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【线性代数】向量组与方程组(一) – Machine World

综述

        方程组的解就是描述一个向量与一组向量的表示系数,一般表示如下:

\(\begin{align} &\begin{cases}a_{11}x_1+a_{12}x_2 + \cdots +a_{1n}x_n = b_1 \\ a_{21}x_1+a_{22}x_2 + \cdots +a_{2n}x_n = b_2 \\ \cdots \cdots \cdots \\ a_{31}x_1+a_{32}x_2 + \cdots +a_{mn}x_n = b_3\end{cases}\\ &还可以看做如下形式:\\& x_1\left(\begin{matrix} a_{11} \\ a_{21} \\ \vdots \\ a_{m1}\end{matrix}\right) + x_2 \left( \begin{matrix}a_{12} \\ a_{22} \\ \vdots \\ a_{m2}  \end{matrix}\right) + \cdots + x_n\left( \begin{matrix} a_{1n} \\ a_{2n} \\ \vdots \\ a_{mn}\end{matrix}\right) = \left( \begin{matrix} b1 \\ b2 \\ \vdots \\ b_m \end{matrix}\right) \\ & 将上述列向量记作:\alpha_1,\alpha_2,\cdots ,\alpha_n 于是将式子记作:\\ & x_1 \alpha_1 + x_2 \alpha_2 + \cdots + x_n \alpha_n = \beta\\ & 即:一个向量与一组向量的表示问题\end{align}\)

学习过程

    总体来说,学习向量组与方程组大致分为两步:

\(\begin{cases}一、定性研究 \begin{cases}①相关性问题(有无多余变量)\\②表示性问题(如何表达出多余变量)\\③代表性问题(极大无关组)\\④等价性问题(等价向量组) \end{cases} \\ 二、定量计算 \rightarrow 求通解(基础解系)\end{cases}\)

    在本片文章接下来的篇幅,只针对定性研究四大问题进行展开,定量计算见下章。

①相关性问题

\( \begin{align}&\alpha_1, \alpha_2, \cdots,\alpha_s有无多余变量\begin{cases}有\\无\end{cases} \Longleftrightarrow \\ &\begin{cases} |\begin{matrix} \alpha_1 & \alpha_2 & \cdots &\alpha_s\end{matrix}| = 0 \\ |\begin{matrix}\alpha_1&\alpha_2&\cdots & \alpha_s \end{matrix}| \neq 0\end{cases} \Longleftrightarrow\\ & \begin{cases} r(\alpha_1, \alpha_2, \cdots,\alpha_s) < s \\ r(\alpha_1, \alpha_2, \cdots,\alpha_s) = s\end{cases}\Longleftrightarrow  \\ & \begin{cases} 存在一组不全为0的数x_1,x_2,\cdots,x_s使得x_1\alpha_1+ x_2\alpha_2+\cdots+x_s\alpha_s = 0成立,\\称\alpha_1, \alpha_2, \cdots,\alpha_s线性相关\\若要有x_1\alpha_1 + x_2\alpha_2+ \cdots + x_s\alpha_s = 0成立,必须有x_1 = x_2 \cdots =x_s = 0 \\ 称\alpha_1, \alpha_2, \cdots,\alpha_s线性无关 \end{cases} \Longleftrightarrow \\ & \begin{cases}\left(\begin{matrix}\alpha_1&\alpha_2&\cdots&\alpha_s\end{matrix}\right)\left( \begin{matrix}x_1\\x_2\\ \cdots \\ x_s\end{matrix}\right) = 0 &有非零解(必定为无穷多个) \\ AX=0 \rightarrow 齐次方程组 \\ \left(\begin{matrix}\alpha_1&\alpha_2&\cdots&\alpha_s\end{matrix}\right)\left( \begin{matrix}x_1\\x_2\\ \cdots \\ x_s\end{matrix}\right) = 0 &有非零解(只有零解)\end{cases}\end{align}  \)

