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Linear Algebra (2ed) Hoffman&Kunze 1.4

2019年8月24日2019年11月24日christangdt留下评论

我已经忘了这个概念中文如何翻译，但是其本质就是在Row-reduced的基础上，作行的对调，使得所有非零行在上面且每一行开头的1以从左至右自上而下的方式排列、所有零行在下面。这也是Theorem 5的结论，即任何 $m\times n$ 矩阵都和一个row-reduced echelon matrix是行等价的。
前两节讨论了如此多的row-reduced问题，目的是为了解方程组，因为row-reduced的形式使得解方程非常方便，并且如果非零行数比未知数的个数少，那么一定有非退化解（non-trivial solution），即不是0向量的解。Theorem 6是一个更强的假设和更弱一些的结论：如果 $A$ 是 $m\times n$ 矩阵且 $m<n$ ，那么 $AX=0$ 一定有non-trivial solution。Theorem 7是一个特殊一点的结论： $n\times n$ 方阵行等价于单位矩阵的充要条件是 $AX=0$ 只有0解。
以上已经讨论了足够多的关于homogeneous system的结论。如果将条件放松，即 $AX=0$ 变为 $AX=Y$ ，那么需要考虑augmented matrix $\begin{bmatrix}A&Y\end{bmatrix}$ , 在inhomogeneous system中，首先要考虑的是有没有解（不像homogeneous system里一定有0解），按照Theorem 4和Theorem 5可以将 $A$ 变为一个row-reduced echelon matrix $A'$ ，对应的会把 $Y$ 变成一个新的列向量 $Z$ , 那么这个system有解的充要条件是 $A'$ 里全为零的最后几行对应的 $Z$ 的最后几个分量也都是0。
最后有一个看似很奇怪但确实正确的结论： $AX=Y$ 如果 $A,Y$ 都是在 $F$ 的一个subfield $F_1$ 上，那么其在 $F$ 中有解可以推出其在 $F_1$ 中有解。

Exercises

1.Find all solutions to the following systems of equations by row-reducing the coefficient matrix:

$\begin{aligned}\frac{1}{3}&x_1+2x_2&-6x_3=0\\-4&x_1&+5x_3=0\\-3&x_1+6x_2&-13x_3=0\\-\frac{7}{3}&x_1+2x_2&-\frac{8}{3}x_3=0\end{aligned}$

Solution: We have

$\begin{bmatrix}1/3&2&-6\\-4&0&5\\-3&6&-13\\-7/3&2&-8/3\end{bmatrix}\rightarrow\begin{bmatrix}1&6&-18\\0&24&-67\\0&24&-67\\-7&6&-8\end{bmatrix}\rightarrow\begin{bmatrix}1&6&-18\\0&24&-67\\0&0&0\\0&48&-134\end{bmatrix}\rightarrow\begin{bmatrix}1&6&-18\\0&24&-67\\0&0&0\\0&0&0\end{bmatrix}$

let $x_3=c$ , then all solutions is of the form:

$x_1=\dfrac{1062}{67} c,\quad x_2=\dfrac{24}{67} c,\quad x_3=c,\quad c\in R$

2. Find a row-reduced echelon matrix which is row-equivalent to
$A=\begin{bmatrix}1&-i\\2&2\\i&1+i\end{bmatrix}$ . What are the solutions of $AX=0$ ?

Solution: We have

$A=\begin{bmatrix}1&-i\\2&2\\i&1+i\end{bmatrix}\rightarrow\begin{bmatrix}1&1\\0&-i-1\\i&1+i\end{bmatrix}\rightarrow\begin{bmatrix}1&1\\0&1\\0&-i\end{bmatrix}\rightarrow\begin{bmatrix}1&0\\0&1\\0&0\end{bmatrix}$

the solutions of $AX=0$ are trivial solutions.

3.Describe explicitly all $2\times 2$ row-reduced echelon matrices.

