双语矩阵论课程笔记

文章目录

1. Definitions and Examples
- 1.1 Number filed（数域）
- 1.2 Algebraic systems（代数系统）
- 1.3 Linear space / Vector space（线性空间 / 向量空间）
- - 1.3.1 Definition
  - 1.3.2 Remark on Linear space
  - 1.3.3 Verify a linear space
2. Linear Dependence and Independence（线性相关和线性无关）
- 2.1 Basic concept
- - 2.1.1 Linear combinations of vectors（线性组合）
  - 2.1.2 linearly represented（线性表示）
  - 2.1.3 linear dependence and Independence（线性相关和线性无关）
- 2.2 Theorem and corollary
3. Basis and Dimension（基和维数）
- 3.1 Dimension（维数）
- 3.2 Basis and coordinate（基和坐标）
- - 3.2.1 Definition
  - 3.2.2 Remark on basis
  - 3.3.3 Transition Matrix（过渡矩阵）
  - 3.3.4 Examples
4. Linear subspace（线性子空间）
- 4.1 Basic concept
- - 4.1.1 linear subspace（线性子空间）
  - 4.1.2 direct sum（直和）
- 4.2 Sum of subspaces
- 4.3 Theorems
- - 4.3.1 Theorem 1
  - 4.3.2 Theorem 2
  - 4.3.3 Theorem 3
  - 4.3.4 Theorem 4 (Dimension Formula 维数公式)
  - 4.3.5 Theorem 5 (Equivalent to direct sum 直和等价于)
  - 4.3.6 Theorem 6 (Direct sum decomposition 直和分解)

1. Definitions and Examples

1.1 Number filed（数域）

A number filed F is a set of numbers containing at least 0 and 1 and is closed under addition, subtraction, multiplication and division, which means following properties are satisfied：
∀a∈F,−a∈F∀b∈F,b≠0,b−1∈F∀a,b∈F,a+b∈F,∀a,b∈F,a⋅b∈F\begin{aligned} &\forall a \in F &&,-a\in F \\ &\forall b \in F, b\neq 0 &&,b^{-1} \in F \\ &\forall a,b \in F &&,a+b\in F, \\ &\forall a,b \in F &&,a·b\in F \end{aligned} ∀a∈F∀b∈F,b=0∀a,b∈F∀a,b∈F,−a∈F,b−1∈F,a+b∈F,,a⋅b∈F
for example, the real number filed R（实数域） is a number filed, it contains real number 0 and 1, and closed under arithmetic（对四则运算封闭）
域、群、环等概念，其实都是具有一些好的性质的集合。数域是针对数的域，其元素是实数、复数、有理数等等各种数

1.2 Algebraic systems（代数系统）

An algebraic system is usually referred to as a set in which some operation together with some rules are defined. To describe an algebraic system, we need
1. a set of elements（Given）
2. operations（Well Defined）
3. rules of operation（deduced）
note：
1. the elements in the set can be anything，number、function、matrix…
2. well defined means the result of any operation is unique. You can view the operation as a mapping, and well define means that any element can only be mapped to a unique element
3. with the rules of operation, we can deduce the properties of the algebraic system
examples：Pay attention to example 4 and example 5, the set of them is set of function, so the addition operation represented as (f+g)(x)(f+g)(x)(f+g)(x)

1.3 Linear space / Vector space（线性空间 / 向量空间）

1.3.1 Definition

Linear space/vector space：A set V(V≠∅)\pmb{V}(\pmb{V} \neq \empty)VVV(VVV=∅) over a number filed F\pmb{F}FFF is a set of elements together with two operations, addition(加法) and and scalar multiplication(数乘), if V\pmb{V}VVV satisfies the following conditions, it’s a linear space/vector space
1. closure properties（封闭性）：
  1. ∀α,β∈V,α+β∈V\forall \alpha,\beta \in \pmb{V}, \alpha+\beta \in \pmb{V}∀α,β∈VVV,α+β∈VVV is unique
  2. ∀k∈F,α∈V,k⋅α∈V\forall k \in \pmb{F}, \alpha\in \pmb{V},\space k·\alpha\in \pmb{V}∀k∈FFF,α∈VVV, k⋅α∈VVV is unique
    
    说明：这里的加法运算 addition 没有什么要求，可以定义任意 V\pmb{V}VVV 上的运算作为 “addition”，但是数乘运算 scalar multiplication 要求必须是从数域 F\pmb{F}FFF 中取一个数和 V\pmb{V}VVV 中元素做数乘，Linear space V\pmb{V}VVV over a filed F\pmb{F}FFF 指的就是这个。封闭性要求这两个运算的结果都在 V\pmb{V}VVV 中
2. The addition axioms（加法公理）：
3. The scalar multiplication axioms（乘法公理）：

特殊元素：

元素	性质
`加法零元 (zero element)` （0∈V\pmb{0} \in \pmb{V}000∈VVV）	∀α∈V,0+α=α\forall \alpha \in \pmb{V}, \space \pmb{0}+\alpha = \alpha∀α∈VVV, 000+α=α
`加法负元 (additive inverse)`（−α∈V-\alpha \in \pmb{V}−α∈VVV）	∀α∈V\forall \alpha \in \pmb{V}∀α∈VVV，∃−α∈V\exists -\alpha\in\pmb{V}∃−α∈VVV，S.t. α+(−α)=0\alpha+(-\alpha) = \pmb{0}α+(−α)=000
`乘法幺元 (unit element)`（ 1∈F1 \in \pmb{F}1∈FFF ）	∀α∈V,1⋅α=α\forall \alpha \in \pmb{V}, \space 1·\alpha = \alpha∀α∈VVV, 1⋅α=α

