《线性代数应该这样学》课内验证1-向量空间

作者: Raow1 | 来源:发表于2021-09-14 17:21 被阅读0次

《线性代数应该这样学》课内验证1-向量空间
《线性代数应该这样学》课内验证2-有限维向量空间
《线性代数应该这样学》课内验证3-线性映射
线性代数（待续）
线性代数的本质（笔记1）
如何生动有趣的入门线性代数
线性代数（二）列空间和零空间。
矩阵分析 (一) 线性空间和线性变换
人工智能学习笔记-Day12
线性代数启示录(一)

《线性代数应该这样学》（Linear Algebra Done Right）中说：You cannot read mathematics the way you read a novel. If you zip through a page in less than an hour, you are probably going too fast. When you encounter the phrase “as you should verify”, you should indeed do the verification, which will usually require some writing on your part. 这个笔记也由此而生。

笔记将会记录一部分书内要求自行验证的题目。出于某些原因，我将试着用英语来进行记录。这是我第一次尝试着用英语写数学证明，因此在格式、单词、语法等地方出错应该是难以避免的，望谅解。

CHAPTER 1 Vector Spaces

With the usual operations of addition and scalar multiplication, $\mathbb{F}^n$ is a vector space over $\mathbb{F}$ , as you should verify.

Proof 1.13 states the commutativity. As for associativity, suppose $x=(x_1, \cdots , x_n)$ , $y=(y_1, \cdots , y_n)$ and $z=(z_1, \cdots , z_n)$ . Then
$\begin{align*} (x+y)+z &= ((x_1, \cdots , x_n) + (y_1, \cdots , y_n)) + (z_1, \cdots , z_n) \\ &= (x_1+y_1+z_1, \cdots , x_n+y_n+z_n) \\ &= (x_1+(y_1+z_1), \cdots , x_n+(y_n+z_n)) \\ &= (x_1, \cdots , x_n)+(y_1+z_1, \cdots , y_n+z_n) \\ &= x+(y+z) \end{align*}$
Suppose $a,b \in \mathbb{F}$ . Then
$\begin{align*} (ab)x &= (abx_1, \cdots , abx_n) \\ &= (a(bx_1), \cdots ,a(bx_n)) \\ &= a(bx) \end{align*}$
For additive identity, let $0=(0, \cdots , 0)$ . Obviously, for all $x \in \mathbb{F}^n$ , suppose $x=(x_1, \cdots , x_n)$
$\begin{align*} x+0 &= (x_1 , \cdots , x_n) + (0, \cdots , 0) \\ &= (x_1+0, \cdots , x_n+0) \\ &= (x_1,\cdots , x_n) \\ &= x \end{align*}$
For additive inverse, suppose $x=(x_1, \cdots , x_n) \in \mathbb{F}^n$ . There exists $y=(-x_1, \cdots ,-x_n) \in \mathbb{F}^n$ such that
$\begin{align*} x+y &= (x_1, \cdots , x_n) + (-x_1, \cdots , -x_n) \\ &= (x_1+(-x_1), \cdots ,x_n+(-x_n)) \\ &=(0, \cdots ,0) \\ &= 0 \end{align*}$
For multiplicative identity, suppose $x=(x_1, \cdots , x_n) \in \mathbb{F}^n$ . Then
$\begin{align*} 1x &=(1x_1, \cdots , 1x_n) \\ &=(x_1, \cdots , x_n) \\ &=x \end{align*}$
For distributive properties, suppose $a,b \in \mathbb{F}$ , $x=(x_1, \cdots ,x_n) \in \mathbb{F}^n$ , $y=(y_1, \cdots ,y_n) \in \mathbb{F}^n$ . Then
$\begin{align*} a(x+y) &= a(x_1 + y_1 , \cdots , x_n+y_n) \\ &= (a(x_1+y_1), \cdots , a(x_n+y_n)) \\ &= (ax_1+ay_1, \cdots , ax_n+ay_n) \\ &=(ax_1, \cdots ,ax_n) + (ay_1,\cdots,ay_n) \\ &=ax+ay \end{align*}$

$\begin{align*} (a+b)x &= ((a+b)x_1, \cdots , (a+b)x_n) \\ &= (ax_1+bx_1, \cdots ,ax_n+bx_n) \\ &=(ax_1,\cdots,ax_n)+(bx_1,\cdots,bx_n) \\ &=ax+bx \end{align*}$

Thus $\mathbb{F}^n$ is a vector space over $\mathbb{F}$ .

