线性空间
性质（判断是否是线性空间的依据）
定义 5.1.1 设 V V V F F F V V V V V V α \alpha α β \beta β V V V γ \gamma γ α \alpha α β \beta β γ = α + β \gamma = \alpha + \beta γ = α + β F F F V V V F F F k k k V V V α \alpha α V V V δ \delta δ k k k α \alpha α δ = k α \delta = k\alpha δ = k α 满足如下性质：
(1) 加法交换律 α + β = β + α \alpha + \beta = \beta + \alpha α + β = β + α
(2) 加法结合律 ( α + β ) + γ = α + ( β + γ ) (\alpha + \beta) + \gamma = \alpha + (\beta + \gamma) ( α + β ) + γ = α + ( β + γ )
(3) 存在“零元 ”，即存在 0 ∈ V \mathbf{0} \in V 0 ∈ V ∀ α ∈ V , 0 + α = α \forall \alpha \in V, \ \mathbf{0} + \alpha = \alpha ∀ α ∈ V , 0 + α = α
(4) 存在负元 ，即 ∀ α ∈ V \forall \alpha \in V ∀ α ∈ V β ∈ V \beta \in V β ∈ V α + β = 0 \alpha + \beta = \mathbf{0} α + β = 0
(5) 1 ⋅ α = α 1 \cdot \alpha = \alpha 1 ⋅ α = α
(6) 数乘结合律 ( k l ) α = k ( l α ) = l ( k α ) (kl)\alpha = k(l\alpha) = l(k\alpha) ( k l ) α = k ( l α ) = l ( k α )
(7) 分配律 ( k + l ) α = k α + l α (k + l)\alpha = k\alpha + l\alpha ( k + l ) α = k α + l α
(8) 分配律 k ( α + β ) = k α + k β k(\alpha + \beta) = k\alpha + k\beta k ( α + β ) = k α + k β
基变换与坐标变换
过渡矩阵
ε 1 , ε 2 , ⋯ , ε n \varepsilon_1, \varepsilon_2, \cdots, \varepsilon_n ε 1 , ε 2 , ⋯ , ε n η 1 , η 2 , ⋯ , η n \eta_1, \eta_2, \cdots, \eta_n η 1 , η 2 , ⋯ , η n V V V
$$\begin{cases} \eta_1 = a_{11}\varepsilon_1 + a_{21}\varepsilon_2 + \cdots + a_{n1}\varepsilon_n, \\ \eta_2 = a_{12}\varepsilon_1 + a_{22}\varepsilon_2 + \cdots + a_{n2}\varepsilon_n, \\ \cdots\cdots \\ \eta_n = a_{1n}\varepsilon_1 + a_{2n}\varepsilon_2 + \cdots + a_{nn}\varepsilon_n. \end{cases} \quad (5.3.1)$$
将式 (5.3.1) 写成矩阵形式
$$(\eta_1, \eta_2, \cdots, \eta_n) = (\varepsilon_1, \varepsilon_2, \cdots, \varepsilon_n) \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix}. \quad (5.3.2)$$
这也就是基变换 公式。
定义 5.3.1 我们称式 (5.3.2) 中的矩阵
$$A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix}$$
为从基 ε 1 , ε 2 , ⋯ , ε n \varepsilon_1, \varepsilon_2, \cdots, \varepsilon_n ε 1 , ε 2 , ⋯ , ε n η 1 , η 2 , ⋯ , η n \eta_1, \eta_2, \cdots, \eta_n η 1 , η 2 , ⋯ , η n 过渡矩阵 .