Solution: They are (due to the nonzero entries of the matrix)

$\begin{bmatrix}0&0\\0&0\end{bmatrix},\quad\begin{bmatrix}1&0\\0&0\end{bmatrix},\quad\begin{bmatrix}1&b\\0&0\end{bmatrix},\quad\begin{bmatrix}1&0\\0&1\end{bmatrix}$

4. Consider the system of equations

$\begin{aligned}&x_1-x_2+&2x_3=1\\2&x_1&+2x_3=1\\&x_1-3x_2&+4x_3=2\end{aligned}$

Does this system have a solution? If so, describe explicitly all solutions.

Solution: We have

$\begin{bmatrix}1&-1&2&1\\2&0&2&1\\1&-3&4&2\end{bmatrix}\rightarrow\begin{bmatrix}1&-1&2&1\\0&2&-2&-1\\0&-2&2&1\end{bmatrix}\rightarrow\begin{bmatrix}1&-1&2&1\\0&1&-1&-\frac{1}{2}\\0&0&0&0\end{bmatrix}\rightarrow\begin{bmatrix}1&0&1&\frac{1}{2}\\0&1&-1&-\frac{1}{2}\\0&0&0&0\end{bmatrix}$

this system has a solution, let $x_3=c$ , then all solutions of the system is

$\left(\dfrac{1}{2}-c,c-\dfrac{1}{2},c\right)$

5. Give an example of a system of two linear equations in two unknowns which has no solution.

Solution: An example may be

$\begin{cases}x_1+x_2&=1\\2x_1+2x_2&=3\end{cases}$

6. Show that the system

$\begin{array}{r}x_1-2x_2+x_3+2x_4=1\\x_1+x_2-x_3+x_4=2\\x_1+7x_2-5x_3-x_4=3\end{array}$

has no solution.

Solution: Use row operation on the augmented matrix of the system we get

$\begin{bmatrix}1&-2&1&2&1\\1&1&-1&1&2\\1&7&-5&-1&3\end{bmatrix}\rightarrow\begin{bmatrix}1&-2&1&2&1\\0&3&-2&-1&1\\0&9&-6&-3&2\end{bmatrix}\rightarrow\begin{bmatrix}1&-2&1&2&1\\0&3&-2&-1&1\\0&0&0&0&-1 \end{bmatrix}$

thus the system has no solution.

7. Find all solutions of

$\begin{aligned}2x_1-3x_2-7x_3+5x_4+2x_5&=-2\\x_1-2x_2-4x_3+3x_4+x_5&=-2\\2x_1-4x_3+2x_4+x_5&=3\\x_1-5x_2-7x_3+6x_4+2x_5&=-7\end{aligned}$

Solution: We have

$\begin{bmatrix}2&-3&-7&5&2&-2\\1&-2&-4&3&1&-2\\2&0&-4&2&1&3\\1&-5&-7&6&2&-7\end{bmatrix}\rightarrow\begin{bmatrix}1&-2&-4&3&1&-2\\0&1&1&-1&0&2\\0&4&4&-4&-1&7\\0&-3&-3&3&1&-5\end{bmatrix}\\ \rightarrow\begin{bmatrix}1&-2&-4&3&1&-2\\0&1&1&-1&0&2\\0&0&0&0&-1&-1\\0&0&0&0&1&1\end{bmatrix}\rightarrow\begin{bmatrix}1&0&-2&1&1&2\\0&1&1&-1&0&2\\0&0&0&0&1&1\\0&0&0&0&0&0\end{bmatrix}$

thus $x_5=1$ , if we let $x_3=c_1,x_4=c_2$ , then all solutions of the system is

$x_1=1+2c_1-c_2,x_2=2-c_1+c_2,x_3=c_1,x_4=c_2,x_5=1$

8. Let $A=\begin{bmatrix}3&-1&2\\2&1&1\\1&-3&0\end{bmatrix}$ . For which triples $(y_1,y_2,y_3)$ does the system $AX=Y$ have a solution?