You should verify all three bullet points in the next example.
- If $S$ is a nonempty set, then $\mathbb{F}^S$ (with the operations of addition and scalar multiplication as defined above) is a vector space over $\mathbb{F}$ .
- The additive identity of $\mathbb{F}^S$ is the function $0 : S \to \mathbb{F}$ defined by $0(x)=0$ for all $x \in S$ .
- For $f \in \mathbb{F}^S$ , the additive inverse of $f$ is the function $-f : S \to \mathbb{F}$ defined by $(-f)(x)=-f(x)$ for all $x \in S$ .

Proof Commutativity. Suppose $f,g \in \mathbb{F}^S$ , then
$\begin{align*} (f+g)(x) &= f(x)+g(x) \\ &= g(x)+f(x) \\ &= (g+f)(x) \end{align*}$
Associativity. Suppose $f,g,h \in \mathbb{F}^S$ , $a,b \in \mathbb{F}$ , then
$\begin{align*} ((f+g)+h)(x) &= (f+g)(x)+h(x) \\ &= f(x)+g(x)+h(x) \\ &= f(x)+ (g(x)+h(x)) \\ &= f(x) +((g+h)(x)) \\ &= (f+(g+h))(x) \end{align*}$

$\begin{align*} ((ab)f)(x) &= abf(x) \\ &= a(bf(x)) \\ &= a(bf)(x) \end{align*}$

Additive identity. Suppose $f \in \mathbb{F}^S$ and $0:S \to \mathbb{F}$ defined by $0(x)=0$ for all $x \in S$ , then
$\begin{align*} (f+0)(x) &= f(x)+0(x) \\ &= f(x)+0 \\ &= f(x) \\ \end{align*}$
Additive inverse. Suppose $f \in \mathbb{F}^S$ and $-f:S \to \mathbb{F}$ defined by $(-f)(x)=-f(x)$ for all $x \in S$ , then
$\begin{align*} (f+(-f))(x) &= f(x)+(-f(x)) =0 \end{align*}$
Multiplicative identity. Suppose $f \in \mathbb{F}^S$ , then
$\begin{align*} (1f)(x) = 1f(x)=f(x) \end{align*}$
Distributive properties. Suppose $a,b \in \mathbb{F}$ and $f,g \in \mathbb{F}^S$ , then
$\begin{align*} a((f+g)(x)) &= a(f(x)+g(x)) =af(x)+ag(x) \\ \end{align*}$

$\begin{align*} ((a+b)f)(x) &= (a+b)f(x)=af(x)+bf(x) \\ \end{align*}$

You should verify all the assertions in the next example.

(a) If $b \in \mathbb{F}$ , then $\{(x_1,x_2,x_3,x_4) \in \mathbb{F}^4 : x_3=5x_4+b\}$ is a subspace of $\mathbb{F}^4$ if and only if $b=0$ .

(b) The set of continuous real-valued functions on the interval $[0,1]$ is a subspace of $\mathbb{R}^{[0,1]}$ .

(c) The set of differentiable real-valued functions on $\mathbb{R}$ is a subspace of $\mathbb{R}^{\mathbb R}$ .

(d) The set of differentiable real-valued functions $f$ on the interval $(0,3)$ such that $f'(2)=b$ is a subspace of $\mathbb{R}^{(0,3)}$ if and only if $b=0$ .

(e) The set of all sequences of complex numbers with limit $0$ is a subspace of $\mathbb{C}^{\infty}$ .

Proof (a) Denote the set by $V$ .

If it is a subspace of $\mathbb{F}^4$ , then $(0,0,0,0)$ in it. Hence $0=0+b$ . This happens if $b=0$ .

If $b=0$ , then $(0,0,0,0)$ belongs to the set. For any $x,y \in V$ and $\lambda \in \mathbb{F}$ , suppose $x=(x_1,x_2,x_3,x_4)$ , $y=(y_1,y_2,y_3,y_4)$ . So we have that $x_3=5x_4$ and $y_3=5y_4$ . Then
$\begin{align*} x+y &= (x_1,x_2,x_3,x_4) +(y_1,y_2,y_3,y_4) \\ &= (x_1+y_1,x_2+y_2,x_3+y_3,x_4+y_4) \\ &= (x_1+y_1,x_2+y_2,5(x_4+y_4),x_4+y_4) \in V \end{align*}$
i.e. $V$ is closed under addition. Similarly,
$\begin{align*} \lambda x &= (\lambda x_1,\lambda x_2 , \lambda x_3,\lambda x_4) \\ &= (\lambda x_1,\lambda x_2, 5\lambda x_4, \lambda x_4) \in V \end{align*}$
i.e. $V$ is closed under scalar multiplication.

Hence if $b=0$ , $V$ is a subspace of $\mathbb{F}^4$ .

(b) Denote the set of continuous real-valued functions on the interval $[0,1]$ by $V$ .