1.3.2 Remark on Linear space

Important things to remember：In the definition of linear linear space, we need
1. a set V≠∅\pmb{V} \neq \emptyVVV=∅
2. a number filed F\pmb{F}FFF
3. two operations (addition & scalar multiplication; well-defined and closed)
4. eight operation rules
when you do scalar multiplication, the scalar must come from the filed F\pmb{F}FFF
there are some rules derived from the definition：If V\pmb{V}VVV is a linear space over the filed F\pmb{F}FFF, and α∈V,k∈F\alpha \in \pmb{V}, k\in\pmb{F}α∈VVV,k∈FFF，then
1. 0\pmb{0}000 is unique
2. 0⋅α=00·\alpha = \pmb{0}0⋅α=000
3. k⋅0=0k·\pmb{0} = \pmb{0}k⋅000=000
4. α+β=0⇒β=−α\alpha +\beta = \pmb{0} \Rightarrow \beta = -\alphaα+β=000⇒β=−α（i.e. the addition inverse is unique）
5. (−1)⋅α=−α(-1)·\alpha = -\alpha(−1)⋅α=−α
6. k⋅α=0,k∈F,α∈V⇒k=0orα=0k·\alpha = \pmb{0}, k\in \pmb{F}, \alpha\in \pmb{V} \Rightarrow k=0 \space\space or \space\space \alpha = \pmb{0}k⋅α=000,k∈FFF,α∈VVV⇒k=0 or α=000
证明一下第2条
∵0⋅α+(−0⋅α)=(0+(−0))⋅α=(−0)⋅α∴0⋅α+(−0⋅α)+(−(−0⋅α))=(−0⋅α)+(−(−0⋅α))∴0⋅α+0=0∴0⋅α=0\begin{aligned} &\because 0·\alpha + (-0·\alpha) = (0+(-0))·\alpha =(-0)·\alpha \\ &\therefore 0·\alpha + (-0·\alpha) +(-(-0·\alpha)) = (-0·\alpha) +(-(-0·\alpha)) \\ &\therefore 0·\alpha +\pmb{0}= \pmb{0} \\ &\therefore 0·\alpha = \pmb{0} \end{aligned} ∵0⋅α+(−0⋅α)=(0+(−0))⋅α=(−0)⋅α∴0⋅α+(−0⋅α)+(−(−0⋅α))=(−0⋅α)+(−(−0⋅α))∴0⋅α+000=000∴0⋅α=000

1.3.3 Verify a linear space

How to verify a linear space?
1. Find out the number field F and the set V
2. What are the two operations?
3. satisfy the Closure property?
4. satisfy the 8 rules?
Example 1
Example 2：c[a.b]c[a.b]c[a.b] is a set of continuous functions on interval [a,b], is it a linear space ?

2. Linear Dependence and Independence（线性相关和线性无关）

2.1 Basic concept

2.1.1 Linear combinations of vectors（线性组合）

V\pmb{V}VVV is a linear space over F\pmb{F}FFF, there are {α1,α2,...,αn∈Vk1,k2,...,kn∈F\left\{ \begin{aligned}\alpha_1,\alpha_2,...,\alpha_n \in \pmb{V} \\ k_1,k_2,...,k_n \in \pmb{F}\end{aligned}\right.{α1,α2,...,αn∈VVVk1,k2,...,kn∈FFF, a linear combination(线性组合) of α1,α2,...,αn\alpha_1,\alpha_2,...,\alpha_nα1,α2,...,αn is
k1α1+k2α2+...+knαnk_1\alpha_1+k_2\alpha_2+...+k_n\alpha_n k1α1+k2α2+...+knαn

2.1.2 linearly represented（线性表示）

Given two elements groups {(I):α1,α2,...,αn∈V(II):β1,β2,...,βs∈V\left\{ \begin{aligned}(\mathrm{I}):\alpha_1,\alpha_2,...,\alpha_n \in \pmb{V} \\ (\mathrm{II}):\beta_1,\beta_2,...,\beta_s \in \pmb{V}\end{aligned}\right.{(I):α1,α2,...,αn∈VVV(II):β1,β2,...,βs∈VVV, if each element in group (I)(\mathrm{I})(I) is a combination of group (II)(\mathrm{II})(II), we call the former can be linearly represented(线性表示) by the later.
If two groups can be represented by each other, they are called equivalent(相抵/等价), which is
1. reflexive（有自反性：自己和自己等价）
2. symmetric（有对称性：(I)(\mathrm{I})(I) 和 (II)(\mathrm{II})(II) 等价 ⇒\Rightarrow⇒ (II)(\mathrm{II})(II) 和 (I)(\mathrm{I})(I) 等价）
3. transitive（有传递性：(I)(\mathrm{I})(I) 和 (II)(\mathrm{II})(II) 等价，(II)(\mathrm{II})(II) 和 (III)(\mathrm{III})(III) 等价 ⇒\Rightarrow⇒ (I)(\mathrm{I})(I) 和 (III)(\mathrm{III})(III) 等价）

2.1.3 linear dependence and Independence（线性相关和线性无关）

有一组不全为0的系数 kkk 使得向量组 α\alphaα 的线性组合为零向量 0\pmb{0}000，则向量组 α\alphaα 线性相关，反之线性无关
注意：线性空间 V\pmb{V}VVV 中的元素不一定是向量，也可能是数、抽象函数…等等，因为它本质是一种特殊的 Algebraic systems 代数系统。后面3.2.1节中，我们会说明这些元素和通常意义上的向量有一一对应关系，因此习惯上称这些元素 α∈V\alpha\in \pmb{V}α∈VVV 为向量（vector）

