Fragment 1 数域
/Define/
定义1 设 K K K C C C K K K K K K K K K
例1 Q ( 2 ) = { a + b 2 ∣ a , b ∈ Q } \mathbb{Q}(\sqrt{2}) = \{ a + b\sqrt{2} \mid a, b \in \mathbb{Q} \} Q ( 2 ) = { a + b 2 ∣ a , b ∈ Q }
/proof/
先证明 a + b 2 = 0 ⇒ a = b = 0 a + b\sqrt{2} = 0 \Rightarrow a = b = 0 a + b 2 = 0 ⇒ a = b = 0
( a + b 2 ) ± ( c + d 2 ) = ( a ± c ) ± ( b ± d ) 2 ∈ Q ( 2 ) ( a + b 2 ) ⋅ ( c + d 2 ) = ( a c + 2 b d ) + ( a d + b c ) 2 ∈ Q ( 2 ) c + d 2 ≠ 0 ⇒ c ≠ 0 or d ≠ 0 ⇒ c − d 2 ≠ 0 a + b 2 c + d 2 = ( a + b 2 ) ( c − d 2 ) ( c + d 2 ) ( c − d 2 ) = a c − 2 b d c 2 − 2 d 2 + b c − a d c 2 − 2 d 2 2 ∈ Q ( 2 ) (a + b\sqrt{2}) \pm (c + d\sqrt{2}) = (a \pm c) \pm (b \pm d)\sqrt{2} \in \mathbb{Q}(\sqrt{2}) \\
(a + b\sqrt{2}) \cdot (c + d\sqrt{2}) = (ac + 2bd) + (ad + bc)\sqrt{2} \in \mathbb{Q}(\sqrt{2}) \\
c + d\sqrt{2} \neq 0 \Rightarrow c \neq 0 \text{ or } d \neq 0 \Rightarrow c - d\sqrt{2} \neq 0 \\
\frac{a + b\sqrt{2}}{c + d\sqrt{2}} = \frac{(a + b\sqrt{2})(c - d\sqrt{2})}{(c + d\sqrt{2})(c - d\sqrt{2})} = \frac{ac - 2bd}{c^2 - 2d^2} + \frac{bc - ad}{c^2 - 2d^2}\sqrt{2} \in \mathbb{Q}(\sqrt{2})
( a + b 2 ) ± ( c + d 2 ) = ( a ± c ) ± ( b ± d ) 2 ∈ Q ( 2 ) ( a + b 2 ) ⋅ ( c + d 2 ) = ( a c + 2 b d ) + ( a d + b c ) 2 ∈ Q ( 2 ) c + d 2 = 0 ⇒ c = 0 or d = 0 ⇒ c − d 2 = 0 c + d 2 a + b 2 = ( c + d 2 ) ( c − d 2 ) ( a + b 2 ) ( c − d 2 ) = c 2 − 2 d 2 a c − 2 b d + c 2 − 2 d 2 b c − a d 2 ∈ Q ( 2 )
例2 $ \mathbb{Q}(\sqrt{3}) = { a + b\sqrt{3} + c\sqrt{4} \mid a, b, c \in \mathbb{Q} } $ 是数域。
/proof/
先证明 a + b 3 + c 4 = 0 ⇒ a 2 + 2 b 2 + 4 c 2 − 2 a b c = 0 a + b\sqrt{3} + c\sqrt{4} = 0 \Rightarrow a^2 + 2b^2 + 4c^2 - 2abc = 0 a + b 3 + c 4 = 0 ⇒ a 2 + 2 b 2 + 4 c 2 − 2 ab c = 0
⇒ a = b = c = 0 \Rightarrow a = b = c = 0
⇒ a = b = c = 0
推广1： Q ( 2 ) = { a 0 + a 1 2 + ⋯ + a n 2 n ∣ a i ∈ Q , 0 ≤ i ≤ n } \mathbb{Q}(\sqrt{2}) = \{ a_0 + a_1\sqrt{2} + \cdots + a_n\sqrt{2^n} \mid a_i \in \mathbb{Q}, 0 \leq i \leq n \} Q ( 2 ) = { a 0 + a 1 2 + ⋯ + a n 2 n ∣ a i ∈ Q , 0 ≤ i ≤ n }
推广2： P P P Q ( p ) = { a + b p ∣ a , b ∈ Q } \mathbb{Q}(\sqrt{p}) = \{ a + b\sqrt{p} \mid a, b \in \mathbb{Q} \} Q ( p ) = { a + b p ∣ a , b ∈ Q }