Solution:

$\begin{bmatrix}3&-1&2&y_1\\2&1&1&y_2\\1&-3&0&y_3 \end{bmatrix}\rightarrow\begin{bmatrix}1&-3&0&y_3\\0&7&1&y_2-2y_3\\0&8&2&y_1-3y_3 \end{bmatrix} \rightarrow \begin{bmatrix}1&-3&0&y_3\\0&7&1&y_2-2y_3\\0&8&2&y_1-3y_3 \end{bmatrix} \rightarrow \begin{bmatrix}1&-3&0&y_3\\0&7&1&y_2-2y_3\\0&1&1&y_1-y_2-y_3\end{bmatrix}$

Since $A$ is row equivalent to the identity matrix, $AX=Y$ has a solution for all $(y_1,y_2,y_3 )$ .

9. Let $A=\begin{bmatrix}3&-6&2&-1\\-2&4&1&3\\0&0&1&1\\1&-2&1&0\end{bmatrix}$ . For which $(y_1,y_2,y_3,y_4)$ does the system $AX=Y$ have a solution?

Solution:

$\begin{bmatrix}3&-6&2&-1&y_1\\-2&4&1&3&y_2\\0&0&1&1&y_3\\1&-2&1&0&y_4\end{bmatrix} \rightarrow \begin{bmatrix}1&-2&1&0&y_4\\0&0&1&1&y_3\\0&0&-1&3&y_2-2y_4\\0&0&-1&-1&y_1-3y_4 \end{bmatrix} \rightarrow \begin{bmatrix}1&-2&1&0&y_4\\0&0&1&1&y_3\\0&0&0&4&*\\0&0&0&0&y_1-3y_4+y_3\end{bmatrix}$

so for triples $(y_1,y_2,y_3,y_4 )$ which satisfies $y_1=3y_4-y_3$ does the system have a solution.

10. Suppose $R$ and $R'$ are $2\times 3$ row-reduced echelon matrices and that the systems $RX=0$ and $R'X=0$ have exactly the same solutions. Prove that $R=R'$ .

Solution: If $R$ and $R'$ are zero matrix then there’s nothing to prove.
If both are nonzero and assume they have different number of rows, let

$R=\begin{bmatrix}1&0&a\\0&1&b\end{bmatrix},\quad R'=\begin{bmatrix}1&c&d\\0&0&0\end{bmatrix}$

then $(c,-1,0)$ is a solution for the system of $R'$ , but not a solution for the system of $R$ , a contradiction, thus $R$ and $R'$ must have the same number of rows.
If both $R$ and $R$ have one row, let

$R=\begin{bmatrix}1&a&b\\0&0&0\end{bmatrix},\quad R'=\begin{bmatrix}1&c&d\\0&0&0\end{bmatrix}$

then $(c,-1,0)$ is a solution for the system of $R'$ , thus a solution for the system of $R$ , so

$c-a+0\cdot b=0 \Rightarrow c=a$

similarly use the solution $(d,0,-1)$ we get $b=d$ , thus $R=R'$ in this case.

If both $R$ and $R$ have two rows, let

$R=\begin{bmatrix}1&0&a\\0&1&b\end{bmatrix},\quad R'=\begin{bmatrix}1&0&c\\0&1&d\end{bmatrix}$

then it’s easy to see $b=d$ , otherwise $(0,b,-1)$ can’t be a solution for the system of $R'$ , also $a=c$ , since $(a,0,-1)$ is a solution for the system of $R'$ and we have $a-c=0$ . The proof is complete.