2.2 Theorem and corollary

Theorems：
1. Suppose that V\pmb{V}VVV is a linear space over PPP. A group of elements α1,α2,...,αr(r≥2)\alpha_1,\alpha_2,...,\alpha_r(r\geq 2)α1,α2,...,αr(r≥2) are linear dependent if and only if there is at least one element that can be written as a linear combination of the others
  若线性相关，则至少有一个向量可以被其他的线性表出
2. Suppose that α1,α2,...,αr\alpha_1,\alpha_2,...,\alpha_rα1,α2,...,αr are linearly independent and can be linearly represented by β1,β2,...,βs\beta_1,\beta_2,...,\beta_sβ1,β2,...,βs, then r≤sr\leq sr≤s
  一组线性无关向量，只能被数目多于等于它的另一组向量线性表出
3. Assume that α1,α2,...,αr\alpha_1,\alpha_2,...,\alpha_rα1,α2,...,αr are linearly independent, while α1,α2,...,αr,β\alpha_1,\alpha_2,...,\alpha_r,\betaα1,α2,...,αr,β are linearly dependent, then β\betaβ can be uniquely and linearly represented by α1,α2,...,αr\alpha_1,\alpha_2,...,\alpha_rα1,α2,...,αr
  一组线性无关的向量，增加一个之后变得线性相关了，多出的一定可以被其他的线性表出
corollary：
1. The number of elements in two equivalent linear independence element groups are same
  两个等价向量组中向量个数相等
examples
- 欲证明某向量组线性无关或相关，可使用定义法，先写出 ∑inkiαi=0\sum_i^n k_i\alpha_i =0∑inkiαi=0 的形式，再说明 kik_iki 是否一定全为 0
- 关于向量组线性相关性的更多性质和证明方法，请参考：线性代数（4）—— 向量与向量组的线性相关性

3. Basis and Dimension（基和维数）

3.1 Dimension（维数）

Definition: If there are n vectors in V\pmb{V}VVV are linearly independent and arbitrary n+1 vectors in V\pmb{V}VVV are linearly dependent, we call the dimension(维数) of V\pmb{V}VVV is n, denoted by dim(V)=n\mathrm{dim}(V) = ndim(V)=n
special case that the dimension equals to 0 and ∞\infin∞
1. If V={0},thendim(V)=0V=\{\pmb{0}\}, then\space \mathrm{dim}(V) = 0V={000},then dim(V)=0
2. If there are mmm linearly independent vectors in V\pmb{V}VVV for any integer mmm, we call V\pmb{V}VVV is infinity dimensional, denoted by dim(V)=+∞\mathrm{dim}(V) = +\infindim(V)=+∞. For example：实多项式集合

3.2 Basis and coordinate（基和坐标）

3.2.1 Definition

Suppose that V\pmb{V}VVV is an n-dimensional linear space, if ε1,ε2,...,εn\varepsilon_1,\varepsilon_2,...,\varepsilon_nε1,ε2,...,εn is a linearly independent vector group in V\pmb{V}VVV, it is a basis(基，复数形式为bases) of V\pmb{V}VVV
∀α∈V\forall \alpha \in \pmb{V}∀α∈VVV can be uniquely linearly represented by the basis ε1,ε2,...,εn\varepsilon_1,\varepsilon_2,...,\varepsilon_nε1,ε2,...,εn that is

where the coefficients（系数） x1,x2,...,xnx_1,x_2,...,x_nx1,x2,...,xn are called coordinates(坐标) over the basis ε1,ε2,...,εn\varepsilon_1,\varepsilon_2,...,\varepsilon_nε1,ε2,...,εn, denoted by (x1,x2,...,xn)(x_1,x_2,...,x_n)(x1,x2,...,xn) or (x1,x2,...,xn)⊤(x_1,x_2,...,x_n)^\top(x1,x2,...,xn)⊤
注意，这里的 coordinates 是由 F\pmb{F}FFF 中的数组成的向量，因此 linear space V\pmb{V}VVV 中的任意一个元素都能通过一组 basis 唯一地映射为一个 vector（有一一对应关系），这就是为什么 linear space 也叫做 vector space

3.2.2 Remark on basis

Bases for a linear space are not unique.
线性空间的基不唯一
- For example, the set of cubic polynomials（三次多项式集合）P4P_4P4 has a basis 1,x,x2,x31,x,x^2,x^31,x,x2,x3, we know this basis can be linearly represented by 1,2x,3x2,4x31,2x,3x^2,4x^31,2x,3x2,4x3, so that 1,2x,3x2,4x31,2x,3x^2,4x^31,2x,3x2,4x3 is also a basis of P4P_4P4
All bases for a linear space have the same number of elements (i.e. dim(V)\mathrm{dim}(V)dim(V))
同一线性空间的所有基，维数一致
If ε1,ε2,...,εn\varepsilon_1,\varepsilon_2,...,\varepsilon_nε1,ε2,...,εn is a linearly independent vector group in V\pmb{V}VVV, and any vector in V\pmb{V}VVV can be uniquely linearly represented by it, then ε1,ε2,...,εn\varepsilon_1,\varepsilon_2,...,\varepsilon_nε1,ε2,...,εn is a basis of V\pmb{V}VVV
如果一组线性无关向量可以线性表示Linear space中的所有向量，则它是一组基
1. 根据2，求线性空间维数，等价于找到线性空间的一组基向量；根据3，求一组基向量，可以先找到一组向量，证明其线性无关，并且能唯一地表示空间中任意向量。2,3两者结合就能求出线性空间的维数