例3 { a 0 + a 1 π + ⋯ + a n π n ∣ a i ∈ Q } \left\{ a_0 + a_1\pi + \cdots + a_n\pi^n \mid a_i \in \mathbb{Q} \right\} { a 0 + a 1 π + ⋯ + a n π n ∣ a i ∈ Q }
/proof/
π \pi π ⟺ d e f \overset{def}\iff ⟺ d e f
则 b 0 = b 1 = ⋯ = b m = 0 b_0 = b_1 = \cdots = b_m = 0 b 0 = b 1 = ⋯ = b m = 0
α ∈ C \alpha \in \mathbb{C} α ∈ C α \alpha α ⟺ d e f ∃ f ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0 \overset{def}\iff \exist f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 ⟺ d e f ∃ f ( x ) = a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0
s t . f ( α ) = 0 st. f(\alpha) = 0 s t . f ( α ) = 0 α \alpha α f ( x ) f(x) f ( x )
例4 S = { a + b 2 ∣ a , b ∈ Z } S = \{ a + b\sqrt{2} \mid a, b \in \mathbb{Z} \} S = { a + b 2 ∣ a , b ∈ Z }
/proof/
用反证法。设 S S S Z ⊆ S \mathbb{Z} \subseteq S Z ⊆ S
1 ∈ S ⇒ 2 ∈ S ⇒ 1 2 ∈ S 1 \in S \Rightarrow 2 \in S \Rightarrow \frac{1}{2} \in S 1 ∈ S ⇒ 2 ∈ S ⇒ 2 1 ∈ S 1 2 = a + b 2 \frac{1}{2} = a + b\sqrt{2} 2 1 = a + b 2 a , b ∈ Z a, b \in \mathbb{Z} a , b ∈ Z
若 $b \neq 0 $ ，则 $ \sqrt{2} = \frac{a - \frac{1}{2}}{b} \in \mathbb{Q}$ 矛盾！
例5 S = { a 2 ∣ a ∈ R } S = \{ a\sqrt{2} \mid a \in \mathbb{R} \} S = { a 2 ∣ a ∈ R }
/proof/
反证法。设 S S S 2 ∈ S ⇒ 2 = 2 ⋅ 2 ∈ S \sqrt{2} \in S \Rightarrow 2 = \sqrt{2} \cdot \sqrt{2} \in S 2 ∈ S ⇒ 2 = 2 ⋅ 2 ∈ S
即 2 = a 2 a ∈ Q ⇒ a = 2 ∉ Q 2 = a\sqrt{2}\quad a \in \mathbb{Q} \Rightarrow a = \sqrt{2} \notin \mathbb{Q} 2 = a 2 a ∈ Q ⇒ a = 2 ∈ / Q
/Theorem/ 定理2：任一数域 K K K Q \mathbb{Q} Q Q \mathbb{Q} Q
/proof/
∀ a ∈ K ⇒ 0 = a − a ∈ K \forall a \in K \Rightarrow 0 = a - a \in K ∀ a ∈ K ⇒ 0 = a − a ∈ K
由定义（数域至少有2个不同元素）再取 K K K b b b 1 = b b ∈ K 1 = \frac{b}{b} \in K 1 = b b ∈ K
{ ∀ m ∈ Z + , m = ( 1 + ⋯ + 1 ) ∈ K , − m = 0 − m ∈ K } ⇒ Z ⊆ K \left\{ \forall m \in \mathbb{Z}^+ ,\quad m = (1 + \cdots + 1) \in K,\quad
-m = 0 - m \in K \right\} \Rightarrow \mathbb{Z} \subseteq K\\
{ ∀ m ∈ Z + , m = ( 1 + ⋯ + 1 ) ∈ K , − m = 0 − m ∈ K } ⇒ Z ⊆ K
任取 m n ∈ Q \frac{m}{n} \in \mathbb{Q} n m ∈ Q n ∈ Z + n \in \mathbb{Z}^+ n ∈ Z + m ∈ Z + m \in \mathbb{Z}^+ m ∈ Z + n ∈ K n \in K n ∈ K m ∈ K ⇒ m n ∈ K m \in K \Rightarrow \frac{m}{n} \in K m ∈ K ⇒ n m ∈ K Q ⊆ K \mathbb{Q} \subseteq K Q ⊆ K
Fragment 2 线性空间首先来定义行向量和列向量
· 行向量 列向量
/Define/
定义1