Linear Algebra (2ed) Hoffman&Kunze 1.3

2019年8月23日2019年11月23日christangdt留下评论

Hoffman这本书有个特点，初读文字有些拗口，英语没点阅读功底容易理解错误他的意思，但是如果能读进去，就能发现这本书最好的地方：视角高。如果说Rudin的分析是大巧不工的大师级，Hoffman的线代就是一个略有些唠叨的大师级。从1.3开始，就有些体会出这样的意思。这一节至少有如下两个拔高：矩阵拔高到函数维度，row-equivalence是一个等价关系。
这一节说Matrix和elementary row operations，我在学北大王萼芳那本高等代数时候也是先学的这一块（当然要在行列式之后），当时就觉得很难理解，把一堆数排成方块形有什么作用，对行进行变换的时候为何选那特定的三种变换。Hoffman是从上一节说的system of linear equations得出的matrix，实际是系数矩阵，但随后正式给的定义又是：matrix是从整数对 $(i,j)$ 到 $F$ 的一个函数，其entry就是这一函数在特定的 $(i,j)$ 上的取值；我们常见的矩阵的样子是矩阵最方便的一种表示。
接下来给出三种elementary row operations：某一行变为原来的 $c$ 倍 $(c\neq 0)$ ；变为自身加上另外一行的 $c$ 倍；调换两行。这三种都可以分别用一个函数 $e$ 精确的表示出来，这个 $e$ 是定义在一个固定的 $m$ （矩阵的行数）上，不用管列是多少。为何选这三种？最重要的是这三种方式可以用同样的方式可逆，也就是Theorem 2的结论。由elementary row operations得到的矩阵和原来矩阵是row-equivalent的，并且揭示了row-equivalent满足self-reflexive,symmetric和transitivity，故而是一个等价关系。由于elementary row operations不会改变以该矩阵作为coefficient matrix的方程组系统的解空间，因此Theorem 3说明：row-equivalent的两个矩阵为coefficient matrix对应的方程组系统有相同的解空间。
最后，有一个row-reduced的定义，即满足一是每一个非零行从左往右先碰到的非零元素（leading non-zero entry）必须是1，二是含有“一是”里这样的1的列，其他元素都是0。Theorem 4说明任何一个matrix都可以被row-reduced。证明是纯说理性的，可能那个时代（20世纪70年代）画图不易，硬是像讲故事一样把证明说清楚了。

Exercises

1. Find all solutions to the system of equations

$(1-i)x_1-ix_2=0\\2x_1+(1-i)x_2=0$

Solution: We have

$\begin{cases}(1-i) x_1-ix_2=0\\2x_1+(1-i) x_2=0\end{cases} \Rightarrow \begin{cases}2x_1-(1+i)ix_2=0\\2x_1+(1-i) x_2=0\end{cases} \Rightarrow \begin{cases}2x_1+(1-i)x_2=0\\2x_1+(1-i) x_2=0\end{cases}$

so let $x_2=c$ , we have $x_1=(i-1)c/2$ , so all solutions to the system is

$x_1=\dfrac{(i-1)c}{2},\quad x_2=c,\quad \forall c\in C$ .

2. If $A=\begin{bmatrix}3&-1&2\\2&1&1\\1&-3&0 \end{bmatrix}$ , find all solutions of $AX=0$ by row-reducing $A$ .

Solution: Since

$\begin{bmatrix}3&-1&2\\2&1&1\\1&-3&0\end{bmatrix}\rightarrow\begin{bmatrix}0&8&2\\0&7&1\\1&-3&0\end{bmatrix}\rightarrow\begin{bmatrix}0&1&1\\0&7&1\\1&-3&0\end{bmatrix}\\ \rightarrow\begin{bmatrix}0&1&1\\0&0&-6\\1&0&3\end{bmatrix}\rightarrow\begin{bmatrix}0&1&1\\0&0&1\\1&0&3\end{bmatrix}\rightarrow\begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix}$

so $AX=0$ has only trivial solutions.

3. If $A=\begin{bmatrix}6&-4&0\\4&-2&0\\-1&0&3 \end{bmatrix}$ , find all solutions of $AX=2X$ and $AX=3X$ .

Solution: It’s equivalent to solving $(A-2I)X=0$ and $(A-3I)X=0$ , use row-reducing we have

$A-2I=\begin{bmatrix}4&-4&0\\4&-4&0\\-1&0&1\end{bmatrix}\rightarrow\begin{bmatrix}1&-1&0\\0&0&0\\0&1&1\end{bmatrix}$

so let $x_2=c$ , we have $x_1=c$ and $x_3=-c$ , so all solutions to the system $(A-2I)X=0$ is $(c,c,-c),c\in C$ .