3.3.3 Transition Matrix（过渡矩阵）

Suppose that vector group ε1,ε2,...,εn\varepsilon_1,\varepsilon_2,...,\varepsilon_nε1,ε2,...,εn is a basis of linear space V\pmb{V}VVV，ε1′,ε2′,...,εn′\varepsilon_1',\varepsilon_2',...,\varepsilon_n'ε1′,ε2′,...,εn′ are n vectors in V\pmb{V}VVV，then there exists a matrix T∈Pn×n\pmb{T}\in P^{n\times n }TTT∈Pn×n such that
(ε1′,ε2′,...,εn′)=(ε1,ε2,...,εn)T(\varepsilon_1',\varepsilon_2',...,\varepsilon_n') = (\varepsilon_1,\varepsilon_2,...,\varepsilon_n)\pmb{T} (ε1′,ε2′,...,εn′)=(ε1,ε2,...,εn)TTT As mentioned above, the column vectors of T\pmb{T}TTT are the coordinates of vectorsε1′,ε2′,...,εn′\varepsilon_1',\varepsilon_2',...,\varepsilon_n'ε1′,ε2′,...,εn′ over the basis ε1,ε2,...,εn\varepsilon_1,\varepsilon_2,...,\varepsilon_nε1,ε2,...,εn
T\pmb{T}TTT is reversible(可逆的) ⇔\Leftrightarrow⇔ ε1′,ε2′,...,εn′\varepsilon_1',\varepsilon_2',...,\varepsilon_n'ε1′,ε2′,...,εn′ is also a basis of V\pmb{V}VVV, cause
(ε1′,ε2′,...,εn′)T−1=(ε1,ε2,...,εn)(\varepsilon_1',\varepsilon_2',...,\varepsilon_n') \pmb{T}^{-1}= (\varepsilon_1,\varepsilon_2,...,\varepsilon_n) (ε1′,ε2′,...,εn′)TTT−1=(ε1,ε2,...,εn) the basis ε1,ε2,...,εn\varepsilon_1,\varepsilon_2,...,\varepsilon_nε1,ε2,...,εn can be linear represented by ε1′,ε2′,...,εn′\varepsilon_1',\varepsilon_2',...,\varepsilon_n'ε1′,ε2′,...,εn′
Especially, when both ε1,ε2,...,εn\varepsilon_1,\varepsilon_2,...,\varepsilon_nε1,ε2,...,εn and ε1′,ε2′,...,εn′\varepsilon_1',\varepsilon_2',...,\varepsilon_n'ε1′,ε2′,...,εn′ are basis of V\pmb{V}VVV, the matrix T\pmb{T}TTT is called the transition Matrix(过渡矩阵) from basis ε1,ε2,...,εn\varepsilon_1,\varepsilon_2,...,\varepsilon_nε1,ε2,...,εn to ε1′,ε2′,...,εn′\varepsilon_1',\varepsilon_2',...,\varepsilon_n'ε1′,ε2′,...,εn′ (第一组基 x 过渡矩阵 = 第二组基)
∀α∈V\forall \alpha \in \pmb{V}∀α∈VVV, suppose the coordinates over the bases ε1,ε2,...,εn\varepsilon_1,\varepsilon_2,...,\varepsilon_nε1,ε2,...,εn and ε1′,ε2′,...,εn′\varepsilon_1',\varepsilon_2',...,\varepsilon_n'ε1′,ε2′,...,εn′ are (x1,x2,...,xn)(x_1,x_2,...,x_n)(x1,x2,...,xn) and (x1′,x2′,...,xn′)(x_1',x_2',...,x_n')(x1′,x2′,...,xn′) respectively, then
这一块就是说，对于空间中的某个向量，当使用不同的基来描述这个空间时，这个向量对应的坐标也不同。由于是同一个向量，不同基在其对应的坐标的下的线性组合都能得到它，即
(ε1′,ε2′,…,εn′)[x1′x2′⋮xn′]=[(ε1,ε2,…,εn)T][T−1[x1x2⋮xn]]=(ε1,ε2,…,εn)[x1x2⋮xn](\varepsilon_1',\varepsilon_2',\dots,\varepsilon_n')\begin{bmatrix}x_1'\\x_2'\\ \vdots \\x_n'\end{bmatrix} = \big[(\varepsilon_1,\varepsilon_2,\dots,\varepsilon_n)\pmb{T}\big]\big[\pmb{T}^{-1}\begin{bmatrix}x_1\\x_2\\ \vdots \\x_n\end{bmatrix} \big] = (\varepsilon_1,\varepsilon_2,\dots,\varepsilon_n)\begin{bmatrix}x_1\\x_2\\ \vdots \\x_n\end{bmatrix} (ε1′,ε2′,…,εn′)⎣⎢⎢⎢⎡x1′x2′⋮xn′⎦⎥⎥⎥⎤=[(ε1,ε2,…,εn)TTT][TTT−1⎣⎢⎢⎢⎡x1x2⋮xn⎦⎥⎥⎥⎤]=(ε1,ε2,…,εn)⎣⎢⎢⎢⎡x1x2⋮xn⎦⎥⎥⎥⎤

3.3.4 Examples

Consider V={A∈R2×2∣A⊤=A}\pmb{V} = \{A\in R^{2\times 2}|A^\top = A\}VVV={A∈R2×2∣A⊤=A}, prove that dim(V)=3\mathrm{dim}(V) = 3dim(V)=3

（这个证明还应当说明一下 ε\varepsilonε 线性无关：当 ∑i=13εiki=0\sum_{i=1}^3\varepsilon_ik_i = 0∑i=13εiki=0 时，kik_iki 必定全为0）
Compute the dimensions of linear spaces
Find the transition matrix and compute the coordinate
Find the transition matrix