基变换公式
η = A ε \mathbf{\eta} = A\mathbf{\varepsilon} η = A ε ( x 1 , x 2 , ⋯ , x n ) = ( ε 1 , ε 2 , ⋯ , ε n ) A (x_{1}, x_{2}, \cdots, x_{n})=(\varepsilon_{1}, \varepsilon_{2}, \cdots, \varepsilon_{n})A ( x 1 , x 2 , ⋯ , x n ) = ( ε 1 , ε 2 , ⋯ , ε n ) A 坐标变换公式
$$\mathbf{x} = A\mathbf{x}'$$
$$(x_{1},x_{2}, \cdots, x_{n})=A(x'_{1}, x'_{2}, \cdots, x'_{n})$$
线性子空间
判断
运算
子空间的交 V 1 ∩ V 2 = { v ∣ v ∈ V 1 且 v ∈ V 2 } V_1 \cap V_2 = \{v \mid v \in V_1 \text{ 且 } v \in V_2\} V 1 ∩ V 2 = { v ∣ v ∈ V 1 且 v ∈ V 2 }
子空间的和 V 1 + V 2 = { v 1 + v 2 ∣ v 1 ∈ V 1 , v 2 ∈ V 2 } V_1 + V_2 = \{v_1 + v_2 \mid v_1 \in V_1, v_2 \in V_2\} V 1 + V 2 = { v 1 + v 2 ∣ v 1 ∈ V 1 , v 2 ∈ V 2 }
子空间的直和 V 1 + V 2 V_1 + V_2 V 1 + V 2 α \mathbf{\alpha} α α = α 1 + α 2 , α 1 ∈ V 1 , α 2 ∈ V 2 \mathbf{\alpha} = \mathbf{\alpha}_1 + \mathbf{\alpha}_2, \quad \mathbf{\alpha}_1 \in V_1, \ \mathbf{\alpha}_2 \in V_2 α = α 1 + α 2 , α 1 ∈ V 1 , α 2 ∈ V 2 V 1 + V 2 V_1 + V_2 V 1 + V 2 直和 ，记为 V 1 ⊕ V 2 V_1 \oplus V_2 V 1 ⊕ V 2
L ( α 1 , α 2 , ⋯ , α s ) + L ( β 1 , β 2 , ⋯ , β t ) = L ( α 1 , α 2 , ⋯ , α s , β 1 , β 2 , ⋯ , β t ) . L(\mathbf{\alpha}_1, \mathbf{\alpha}_2, \cdots, \mathbf{\alpha}_s) + L(\beta_1, \beta_2, \cdots, \beta_t) = L(\mathbf{\alpha}_1, \mathbf{\alpha}_2, \cdots, \mathbf{\alpha}_s, \beta_1, \beta_2, \cdots, \beta_t). L ( α 1 , α 2 , ⋯ , α s ) + L ( β 1 , β 2 , ⋯ , β t ) = L ( α 1 , α 2 , ⋯ , α s , β 1 , β 2 , ⋯ , β t ) . 直和判断的充分必要条件
定理 5.4.6 设 V 1 V_1 V 1 V 2 V_2 V 2 V V V W = V 1 + V 2 W = V_1 + V_2 W = V 1 + V 2
(1) W = V 1 + V 2 W = V_1 + V_2 W = V 1 + V 2
(2) 如果 α 1 + α 2 = 0 , α i ∈ V i ( i = 1 , 2 ) \mathbf{\alpha}_1 + \mathbf{\alpha}_2 = \mathbf{0}, \ \mathbf{\alpha}_i \in V_i (i = 1,2) α 1 + α 2 = 0 , α i ∈ V i ( i = 1 , 2 ) α 1 = α 2 = 0 \mathbf{\alpha}_1 = \mathbf{\alpha}_2 = \mathbf{0} α 1 = α 2 = 0
(3) V 1 ∩ V 2 = { 0 } V_1 \cap V_2 = \{\mathbf{0}\} V 1 ∩ V 2 = { 0 }
(4) 如果 α 1 , α 2 , ⋯ , α s \mathbf{\alpha}_1, \mathbf{\alpha}_2, \cdots, \mathbf{\alpha}_s α 1 , α 2 , ⋯ , α s V 1 V_1 V 1 β 1 , β 2 , ⋯ , β t \beta_1, \beta_2, \cdots, \beta_t β 1 , β 2 , ⋯ , β t V 2 V_2 V 2 α 1 , α 2 , ⋯ , α s , β 1 , β 2 , ⋯ , β t \mathbf{\alpha}_1, \mathbf{\alpha}_2, \cdots, \mathbf{\alpha}_s, \beta_1, \beta_2, \cdots, \beta_t α 1 , α 2 , ⋯ , α s , β 1 , β 2 , ⋯ , β t W W W
(5) dim W = dim V 1 + dim V 2 \dim W = \dim V_1 + \dim V_2 dim W = dim V 1 + dim V 2
维数与基
L ( α 1 , α 2 , ⋯ , α r ) L(\mathbf{\alpha}_1, \mathbf{\alpha}_2, \cdots, \mathbf{\alpha}_r) L ( α 1 , α 2 , ⋯ , α r ) 维数 等于向量组 α 1 , α 2 , ⋯ , α r \mathbf{\alpha}_1, \mathbf{\alpha}_2, \cdots, \mathbf{\alpha}_r α 1 , α 2 , ⋯ , α r 秩 .
