内积

You are currently browsing articles tagged 内积.

此文写于2014年8月13日。原文里涉及n阶情形.现认为没有必要呈现一般的n阶情形,否则叙述过于复杂,只用写到4阶就够了.根据从2阶到4阶的情形完全可以脑补一般情形是怎样的.

1. {2\times 2} 线性方程组

我们来看线性方程组

\displaystyle \begin{cases} a_{11}x_1+a_{12}x_2=b_1\\ a_{21}x_1+a_{22}x_2=b_2. \end{cases} \ \ \ \ \ (1)

其中涉及到的数都是实数.把

\displaystyle \mathbf{a}_{1}=\begin{pmatrix} a_{11}\\ a_{21} \end{pmatrix},\mathbf{a}_2=\begin{pmatrix} a_{12}\\ a_{22} \end{pmatrix},\mathbf{b}=\begin{pmatrix} b_1\\ b_2 \end{pmatrix}

看作 {\mathbf{R}^2} 里的向量,且{\mathbf{a}_1,\mathbf{a}_2} 线性无 关({\mathbf{a}_1}{\mathbf{a}_2} 线性相关的情形,{x_1,x_2} 没有唯一 解.).则方程组 (1) 可以化为

\displaystyle x_1\mathbf{a}_1+x_2\mathbf{a}_2=\mathbf{b}. \ \ \ \ \ (2)

{\mathbf{R}^2} 里找到与 {\mathbf{a}_2} 正交的非零向 量 {\mathbf{a}_2'}.显然,{\mathbf{a}_2'} 可以是 {\begin{pmatrix} a_{22}\\ -a_{12} \end{pmatrix}}.在式(2) 两边点乘 {\mathbf{a}_2'} ,得到

\displaystyle x_1\mathbf{a}_1\cdot \mathbf{a}_2'=\mathbf{b}\cdot \mathbf{a}_2'. \ \ \ \ \ (3)

(3)可得

\displaystyle x_1=\frac{\mathbf{b}\cdot \mathbf{a}_2'}{\mathbf{a}_1\cdot \mathbf{a}_2'}=\frac{b_1a_{22}-b_2a_{12}}{a_{11}a_{22}-a_{21}a_{12}} =\frac{\begin{vmatrix} b_1&a_{12}\\ b_{2}&a_{22}\\ \end{vmatrix}}{ \begin{vmatrix} a_{11}&a_{12}\\ a_{21}&a_{22} \end{vmatrix}}.

下面我们来求{x_2}.把方程 (2) 写成

\displaystyle x_2\mathbf{a}_2+x_1\mathbf{a}_1=\mathbf{b}.

于是

\displaystyle x_2=\frac{ \begin{vmatrix} b_1&a_{11}\\ b_2&a_{21} \end{vmatrix} }{ \begin{vmatrix} a_{12}&a_{11}\\ a_{22}&a_{21} \end{vmatrix}}=\frac{ \begin{vmatrix} a_{11}&b_1\\ a_{21}&b_2 \end{vmatrix} }{ \begin{vmatrix} a_{11}&a_{12}\\ a_{21}&a_{22} \end{vmatrix} }.

2. {3\times 3} 线性方程组

我们来看

\displaystyle x_1 \mathbf{a}_1+x_2 \mathbf{a}_2+x_3\mathbf{a}_3=\mathbf{b}, \ \ \ \ \ (4)

其中

\displaystyle \mathbf{a}_1=\begin{pmatrix} a_{11}\\ a_{21}\\ a_{31} \end{pmatrix},\mathbf{a}_2=\begin{pmatrix} a_{12}\\ a_{22}\\ a_{32} \end{pmatrix},\mathbf{a}_{3}= \begin{pmatrix} a_{13}\\ a_{23}\\ a_{33} \end{pmatrix}, \mathbf{b}=\begin{pmatrix} b_1\\ b_2\\ b_3 \end{pmatrix}.

{\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3} 线性无关.以上涉及到的数都是 实数.设 {\mathbf{a}_{2,3}'} 是与 {\mathbf{a}_2,\mathbf{a}_3} 都正交的向 量,不妨设 {\mathbf{a}_{2,3}'= \begin{pmatrix} u_1\\ u_2\\ u_{3} \end{pmatrix}, }

\displaystyle \begin{cases} a_{12}u_1+a_{22}u_2+a_{32}u_{3}=0,\\ a_{13}u_1+a_{23}u_2+a_{33}u_{3}=0. \end{cases}

上式可以化为

\displaystyle \begin{cases} a_{12}u_1+a_{22}u_2=-u_3a_{32}\\ a_{13}u_1+a_{23}u_2=-u_3a_{33} \end{cases},