4. Linear subspace（线性子空间）

4.1 Basic concept

4.1.1 linear subspace（线性子空间）

Suppose that V\pmb{V}VVV is a vector space on PPP, W⊆VW \subseteq VW⊆V is not empty. If WWW is a vector space for the same two algebraic operation, we call WWW a linear subspace of V\pmb{V}VVV
1. W⊆VW \subseteq VW⊆V and W≠∅W \neq \emptyW=∅
2. WWW is a vector space for the same algebraic operation
trivial subspace (平凡子空间) of V\pmb{V}VVV
1. V\pmb{V}VVV itself
2. zero space V={0}\pmb{V} = \{\pmb{0}\}VVV={000}

4.1.2 direct sum（直和）

二分解：If V1\pmb{V}_1VVV1 and V2\pmb{V}_2VVV2 are two subspaces of a vector space V\pmb{V}VVV, such that each α∈V1+V2\alpha \in \pmb{V}_1+\pmb{V}_2α∈VVV1+VVV2 can be uniquely decomposed as
α=α1+α2,α1∈V1,α2∈V2\alpha = \alpha_1+\alpha_2, \space \space \alpha_1 \in \pmb{V}_1,\alpha_2 \in \pmb{V}_2 α=α1+α2, α1∈VVV1,α2∈VVV2 then the sum V1+V2\pmb{V}_1+\pmb{V}_2VVV1+VVV2 is called a direct sum(直和) of V1\pmb{V}_1VVV1 and V2\pmb{V}_2VVV2, denoted by V1+˙V2\pmb{V}_1 \dot{+} \pmb{V}_2VVV1+˙VVV2
多分解：Suppose that V1,V2,...,Vs\pmb{V}_1,\pmb{V}_2,...,\pmb{V}_sVVV1,VVV2,...,VVVs are finite subspaces of V\pmb{V}VVV, if ∀α∈V1+V2+...+Vs\forall \alpha \in \pmb{V}_1+\pmb{V}_2+...+\pmb{V}_s∀α∈VVV1+VVV2+...+VVVs can be uniquely decomposed as
α=α1+α2+...+αs,αi∈Vi(i=1,....,s)\alpha = \alpha_1+\alpha_2 +...+\alpha_s ,\space\space \alpha_i \in \pmb{V}_i(i=1,....,s) α=α1+α2+...+αs, αi∈VVVi(i=1,....,s) then call V1+V2+...+Vs\pmb{V}_1+\pmb{V}_2+...+\pmb{V}_sVVV1+VVV2+...+VVVs is a direct sum(直和), denoted by V1+˙V2+˙...+˙Vs\pmb{V}_1 \dot{+} \pmb{V}_2\dot{+}...\dot{+} \pmb{V}_sVVV1+˙VVV2+˙...+˙VVVs

4.2 Sum of subspaces

V1+V2={α+β∣α∈V1,β∈V2}\pmb{V}_1 + \pmb{V}_2 = \{\alpha+\beta|\alpha\in \pmb{V}_1, \beta\in \pmb{V}_2\}VVV1+VVV2={α+β∣α∈VVV1,β∈VVV2}
subspaces satisfying the following conditions:
1. {V1∩V2=V2∩V1V1+V2=V2+V1\left\{\begin{aligned} &\pmb{V}_1\cap \pmb{V}_2 = \pmb{V}_2\cap \pmb{V}_1 \\ &\pmb{V}_1+\pmb{V}_2 = \pmb{V}_2+\pmb{V}_1\end{aligned}\right.{VVV1∩VVV2=VVV2∩VVV1VVV1+VVV2=VVV2+VVV1
2. {(V1∩V2)∩V3=V1∩(V2∩V3)(V1+V2)+V3=V1+(V2+V3)\left\{\begin{aligned} &(\pmb{V}_1\cap \pmb{V}_2)\cap \pmb{V}_3 = \pmb{V}_1 \cap (\pmb{V}_2\cap \pmb{V}_3) \\ &(\pmb{V}_1+ \pmb{V}_2)+ \pmb{V}_3 = \pmb{V}_1 + (\pmb{V}_2+ \pmb{V}_3)\end{aligned}\right.{(VVV1∩VVV2)∩VVV3=VVV1∩(VVV2∩VVV3)(VVV1+VVV2)+VVV3=VVV1+(VVV2+VVV3)
Generally, if V1,V2,...,Vs\pmb{V}_1,\pmb{V}_2,...,\pmb{V}_sVVV1,VVV2,...,VVVs are subspaces of V\pmb{V}VVV, then
1. V1∩V2∩...∩Vs=⋂i=1sVi\pmb{V}_1\cap \pmb{V}_2\cap...\cap \pmb{V}_s = \bigcap_{i=1}^s\pmb{V}_iVVV1∩VVV2∩...∩VVVs=⋂i=1sVVVi
2. V1+V2+...+Vs=∑i=1sVi\pmb{V}_1+\pmb{V}_2+...+\pmb{V}_s = \sum_{i=1}^s\pmb{V}_iVVV1+VVV2+...+VVVs=∑i=1sVVVi
are subspaces of V\pmb{V}VVV

