Chapter 1 (Linear Equations in Linear Algebra): Solution sets of linear systems

本文为《Linear algebra and its applications》的读书笔记

Homogeneous Linear Systems

齐次线性方程组

• A system of linear equations is said to be homogeneous if it can be written in the form $A\mathbf{x}=0$, where $A$ is an $m\times n$ matrix and $0$ is the zero vector in ${\mathbb{R}}^m$.
• Such a system $A\mathbf{x}=0$ always has at least one solution, namely, $\mathbf{x}=0$ (the zero vector in ${\mathbb{R}}^n$. This zero solution is usually called the trivial solution (平凡解). For a given equation $A\mathbf{x}=0$; the important question is whether there exists a nontrivial solution (非平凡解), that is, a nonzero vector $\mathbf{x}$ that satisfies $A\mathbf{x}=0$

• The Existence and Uniqueness Theorem in Section 1.2 leads immediately to the following fact:
  • The homogeneous equation $A\mathbf{x}=0$ has a nontrivial solution if and only if the equation has at least one free variable.

区别：
当 $A$ 的每个列都是主元列时，$A\mathbf{x}=0$ 或 $A\mathbf{x}=\mathbf{b}$ 有唯一解 (假设 $A\mathbf{x}=\mathbf{b}$ 有解)
当 $A$ 的每行都有主元时，$A\mathbf{x}=\mathbf{b}$ 对任意 $\mathbf{b}$ 均有解

补充 ($A$ 为 $m\times n$ 的矩阵)：

• $A\mathbf{x}=\mathbf{b}$ 有解 $\iff$ $rank([A\mathbf{b}])=rank(A)$
• $A\mathbf{x}=\mathbf{b}$ 有唯一解 $\iff$ $rank([A\mathbf{b}])=rank(A)=n$
• $A\mathbf{x}=0$ 有非零解 (只有零解) $\iff$ $rank(A)<n$ ($rank(A)=n$)

EXAMPLE 1

Determine if the following homogeneous system has a nontrivial solution. Then describe the solution set.

SOLUTION

• As a vector, the general solution of $A\mathbf{x}=0$ has the form
  
• This shows that every solution of $A\mathbf{x}=0$ in this case is a scalar multiple of $\mathbf{v}$. The trivial solution is obtained by choosing $x_3=0$: Geometrically, the solution set is a line through $0$ in ${\mathbb{R}}^3$.
  

EXAMPLE 2

A single linear equation can be treated as a very simple system of equations. Describe all solutions of the homogeneous “system”
$$10x_1-3x_2-2x_3=0(1)$$
SOLUTION

• The general solution is $x_1=.3x_2+.2x_3$, with $x_2$ and $x_3$ free. As a vector, the general solution is
  
• This calculation shows that every solution is a linear combination of the vectors $\mathbf{u}$ and $\mathbf{v}$. That is, the solution set is Span{$\mathbf{u},\mathbf{v}$}. Since neither $\mathbf{u}$ nor $\mathbf{v}$ is a scalar multiple of the other, the solution set is a plane through the origin. See Figure 2.
  
• Examples 1 and 2 illustrate the fact that the solution set of a homogeneous equation $A\mathbf{x}=0$ can always be expressed explicitly as Span{${\mathbf{v}}_1,...,{\mathbf{v}}_p$} for suitable vectors ${\mathbf{v}}_1,...,{\mathbf{v}}_p$.
  • If the only solution is the zero vector, then the solution set is Span{$0$}.
  • If the equation $A\mathbf{x}=0$ has only one free variable, the solution set is a line through the origin, as in Figure 1.
  • A plane through the origin, as in Figure 2, provides a good mental image for the solution set of $A\mathbf{x}=0$ when there are two or more free variables.

Parametric Vector Form

参数向量形式

• The original equation (1) for the plane in Example 2 is an $implicit$ description of the plane. Solving this equation amounts to finding an $explicit$ description of the plane as the set spanned by $\mathbf{u}$ and $\mathbf{v}$. Equation (2) is called a parametric vector equation of the plane. Sometimes such an equation is written as
  $$\mathbf{x}=s\mathbf{u}+t\mathbf{v}(s,tin\mathbb{R})$$
  • In Example 1, the equation $\mathbf{x}=t\mathbf{v}$ (with $t$ in $\mathbb{R}$), is a parametric vector equation of a line.
• Whenever a solution set is described explicitly with vectors as in Examples 1 and 2, we say that the solution is in parametric vector form.

通过将解写为参数向量形式，可以清楚的描述出解的几何分布情况 (直线、平面、点)

Solutions of Nonhomogeneous Systems

• When a nonhomogeneous linear system has many solutions, the general solution can be written in parametric vector form as one vector plus an arbitrary linear combination of vectors that satisfy the corresponding homogeneous system.