设 K K K a 1 , a 2 , … , a n ∈ K a_1, a_2, \ldots, a_n \in K a 1 , a 2 , … , a n ∈ K
1 × n 1 \times n 1 × n [ a 1 , a 2 , … , a n ] [a_1, a_2, \ldots, a_n] [ a 1 , a 2 , … , a n ] K K K n n n n × 1 n \times 1 n × 1 [ a 1 ⋮ a n ] \begin{bmatrix} a_1 \\ \vdots \\ a_n \end{bmatrix} a 1 ⋮ a n K K K n n n
记 K n = { ( a 1 , … , a n ) ∣ a i = k , i ∈ N } K_n = \{ (a_1, \ldots, a_n) \mid a_i = k, i \in \mathbb{N} \} K n = {( a 1 , … , a n ) ∣ a i = k , i ∈ N } K K K n n n
K n = { [ a 1 ⋮ a n ] ∣ a i ∈ K , i ∈ N } K^n = \{ \begin{bmatrix} a_1 \\ \vdots \\ a_n \end{bmatrix} \mid a_i \in K, i \in \mathbb{N} \} K n = { a 1 ⋮ a n ∣ a i ∈ K , i ∈ N } K K K n n n
α = ( a 1 a 2 ⋮ a n ) , β = ( b 1 b 2 ⋮ b n ) ∈ K n , c ∈ K \alpha = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix}, \quad \beta = \begin{pmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{pmatrix} \in K^n, \quad c \in K
α = a 1 a 2 ⋮ a n , β = b 1 b 2 ⋮ b n ∈ K n , c ∈ K
α − β = def ( a 1 − b 1 ⋮ a i − b i ) , ∀ i ∈ N α + β = def ( a 1 + b 1 ⋮ a n + b n ) c ⋅ α = def ( c a 1 ⋮ c a n ) \alpha - \beta \overset{\text{def}}{=} \begin{pmatrix} a_1 - b_1 \\ \vdots \\ a_i - b_i \end{pmatrix}, \quad \forall i \in \mathbb{N} \quad \quad
\alpha + \beta \overset{\text{def}}{=} \begin{pmatrix} a_1 + b_1 \\ \vdots \\ a_n + b_n \end{pmatrix}\quad \quad
c \cdot \alpha \overset{\text{def}}{=} \begin{pmatrix} ca_1 \\ \vdots \\ ca_n \end{pmatrix}
α − β = def a 1 − b 1 ⋮ a i − b i , ∀ i ∈ N α + β = def a 1 + b 1 ⋮ a n + b n c ⋅ α = def c a 1 ⋮ c a n
行向量运算规则：α , β , γ ∈ K n ( K n ) , k ∈ K \alpha, \beta, \gamma \in K^n (K_n), k \in K α , β , γ ∈ K n ( K n ) , k ∈ K
加法交换律：α + β = β + α \alpha + \beta = \beta + \alpha α + β = β + α
加法结合律：α + ( β + γ ) = ( α + β ) + γ \alpha + (\beta + \gamma) = (\alpha + \beta) + \gamma α + ( β + γ ) = ( α + β ) + γ
零向量：α + 0 = α \alpha + 0 = \alpha α + 0 = α
负向量：α + ( − α ) = 0 \alpha + (-\alpha) = 0 α + ( − α ) = 0
乘法单位元：1 ⋅ α = α 1 \cdot \alpha = \alpha 1 ⋅ α = α
数乘分配律（左）：( k + l ) α = k α + l α (k + l) \alpha = k\alpha + l\alpha ( k + l ) α = k α + l α
数乘分配律（右）：k ( α + β ) = k α + k β k(\alpha + \beta) = k\alpha + k\beta k ( α + β ) = k α + k β