$A-3I=\begin{bmatrix}3&-4&0\\4&-5&0\\-1&0&0\end{bmatrix}\rightarrow\begin{bmatrix}0&0&0\\0&1&0\\1&0&0\end{bmatrix}$

so all solutions to the system $(A-3I)X=0$ is $(0,0,c),c\in C$ .

4. Find a row-reduced matrix which is row-equivalent to $A=\begin{bmatrix}i&-(1+i)&0\\1&-2&1\\1&2i&-1 \end{bmatrix}$

Solution:

$\begin{bmatrix}i&-(1+i)&0\\1&-2&1\\1&2i&-1\end{bmatrix}\rightarrow\begin{bmatrix}0&-1+i&-i\\1&-2&1\\0&2i+2&-2\end{bmatrix}\\ \rightarrow\begin{bmatrix}0&-1+i&-i\\1&-2&1\\0&i+1&-1\end{bmatrix}\rightarrow\begin{bmatrix}0&0&0\\1&-2&1\\0&1&\frac{i-1}{2}\end{bmatrix}\rightarrow\begin{bmatrix}0&0&0\\1&0&i\\0&1&\frac{i-1}{2}\end{bmatrix}$

5. Prove that the following two matrices are not row-equivalent:

$\begin{bmatrix}2&0&0\\a&-1&0\\b&c&3 \end{bmatrix},\quad \begin{bmatrix}1&1&2\\-2&0&-1\\1&3&5 \end{bmatrix}$

Solution: The first matrix can be row reduced to

$\begin{bmatrix}2&0&0\\a&-1&0\\b&c&3\end{bmatrix}\rightarrow\begin{bmatrix}1&0&0\\0&-1&0\\0&c&3\end{bmatrix}\rightarrow\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$

the second matrix can be row-reduced to

$\begin{bmatrix}1&1&2\\-2&0&-1\\1&3&5\end{bmatrix}\rightarrow\begin{bmatrix}1&1&2\\0&2&3\\0&2&3\end{bmatrix}\rightarrow\begin{bmatrix}1&1&2\\0&1&1/2\\0&0&0\end{bmatrix}\rightarrow\begin{bmatrix}1&0&2\\0&1&1/2\\0&0&0\end{bmatrix}$

thus they are not row-equivalent.

6. Let $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ be a $2\times 2$ matrix with complex entries. Suppose that $A$ is row-reduced and also that $a+b+c+d=0$ . Prove that there are exactly three such matrices.

Solution: The zero matrix $\begin{bmatrix}0&0\\0&0\end{bmatrix}$ is row-reduced and satisfy $a+b+c+d=0$ , suppose there’re nonzero entries, then at least two entries shall be nonzero, otherwise contradicting $a+b+c+d=0$ , consider the case when there’re exactly $2$ nonzero entries, then they can’t belong to different rows, since then the matrix would be $\begin{bmatrix}1&0\\0&1\end{bmatrix}$ or $\begin{bmatrix}0&1\\1&0\end{bmatrix}$ , under the restriction of row-reduced, but then contradicting $a+b+c+d=0$ , in the case when they belong to the same row, we get the matrix $\begin{bmatrix}1&-1\\0&0\end{bmatrix}$ or $\begin{bmatrix}0&0\\1&-1\end{bmatrix}$ satisfy the condition.
Now consider the case when there’re exactly $3$ nonzero entries, then the matrix is of the form $\begin{bmatrix}1&b\\0&d\end{bmatrix}$ or $\begin{bmatrix}0&b\\1&d\end{bmatrix}$ , but notice then either $b=0,d=1$ or $b=1,d=0$ , under the restriction of row-reduced, this contradicting $a+b+c+d=0$ .
Finally the case when all $4$ entries is nonzero contradicts the restriction of row-reduced.
Thus there’re exactly three such matrices: $\begin{bmatrix}0&0\\0&0\end{bmatrix}$ , $\begin{bmatrix}1&-1\\0&0\end{bmatrix}$ and $\begin{bmatrix}0&0\\1&-1\end{bmatrix}$ .

