Linear Algebra (2ed) Hoffman&Kunze 1.2

christangdt

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Theorem1 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\in Q\}, then 0=0+0\sqrt{2},1=1+0\sqrt{2}, thus 0,1\in F, let x,y \in F , then x=a+b \sqrt{2},y=c+d \sqrt{2},a,b,c,d \in 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\cdot 0=0\cdot 0=0,0\cdot 1=1\cdot 0=0,1\cdot 1=1\cdot 1=1

6) Multiplication is associative

0\cdot (0\cdot 0)=0\cdot 0=0=0\cdot 0=(0\cdot 0)\cdot 0 \\ 0\cdot (1\cdot 0)=0\cdot 0=0=0\cdot 0=(0\cdot 1)\cdot 0 \\ 0\cdot (0\cdot 1)=0\cdot 0=0=0\cdot 0=(0\cdot 0)\cdot 1 \\ 0\cdot (1\cdot 1)=0\cdot 1=0=0\cdot 1=(0\cdot 1)\cdot 1 \\ 1\cdot (0\cdot 0)=1\cdot 0=0=0\cdot 0=(1\cdot 0)\cdot 0 \\ 1\cdot (1\cdot 0)=1\cdot 0=0=1\cdot 0=(1\cdot 1)\cdot 0 \\ 1\cdot (0\cdot 1)=1\cdot 0=0=0\cdot 1=(1\cdot 0)\cdot 1 \\ 1\cdot (1\cdot 1)=1\cdot 1=1=1\cdot 1=(1\cdot 1)\cdot 1

7) The 1 is the element 1 in F since 0\cdot 1=0 and 1\cdot 1=1

8) To 1 there’s 1^{-1}=1, since 1\cdot 1=1

9) Multiplication distributes over addition

0\cdot (0+0)=0\cdot 0=0=0+0=0\cdot 0+0\cdot 0 \\ 0\cdot (1+0)=0\cdot 1=0=0+0=0\cdot 1+0\cdot 0 \\ 0\cdot (0+1)=0\cdot 0=0=0+0= 0\cdot 0+0\cdot 1 \\ 0\cdot (1+1)=0\cdot 0=0=0+0=0\cdot 1+0\cdot 1 \\ 1\cdot (0+0)=1\cdot 0=0=0+0=1\cdot 0+1\cdot 0 \\ 1\cdot (1+0)=1\cdot 1=1=1+0=1\cdot 1+1\cdot 0 \\ 1\cdot (0+1)=1\cdot 1=1=0+1=1\cdot 0+1\cdot 1 \\ 1\cdot (1+1)=1\cdot 0=0=1+1=1\cdot 1+1\cdot 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\in 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 := 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=\dfrac{b_1}{d_1} d_1 (x_1+mx_2 )=\dfrac{b_1}{d_1} (d_1 x_1+d_2 x_2 )

and also we have

d_1 x_1+d_2 x_2=\dfrac{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\in F and 0\in F, also from x^{-1}\in F we have

A:=\left\{\dfrac{1}{n}:n\in Z \backslash \{0\}\right\}\subset F

finally from xy\in F we see Q=Z\times A\subset F since Q=\{xy:x\in Z,y\in 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 \\ \quad \cdots \\ n:= \underbrace{1+1+\cdots+1}_n

and form a copy of N^+.
Use the property (x\in F)\Rightarrow(-x\in F) we can get a copy of Z.
Use the property (x\in F,x\neq 0)\Rightarrow(x^{-1}\in F) and (x\in F,y\in F)\Rightarrow(xy\in F) we can finally get a copy of Q.

《Linear Algebra (2ed) Hoffman&Kunze 1.2》有2条评论

  1. 未知 的头像 匿名 2019年9月11日 / 上午8:25

    你好，请教一下。线性代数“1.4”小节最后一段文字：方程在域F中有解，则一定在子域F1中有解。觉得不能接受。在增广矩阵行初等变换过程中，那些系数还一直在子域F1中吗？？

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  2. christangdt 的头像 christangdt 2019年9月16日 / 下午9:25

    你说的是1.4正文里最后的内容吧？初等行变换实质是对系数进行加或者乘，这两个运算在field里都是封闭的（所以在subfield里也是），如果F1是一个子域，那么让y1,…,ym在F1中满足的条件和在F中满足的条件是等价的

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