数乘结合律：( k l ) α = k ( l α ) (kl)\alpha = k(l\alpha) ( k l ) α = k ( l α )
· 线性空间
然后我们可以给出代数学中的一个核心定义：
/Define/
定义2
设 K K K V V V V V V + : V × V → V + : V \times V \rightarrow V + : V × V → V ( α , β ) ↦ α + β ( \alpha, \beta ) \mapsto \alpha + \beta ( α , β ) ↦ α + β
K K K V V V K × V → V K \times V \rightarrow V K × V → V ( k , α ) ↦ k α (k, \alpha) \mapsto k\alpha ( k , α ) ↦ k α
若加法运算和数乘运算满足八条性质，则称 V V V K K K
请注意，线性空间中的加法和数乘已经不再局限于数的加法和乘法的概念中了。也就是说，1+1等于几是由你自己定义的，只要你所定义的满足八条运算法则。
正是因为这个，对于零元，单位元，负元也是要根据八条法则确定。比如零元并不一定就是0.我们可以根据线性空间的性质得出。
此处本喵给出一个想法：
f ( α + β ) = f ( α ) + f ( β ) f ( k α ) = k f ( α ) f(\alpha+\beta)=f(\alpha)+f(\beta)\quad \quad f(k\alpha)=kf(\alpha)
f ( α + β ) = f ( α ) + f ( β ) f ( k α ) = k f ( α )
/question/ 我们是否可以说集合 V V V α \alpha α β \beta β k ∈ K k\in K k ∈ K
例1： K n K^n K n K n K_n K n K K K
例2： K [ x ] = { a 0 x n + ⋯ + a n ∣ x K[x] = \{ a_0x^n + \cdots + a_n \mid x K [ x ] = { a 0 x n + ⋯ + a n ∣ x , a i ∈ K , i ∈ N } , a_i \in K, i \in \mathbb{N} \} , a i ∈ K , i ∈ N }
加法：
f ( x ) = a 0 x n + ⋯ + a n g ( x ) = b 0 x m + ⋯ + b m n ≥ m f ( x ) + g ( x ) = def a 0 x n + ⋯ + ( a m + b m ) x m + ⋯ + ( a 0 + b 0 ) f(x) = a_0x^n + \cdots + a_n\quad \quad
g(x) = b_0x^m + \cdots + b_m \quad n \geq m\\
f(x) + g(x) \overset{\text{def}}{=} a_0x^n + \cdots + (a_m + b_m)x^m + \cdots + (a_0 + b_0)
f ( x ) = a 0 x n + ⋯ + a n g ( x ) = b 0 x m + ⋯ + b m n ≥ m f ( x ) + g ( x ) = def a 0 x n + ⋯ + ( a m + b m ) x m + ⋯ + ( a 0 + b 0 )
数乘：k ⋅ f ( x ) = def k a 0 x n + ⋯ + k a n x + k a n k \cdot f(x) \overset{\text{def}}{=} ka_0x^n + \cdots + ka_nx + ka_n k ⋅ f ( x ) = def k a 0 x n + ⋯ + k a n x + k a n
K [ x ] K[x] K [ x ] K K K
例3： C [ 0 , 1 ] C[0,1] C [ 0 , 1 ] R \mathbb{R} R
加法：( f + g ) ( x ) = def f ( x ) + g ( x ) (f+g)(x) \overset{\text{def}}{=} f(x) + g(x) ( f + g ) ( x ) = def f ( x ) + g ( x ) ( k f ) ( x ) = def k f ( x ) (kf)(x) \overset{\text{def}}{=} kf(x) ( k f ) ( x ) = def k f ( x )
→ 成立八条性质 → C [ 0 , 1 ] C[0,1] C [ 0 , 1 ] R \mathbb{R} R
例4： M m n ( K ) M_{mn}(K) M mn ( K ) K K K m × n m \times n m × n
加法：A + B A + B A + B k ⋅ A k \cdot A k ⋅ A
→ 成立八条性质 → M m n ( K ) M_{mn}(K) M mn ( K ) K K K
