《線性代數(shù)應(yīng)該這樣學(xué)》課內(nèi)驗(yàn)證1-向量空間

《線性代數(shù)應(yīng)該這樣學(xué)》（Linear Algebra Done Right）中說：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. 這個(gè)筆記也由此而生。

筆記將會(huì)記錄一部分書內(nèi)要求自行驗(yàn)證的題目晚树。出于某些原因想邦，我將試著用英語來進(jìn)行記錄。這是我第一次嘗試著用英語寫數(shù)學(xué)證明击碗，因此在格式、單詞们拙、語法等地方出錯(cuò)應(yīng)該是難以避免的稍途，望諒解。

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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《線性代數(shù)應(yīng)該這樣學(xué)》課內(nèi)驗(yàn)證1-向量空間

CHAPTER 1 Vector Spaces

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