{2\times 2}方程组的情形,解得

\displaystyle u_1=\frac{ \begin{vmatrix} -u_3a_{32}&a_{22}\\ -u_3a_{33}&a_{23} \end{vmatrix} }{ \begin{vmatrix} a_{12}&a_{22}\\ a_{13}&a_{23} \end{vmatrix} }=-u_3 \frac{ \begin{vmatrix} a_{32}&a_{22}\\ a_{33}&a_{23} \end{vmatrix} }{ \begin{vmatrix} a_{12}&a_{22}\\ a_{13}&a_{23} \end{vmatrix} }=u_{3}\frac{ \begin{vmatrix} a_{22}&a_{32}\\ a_{23}&a_{33} \end{vmatrix} }{ \begin{vmatrix} a_{12}&a_{22}\\ a_{13}&a_{23} \end{vmatrix} },

\displaystyle u_2=\frac{ \begin{vmatrix} a_{12}&-u_3a_{32}\\ a_{13}&-u_3a_{33} \end{vmatrix} }{ \begin{vmatrix} a_{12}&a_{22}\\ a_{13}&a_{23} \end{vmatrix} }=-u_3 \frac{ \begin{vmatrix} a_{12}&a_{32}\\ a_{13}&a_{33} \end{vmatrix} }{ \begin{vmatrix} a_{12}&a_{22}\\ a_{13}&a_{23} \end{vmatrix} }

不妨设{u_3= \begin{vmatrix} a_{12}&a_{22}\\ a_{13}&a_{23} \end{vmatrix} },则{ u_{1}=\begin{vmatrix} a_{22}&a_{32}\\ a_{23}&a_{33} \end{vmatrix} },{u_{2}= - \begin{vmatrix} a_{12}&a_{32}\\ a_{13}&a_{33} \end{vmatrix} },于是

\displaystyle \mathbf{a}_{2,3}'= \begin{pmatrix} \begin{vmatrix} a_{22}&a_{32}\\ a_{23}&a_{33} \end{vmatrix}\\ -\begin{vmatrix} a_{12}&a_{32}\\ a_{13}&a_{33} \end{vmatrix}\\ \begin{vmatrix} a_{12}&a_{22}\\ a_{13}&a_ {23} \end{vmatrix} \end{pmatrix}= \begin{pmatrix} \begin{vmatrix} a_{22}&a_{23}\\ a_{32}&a_{33} \end{vmatrix}\\ - \begin{vmatrix} a_{12}&a_{13}\\ a_{32}&a_{33} \end{vmatrix}\\ \begin{vmatrix} a_{12}&a_{13}\\ a_{22}&a_{23} \end{vmatrix} \end{pmatrix}.

将向量 {\mathbf{a}_{2,3}'} 点乘方 程 (4) 两边,得

\displaystyle x_1\mathbf{a}_1\cdot \mathbf{a}_{2,3}'=\mathbf{b}\cdot \mathbf{a}_{2,3}'. \ \ \ \ \ (5)

 

由式(5)可得

\displaystyle x_1=\frac{\mathbf{b}\cdot \mathbf{a}_{2,3}'}{\mathbf{a}_1\cdot \mathbf{a}_{2,3}'}=\frac{b_1 \begin{vmatrix} a_{22}&a_{23}\\ a_{32}&a_{33} \end{vmatrix}-b_2 \begin{vmatrix} a_{12}&a_{13}\\ a_{32}&a_{33} \end{vmatrix}+b_3 \begin{vmatrix} a_{12}&a_{13}\\ a_{22}&a_{23} \end{vmatrix} }{a_{11} \begin{vmatrix} a_{22}&a_{23}\\ a_{32}&a_{33} \end{vmatrix}-a_{21} \begin{vmatrix} a_{12}&a_{13}\\ a_{32}&a_{33} \end{vmatrix}+a_{31} \begin{vmatrix} a_{12}&a_{13}\\ a_{22}&a_{23} \end{vmatrix}}=\frac{ \begin{vmatrix} b_{1}&a_{12}&a_{13}\\ b_{2}&a_{22}&a_{23}\\ b_{3}&a_{32}&a_{33} \end{vmatrix}}{ \begin{vmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{vmatrix} }.

将方程 (4) 化为

\displaystyle x_2\mathbf{a}_2+x_1\mathbf{a}_1+x_3\mathbf{a}_3=\mathbf{b}.

\displaystyle x_2=\frac{ \begin{vmatrix} b_{1}&a_{11}&a_{13}\\ b_{2}&a_{21}&a_{23}\\ b_{3}&a_{31}&a_{33} \end{vmatrix}}{ \begin{vmatrix} a_{12}&a_{11}&a_{13}\\ a_{22}&a_{21}&a_{23}\\ a_{32}&a_{31}&a_{33} \end{vmatrix} }=\frac{ \begin{vmatrix} a_{11}&b_{1}&a_{13}\\ a_{21}&b_{2}&a_{23}\\ a_{31}&b_3&a_{33} \end{vmatrix}}{ \begin{vmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{vmatrix} }.

