1.2 System of Linear Equations

这一节与1.1 Fields一起，是一个基础知识介绍，主要概念包括field和subfield，system of m linear equations in n unknowns，以及什么是solution，什么是homogeneous system。
field需要满足9个条件，subfield则重点介绍的是复数域C上自成一域的集合。域的特征值是挺有意思的一个新概念，一般能遇到的field都是characteristic zero的。
在介绍方程组系统时，很快就引入了linear combination的概念，如果用线性空间的思维来看，equations in n unknowns with coefficients in F可以看作一个线性空间V，如果说一个方程是某个系统中方程的linear combination，那么这一方程处于这个系统在V中span的subfield，如果两个系统等价，按照文中定义，就是他们span成相同的subfield，因此解相同，这也是Theorem 1 的内容。并且，V中的任何一个subfield都对应着 $F^n$里的一个subfield（即解空间solution space），当然按照1.2的内容还没法证明 $V$和 $F^n$是不是同构（isomorphism）的。

Exercises

1.Verify that the set of complex numbers described in Example 4 is a subfield of $C$.

Solution:Let $F=\{x+y\sqrt{2}:x,y∈Q\}$, then $0=0+0\sqrt{2},1=1+0\sqrt{2}$, thus $0,1∈F$, let $x,y∈F$, then $x=a+b\sqrt{2},y=c+d\sqrt{2},a,b,c,d∈Q$, so we have
$x+y=a+c+(b+d)\sqrt{2}\in F\\-x=-a-b\sqrt{2}\in F \\ xy=ac+2bd+(ad+bc)\sqrt{2}\in F\\x^{-1}=\frac{1}{a+b\sqrt{2}}=\frac{a-b\sqrt{2}}{a^2-2b^2 }\in F$

2.Let $F$ be the field of complex numbers. Are the following two systems of linear equations equivalent? If so, express each equation in each system as a linear combination of the equations in the other system.

$\begin{aligned}x_1-x_2&=0\quad\quad 3x_1+x_2&=0\\2x_1+x_2&=0\quad\quad x_1+x_2&=0\end{aligned}$
Solution:They are equivalent, since
$3x_1+x_2=1/3 (x_1-x_2 )+4/3 (2x_1+x_2 )\\x_1+x_2=-1/3 (x_1-x_2 )+2/3 (2x_1+x_2 )$
and
$x_1-x_2=(3x_1+x_2 )-2(x_1+x_2)\\2x_1+x_2=1/2 (3x_1+x_2 )+1/2 (x_1+x_2 )$

3.Test the following systems of equations as in Exercise 2.

$\begin{aligned}-x_1+x_2+4x_3&=0\quad\quad x_1-&&x_3&=0\\x_1+3x_2+8x_3&=0\quad\quad &x_2+3&x_3&=0\\ \frac{1}{2}x_1+x_2+\frac{5}{2}x_3&=0\end{aligned}$
Solution: They are equivalent, since
$-x_1+x_2+4x_3=-(x_1-x_3 )+(x_2+3x_3)\\ x_1+3x_2+8x_3=(x_1-x_3 )+3(x_2+3x_3)\\ 1/2 x_1+x_2+5/2 x_3=1/2 (x_1-x_3 )+(x_2+3x_3)$
and
$x_1-x_3=-2/3 (-x_1+x_2+4x_3 )+2/3 (1/2 x_1+x_2+5/2 x_3 )\\ x_2+3x_3=1/4 (-x_1+x_2+4x_3 )+1/4(x_1+3x_2+8x_3)$

4.Test the following systems of equations as in Exercise 2.

$\begin{aligned}2x_1+(-1+i)&x_2+&x_4&=0\quad\quad &\left(1+\frac{i}{2}\right)x_1+8x_2-ix_3-x_4&=0\\&3x_2-2ix_3+&5x_4&=0\quad\quad &\frac{2}{3}x_1-\frac{1}{2}x_2+x_3+7x_4&=0\end{aligned}$
Solution: They are not equivalent, suppose
$a(2x_1+(-1+i) x_2+x_4 )+b(3x_2-2ix_3+5x_4 )\\=(1+i/2) x_1+8x_2-ix_3-x_4$
then compare $x_1$ we have $a=1/2+i/4$, compare $x_3$ we have $b=1/2$, but then $a+5b=3+i/4\neq -1$

5. Let F be a set which contains exactly two elements, 0 and 1. Define an addition and multiplication by the tables:

$\begin{array}{lr} \begin{array}{c|lr} {+} & 0 & 1 \\ \hline 0 & 0 & 1 \\ 1 & 1 & 0 \end{array} &\quad \begin{array}{c|lr} {\cdot} & 0 & 1 \\ \hline 0 & 0 & 0 \\ 1 & 0 & 1 \end{array} \end{array}$

Verify that the set $F$, together with these two operations, is a field.