4.3 Theorems

4.3.1 Theorem 1

A nonempty subset W\pmb{W}WWW in V\pmb{V}VVV is a subspace of V\pmb{V}VVV ⇔\Leftrightarrow⇔ W\pmb{W}WWW satisfies the closure properties of addition and scalar multiplication
原线性空间的非空子集，只要满足加法和数乘封闭性，其他8条性质一定满足，即为线性子空间
{∀α,β∈W,α+β∈W∀α∈W,k∈P,kα∈W\left\{\begin{aligned} &\forall \alpha,\beta \in \pmb{W}, \alpha +\beta \in \pmb{W} \\ &\forall \alpha \in \pmb{W}, k\in P, k\alpha \in \pmb{W} \end{aligned} \right. {∀α,β∈WWW,α+β∈WWW∀α∈WWW,k∈P,kα∈WWW 说明：kkk 可以是数0，这样 kα=0∈Wk\alpha = \pmb{0}\in Wkα=000∈W 保证加法零元存在；kkk 可以是 -1，对任意 α∈W\alpha \in Wα∈W 有 −1⋅α=−α∈W-1·\alpha =-\alpha \in W−1⋅α=−α∈W，加法逆元存在；kkk 可以是1，保证乘法幺元存在，可以保证8条性质
Remark：
1. W⊆V\pmb{W}\subseteq \pmb{V}WWW⊆VVV is a subspace ⇔W\Leftrightarrow \pmb{W}⇔WWW is closed with the two operations defined on V\pmb{V}VVV
2. W⊆V\pmb{W}\subseteq \pmb{V}WWW⊆VVV is a subspace ⇒\Rightarrow⇒ dim(W\pmb{W}WWW) ≤\leq≤ dim(V\pmb{V}VVV)
  W\pmb{W}WWW 的基能被 V\pmb{V}VVV 的基线性表出，根据2.3.2 theorems 2，说明 V\pmb{V}VVV 的基中向量个数肯定更多

4.3.2 Theorem 2

suppose that α1,α2,...,αs\alpha_1,\alpha_2,...,\alpha_sα1,α2,...,αs is a group of vectors in V\pmb{V}VVV. Denote
W=span(α1,α2,...,αs)={k1α1+k2α2+....+ksαs∣ki∈P,i=1,...,s}\pmb{W} = span(\alpha_1,\alpha_2,...,\alpha_s) = \{k_1\alpha_1+k_2\alpha_2+....+k_s\alpha_s|k_i\in P,i=1,...,s\} WWW=span(α1,α2,...,αs)={k1α1+k2α2+....+ksαs∣ki∈P,i=1,...,s} Then W\pmb{W}WWW is a subspace of V\pmb{V}VVV. We call W\pmb{W}WWW the spanned subspace by α1,α2,...,αs\alpha_1,\alpha_2,...,\alpha_sα1,α2,...,αs
线性空间中一组向量生成（span）的空间一定是线性子空间