EXAMPLE 3

Describe all solutions of $A\mathbf{x}=\mathbf{b}$, where

SOLUTION

• Here $A$ is the matrix of coefficients from Example 1.
  
• The equation
  $$\mathbf{x}=\mathbf{p}+t\mathbf{v}(tin\mathbb{R})(3)$$describes the solution set of $A\mathbf{x}=\mathbf{b}$ in parametric vector form. Recall from Example 1 that the solution set of $A\mathbf{x}=0$ has the parametric vector equation
  $$\mathbf{x}=t\mathbf{v}(tin\mathbb{R})(4)$$[with the same $\mathbf{v}$]. Thus the solutions of $A\mathbf{x}=\mathbf{b}$ are obtained by adding the vector $\mathbf{p}$ to the solutions of $A\mathbf{x}=0$. The vector $\mathbf{p}$ itself is just one particular solution of $A\mathbf{x}=\mathbf{b}$ [corresponding to $t=0$ in (3)].

• To describe the solution set of $A\mathbf{x}=\mathbf{b}$ geometrically, we can think of vector addition as a $translation(平移)$. Given $\mathbf{v}$ and $\mathbf{p}$ in ${\mathbb{R}}^2$ or ${\mathbb{R}}^3$, the effect of adding $\mathbf{p}$ to $\mathbf{v}$ is to move $\mathbf{v}$ in a direction parallel to the line through $\mathbf{p}$ and $0$. We say that $\mathbf{v}$ is translated by $\mathbf{p}$ to $\mathbf{v}$ $+$ $\mathbf{p}$. See Figure 3.
  
• If each point on a line $L$ in ${\mathbb{R}}^2$ or ${\mathbb{R}}^3$ is $translated$ by a vector $\mathbf{p}$, the result is a line parallel to $L$. See Figure 4.
  
• Suppose $L$ is the line through $0$ and $\mathbf{v}$, described by equation (4). Adding $\mathbf{p}$ to each point on $L$ produces the translated line described by equation (3). Note that $\mathbf{p}$ is on the line in equation (3). We call (3) the equation of the line through $\mathbf{p}$ parallel to $\mathbf{v}$ (通过 $\mathbf{p}$ 平行于 $\mathbf{v}$ 的直线方程). Thus the solution set of $A\mathbf{x}=\mathbf{b}$ is a line through $\mathbf{p}$ parallel to the solution set of $A\mathbf{x}=0$. Figure 5 illustrates this case.
  
• The relation between the solution sets of $A\mathbf{x}=\mathbf{b}$ and $A\mathbf{x}=0$ shown in Figure 5 generalizes to any consistent equation $A\mathbf{x}=\mathbf{b}$, although the solution set will be larger than a line when there are several free variables. The following theorem gives the precise statement.

$Warning$: Theorem 6 and Figure 6 apply only to an equation $A\mathbf{x}=\mathbf{b}$ that has at least one nonzero solution $\mathbf{p}$. When $A\mathbf{x}=\mathbf{b}$ has no solution, the solution set is empty.

PROOF

• Suppose $\mathbf{p}$ satisfies $A\mathbf{x}=\mathbf{b}$. Then $A\mathbf{p}=\mathbf{b}$. Theorem 6 says that the solution set of $A\mathbf{x}=\mathbf{b}$ equals the set S = { w : w = p + v h f o r s o m e v h s u c h t h a t A v h = 0 } S = \{\boldsymbol w : \boldsymbol w = \boldsymbol p +\boldsymbol v_h\ for\ some\ \boldsymbol v_h\ such\ that\ A\boldsymbol v_h = \boldsymbol 0\} S={w:w=p+vh​ for some vh​ such that Avh​=0}. There are two things to prove: (a) every vector in $S$ satisfies $A\mathbf{x}=\mathbf{b}$, (b) every vector that satisfies $A\mathbf{x}=\mathbf{b}$ is in $S$.
  

• Theorem 6 says that if $A\mathbf{x}=\mathbf{b}$ has a solution, then the solution set is obtained by translating the solution set of $A\mathbf{x}=0$, using any particular solution $\mathbf{p}$ of $A\mathbf{x}=\mathbf{b}$ for the translation. Figure 6 illustrates the case in which there are two free variables.
  
• Theorem 6 also indicates that the number of free variables in $A\mathbf{x}=\mathbf{b}$ depends only on $A$, not on $\mathbf{b}$.

$Checkpoint$:
Let $A$ be a $2\times 2$ matrix. Answer True or False: If the solution set of $A\mathbf{x}=0$ is a line through the origin in ${\mathbb{R}}^2$ and if $b\ne 0$, then the solution set of $A\mathbf{x}=\mathbf{b}$ is a line not through the origin

SOLUTION

• False. The solution set could be empty. In this case, the solution set of $A\mathbf{x}=\mathbf{b}$ is not produced by translating the (nonempty) solution set of $A\mathbf{x}=0$. See the Warning after Theorem 6.