类似地,可得

\displaystyle x_3=\frac{ \begin{vmatrix} a_{11}&a_{12}&b_{1}\\ a_{21}&a_{22}&b_{2}\\ a_{31}&a_{32}&b_{3} \end{vmatrix}}{ \begin{vmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{vmatrix} }

3. {4\times 4}线性方程组

继续看

\displaystyle x_1\mathbf{a}_1+x_2\mathbf{a}_2+x_3\mathbf{a}_3+x_4\mathbf{a}_4=\mathbf{b}, \ \ \ \ \ (6)

其中

\displaystyle \mathbf{a}_1= \begin{pmatrix} a_{11}\\ a_{21}\\ a_{31}\\ a_{41} \end{pmatrix},\mathbf{a}_{2}= \begin{pmatrix} a_{12}\\ a_{22}\\ a_{32}\\ a_{42} \end{pmatrix},\mathbf{a}_3= \begin{pmatrix} a_{13}\\ a_{23}\\ a_{33}\\ a_{43} \end{pmatrix},\mathbf{a}_4= \begin{pmatrix} a_{14}\\ a_{24}\\ a_{34}\\ a_{44} \end{pmatrix},\mathbf{b}= \begin{pmatrix} b_1\\ b_2\\ b_3\\ b_4 \end{pmatrix}.

而且向量{\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3,\mathbf{a}_4}线性无关.

接下来寻找与向量{\mathbf{a}_2,\mathbf{a}_3,\mathbf{a}_4}都正交的某个非 零向量,设这个向量为{\mathbf{a}_{2,3,4}'= \begin{pmatrix} u_1\\ u_2\\ u_3\\ u_4 \end{pmatrix} }.则有

\displaystyle \begin{pmatrix} a_{12}&a_{22}&a_{32}&a_{42}\\ a_{13}&a_{23}&a_{33}&a_{43}\\ a_{14}&a_{24}&a_{34}&a_{44} \end{pmatrix} \begin{pmatrix} u_1\\ u_2\\ u_3\\ u_4 \end{pmatrix}= \begin{pmatrix} 0\\ 0\\ 0\\ \end{pmatrix},

上式化为

\displaystyle \begin{pmatrix} a_{12}&a_{22}&a_{32}\\ a_{13}&a_{23}&a_{33}\\ a_{14}&a_{24}&a_{34} \end{pmatrix} \begin{pmatrix} u_1\\ u_2\\ u_3 \end{pmatrix}=-u_4 \begin{pmatrix} a_{42}\\ a_{43}\\ a_{44} \end{pmatrix}.

{u_4= -\begin{vmatrix} a_{12}&a_{22}&a_{32}\\ a_{13}&a_{23}&a_{33}\\ a_{14}&a_{24}&a_{34} \end{vmatrix} },则

\displaystyle u_1= \begin{vmatrix} a_{42}&a_{22}&a_{32}\\ a_{43}&a_{23}&a_{33}\\ a_{44}&a_{24}&a_{34} \end{vmatrix}= \begin{vmatrix} a_{22}&a_{32}&a_{42}\\ a_{23}&a_{33}&a_{43}\\ a_{24}&a_{34}&a_{44} \end{vmatrix},

\displaystyle u_2= \begin{vmatrix} a_{12}&a_{42}&a_{32}\\ a_{13}&a_{43}&a_{33}\\ a_{14}&a_{44}&a_{34} \end{vmatrix}=- \begin{vmatrix} a_{12}&a_{32}&a_{42}\\ a_{13}&a_{33}&a_{43}\\ a_{14}&a_{34}&a_{44} \end{vmatrix},

\displaystyle u_3= \begin{vmatrix} a_{12}&a_{22}&a_{42}\\ a_{13}&a_{23}&a_{43}\\ a_{14}&a_{24}&a_{44} \end{vmatrix},