4.3.3 Theorem 3

consider the two vector groups in V\pmb{V}VVV, α1,α2,...,αs\alpha_1,\alpha_2,...,\alpha_sα1,α2,...,αs and β1,β2,...,βt\beta_1,\beta_2,...,\beta_tβ1,β2,...,βt, then span((α1,α2,...,αs)=span(β1,β2,...,βt)⇔span((\alpha_1,\alpha_2,...,\alpha_s) = span(\beta_1,\beta_2,...,\beta_t) \Leftrightarrowspan((α1,α2,...,αs)=span(β1,β2,...,βt)⇔ two vector groups are equivalent.
两个向量组等价(可以互相线性表出) <=> 这两个向量组生成的线性子空间相等
Remark
1. dim(span(α1,α2,...,αs))≤s\mathrm{dim}(span(\alpha_1,\alpha_2,...,\alpha_s))\leq sdim(span(α1,α2,...,αs))≤s and the equality holds if and only if α1,α2,...,αs\alpha_1,\alpha_2,...,\alpha_sα1,α2,...,αs are linearly independent
  对于 span(α1,α2,...,αs)span(\alpha_1,\alpha_2,...,\alpha_s)span(α1,α2,...,αs) 的任意一组基 ε1,ε2,...,εr\varepsilon_1,\varepsilon_2,...,\varepsilon_rε1,ε2,...,εr，一定有 εi∈span(α1,α2,...,αs)\varepsilon_i\in span(\alpha_1,\alpha_2,...,\alpha_s)εi∈span(α1,α2,...,αs) ，也就是说这组线性无关的基向量可以由 α1,α2,...,αs\alpha_1,\alpha_2,...,\alpha_sα1,α2,...,αs 线性表出，根据2.3.2 theorems 2，有 s≥rs\geq rs≥r。当 α1,α2,...,αs\alpha_1,\alpha_2,...,\alpha_sα1,α2,...,αs 线性无关时，其本身也成为 span(α1,α2,...,αs)span(\alpha_1,\alpha_2,...,\alpha_s)span(α1,α2,...,αs) 的一组基，此时等号成立
Examples
1. A选项对：设 x,y∈V1⋂V2,k∈Px,y \in V_1 \bigcap V_2, k \in Px,y∈V1⋂V2,k∈P，有
  x,y∈V1⇒x+y∈V1x,y∈V2⇒x+y∈V2∴x+y∈V1∩V2x,y∈V1,V2⇒kx,ky∈V1,V2∴kx,ky∈V1∩V2∵{0}∈V1,V2∴V1∩V2≠∅\begin{aligned} &x,y\in V_1 \Rightarrow x+y\in V_1 \\ &x,y\in V_2 \Rightarrow x+y\in V_2 \\ &\therefore x+y \in V_1\cap V_2 \\ &x,y \in V_1, V_2 \Rightarrow kx,ky \in V_1,V_2 \\ &\therefore kx,ky \in V_1\cap V_2 \\ &\because \{0\} \in V_1,V_2 \\ &\therefore V_1\cap V_2 \neq \empty \end{aligned} x,y∈V1⇒x+y∈V1x,y∈V2⇒x+y∈V2∴x+y∈V1∩V2x,y∈V1,V2⇒kx,ky∈V1,V2∴kx,ky∈V1∩V2∵{0}∈V1,V2∴V1∩V2=∅
2. B选项错：设 (α1,α2)=(0,1),(β1,β2)=(1,0)(\alpha_1,\alpha_2) = (0,1), \space (\beta_1,\beta_2) = (1,0)(α1,α2)=(0,1), (β1,β2)=(1,0), 则 V1+V2=span(1,1)V_1+V_2=span(1,1)V1+V2=span(1,1)，这个既不在 V1V_1V1 也不在 V2V_2V2 因此加法不封闭，此为反例
3. C选项错：设 (α1,α2...αs)=(β1,β2,...,βr)(\alpha_1,\alpha_2...\alpha_s) = (\beta_1,\beta_2,...,\beta_r)(α1,α2...αs)=(β1,β2,...,βr) 则 V1⋃V2=V1=V2V_1 \bigcup V_2 = V_1 = V_2V1⋃V2=V1=V2 这时是 subspace，此为反例
4. D选项对：设 x,y∈V1+V2x,y \in V_1+V_2x,y∈V1+V2, 则
  x=α′+β′(α′∈V1,β′∈V2)y=α′′+β′′(α′′∈V1,β′′∈V2)x+y=(α′+α′′)+(β′+β′′)α′+α′′∈V1,β′+β′′∈V2⇒x+y∈V1+V2同理,kx=kα′+kβ′∈V1+V2\begin{aligned} &x = \alpha'+\beta' \space\space(\alpha'\in V_1,\beta'\in V_2)\\ &y = \alpha''+\beta'' \space\space(\alpha''\in V_1,\beta''\in V_2)\\ &x + y = (\alpha'+\alpha'') +(\beta'+\beta'') \\ & \alpha'+\alpha'' \in V_1, \beta'+\beta''\in V_2 \Rightarrow x + y \in V_1+V_2 \\ &同理, kx = k\alpha'+k\beta' \in V_1+V_2 \end{aligned} x=α′+β′ (α′∈V1,β′∈V2)y=α′′+β′′ (α′′∈V1,β′′∈V2)x+y=(α′+α′′)+(β′+β′′)α′+α′′∈V1,β′+β′′∈V2⇒x+y∈V1+V2同理,kx=kα′+kβ′∈V1+V2
5. E选项对：设 α∈V1=k1α1+...+ksαs\alpha \in V_1 = k_1\alpha_1+...+k_s\alpha_sα∈V1=k1α1+...+ksαs， β∈V2=k1′β1+...+kr′βr\beta \in V_2 = k_1'\beta_1+...+k_r'\beta_rβ∈V2=k1′β1+...+kr′βr, 则
  ∵α+β=∑i=1skiαi+∑i=1rki′αi∈span(α1,...,αs,β1,...,βr)∴V1+V2⊆span(α1,...,αs,β1,...,βr)同理span(α1,...,αs,β1,...,βr)中任意向量可以拆解为span(α1,...,αs)+span(β1,...,βr)，即属于V1+V2∴span(α1,...,αs,β1,...,βr)⊆V1+V2∴V1+V2=span(α1,...,αs,β1,...,βr)\begin{aligned} &\because \alpha + \beta = \sum_{i=1}^s k_i\alpha_i + \sum_{i=1}^r k_i'\alpha_i \in span(\alpha_1,...,\alpha_s,\beta_1,...,\beta_r) \\ &\therefore V_1+V_2 \subseteq span(\alpha_1,...,\alpha_s,\beta_1,...,\beta_r) \\ & 同理 span(\alpha_1,...,\alpha_s,\beta_1,...,\beta_r) 中任意向量可以拆解为 span(\alpha_1,...,\alpha_s) + span(\beta_1,...,\beta_r)，即属于 V_1 + V_2 \\ &\therefore span(\alpha_1,...,\alpha_s,\beta_1,...,\beta_r) \subseteq V_1+V_2\\ &\therefore V_1 + V_2 = span(\alpha_1,...,\alpha_s,\beta_1,...,\beta_r) \end{aligned} ∵α+β=i=1∑skiαi+i=1∑rki′αi∈span(α1,...,αs,β1,...,βr)∴V1+V2⊆span(α1,...,αs,β1,...,βr)同理span(α1,...,αs,β1,...,βr)中任意向量可以拆解为span(α1,...,αs)+span(β1,...,βr)，即属于V1+V2∴span(α1,...,αs,β1,...,βr)⊆V1+V2∴V1+V2=span(α1,...,αs,β1,...,βr)