例5 设 k 1 ≤ k 2 k_1 \leq k_2 k 1 ≤ k 2 Q ⊆ R ⊆ C \mathbb{Q} \subseteq \mathbb{R} \subseteq \mathbb{C} Q ⊆ R ⊆ C
加法：a , b ∈ k 2 a, b \in k_2 a , b ∈ k 2 a + b a + b a + b k 1 ∈ k 2 k_1 \in k_2 k 1 ∈ k 2 a ∈ k 2 a \in k_2 a ∈ k 2 k 1 ⋅ a k_1 \cdot a k 1 ⋅ a
→ 成立 → k 2 k_2 k 2 k 1 k_1 k 1
特别：k k k k k k
命题3 设 V k V_k V k
零向量唯一
设 0 1 , 0 2 0_1, 0_2 0 1 , 0 2 0 1 = 0 1 + 0 2 = 0 2 0_1 = 0_1 + 0_2 = 0_2 0 1 = 0 1 + 0 2 = 0 2
负向量唯一
设 β \beta β α \alpha α α + β = 0 \alpha + \beta = 0 α + β = 0 α + γ = 0 \alpha + \gamma = 0 α + γ = 0 β = β + 0 = β + ( α + γ ) = ( β + α ) + γ = 0 + γ = γ \beta = \beta + 0 = \beta + (\alpha + \gamma) = (\beta + \alpha) + \gamma = 0 + \gamma = \gamma β = β + 0 = β + ( α + γ ) = ( β + α ) + γ = 0 + γ = γ
加法消去律：α + β = α + γ ⇒ β = γ \alpha + \beta = \alpha + \gamma \Rightarrow \beta = \gamma α + β = α + γ ⇒ β = γ
α − α + α + β = α + α + γ ⇒ β = γ \alpha - \alpha + \alpha + \beta = \alpha + \alpha + \gamma \Rightarrow \beta = \gamma α − α + α + β = α + α + γ ⇒ β = γ
0 ⋅ α = 0 0\cdot \alpha = 0 0 ⋅ α = 0
α + 0 ⋅ α = ( 0 + 0 ) α = 0 ⋅ α + 0 ⋅ α = 0 \alpha + 0 \cdot \alpha = (0 + 0) \alpha = 0 \cdot \alpha + 0 \cdot \alpha = 0 α + 0 ⋅ α = ( 0 + 0 ) α = 0 ⋅ α + 0 ⋅ α = 0
k ⋅ 0 = 0 k \cdot 0 = 0 k ⋅ 0 = 0
k ( 0 + 0 ) = k ⋅ 0 + k ⋅ 0 = 0 k(0 + 0) = k \cdot 0 + k \cdot 0 = 0 k ( 0 + 0 ) = k ⋅ 0 + k ⋅ 0 = 0
− α = ( − 1 ) α -\alpha = (-1) \alpha − α = ( − 1 ) α
α + ( − 1 ) α = ( 1 − 1 ) α = 0 ⋅ α = 0 \alpha + (-1) \alpha = (1 - 1) \alpha = 0 \cdot \alpha = 0 α + ( − 1 ) α = ( 1 − 1 ) α = 0 ⋅ α = 0
若 k α = 0 k \alpha = 0 k α = 0 k = 0 k = 0 k = 0 α = 0 \alpha = 0 α = 0
k = 0 k = 0 k = 0 α ≠ 0 \alpha \neq 0 α = 0 k = 0 k = 0 k = 0
注意：
(1) 加法消去律 ⇒ 可移项
α = β + γ ⇒ { α − β = γ α − β − γ = 0 \alpha = \beta + \gamma \Rightarrow
\begin{cases}
\alpha - \beta = \gamma\\
\alpha - \beta - \gamma = 0
\end{cases}
α = β + γ ⇒ { α − β = γ α − β − γ = 0
(2) α + ⋯ + 0 \alpha + \cdots + 0 α + ⋯ + 0
Fragment 3 向量组线性关系
/Define/
定义1
设 V V V K K K α 1 , … , α n , β ∈ V \alpha_1, \ldots, \alpha_n, \beta \in V α 1 , … , α n , β ∈ V k 1 , … , k n ∈ K k_1, \ldots, k_n \in K k 1 , … , k n ∈ K β = k 1 α 1 + ⋯ + k n α n \beta = k_1\alpha_1 + \cdots + k_n\alpha_n β = k 1 α 1 + ⋯ + k n α n β \beta β α 1 , … , α n \alpha_1, \ldots, \alpha_n α 1 , … , α n β \beta β α 1 , … , α n \alpha_1, \ldots, \alpha_n α 1 , … , α n β \beta β α 1 , … , α n \alpha_1, \ldots, \alpha_n α 1 , … , α n