②表示性问题

\(\begin{align}&\beta能否由 \alpha_1,\alpha_2\cdots\alpha_s表示\begin{cases}能\\不能\end{cases} \Longleftrightarrow \\ &方形 \begin{cases} |\begin{matrix}\alpha_1&\alpha_2\cdots&\alpha&\beta \end{matrix}| = 0 \\  |\begin{matrix}\alpha_1&\alpha_2\cdots&\alpha&\beta \end{matrix} | \neq 0\end{cases} \Longleftrightarrow \\ & \begin{cases}r(\begin{matrix}\alpha_1&\alpha_2\cdots&\alpha&\beta \end{matrix} ) = r(\begin{matrix}\alpha_1&\alpha_2\cdots&\alpha \end{matrix} ) \\ r(\begin{matrix}\alpha_1&\alpha_2\cdots&\alpha&\beta \end{matrix} ) = r(\begin{matrix}\alpha_1&\alpha_2\cdots&\alpha \end{matrix} ) +1\end{cases}  \\&\begin{cases}存在一组不全为0的数x_1,x_2,\cdots,x_s使得x_1\alpha_1+ x_2\alpha_2+\cdots+x_s\alpha_s = \beta成立\\称\beta 可由\alpha_1, \alpha_2, \cdots,\alpha_s线性表出 \\不存在任何一组数x_1,x_2,\cdots,x_s使得x_1\alpha_1+ x_2\alpha_2+\cdots+x_s\alpha_s = \beta成立\\称\beta 不可由\alpha_1, \alpha_2, \cdots,\alpha_s线性表出\end{cases}\Longleftrightarrow \\ &\\ & \begin{cases}\left(\begin{matrix}\alpha_1&\alpha_2&\cdots&\alpha_s\end{matrix}\right)\left( \begin{matrix}x_1\\x_2\\ \cdots \\ x_s\end{matrix}\right) = \beta  &有解 \\ AX=\beta \rightarrow 非齐次方程组&(A|B)叫增广矩阵 \\ \left(\begin{matrix}\alpha_1&\alpha_2&\cdots&\alpha_s\end{matrix}\right)\left( \begin{matrix}x_1\\x_2\\ \cdots \\ x_s\end{matrix}\right) = \beta &无解\end{cases} \end{align}\)

③代表性问题

\(\begin{align} &定义:若\alpha_{i_1},\alpha_{i_2},\cdots,\alpha{i_r}满足:\\&①取自\alpha_1,\alpha_2,\cdots,\alpha_s \\&②线性无关\\&③ \alpha_1,\alpha_2,\cdots,\alpha_s中任一\alpha_i均可由其线性表出\\ &则称\alpha_{i_1},\alpha_{i_2},\cdots,\alpha{i_r}为原向量组 \alpha_1,\alpha_2,\cdots,\alpha_s的极大线性无关组(不唯一)\end{align}\)

④等价性问题(同维)

\(\begin{align}&设(Ⅰ)\alpha_1,\alpha2,\cdots,\alpha_s (Ⅱ)\beta_1,\beta_2,\cdots,\beta_t ,其中s并不需要与t相同 \\ &1)若\begin{cases} \alpha_1 = k_{11}\beta_1+ \cdots+k_{1t}\beta{t} \\ \vdots \\\alpha_s = k_{s1}\beta_1 + \cdots + k_{st}\beta_t\end{cases} 成立,则称(Ⅰ)可由(Ⅱ)线性表出 \\ & 2)若\begin{cases}\beta_1 = l_{11}\alpha_1 + \cdots + l_{ds}d_s \\ \vdots \\ \beta_t = l_{t1}+ \cdots+ l_{ts}d_s\end{cases}成立,则称(Ⅱ)可由(Ⅰ)线性表出 \\ &若1),2)同时成立\Rightarrow (Ⅰ)与(Ⅱ)等价\end{align}\)

参考文献:

  • 《工程数学:线性代数》同济大学出版社

作者 WellLee

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