rangle}{\|v_j\|^2}v_j, w ℓ = v ℓ + j = 1 ∑ ℓ − 1 ∥ v j ∥ 2 ⟨ w ℓ , v j ⟩ v j , 便得到反向包含，兩個部分張成空間相等。最後，若 v ℓ = 0 v_\ell=0 v ℓ = 0 w ℓ w_\ell w ℓ span { v 1 , … , v ℓ − 1 } = span { w 1 , … , w ℓ − 1 } \operatorname{span}\{v_1,\dots,v_{\ell-1}\}
=\operatorname{span}\{w_1,\dots,w_{\ell-1}\} span { v 1 , … , v ℓ − 1 } = span { w 1 , … , w ℓ − 1 }
因此正交性不只是若干個點積為零；正交集還必須不含零向量。這個非零檢查令後面的正規化
有合法分母。
在三維空間中扣除已有方向
例題
R 3 \mathbb R^3 R 3 取有序向量組
w 1 = [ 1 0 1 ] , w 2 = [ 1 1 1 ] , w 3 = [ 0 1 2 ] . w_1=\begin{bmatrix}1\\0\\1\end{bmatrix},\qquad
w_2=\begin{bmatrix}1\\1\\1\end{bmatrix},\qquad
w_3=\begin{bmatrix}0\\1\\2\end{bmatrix}. w 1 = 1 0 1 , w 2 = 1 1 1 , w 3 = 0 1 2 . 以它們為欄的矩陣行列式是 2 2 2
v 1 = w 1 = [ 1 0 1 ] , ∥ v 1 ∥ 2 = 2. v_1=w_1=\begin{bmatrix}1\\0\\1\end{bmatrix},
\qquad \|v_1\|^2=2. v 1 = w 1 = 1 0 1 , ∥ v 1 ∥ 2 = 2. 第二步的係數為
⟨ w 2 , v 1 ⟩ ∥ v 1 ∥ 2 = 2 2 = 1 , v 2 = w 2 − v 1 = [ 0 1 0 ] . \frac{\langle w_2,v_1\rangle}{\|v_1\|^2}
=\frac{2}{2}=1,\qquad
v_2=w_2-v_1=\begin{bmatrix}0\\1\\0\end{bmatrix}. ∥ v 1 ∥ 2 ⟨ w 2 , v 1 ⟩ = 2 2 = 1 , v 2 = w 2 − v 1 = 0 1 0 . 第三步必須同時處理兩個已構造的方向：
⟨ w 3 , v 1 ⟩ ∥ v 1 ∥ 2 = 2 2 = 1 , ⟨ w 3 , v 2 ⟩ ∥ v 2 ∥ 2 = 1 1 = 1 , \frac{\langle w_3,v_1\rangle}{\|v_1\|^2}
=\frac{2}{2}=1,\qquad
\frac{\langle w_3,v_2\rangle}{\|v_2\|^2}
=\frac{1}{1}=1, ∥ v 1 ∥ 2 ⟨ w 3 , v 1 ⟩ = 2 2 = 1 , ∥ v 2 ∥ 2 ⟨ w 3 , v 2 ⟩ = 1 1 = 1 , 所以
v 3 = w 3 − v 1 − v 2 = [ − 1 0 1 ] . v_3=w_3-v_1-v_2=\begin{bmatrix}-1\\0\\1\end{bmatrix}. v 3 = w 3 − v 1 − v 2 = − 1 0 1 . 三個平方範數是 2 , 1 , 2 2,1,2 2 , 1 , 2
⟨ v 1 , v 2 ⟩ = ⟨ v 1 , v 3 ⟩ = ⟨ v 2 , v 3 ⟩ = 0. \langle v_1,v_2\rangle=\langle v_1,v_3\rangle
=\langle v_2,v_3\rangle=0. ⟨ v 1 , v 2 ⟩ = ⟨ v 1 , v 3 ⟩ = ⟨ v 2 , v 3 ⟩ = 0. 張成空間保持也可直接看出：
w 2 = v 1 + v 2 , w 3 = v 1 + v 2 + v 3 . w_2=v_1+v_2,\qquad w_3=v_1+v_2+v_3. w 2 = v 1 + v 2 , w 3 = v 1 + v 2 + v 3 . 因此輸出與輸入張成完全相同的 R 3 \mathbb{R}^3 R 3
