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MA&ALA2

2024-07-15 23:08:00| 来源: 网络整理| 查看: 265

注:本文是对Matrix Analysis and Applied Linear Algebra一书第二章Rectangular Systems and Echelon Forms的学习笔记

内容结构 行阶梯矩阵 (Row Echelon Form)最简行阶梯矩阵 (Reduced Row Echelon Form)矩阵的秩 (Rank)线性方程组的解 (Consistency of Linear Systems)齐次/非齐次方程组 (Homogeneous / Nonhomogeneous System)

行阶梯矩阵 (Row Echelon Form)

An m × n m \times n m×n matrix E \mathbf{E} E with rows E i ∗ \mathbf{E}_{i*} Ei∗​ and columns E ∗ j \mathbf{E}_{*j} E∗j​ is said to be in row echelon form provided the following two conditions hold.

If E i ∗ \mathbf{E}_{i*} Ei∗​ consists entirely of zeros, then all rows below E i ∗ \mathbf{E}_{i*} Ei∗​ are also entirely zero; i.e., all zero rows are at the bottom.If the first nonzero entry in E i ∗ \mathbf{E}_{i*} Ei∗​ lies in the j t h j^{th} jth position, then all entries below the i t h i^{th} ith position in columns E ∗ 1 , E ∗ 2 , ⋯   , E ∗ j \mathbf{E}_{*1},\mathbf{E}_{*2},\cdots,\mathbf{E}_{*j} E∗1​,E∗2​,⋯,E∗j​ are zero. These two conditions say that the nonzero entries in an echelon form must lie on or above a stair-step line that emanates from the upper-left-hand cornor and slopes down and to the right. The pivots are the first nonzero entries in each row. A typical structure for a matrix in row echelon form is illustrated below with the pivots circled. 在这里插入图片描述

基本列 (basic columns): 假设原矩阵 A m × n \mathbf{A}_{m \times n} Am×n​被化为行阶梯矩阵 E \mathbf E E, E \mathbf E E取值不唯一。 A \mathbf A A的基本列定义为 A \mathbf A A中包含pivots的那些列。

基本列为什么“基本”?也即,它有怎样的性质? 下面将证明,矩阵的所有列都可以通过基本列的线性组合表示出来。

最简行阶梯矩阵 (Reduced Row Echelon Form)

A matrix E m × n \mathbf{E}_{m \times n} Em×n​ is said to be in reduced row echelon form provided that the following three conditions hold.

E \mathbf E E is in row echelon form.The first nonzero entry in each row (i.e., each pivot) is 1.All entries above each pivot are 0. A typical structure for a matrix in reduced row echelon form is illustrated below, where entries marked * can be either zero or nonzero numbers: 在这里插入图片描述

定义 E A \mathbf{E}_\mathbf A EA​表示 A \mathbf A A的最简行阶梯矩阵, E A \mathbf{E}_\mathbf A EA​是由 A \mathbf A A唯一确定的。

下面我们来探究一下 E A \mathbf{E}_\mathbf A EA​和 A \mathbf A A列之间的关系(Column Relationships in E A \mathbf{E}_\mathbf A EA​ and A \mathbf A A). 对于 E A \mathbf{E}_\mathbf A EA​来说,以下结论是显然的:

Each nonbasic column E ∗ k \mathbf{E}_{∗k} E∗k​ in E A \mathbf{E}_\mathbf A EA​ is a combination (a sum of multiples) of the basic columns in E A \mathbf{E}_\mathbf A EA​ to the left of E ∗ k \mathbf{E}_{∗k} E∗k​. That is, E ∗ k = μ 1 E ∗ b 1 + μ 2 E ∗ b 2 + ⋯ + μ j E ∗ b j = μ 1 ( 1 0 ⋮ 0 ⋮ 0 ) + μ 2 ( 0 1 ⋮ 0 ⋮ 0 ) + ⋯ + μ j ( 0 0 ⋮ 1 ⋮ 0 ) = ( μ 1 μ 2 ⋮ μ j ⋮ 0 ) \begin{aligned} \mathbf{E}_{∗k}&=\mu_1\mathbf{E}_{∗b_1}+\mu_2\mathbf{E}_{∗b_2}+\cdots +\mu_j\mathbf{E}_{∗b_j} \\ &=\mu_1\left(\begin{matrix} 1 \\ 0\\ \vdots \\ 0\\ \vdots \\ 0 \end{matrix}\right)+\mu_2\left(\begin{matrix} 0 \\ 1\\ \vdots \\ 0\\ \vdots \\ 0 \end{matrix}\right)+\cdots+\mu_j\left(\begin{matrix} 0 \\ 0\\ \vdots \\ 1\\ \vdots \\ 0 \end{matrix}\right)=\left(\begin{matrix} \mu_1 \\ \mu_2 \\ \vdots \\ \mu_j \\ \vdots \\ 0 \end{matrix}\right) \end{aligned} E∗k​​=μ1​E∗b1​​+μ2​E∗b2​​+⋯+μj​E∗bj​​=μ1​⎝⎜⎜⎜⎜⎜⎜⎜⎜⎛​10⋮0⋮0​⎠⎟⎟⎟⎟⎟⎟⎟⎟⎞​+μ2​⎝⎜⎜⎜⎜⎜⎜⎜⎜⎛​01⋮0⋮0​⎠⎟⎟⎟⎟⎟⎟⎟⎟⎞​+⋯+μj​⎝⎜⎜⎜⎜⎜⎜⎜⎜⎛​00⋮1⋮0​⎠⎟⎟⎟⎟⎟⎟⎟⎟⎞​=⎝⎜⎜⎜⎜⎜⎜⎜⎜⎛​μ1​μ2​⋮μj​⋮0​⎠⎟⎟⎟⎟⎟⎟⎟⎟⎞​​ where the E ∗ b i \mathbf{E}_{∗b_i} E∗bi​​’s are the basic columns to the left of E ∗ k \mathbf{E}_{∗k} E∗k​ and where the multipliers μ i \mu_i μi​ are the first j j j entries in E ∗ k \mathbf{E}_{∗k} E∗k​.

下面我们证明:

The relationships that exist among the columns of A \mathbf A A are exactly the same as the relationships that exist among the columns of E A \mathbf{E}_\mathbf A EA​. In particular, if A ∗ k \mathbf{A}_{∗k} A∗k​ is a nonbasic column in A \mathbf A A, then A ∗ k = μ 1 A ∗ b 1 + μ 2 A ∗ b 2 + ⋯ + μ j A ∗ b j , \mathbf{A}_{∗k} = \mu_1\mathbf{A}_{∗b_1} + \mu_2\mathbf{A}_{∗b_2} + \cdots + \mu_j\mathbf{A}_{∗b_j} , A∗k​=μ1​A∗b1​​+μ2​A∗b2​​+⋯+μj​A∗bj​​, where the A ∗ b i \mathbf{A}_{∗b_i} A∗bi​​’s are the basic columns to the left of A ∗ k \mathbf{A}_{∗k} A∗k​, and where the multipliers μ i \mu_i μi​ are as described above—the first j j j entries in E ∗ k \mathbf{E}_{∗k} E∗k​.

首先,矩阵 A \mathbf A A经过一系列初等变换化为最简行阶梯矩阵 E A \mathbf E_\mathbf A EA​,该过程可以等价为给 A \mathbf A A左乘一个非奇异矩阵 P \mathbf P P,也即 E A = P A . \mathbf E_\mathbf A=\mathbf P\mathbf A. EA​=PA. 根据 E ∗ k = ( P A ) ∗ k = P A ∗ k \mathbf{E}_{∗k}=(\mathbf P\mathbf A)_{*k}=\mathbf P\mathbf A_{*k} E∗k​=(PA)∗k​=PA∗k​,则若有 E ∗ k = ∑ i = 1 j μ i E ∗ b i , \mathbf{E}_{∗k}=\sum_{i=1}^{j} \mu_i\mathbf{E}_{∗b_i}, E∗k​=i=1∑j​μi​E∗bi​​, 等式两边左乘 P − 1 \mathbf P^{-1} P−1即可得到 A ∗ k = ∑ i = 1 j μ i A ∗ b i . \mathbf{A}_{∗k}=\sum_{i=1}^{j} \mu_i\mathbf{A}_{∗b_i}. A∗k​=i=1∑j​μi​A∗bi​​.

则基本列的性质已经得到证明,即矩阵的所有列都可以通过基本列的线性组合表示出来。

矩阵的秩 (Rank)

Suppose A m × n \mathbf A_{m\times n} Am×n​ is reduced by row operations to an echelon form E \mathbf E E. The rank of A \mathbf A A is defined to be the number r a n k ( A ) = number of pivots = number of nonzero rows in  E = number of basic columns in  A . \begin{aligned} rank (\mathbf A) &= \text{number of pivots} \\ &=\text{number of nonzero rows in }\mathbf E \\ &=\text{number of basic columns in }\mathbf A. \end{aligned} rank(A)​=number of pivots=number of nonzero rows in E=number of basic columns in A.​

线性方程组的解 (Consistency of Linear Systems)

A system of m m m linear equations in n n n unknowns is said to be a consistent system if it possesses at least one solution. If there are no solutions, then the system is called inconsistent. Each of the following is equivalent to saying that [ A ∣ b ] [\mathbf A|\mathbf b] [A∣b] is consistent.