例1： V = K 3 V = K_3 V = K 3 β = ( 1 , 1 , 1 ) \beta = (1, 1, 1) β = ( 1 , 1 , 1 ) α 1 = ( 1 , 1 , 1 ) \alpha_1 = (1, 1, 1) α 1 = ( 1 , 1 , 1 ) α 2 = ( 1 , 0 , 1 ) \alpha_2 = (1, 0, 1) α 2 = ( 1 , 0 , 1 ) α 3 = ( 1 , 1 , 0 ) \alpha_3 = (1, 1, 0) α 3 = ( 1 , 1 , 0 )
例2： V = K n V = K_n V = K n e i = ( 0 , … , 1 , … , 0 ) e_i = (0, \ldots, 1, \ldots, 0) e i = ( 0 , … , 1 , … , 0 ) i ∈ N i \in \mathbb{N} i ∈ N { e 1 , … , e n } \{e_1, \ldots, e_n\} { e 1 , … , e n } K n K_n K n
例3：任一向量是标准单位行向量的线性组合。
/proof/
任取 α ∈ K n \alpha \in K_n α ∈ K n α = ( a 1 , … , a n ) , α 1 ∈ K \alpha = (a_1, \ldots, a_n),\alpha_1 \in K α = ( a 1 , … , a n ) , α 1 ∈ K
⇒ α = a 1 e 1 + ⋯ + a n e n \Rightarrow \alpha = a_1e_1 + \cdots + a_ne_n ⇒ α = a 1 e 1 + ⋯ + a n e n
接着给出线性相关性定义
/Define/
定义2
设 V k V_k V k α 1 , … , α n ∈ V \alpha_1, \ldots, \alpha_n \in V α 1 , … , α n ∈ V K K K c 1 , … , c n c_1, \ldots, c_n c 1 , … , c n c 1 α 1 + ⋯ + c n α n = 0 c_1\alpha_1 + \cdots + c_n\alpha_n = 0 c 1 α 1 + ⋯ + c n α n = 0 α 1 , … , α n \alpha_1, \ldots, \alpha_n α 1 , … , α n
自然引出问题：如何判断向量组线性相关性？
法则1：线性无关的等价定义
若 k 1 , … , k n ∈ K k_1, \ldots, k_n \in K k 1 , … , k n ∈ K k 1 α 1 + ⋯ + k n α n = 0 k_1\alpha_1 + \cdots + k_n\alpha_n = 0 k 1 α 1 + ⋯ + k n α n = 0 k 1 = k 2 = ⋯ = k n = 0 k_1 = k_2 = \cdots = k_n = 0 k 1 = k 2 = ⋯ = k n = 0
注意：线性相关性和线性无关性依赖于基域 K K K k 1 , … , k n ∈ K k_1, \ldots, k_n \in K k 1 , … , k n ∈ K
/example/ R ⊆ C \mathbb{R} \subseteq \mathbb{C} R ⊆ C C \mathbb{C} C R \mathbb{R} R
断言：{ i , i 2 } \{i, i^2\} { i , i 2 } R \mathbb{R} R a 1 + b 1 i = 0 a_1 + b_1i = 0 a 1 + b 1 i = 0 ( a , b ∈ R ) (a, b \in \mathbb{R}) ( a , b ∈ R ) a + b i = 0 a + bi = 0 a + bi = 0 ⇒ \Rightarrow ⇒ a = b = 0 a = b = 0 a = b = 0
/example/ C ⊆ C \mathbb{C} \subseteq \mathbb{C} C ⊆ C C \mathbb{C} C C \mathbb{C} C
断言：{ i , i 2 } \{i, i^2\} { i , i 2 } C \mathbb{C} C 1 + i + i 2 = 0 1 + i + i^2 = 0 1 + i + i 2 = 0
自然得出结论：向量组 = 有限个向量构成的集合。
例4：包含零向量的向量组必线性相关。