第三步的算術可在接受答案前逐項核對：⟨ w 3 , v 1 ⟩ = 0 ⋅ 1 + 1 ⋅ 0 + 2 ⋅ 1 = 2 \langle w_3,v_1\rangle=0\cdot1+1\cdot0+2\cdot1=2 ⟨ w 3 , v 1 ⟩ = 0 ⋅ 1 + 1 ⋅ 0 + 2 ⋅ 1 = 2 ⟨ w 3 , v 2 ⟩ = 0 ⋅ 0 + 1 ⋅ 1 + 2 ⋅ 0 = 1 \langle w_3,v_2\rangle=0\cdot0+1\cdot1+2\cdot0=1 ⟨ w 3 , v 2 ⟩ = 0 ⋅ 0 + 1 ⋅ 1 + 2 ⋅ 0 = 1 ( 1 , 0 , 1 ) T + ( 0 , 1 , 0 ) T (1,0,1)^T+(0,1,0)^T ( 1 , 0 , 1 ) T + ( 0 , 1 , 0 ) T ( − 1 , 0 , 1 ) T (-1,0,1)^T ( − 1 , 0 , 1 ) T
反例模式
為甚麼不能用原來的 w 向量代替 v 向量 若錯誤地用原向量計算第三步，會得到
w 3 − ⟨ w 3 , w 1 ⟩ ∥ w 1 ∥ 2 w 1 − ⟨ w 3 , w 2 ⟩ ∥ w 2 ∥ 2 w 2 = w 3 − w 1 − w 2 = [ − 2 0 0 ] . w_3-
\frac{\langle w_3,w_1\rangle}{\|w_1\|^2}w_1-
\frac{\langle w_3,w_2\rangle}{\|w_2\|^2}w_2
=w_3-w_1-w_2=\begin{bmatrix}-2\\0\\0\end{bmatrix}. w 3 − ∥ w 1 ∥ 2 ⟨ w 3 , w 1 ⟩ w 1 − ∥ w 2 ∥ 2 ⟨ w 3 , w 2 ⟩ w 2 = w 3 − w 1 − w 2 = − 2 0 0 . 這個向量與 w 1 w_1 w 1 − 2 -2 − 2 v 1 , v 2 v_1,v_2 v 1 , v 2 v 3 = ( − 1 , 0 , 1 ) T v_3=(-1,0,1)^T v 3 = ( − 1 , 0 , 1 ) T
準確保留投影係數中的分數
這裏的投影係數是分數。每次相減時保留精確分數，再核對正交性與張成空間相等。
例題
三維空間中的分數投影係數 從
w 1 = ( 1 , 1 , 0 ) T , w 2 = ( 1 , 0 , 1 ) T , w 3 = ( 0 , 1 , 1 ) T w_1=(1,1,0)^T,\qquad w_2=(1,0,1)^T,\qquad w_3=(0,1,1)^T w 1 = ( 1 , 1 , 0 ) T , w 2 = ( 1 , 0 , 1 ) T , w 3 = ( 0 , 1 , 1 ) T 開始。
以這些向量為欄的行列式是 − 2 -2 − 2 ∥ v 2 ∥ 2 = 3 / 2 \|v_2\|^2=3/2 ∥ v 2 ∥ 2 = 3/2 ⟨ w 3 , v 2 ⟩ = 0 ( 1 / 2 ) + 1 ( − 1 / 2 ) + 1 ( 1 ) = 1 / 2 \langle w_3,v_2\rangle=0(1/2)+1(-1/2)+1(1)=1/2 ⟨ w 3 , v 2 ⟩ = 0 ( 1/2 ) + 1 ( − 1/2 ) + 1 ( 1 ) = 1/2 ( 1 / 2 ) / ( 3 / 2 ) = 1 / 3 (1/2)/(3/2)=1/3 ( 1/2 ) / ( 3/2 ) = 1/3 v 3 v_3 v 3 4 / 3 4/3 4/3 v 1 ⋅ v 2 = v 1 ⋅ v 3 = v 2 ⋅ v 3 = 0 v_1\cdot v_2=v_1\cdot v_3=v_2\cdot v_3=0 v 1 ⋅ v 2 = v 1 ⋅ v 3 = v 2 ⋅ v 3 = 0
正規化與邊界情況
定理
由正交基得到標準正交基 若 v 1 , … , v k v_1,\dots,v_k v 1 , … , v k
u i = v i ∥ v i ∥ . u_i=\frac{v_i}{\|v_i\|}. u i = ∥ v i ∥ v i . 每個 u i u_i u i 1 1 1
例題