The additive identity of $\mathbb{R}^{[0,1]}$ is the constant function $f \equiv 0$ on $[0,1]$ , and it obviously belongs to the set.

The sum of tow continuous functions is continuous, i.e. $V$ is closed under addition.

The product of constant $a$ with continuous function is continuous, i.e. $V$ is closed under scalar multiplication.

Thus $V$ is a subspace of $\mathbb{R}^{[0,1]}$ .

(d) Denote the set by $V$ .

If $V$ is a subspace of $\mathbb{R}^{[0,3]}$ , then the additive identity $f \equiv 0 \in V$ . Hence $f'(2)=0=b$ .

If $b=0$ , it is clear that the additive identity $f \equiv 0$ is contained in $V$ .

Closed under addition: suppose $f,g \in V$ , then $f,g$ are differentiable real-valued functions. So is $f+g$ . Moreover,
$\begin{align*} (f+g)'(2) &= f'(2)+g'(2) \\ &= 0+0 \\ &=0 \end{align*}$
Closed under scalar multiplication: suppose $f \in V$ and $a \in \mathbb{R}$ , then $f$ is differentiable read-valued function. So is $af$ . Moreover,
$\begin{align*} (af)'(2)=af'(2)=0 \end{align*}$
(e) Denote the set by $V$ .

Additive identity: it is clear that $(0,0,\cdots) \in V$ .

Closed under addition: suppose $a=(a_1,a_2,\cdots) \in V$ and $b=(b_1,b_2, \cdots) \in V$ , so $\lim\limits_{n \to \infty} a_n=0$ and $\lim\limits_{n \to \infty} b_n=0$ .
$\begin{align*} \lim\limits_{n \to \infty} (a_n+b_n)=\lim\limits_{n \to \infty} a_n +\lim\limits_{n \to \infty} b_n=0+0 =0 \end{align*}$
Hence, $a+b=(a_1+b_1,a_2+b_2,\cdots) \in V$ .

Closed under scalar multiplication: suppose $a=(a_1,a_2,\cdots) \in V$ and $\lambda \in \mathbb{C}$ , then $\lim\limits_{n \to \infty} a_n=0$ .
$\begin{align*} \lim\limits_{n \to \infty} \lambda a_n=\lambda \lim\limits_{n \to \infty} a_n= \lambda 0 =0 \end{align*}$
Hence, $\lambda a=(\lambda a_1,\lambda a_2,\cdots) \in V$ .

Suppose $U$ is the set of all elements of $\mathbb{F}^3$ whose second and third coordinates equal $0$ , and $W$ is the set of all elements of $\mathbb{F}^3$ whose first and third coordinates equal $0$ :
$U=\{(x,0,0) \in \mathbb{F}^3:x \in \mathbb{F} \} \quad \text{and} \quad W=\{(0,y,0)\in \mathbb{F}^3 : y\in \mathbb{F} \}.$
Then
$U+W =\{(x,y,0): x,y \in \mathbb{F} \},$
as you should verify.

Proof Suppose $u=(x_1,0,0) \in U$ and $w=(0,y_1,0) \in W$ , then
$u+w=(x_1,0,0)+(0,y_1,0) =(x_1,y_1,0) \in \{(x,y,0):x,y\in \mathbb{F} \}$
Hence $U+W \subset \{(x,y,0):x,y\in \mathbb{F} \}$ .

Every vector in $\{(x,y,0):x,y \in \mathbb{F} \}$ , can be written as
$(x,y,0)=(x,0,0)+(y,0,0),$
where the first vector on the right side is in $U$ , the second vector is in $W$ .

Hence $\{(x,y,0):x,y \in \mathbb{F} \} \subset U+W$ .

Thus $U+W=\{(x,y,0):x,y \in \mathbb{F} \}$ .

Suppose $U$ is the subspace of $\mathbb{F}^3$ of those vectors whose last coordinate equals $0$ , and $W$ is the subspace of $\mathbb{F}^3$ of those vectors whose first two coordinates equal $0$ :
$U=\{ (x,y,0)\in \mathbb{F}^3:x,y\in \mathbb{F} \} \quad \text{and} \quad W=\{(0,0,z) \in \mathbb{F}^3 :z\in \mathbb{F} \}.$
Then $\mathbb{F}^3 = U \oplus W$ , as you should verify.

Proof $U \cap W={0}$ , hence $U+W$ is direct sum.

Clearly $\mathbb{F}^3=U+W$ , because every vector $(x,y,z) \in \mathbb{F}^3$ can be written as
$(x,y,z)=(x,y,0)+(0,0,z),$
where the first vector on the right side is in $U$ , the second vector is in $W$ .

Thus $\mathbb{F}^3=U \oplus W$ .

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