7. Prove that the interchange of two rows of a matrix can be accomplished by a finite sequence of elementary row operations of the other two types.

Solution: It’s enough to show the process for a two-row matrix:

$\begin{aligned}\begin{bmatrix}a_1&\cdots&a_n\\b_1&\cdots&b_n\end{bmatrix} \xrightarrow {\text{add row1 to row2}}\begin{bmatrix}a_1&\cdots&a_n\\b_1+a_1&\cdots&b_n+a_n\end{bmatrix} \\ \xrightarrow{\text{add (-1)*row2 to row1}} \begin{bmatrix}-b_1&\cdots&-b_n\\b_1+a_1&\cdots&b_n+a_n\end{bmatrix}\\ \xrightarrow{\text{add row1 to row2}} \begin{bmatrix}-b_1&\cdots&-b_n\\a_1&\cdots&a_n\end{bmatrix}\\ \xrightarrow{\text{multiply row1 by -1}}\begin{bmatrix}b_1&\cdots&b_n\\a_1&\cdots&a_n\end{bmatrix}\end{aligned}$

8. Consider the system of equations $AX=0$ where $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ is a $2\times 2$ matrix over the field $F$ . Prove the following:
(a) If every entry of $A$ is $0$ , then every pair $(x_1,x_2)$ is a solution of $AX=0$ .
(b) If $ad-bc\neq 0$ , the system $AX=0$ has only the trivial solution $x_1=x_2=0$ .
(c) If $ad-bc= 0$ , and some entry of $A$ is different from $0$ , then there is a solution $(x_1^0,x_2^0)$ such that $(x_1,x_2)$ is a solution if and only if there is a scalar $y$ such that $x_1=yx_1^0,x_2=yx_2^0$ .

Solution:
(a) We have $0X=0$ for every $X$ .
(b) Since $a$ and $c$ can’t both be $0$ , we suppose $a\neq 0$ , use row reduction, we have

$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\rightarrow\begin{bmatrix}a&b\\ca&da\end{bmatrix}\rightarrow\begin{bmatrix}a&b\\0&ad-bc\end{bmatrix}\rightarrow\begin{bmatrix}1&0\\0&1 \end{bmatrix}$

and the conclusion follows.
(c) If both $a$ and $c$ is $0$ , then $A$ is row-equivalent to $\begin{bmatrix}0&1\\0&0\end{bmatrix}$ or $\begin{bmatrix}0&0\\0&1\end{bmatrix}$ . If not, then $A$ is row-equivalent to $\begin{bmatrix}a&b\\0&ad-bc\end{bmatrix}$ or $\begin{bmatrix}0&bc-ad\\c&d\end{bmatrix}$ , in all cases $A$ is row-equivalent to a matrix that has nonzero entries in only one row, thus the system $AX=0$ has the same solution with

$\alpha x_1+\beta x_2=0,\quad (\alpha \neq 0)\vee (\beta \neq 0)$

let $(x_1^0,x_2^0)$ be a nonzero solution, then $\alpha x_1^0=-\beta x_2^0$ , thus either $\alpha\neq 0$ , which leads to $x_1^0=kx_2^0,k=-\beta/\alpha$ , or $\beta\neq 0$ , which leads to $x_2^0=lx_1^0,l=-\alpha/\beta$ . If there’s any $(x_1,x_2)$ which satisfies $\alpha x_1+\beta x_2=0$ , then we must also have $x_1=kx_2$ or $x_2=lx_1$ , and the conclusion follows.
In fact, $(x_1^0,x_2^0)$ is a basis for the solution space in this case, so any $(x_1,x_2)$ being a solution, we must have $(x_1,x_2 )=y(x_1^0,x_2^0)$ .

Linear Algebra (2ed) Hoffman&Kunze 1.2

2019年8月21日2019年11月23日christangdt2条评论

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 $Q$ .

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