正規化第一個 R 3 \mathbb R^3 R 3 對第一個 R 3 \mathbb R^3 R 3
∥ v 1 ∥ = 2 , ∥ v 2 ∥ = 1 , ∥ v 3 ∥ = 2 . \|v_1\|=\sqrt2,\qquad \|v_2\|=1,\qquad \|v_3\|=\sqrt2. ∥ v 1 ∥ = 2 , ∥ v 2 ∥ = 1 , ∥ v 3 ∥ = 2 . 因此
u 1 = 1 2 [ 1 0 1 ] , u 2 = [ 0 1 0 ] , u 3 = 1 2 [ − 1 0 1 ] . u_1=\frac1{\sqrt2}\begin{bmatrix}1\\0\\1\end{bmatrix},\qquad
u_2=\begin{bmatrix}0\\1\\0\end{bmatrix},\qquad
u_3=\frac1{\sqrt2}\begin{bmatrix}-1\\0\\1\end{bmatrix}. u 1 = 2 1 1 0 1 , u 2 = 0 1 0 , u 3 = 2 1 − 1 0 1 . 這裏沿用第一個輸出，而不是靜態表格的另一組輸入。
因為範數為正，u i u_i u i v i v_i v i ∥ v i ∥ \|v_i\| ∥ v i ∥
單位長度可直接核對：∥ u 1 ∥ 2 = ( 1 / 2 ) ∥ v 1 ∥ 2 = 1 \|u_1\|^2=(1/2)\|v_1\|^2=1 ∥ u 1 ∥ 2 = ( 1/2 ) ∥ v 1 ∥ 2 = 1 ∥ u 2 ∥ 2 = 1 \|u_2\|^2=1 ∥ u 2 ∥ 2 = 1 ∥ u 3 ∥ 2 = ( 1 / 2 ) ∥ v 3 ∥ 2 = 1 \|u_3\|^2=(1/2)\|v_3\|^2=1 ∥ u 3 ∥ 2 = ( 1/2 ) ∥ v 3 ∥ 2 = 1 v i = ∥ v i ∥ u i v_i=\|v_i\|u_i v i = ∥ v i ∥ u i u i u_i u i v i v_i v i
例題
由 R 4 \mathbb R^4 R 4 令
V = N [ 1 1 1 1 ] = { x ∈ R 4 : x 1 + x 2 + x 3 + x 4 = 0 } . V=N\begin{bmatrix}1&1&1&1\end{bmatrix}
=\{x\in\mathbb{R}^4:x_1+x_2+x_3+x_4=0\}. V = N [ 1 1 1 1 ] = { x ∈ R 4 : x 1 + x 2 + x 3 + x 4 = 0 } . 寫成 x = ( − a − b − c , a , b , c ) T x=(-a-b-c,a,b,c)^T x = ( − a − b − c , a , b , c ) T V V V
w 1 = [ − 1 1 0 0 ] , w 2 = [ − 1 0 1 0 ] , w 3 = [ − 1 0 0 1 ] . w_1=\begin{bmatrix}-1\\1\\0\\0\end{bmatrix},\quad
w_2=\begin{bmatrix}-1\\0\\1\\0\end{bmatrix},\quad
w_3=\begin{bmatrix}-1\\0\\0\\1\end{bmatrix}. w 1 = − 1 1 0 0 , w 2 = − 1 0 1 0 , w 3 = − 1 0 0 1 . 這些向量線性無關：若 a 1 w 1 + a 2 w 2 + a 3 w 3 = 0 a_1w_1+a_2w_2+a_3w_3=0 a 1 w 1 + a 2 w 2 + a 3 w 3 = 0 a 1 = a 2 = a 3 = 0 a_1=a_2=a_3=0 a 1 = a 2 = a 3 = 0 V V V
Gram-Schmidt 給出
v 1 = w 1 , v 2 = w 2 − 1 2 v 1 = [ − 1 / 2 − 1 / 2 1 0 ] , v_1=w_1,\qquad
v_2=w_2-\frac12v_1=