Solution:The verification is as follows:

1. Addition is commutative
   $0+0=0+0,0+1=1=1+0,1+1=0=1+1$
2. Addition is associative
   $0+(0+0)=0=(0+0)+0 \\ 0+(1+0)=1=(0+1)+0 \\ 0+(0+1)=1=(0+0)+1 \\ 0+(1+1)=0+0=0=(0+1)+1 \\ 1+(0+0)=1=1+0=(1+0)+0 \\ 1+(1+0)=1+1=0=(1+1)+0 \\ 1+(0+1)=0=(1+0)+1 \\ 1+(1+1)=1=0+1=(1+1)+1$
3. The $0$ is the element $0$ in $F$, since $0+0=0,1+0=1$
4. To $0$, we have $-0=0$, to $1$, we have $-1=1$ since $1+1=0$
5. Multiplication is commutative
   $0⋅0=0⋅0=0,0⋅1=1⋅0=0,1⋅1=1⋅1=1$
6. Multiplication is associative
   $0⋅(0⋅0)=0⋅0=0=0⋅0=(0⋅0)⋅0 \\ 0⋅(1⋅0)=0⋅0=0=0⋅0=(0⋅1)⋅0 \\ 0⋅(0⋅1)=0⋅0=0=0⋅0=(0⋅0)⋅1 \\ 0⋅(1⋅1)=0⋅1=0=0⋅1=(0⋅1)⋅1 \\ 1⋅(0⋅0)=1⋅0=0=0⋅0=(1⋅0)⋅0 \\ 1⋅(1⋅0)=1⋅0=0=1⋅0=(1⋅1)⋅0 \\ 1⋅(0⋅1)=1⋅0=0=0⋅1=(1⋅0)⋅1 \\ 1⋅(1⋅1)=1⋅1=1=1⋅1=(1⋅1)⋅1$
7. The $1$ is the element $1$ in $F$ since $0⋅1=0$ and $1⋅1=1$
8. To $1$ there’s $1^{-1}=1$, since $1⋅1=1$
9. Multiplication distributes over addition
   $0⋅(0+0)=0⋅0=0=0+0=0⋅0+0⋅0 \\ 0⋅(1+0)=0⋅1=0=0+0=0⋅1+0⋅0 \\ 0⋅(0+1)=0⋅0=0=0+0= 0⋅0+0⋅1 \\ 0⋅(1+1)=0⋅0=0=0+0=0⋅1+0⋅1 \\ 1⋅(0+0)=1⋅0=0=0+0=1⋅0+1⋅0 \\ 1⋅(1+0)=1⋅1=1=1+0=1⋅1+1⋅0 \\ 1⋅(0+1)=1⋅1=1=0+1=1⋅0+1⋅1 \\ 1⋅(1+1)=1⋅0=0=1+1=1⋅1+1⋅1$

6. Prove that if two homogeneous systems of linear equations in two unknowns have the same solutions, then they are equivalent.

Solution:Let
$\begin{cases}a_1 x_1+a_2 x_2=0\\ b_1 x_1+b_2 x_2=0\end{cases},\quad \begin{cases}c_1 x_1+c_2 x_2=0\\d_1 x_1+d_2 x_2=0 \end{cases}$
be two homogeneous system of linear equations, if they have only zero solutions, then $(a_1,a_2 )$ and $(b_1,b_2 )$ are linearly independent, thus a basis of $R^2$, thus they can represent $(c_1,c_2 )$ and $(d_1,d_2 )$, vice versa.
If they have non-zero solutions, then either all $a_i,b_i,c_i,d_i$ are $0$, or in each of the set $(a_1,a_2,b_1,b_2 )$ and $(c_1,c_2,d_1,d_2 )$ we have at least one nonzero element. Without loss of generality we suppose $(a_1,a_2 )=k(b_1,b_2 ),(c_1,c_2 )=l(d_1,d_2 )$ for some $k,l∈R$, in which one of $b_1,b_2$ and one of $d_1,d_2$ is nonzero, we further suppose $b_1\neq 0,d_1\neq 0$, then let $x_1,x_2$ be a common solution, we have
$b_2/b_1 =-x_2/x_1 =d_2/d_1 \coloneqq m$
then $b_2=b_1 m,d_2=d_1 m$, thus
$b_1 x_1+b_2 x_2=b_1 x_1+b_1 mx_2=\frac{b_1}{d_1} d_1 (x_1+mx_2 )=\frac{b_1}{d_1} (d_1 x_1+d_2 x_2 )$
and also we have
$d_1 x_1+d_2 x_2=\frac{d_1}{b_1} (b_1 x_1+b_2 x_2 )$
thus the equivalence is proved.

7. Prove that each subfield of the field of complex numbers contains every rational number.

Solution:Let $F$ be a subfield of $C$, then $1\in F$, thus $N^+\subset F$ from the fact that $x+y\in F$, and $Z\subset F$ since $-x∈F$ and $0∈F$, also from $x^{-1}\in F$ we have
$A≔\left\{\frac{1}{n}:n\in Z\backslash \{0\}\right\}⊂F$
finally from $xy∈F$ we see $Q=Z×A⊂F$ since $Q=\{xy:x∈Z,y∈A\}$.

8. Prove that each field of characteristic zero contains a copy of the rational number field.

Solution:In a field of characteristic zero, we can successfully define
$2≔1+1 \\ 3≔1+1+1 \\ \cdots \\ n\coloneqq \underbrace{1+1+⋯+1}_n$
and form a copy of $N^+$.
Use the property $(x∈F)⇒(-x∈F)$ we can get a copy of $Z$.
Use the property $(x∈F,x≠0)⇒(x^{-1}∈F)$ and $(x∈F,y∈F)⇒(xy∈F)$ we can finally get a copy of $Q$.