设 x 1 = 0 , x 2 , … , x n ∈ V x_1 = 0, x_2, \ldots, x_n \in V x 1 = 0 , x 2 , … , x n ∈ V ( x 1 + x 2 + ⋯ + x n = 0 ⇒ { x 1 , … , x n } (x_1 + x_2 + \cdots + x_n = 0 \Rightarrow \{ x_1, \ldots, x_n \} ( x 1 + x 2 + ⋯ + x n = 0 ⇒ { x 1 , … , x n }
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 ## 定理与证明 ### 定理5 n维标准单位向量必线性无关。 **证明：** 设 $ k_1e_1 + \cdots + k_ne_n = 0 $ $ e_i \in K $ 。 即 $ 0 = (k_1, \cdots, k_n) $ 从而 $ k_1 = \cdots = k_n = 0 $ 从而 $ e_1, e_2, \cdots, e_n $ 线性无关。 ### 定理3 设 $ l \leq m \leq n $ 则若 $ \{\alpha_1, \ldots, \alpha_m\} $ 线性相关，则 $ \{\alpha_1, \ldots, \alpha_n\} $ 也线性相关。 若 $ \{\alpha_1, \ldots, \alpha_n\} $ 线性无关，则 $ \{\alpha_1, \ldots, \alpha_m\} $ 也线性无关。 **证明：** 1) 由定义存在 $ K $ 中不全为 0 的数 $ k_1, \ldots, k_m $ 使 $ k_1\alpha_1 + \cdots + k_m\alpha_m = 0 $ 从而 $ k_1\alpha_1 + \cdots + k_m\alpha_m + 0\alpha_{m+1} + \cdots + 0\cdot \alpha_n = 0 $ 。 这些系数不全为 0，故 $ \alpha_1, \ldots, \alpha_n $ 线性相关。 2) 是 (1) 的逆否命题 # ### 定理4 设 $ V_k $ 是线性空间， $ \alpha_1, \ldots, \alpha_n \in V $ 则 $ \alpha_1, \ldots, \alpha_n $ 线性相关 $ \Leftrightarrow \exists 1 \leq i \leq n $ st. $ \alpha_i $ 是 $ \alpha_1, \ldots, \alpha_{i-1}, \alpha_{i+1}, \ldots, \alpha_n $ 的线性组合。 **证明：** 充分性：不妨设 $ \alpha_1 $ 是 $ \alpha_2, \ldots, \alpha_n $ 的线性组合，则存在不全为 0 的数 $ k_2, \ldots, k_n \in K $ 使 $ \alpha_1 = k_2\alpha_2 + \cdots + k_n\alpha_n $ 即 $ -\alpha_1 + k_2\alpha_2 + \cdots + k_n\alpha_n = 0 $ 必要性：设 $ \alpha_1, \ldots, \alpha_n $ 线性相关，即存在不全为 0 的数 $ k_1, \ldots, k_n \in K $ 使 $ k_1\alpha_1 + \cdots + k_n\alpha_n = 0 $ 不妨设 $ k_1 \neq 0 $ 则 $ \alpha_1 = -\frac{k_2}{k_1}\alpha_2 - \cdots - \frac{k_n}{k_1}\alpha_n $ # ### 定理5 设 $ V_k $ 是线性空间， $ \alpha_1, \ldots, \alpha_n, \beta \in V $ 若 $ \alpha_1, \ldots, \alpha_n $ 线性无关， 则或者 $ \alpha_1, \ldots, \alpha_n, \beta $ 线性无关，或者 $ \beta $ 是 $ \alpha_1, \ldots, \alpha_n $ 的线性组合。 **证明：** 1° $ \alpha_1, \ldots, \alpha_n, \beta $ 线性无关。 2° $ \alpha_1, \ldots, \alpha_n, \beta $ 线性相关，即存在不全为 0 的数 $ k_1, \ldots, k_n \in K $ st $ k_1\alpha_1 + \cdots + k_n\alpha_n + k_{n+1}\beta = 0 $ 断言 $ k_{n+1} \neq 0 $ ① 用反证法 设 $ k_{n+1} = 0 $ 则 $ k_1\alpha_1 + \cdots + k_n\alpha_n = 0 $ 又由 $ \alpha_1, \ldots, \alpha_n $ 无关 ⇒ $ k_1 = \cdots = k_n = 0 $ 矛盾 #