\begin{bmatrix}-1/2\\-1/2\\1\\0\end{bmatrix}, v 1 = w 1 , v 2 = w 2 − 2 1 v 1 = − 1/2 − 1/2 1 0 , 這裏 ⟨ w 2 , v 1 ⟩ = 1 \langle w_2,v_1\rangle=1 ⟨ w 2 , v 1 ⟩ = 1 ∥ v 1 ∥ 2 = 2 \|v_1\|^2=2 ∥ v 1 ∥ 2 = 2 1 / 2 1/2 1/2 ⟨ w 3 , v 1 ⟩ = 1 \langle w_3,v_1\rangle=1 ⟨ w 3 , v 1 ⟩ = 1 ⟨ w 3 , v 2 ⟩ = 1 / 2 \langle w_3,v_2\rangle=1/2 ⟨ w 3 , v 2 ⟩ = 1/2 ∥ v 2 ∥ 2 = 3 / 2 \|v_2\|^2=3/2 ∥ v 2 ∥ 2 = 3/2 ( 1 / 2 ) / ( 3 / 2 ) = 1 / 3 (1/2)/(3/2)=1/3 ( 1/2 ) / ( 3/2 ) = 1/3
v 3 = w 3 − 1 2 v 1 − 1 3 v 2 = [ − 1 / 3 − 1 / 3 − 1 / 3 1 ] . v_3=w_3-\frac12v_1-\frac13v_2
=\begin{bmatrix}-1/3\\-1/3\\-1/3\\1\end{bmatrix}. v 3 = w 3 − 2 1 v 1 − 3 1 v 2 = − 1/3 − 1/3 − 1/3 1 . 平方範數為 2 , 3 / 2 , 4 / 3 2,3/2,4/3 2 , 3/2 , 4/3
u 1 = 1 2 [ − 1 1 0 0 ] , u 2 = 1 6 [ − 1 − 1 2 0 ] , u 3 = 1 12 [ − 1 − 1 − 1 3 ] . u_1=\frac1{\sqrt2}\begin{bmatrix}-1\\1\\0\\0\end{bmatrix},\quad
u_2=\frac1{\sqrt6}\begin{bmatrix}-1\\-1\\2\\0\end{bmatrix},\quad
u_3=\frac1{\sqrt{12}}\begin{bmatrix}-1\\-1\\-1\\3\end{bmatrix}. u 1 = 2 1 − 1 1 0 0 , u 2 = 6 1 − 1 − 1 2 0 , u 3 = 12 1 − 1 − 1 − 1 3 . 例如 ∥ v 2 ∥ 2 = 1 / 4 + 1 / 4 + 1 = 3 / 2 \|v_2\|^2=1/4+1/4+1=3/2 ∥ v 2 ∥ 2 = 1/4 + 1/4 + 1 = 3/2 ∥ v 3 ∥ 2 = 1 / 9 + 1 / 9 + 1 / 9 + 1 = 4 / 3 \|v_3\|^2=1/9+1/9+1/9+1=4/3 ∥ v 3 ∥ 2 = 1/9 + 1/9 + 1/9 + 1 = 4/3 1 1 1 v i = ∥ v i ∥ u i v_i=\|v_i\|u_i v i = ∥ v i ∥ u i u i u_i u i v i v_i v i 3 3 3 V V V
R 4 \mathbb R^4 R 4
w 2 = v 2 + 1 2 v 1 , w 3 = v 3 + 1 2 v 1 + 1 3 v 2 . w_2=v_2+\frac12v_1,\qquad
w_3=v_3+\frac12v_1+\frac13v_2. w 2 = v 2 + 2 1 v 1 , w 3 = v 3 + 2 1 v 1 + 3 1 v 2 . 反過來，每個 v i v_i v i w i w_i w i 1 1 1 V V V u 1 u_1 u 1 ( − 1 + 1 ) / 2 = 0 (-1+1)/\sqrt2=0 ( − 1 + 1 ) / 2 = 0 u 2 u_2 u 2 ( − 1 − 1 + 2 ) / 6 = 0 (-1-1+2)/\sqrt6=0 ( − 1 − 1 + 2 ) / 6 = 0 u 3 u_3 u 3 ( − 1 − 1 − 1 + 3 ) / 12 = 0 (-1-1-1+3)/\sqrt{12}=0 ( − 1 − 1 − 1 + 3 ) / 12 = 0 2 / 2 2/2 2/2 ( 1 + 1 + 4 ) / 6 (1+1+4)/6 ( 1 + 1 + 4 ) /6 ( 1 + 1 + 1 + 9 ) / 12 (1+1+1+9)/12 ( 1 + 1 + 1 + 9 ) /12 1 1 1