所以

\displaystyle \mathbf{a}_{2,3,4}'= \begin{pmatrix} \begin{vmatrix} a_{22}&a_{32}&a_{42}\\ a_{23}&a_{33}&a_{43}\\ a_{24}&a_{34}&a_{44} \end{vmatrix}\\ -\begin{vmatrix} a_{12}&a_{32}&a_{42}\\ a_{13}&a_{33}&a_{43}\\ a_{14}&a_{34}&a_{44} \end{vmatrix}\\ \begin{vmatrix} a_{12}&a_{22}&a_{42}\\ a_{13}&a_{23}&a_{43}\\ a_{14}&a_{24}&a_{44} \end{vmatrix}\\ -\begin{vmatrix} a_{12}&a_{22}&a_{32}\\ a_{13}&a_{23}&a_{33}\\ a_{14}&a_{24}&a_{34} \end{vmatrix} \end{pmatrix}= \begin{pmatrix} \begin{vmatrix} a_{22}&a_{23}&a_{24}\\ a_{32}&a_{33}&a_{34}\\ a_{42}&a_{43}&a_{44} \end{vmatrix}\\ - \begin{vmatrix} a_{12}&a_{13}&a_{14}\\ a_{32}&a_{33}&a_{34}\\ a_{42}&a_{43}&a_{44} \end{vmatrix}\\ \begin{vmatrix} a_{12}&a_{13}&a_{14}\\ a_{22}&a_{23}&a_{24}\\ a_{42}&a_{43}&a_{44} \end{vmatrix}\\ - \begin{vmatrix} a_{12}&a_{13}&a_{14}\\ a_{22}&a_{23}&a_{24}\\ a_{32}&a_{33}&a_{34} \end{vmatrix} \end{pmatrix}.

在式(6)两边同时点乘向量{\mathbf{a}_{2,3,4}'},可得 {x_1\mathbf{a}_1\cdot \mathbf{a}_{2,3,4}'=\mathbf{b}\cdot \mathbf{a}_{2,3,4}'}, 因此

\displaystyle \begin{array}{rcl} x_1&=&\frac{b_1 \begin{vmatrix} a_{22}&a_{23}&a_{24}\\ a_{32}&a_{33}&a_{34}\\ a_{42}&a_{43}&a_{44} \end{vmatrix}-b_2\begin{vmatrix} a_{12}&a_{13}&a_{14}\\ a_{32}&a_{33}&a_{34}\\ a_{42}&a_{43}&a_{44} \end{vmatrix}+b_3 \begin{vmatrix} a_{12}&a_{13}&a_{14}\\ a_{22}&a_{23}&a_{24}\\ a_{42}&a_{43}&a_{44} \end{vmatrix}-b_4\begin{vmatrix} a_{12}&a_{13}&a_{14}\\ a_{22}&a_{23}&a_{24}\\ a_{32}&a_{33}&a_{34} \end{vmatrix} }{a_{11} \begin{vmatrix} a_{22}&a_{23}&a_{24}\\ a_{32}&a_{33}&a_{34}\\ a_{42}&a_{43}&a_{44} \end{vmatrix}-a_{21}\begin{vmatrix} a_{12}&a_{13}&a_{14}\\ a_{32}&a_{33}&a_{34}\\ a_{42}&a_{43}&a_{44} \end{vmatrix}+a_{31} \begin{vmatrix} a_{12}&a_{13}&a_{14}\\ a_{22}&a_{23}&a_{24}\\ a_{42}&a_{43}&a_{44} \end{vmatrix}-a_{41}\begin{vmatrix} a_{12}&a_{13}&a_{14}\\ a_{22}&a_{23}&a_{24}\\ a_{32}&a_{33}&a_{34} \end{vmatrix}} \\&=&\frac{ \begin{vmatrix} b_1&a_{12}&a_{13}&a_{14}\\ b_{2}&a_{22}&a_{23}&a_{24}\\ b_{3}&a_{32}&a_{33}&a_{34}\\ b_{4}&a_{42}&a_{43}&a_{44} \end{vmatrix} }{ \begin{vmatrix} a_{11}&a_{12}&a_{13}&a_{14}\\ a_{21}&a_{22}&a_{23}&a_{24}\\ a_{31}&a_{32}&a_{33}&a_{34}\\ a_{41}&a_{42}&a_{43}&a_{44} \end{vmatrix}} \end{array}

同理可得

\displaystyle x_2=\frac{ \begin{vmatrix} b_1&a_{11}&a_{13}&a_{14}\\ b_{2}&a_{21}&a_{23}&a_{24}\\ b_{3}&a_{31}&a_{33}&a_{34}\\ b_{4}&a_{41}&a_{43}&a_{44} \end{vmatrix} }{ \begin{vmatrix} a_{12}&a_{11}&a_{13}&a_{14}\\ a_{22}&a_{21}&a_{23}&a_{24}\\ a_{32}&a_{31}&a_{33}&a_{34}\\ a_{42}&a_{41}&a_{43}&a_{44} \end{vmatrix}}=\frac{ \begin{vmatrix} a_{11}&b_1&a_{13}&a_{14}\\ a_{21}&b_2&a_{23}&a_{24}\\ a_{31}&b_3&a_{33}&a_{34}\\ a_{41}&b_4&a_{43}&a_{44} \end{vmatrix} }{ \begin{vmatrix} a_{11}&a_{12}&a_{13}&a_{14}\\ a_{21}&a_{22}&a_{23}&a_{24}\\ a_{31}&a_{32}&a_{33}&a_{34}\\ a_{41}&a_{42}&a_{43}&a_{44} \end{vmatrix} },\cdots

Tags: ,