In row reducing [ A ∣ b ] [\mathbf A|\mathbf b] [A∣b], a row of the following form never appears: ( 0   0   ⋯   0   ∣   α ) ,  where  α ≠ 0. (0~0~\cdots~0~|~\alpha),\text{ where } \alpha \ne 0. (0 0 ⋯ 0 ∣ α), where α​=0. b \mathbf b b is a nonbasic column in [ A ∣ b ] [\mathbf A|\mathbf b] [A∣b]. r a n k [ A ∣ b ] = r a n k ( A ) rank[\mathbf A|\mathbf b]=rank(\mathbf A) rank[A∣b]=rank(A). b \mathbf b b is a combination of the basic columns in A \mathbf A A.

第一点:显然; 第二点:若 b \mathbf b b是基本列,则 α \alpha α是一个pivot,与第一点矛盾; 第三点:根据矩阵的秩的定义,秩=基本列列数。由第二点 b \mathbf b b不是基本列,有 r a n k [ A ∣ b ] = r a n k ( A ) rank[\mathbf A|\mathbf b]=rank(\mathbf A) rank[A∣b]=rank(A)成立; 第四点:所有非基本列都可以表示成基本列的线性组合。

齐次/非齐次方程组 (Homogeneous / Nonhomogeneous System)

A system of m m m linear equations in n n n unknowns a 11 x 1 + a 12 x 2 + ⋯ + a 1 n x n = b 1 , a 21 x 1 + a 22 x 2 + ⋯ + a 2 n x n = b 2 , ⋮ a m 1 x 1 + a m 2 x 2 + ⋯ + a m n x n = b m , 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,\\ \vdots\\ a_{m1}x_1+a_{m2}x_2+\cdots+a_{mn}x_n=b_m,\\ a11​x1​+a12​x2​+⋯+a1n​xn​=b1​,a21​x1​+a22​x2​+⋯+a2n​xn​=b2​,⋮am1​x1​+am2​x2​+⋯+amn​xn​=bm​, in which the right-hand side consists entirely of 0 0 0’s is said to be a homogeneous system. If there is at least one nonzero number on the right-hand side, then the system is called nonhomogeneous.

对于齐次方程组,consistency不是需要考虑的问题,因为一定存在一个平凡解(trival solution) x 1 = x 2 = ⋯ = x n = 0 x_1=x_2=\cdots=x_n=0 x1​=x2​=⋯=xn​=0. 因此,需要考虑的问题是,除了平凡解之外,有没有其它的解,以及如何表示它们。

Let A m × n \mathbf A_{m\times n} Am×n​ be the coefficient matrix for a homogeneous system of m m m linear equations in n n n unknowns, and suppose r a n k ( A ) = r rank (\mathbf A)= r rank(A)=r.

The unknowns that correspond to the positions of the basic columns (i.e., the pivotal positions) are called the basic variables, and the unknowns corresponding to the positions of the nonbasic columns are called the free variables.There are exactly r r r basic variables and n − r n − r n−r free variables.To describe all solutions, reduce A \mathbf A A to a row echelon form using Gaussian elimination, and then use back substitution to solve for the basic variables in terms of the free variables. This produces the general solution that has the form x = x f 1 h 1 + x f 2 h 2 + ⋯ + x f n − r h n − r , \mathbf x=x_{f_1}\mathbf h_1+x_{f_2}\mathbf h_2+\cdots+x_{f_{n-r}}\mathbf h_{n-r}, x=xf1​​h1​+xf2​​h2​+⋯+xfn−r​​hn−r​, where the terms x f 1 , x f 2 , ⋯   , x f n − r x_{f_1},x_{f_2},\cdots,x_{f_{n−r}} xf1​​,xf2​​,⋯,xfn−r​​ are the free variables and where h 1 , h 2 , ⋯   , h n − r \mathbf h_1, \mathbf h_2,\cdots, \mathbf h_{n−r} h1​,h2​,⋯,hn−r​ are n × 1 n \times 1 n×1 columns that represent particular solutions of the homogeneous system. The h i \mathbf h_i hi​’s are independent of which row echelon form is used in the back substitution process. As the free variables x f i x_{f_i} xfi​​ range over all possible values, the general solution generates all possible solutions.A homogeneous system possesses a unique solution (the trivial solution) if and only if r a n k ( A ) = n rank (\mathbf A)= n rank(A)=n — i.e., if and only if there are no free variables.