定理
每個有限維子空間都有標準正交基 R m \mathbb{R}^m R m
常見錯誤
相關輸入會產生零餘量 若 w 1 = ( 1 , 0 ) T w_1=(1,0)^T w 1 = ( 1 , 0 ) T w 2 = ( 2 , 0 ) T w_2=(2,0)^T w 2 = ( 2 , 0 ) T
v 2 = w 2 − ⟨ w 2 , v 1 ⟩ ∥ v 1 ∥ 2 v 1 = ( 2 , 0 ) T − 2 ( 1 , 0 ) T = 0. v_2=w_2-\frac{\langle w_2,v_1\rangle}{\|v_1\|^2}v_1
=(2,0)^T-2(1,0)^T=0. v 2 = w 2 − ∥ v 1 ∥ 2 ⟨ w 2 , v 1 ⟩ v 1 = ( 2 , 0 ) T − 2 ( 1 , 0 ) T = 0. 下一步正規化便沒有定義。應先按線性無關性檢查並刪去冗餘向量；Gram-Schmidt 的輸入應是
線性無關的有序向量組。
快速檢查
思考檢查
Gram-Schmidt 的第一個向量 v 1 v_1 v 1 解答 · 答案 v 1 = w 1 v_1=w_1 v 1 = w 1
思考檢查
Gram-Schmidt 除了正交性還保持甚麼性質？ 解答 · 答案 它保持每個部分張成空間：
span { w 1 , … , w ℓ } = span { v 1 , … , v ℓ } \operatorname{span}\{w_1,\dots,w_\ell\}=\operatorname{span}\{v_1,\dots,v_\ell\} span { w 1 , … , w ℓ } = span { v 1 , … , v ℓ }
解答 · 答案 對每個非零 v i v_i v i u i = v i / ∥ v i ∥ u_i=v_i/\|v_i\| u i = v i /∥ v i ∥
思考檢查
對 w 1 = ( 1 , 2 , 0 ) w_1=(1,2,0) w 1 = ( 1 , 2 , 0 ) w 2 = ( 2 , 0 , 1 ) w_2=(2,0,1) w 2 = ( 2 , 0 , 1 ) 解答 · 引導解答 v 1 = ( 1 , 2 , 0 ) T v_1=(1,2,0)^T v 1 = ( 1 , 2 , 0 ) T ⟨ w 2 , v 1 ⟩ / ∥ v 1 ∥ 2 = 2 / 5 \langle w_2,v_1\rangle/\|v_1\|^2=2/5 ⟨ w 2 , v 1 ⟩ /∥ v 1 ∥ 2 = 2/5
v 2 = ( 2 , 0 , 1 ) T − 2 5 ( 1 , 2 , 0 ) T = ( 8 / 5 , − 4 / 5 , 1 ) T . v_2=(2,0,1)^T-\frac25(1,2,0)^T=(8/5,-4/5,1)^T. v 2 = ( 2 , 0 , 1 ) T − 5 2 ( 1 , 2 , 0 ) T = ( 8/5 , − 4/5 , 1 ) T . 核對 v 1 ⋅ v 2 = 8 / 5 − 8 / 5 = 0 v_1\cdot v_2=8/5-8/5=0 v 1 ⋅ v 2 = 8/5 − 8/5 = 0 w 2 = v 2 + 2 5 v 1 w_2=v_2+\frac25v_1 w 2 = v 2 + 5 2 v 1
思考檢查
為甚麼獨立輸入列表出現零餘量就表示有問題？ 解答 · 引導解答 餘量為零當且僅當目前向量在先前向量的張成空間中，這與線性無關性矛盾。
思考檢查
若正交向量 v 1 , v 2 v_1,v_2 v 1 , v 2 解答 · 引導解答 它們是 u 1 = v 1 / 3 u_1=v_1/3 u 1 = v 1 /3 u 2 = v 2 / 4 u_2=v_2/4 u 2 = v 2 /4 1 1 1
思想史
Gram–Schmidt 這個名稱之前的正交化 相關筆記
先閱讀
9.2 正交集與標準正交基 ，
因為 Gram-Schmidt 建立在正交係數公式上。約束空間例子也使用
6.2 子空間 與
6.5 基底與維數 的零空間與基方法。
下一節
9.4 Cauchy-Schwarz 與三角不等式
會發展控制這些內積計算的估計。