与运算结合 :
定理 5.4.5（维数公式） 设 V V V V 1 , V 2 V_1, V_2 V 1 , V 2 V V V
dim ( V 1 + V 2 ) = dim V 1 + dim V 2 − dim ( V 1 ∩ V 2 ) . ( 5.4.4 ) \dim(V_1 + V_2) = \dim V_1 + \dim V_2 - \dim(V_1 \cap V_2). \quad (5.4.4) dim ( V 1 + V 2 ) = dim V 1 + dim V 2 − dim ( V 1 ∩ V 2 ) . ( 5.4.4 ) 如果是直和 dim ( W ) = dim V 1 + dim V 2 \dim(W) = \dim V_1 + \dim V_2 dim ( W ) = dim V 1 + dim V 2 其中W = V 1 + V 2 W = V_1 + V_2 W = V 1 + V 2
例题
例 1 已知 R 4 \mathbb{R}^4 R 4
α 1 = ( 1 , 2 , 1 , 0 ) , α 2 = ( − 1 , 1 , 1 , 1 ) ; β 1 = ( 2 , − 1 , 0 , 1 ) , β 2 = ( 1 , − 1 , 3 , 7 ) , \mathbf{\alpha}_1 = (1, 2, 1, 0), \ \mathbf{\alpha}_2 = (-1, 1, 1, 1); \ \mathbf{\beta}_1 = (2, -1, 0, 1), \ \mathbf{\beta}_2 = (1, -1, 3, 7), α 1 = ( 1 , 2 , 1 , 0 ) , α 2 = ( − 1 , 1 , 1 , 1 ) ; β 1 = ( 2 , − 1 , 0 , 1 ) , β 2 = ( 1 , − 1 , 3 , 7 ) , 子空间 V 1 = L ( α 1 , α 2 ) , V 2 = L ( β 1 , β 2 ) , V_1 = L(\mathbf{\alpha}_1, \mathbf{\alpha}_2), \ V_2 = L(\mathbf{\beta}_1, \mathbf{\beta}_2), V 1 = L ( α 1 , α 2 ) , V 2 = L ( β 1 , β 2 ) ,
(1) 求 V 1 + V 2 V_1 + V_2 V 1 + V 2
步骤：
V 1 + V 2 = L ( α 1 , α 2 , β 1 , β 2 ) V_{1}+V_{2}=L(\alpha_{1},\alpha_{2},\beta_{1},\beta_{2}) V 1 + V 2 = L ( α 1 , α 2 , β 1 , β 2 )
写出以 α 1 , α 2 , β 1 β 2 \alpha_{1},\alpha_{2},\beta_{1}\beta_{2} α 1 , α 2 , β 1 β 2
对矩阵进行实行行变换，化成行最简形
秩=维数，主元列为基（与求极大线性无关组一致）
解：(1) V 1 + V 2 = L ( α 1 , α 2 , β 1 , β 2 ) V_1 + V_2 = L(\mathbf{\alpha}_1, \mathbf{\alpha}_2, \mathbf{\beta}_1, \mathbf{\beta}_2) V 1 + V 2 = L ( α 1 , α 2 , β 1 , β 2 )
对以 α 1 , α 2 , β 1 , β 2 \mathbf{\alpha}_1, \mathbf{\alpha}_2, \mathbf{\beta}_1, \mathbf{\beta}_2 α 1 , α 2 , β 1 , β 2 A \mathbf{A} A
$$\mathbf{A} = \begin{pmatrix} 1 & -1 & 2 & 1 \\ 2 & 1 & -1 & -1 \\ 1 & 1 & 0 & 3 \\ 0 & 1 & 1 & 7 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & -1 & 2 & 1 \\ 0 & 3 & -5 & -3 \\ 0 & 2 & -2 & 2 \\ 0 & 1 & 1 & 7 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & -1 & 2 & 1 \\ 0 & 0 & -2 & -6 \\ 0 & 1 & -1 & 1 \\ 0 & 0 & 2 & 6 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & 4 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 \end{pmatrix} = \mathbf{B}$$
由 B \mathbf{B} B α 1 , α 2 , β 1 \mathbf{\alpha}_1, \mathbf{\alpha}_2, \mathbf{\beta}_1 α 1 , α 2 , β 1 α 1 , α 2 , β 2 \mathbf{\alpha}_1, \mathbf{\alpha}_2, \mathbf{\beta}_2 α 1 , α 2 , β 2 α 1 , α 2 , β 1 , β 2 \mathbf{\alpha}_1, \mathbf{\alpha}_2, \mathbf{\beta}_1, \mathbf{\beta}_2 α 1 , α 2 , β 1 , β 2
V 1 + V 2 = L ( α 1 , α 2 , β 1 ) V_1 + V_2 = L(\mathbf{\alpha}_1, \mathbf{\alpha}_2, \mathbf{\beta}_1) V 1 + V 2 = L ( α 1 , α 2 , β 1 ) α 1 , α 2 , β 1 \mathbf{\alpha}_1, \mathbf{\alpha}_2, \mathbf{\beta}_1 α 1 , α 2 , β 1 α 1 , α 2 , β 2 \mathbf{\alpha}_1, \mathbf{\alpha}_2, \mathbf{\beta}_2 α 1 , α 2 , β 2
(2) 求 V 1 ∩ V 2 V_1 \cap V_2 V 1 ∩ V 2
步骤：
γ = k 1 α 1 + k 2 α 2 = l 1 β 1 + l 2 β 2 \mathbf{\gamma} = k_1\mathbf{\alpha}_1 + k_2\mathbf{\alpha}_2 = l_1\mathbf{\beta}_1 + l_2\mathbf{\beta}_2 γ = k 1 α 1 + k 2 α 2 = l 1 β 1 + l 2 β 2
返回总目录
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