对于非齐次方程组,可能会是inconsistent的,需要用上一节的方法判断。在下面的讨论中,我们假设它们是consistent的。

Let [ A ∣ b ] [\mathbf A|\mathbf b] [A∣b] be the augmented matrix for a consistent m × n m\times n m×n nonhomogeneous system in which r a n k ( A ) = r rank (\mathbf A)= r rank(A)=r.

Reducing [ A ∣ b ] [\mathbf A|\mathbf b] [A∣b] to a row echelon form using Gaussian elimination and then solving for the basic variables in terms of the free variables leads to the general solution x = p + x f 1 h 1 + x f 2 h 2 + ⋯ + x f n − r h n − r . \mathbf x=\mathbf p+x_{f_1}\mathbf h_1+x_{f_2}\mathbf h_2+\cdots+x_{f_{n-r}}\mathbf h_{n-r}. x=p+xf1​​h1​+xf2​​h2​+⋯+xfn−r​​hn−r​. As the free variables x f i x_{f_i} xfi​​ range over all possible values, this general solution generates all possible solutions of the system.Column p \mathbf p p is a particular solution of the nonhomogeneous system.The expression x f 1 h 1 + x f 2 h 2 + ⋯ + x f n − r h n − r x_{f_1}\mathbf h_1+x_{f_2}\mathbf h_2+\cdots+x_{f_{n-r}}\mathbf h_{n-r} xf1​​h1​+xf2​​h2​+⋯+xfn−r​​hn−r​ is the general solution of the associated homogeneous system.Column p \mathbf p p as well as the columns h i \mathbf h_i hi​ are independent of the row echelon form to which [ A ∣ b ] [\mathbf A|\mathbf b] [A∣b] is reduced.The system possesses a unique solution if and only if any of the following is true: r a n k ( A ) = n = number of unknowns rank (\mathbf A)= n = \text{number of unknowns} rank(A)=n=number of unknowns; there are no free variables; the associated homogeneous system possesses only the trivial solution.

非齐次通解=特解+齐次通解,其中特解是怎么来的? 首先,记线性方程组的增广矩阵为 [ A ∣ b ] [\mathbf A|\mathbf b] [A∣b],其中 r a n k ( A ) = r rank(\mathbf A)=r rank(A)=r. 将增广矩阵化成最简行阶梯矩阵: E [ A ∣ b ] = [ E A ∣ c ] \mathbf E_{[\mathbf A|\mathbf b]}=[\mathbf E_\mathbf A|\mathbf c] E[A∣b]​=[EA​∣c],则 c = [ ξ 1 ⋯ ξ r 0 ⋯ 0 ] T \mathbf c=[\xi_1 \cdots \xi_r 0 \cdots 0]^T c=[ξ1​⋯ξr​0⋯0]T. 若 b = 0 \mathbf b=\mathbf 0 b=0,也就是齐次方程组的情况,基本变量 x b i x_{b_i} xbi​​可以被自由变量 x f i , x f i + 1 , ⋯   , x f n − r x_{f_i},x_{f_{i+1}},\cdots,x_{f_{n-r}} xfi​​,xfi+1​​,⋯,xfn−r​​表示成: x b i = α i x f i + α i + 1 x f i + 1 + ⋯ + α n − r x f n − r , x_{b_i}=\alpha_ix_{f_i}+\alpha_{i+1}x_{f_{i+1}}+\cdots+\alpha_{n-r}x_{f_{n-r}}, xbi​​=αi​xfi​​+αi+1​xfi+1​​+⋯+αn−r​xfn−r​​, 若 b ≠ 0 \mathbf b\ne\mathbf 0 b​=0,也就是非齐次方程组的情况,基本变量和自由变量在上式的基础上还要加一项常数: x b i = ξ i + α i x f i + α i + 1 x f i + 1 + ⋯ + α n − r x f n − r . x_{b_i}=\xi_i+\alpha_ix_{f_i}+\alpha_{i+1}x_{f_{i+1}}+\cdots+\alpha_{n-r}x_{f_{n-r}}. xbi​​=ξi​+αi​xfi​​+αi+1​xfi+1​​+⋯+αn−r​xfn−r​​. 因此可以看出所谓特解其实就是由 r r r个 ξ i \xi_i ξi​和 n − r n-r n−r个 0 0 0组成的向量, n − r n-r n−r个 0 0 0对应的是自由变量的位置。



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