4.3.4 Theorem 4 (Dimension Formula 维数公式)

V1,V2V_1,V_2V1,V2 are two subspaces of finite dimensional vector space V\pmb{V}VVV，则
dim(V1)+dim(V2)=dim(V1+V2)+dim(V1∩V2)dim(\pmb{V}_1) + dim(\pmb{V}_2) = dim(\pmb{V}_1+\pmb{V}_2) + dim(\pmb{V}_1\cap \pmb{V}_2)dim(VVV1)+dim(VVV2)=dim(VVV1+VVV2)+dim(VVV1∩VVV2) 维数公式 dim(V1)+dim(V2) = dim(V1+V2)+dim(V1∩V2)
证明思路：设 dim(V1\pmb{V}_1VVV1) = s，dim(V2\pmb{V}_2VVV2) = r
1. 对于 V1∩V2V_1\cap V_2V1∩V2 的一组基 ε1,...εk\varepsilon_1,...\varepsilon_kε1,...εk，把 V1V_1V1 中不能用它线性表出的向量 α\alphaα 加入这组基，直到这组基的基向量数目为 sss 时，ε1,...εk,αk+1,...,αs\varepsilon_1,...\varepsilon_k, \alpha_{k+1},...,\alpha_sε1,...εk,αk+1,...,αs 一定是 V1V_1V1 的一组基
2. 同理可以扩展 ε1,...εk,βk+1,...,βr\varepsilon_1,...\varepsilon_k, \beta_{k+1},...,\beta_rε1,...εk,βk+1,...,βr 为 V2V_2V2 的一组基
3. 合在一起 ε1,...εk,βk+1,...,βr,αk+1,...,αs\varepsilon_1,...\varepsilon_k, \beta_{k+1},...,\beta_r, \alpha_{k+1},...,\alpha_sε1,...εk,βk+1,...,βr,αk+1,...,αs 一共有 s+r−ks+r-ks+r−k 个，可以用定义证明他们线性无关
4. 易证，V1+V2V_1+V_2V1+V2 可以用这一组向量线性表出，因此这是 V1+V2V_1+ V_2V1+V2 的一组基，得证
当 V1∩V2={0}\pmb{V}_1 \cap \pmb{V}_2 = \{\pmb{0}\}VVV1∩VVV2={000} 时，dim(V1∩V2)=0dim(\pmb{V}_1\cap \pmb{V}_2) = 0dim(VVV1∩VVV2)=0，此时有
dim(V1)+dim(V2)=dim(V1+V2)\mathrm{dim}(\pmb{V}_1) + \mathrm{dim}(\pmb{V}_2) = \mathrm{dim}(\pmb{V}_1+\pmb{V}_2) dim(VVV1)+dim(VVV2)=dim(VVV1+VVV2)
Examples
1. 证明 subspace & 求维数 & 求一组基
  
  这个题第一问的证明有点问题，证明 subspace 应该说明非空，比如这题应该补一句关于 0∈W\pmb{0}\in W000∈W 的证明
2. 求基和维数
  1. 注意 4.3.3 例子中E选项的论证，说明 V1+V2=span(α1,α2,α3,β1,β2)V_1+V_2 = span(\alpha_1,\alpha_2,\alpha_3,\beta_1,\beta_2)V1+V2=span(α1,α2,α3,β1,β2)，这5个向量构成一组基，所以通过化阶梯求向量组的秩即得到 dim(V1+V2)\text{dim}(V_1+V_2)dim(V1+V2)
  2. V1∩V2V_1 \cap V_2V1∩V2 这个空间中的向量可以同时被 V1V_1V1 和 V2V_2V2 对应的两组基线性表示，因此这里联立解一个齐次线性方程组。通解为
    k=a1[−1−1100]+a2[−1−10−11]=[−a1−a2−a1−a2a1−a2a2]\pmb{k} = a_1\begin{bmatrix}-1 \\-1\\1\\0\\0\end{bmatrix} + a_2\begin{bmatrix}-1 \\-1\\0\\-1\\1\end{bmatrix} = \begin{bmatrix}-a_1-a_2 \\-a_1-a_2\\a_1\\-a_2\\a_2\end{bmatrix} kkk=a1⎣⎢⎢⎢⎢⎡−1−1100⎦⎥⎥⎥⎥⎤+a2⎣⎢⎢⎢⎢⎡−1−10−11⎦⎥⎥⎥⎥⎤=⎣⎢⎢⎢⎢⎡−a1−a2−a1−a2a1−a2a2⎦⎥⎥⎥⎥⎤ 回代，就会发现 V1∩V2V_1 \cap V_2V1∩V2 中的任意向量在 V2V_2V2 的基下表示为 a2(β1−β2)a_2(\beta_1-\beta_2)a2(β1−β2)，所以 dim(V1∩V2)=1\text{dim}(V_1 \cap V_2)=1dim(V1∩V2)=1，V1∩V2V_1 \cap V_2V1∩V2 的基可以是 β1−β2\beta_1-\beta_2β1−β2

4.3.5 Theorem 5 (Equivalent to direct sum 直和等价于)

以下表述等价

4.3.6 Theorem 6 (Direct sum decomposition 直和分解)

Suppose that U\pmb{U}UUU is a subspace of dimensional vector space V\pmb{V}VVV, there exists another subspace W\pmb{W}WWW of V\pmb{V}VVV such that
一个向量空间可以拆成两个（多个）线性子空间的直和
V=U+˙W\pmb{V} = \pmb{U} \dot{+} \pmb{W} VVV=UUU+˙WWW
证明思路：设 dim(V)=ndim(\pmb{V}) = ndim(VVV)=n
1. 设 ε1,...,εs\varepsilon_1,...,\varepsilon_sε1,...,εs 是 U\pmb{U}UUU 的一组基，将其扩展为 V\pmb{V}VVV 的一组基 ε1,...,εs,βs+1,...,βn\varepsilon_1,...,\varepsilon_s,\beta_{s+1},...,\beta_nε1,...,εs,βs+1,...,βn，则 V=span(ε1,...,εs,βs+1,...,βn)\pmb{V} = span(\varepsilon_1,...,\varepsilon_s,\beta_{s+1},...,\beta_n)VVV=span(ε1,...,εs,βs+1,...,βn)，设 W=span(βs+1,...,βn)\pmb{W} = span(\beta_{s+1},...,\beta_n)WWW=span(βs+1,...,βn)，此时有 V=U+W\pmb{V} = \pmb{U} + \pmb{W}VVV=UUU+WWW
2. 再证明 U+W\pmb{U} + \pmb{W}UUU+WWW 是直和，证明等价的维数公式即可
  ∵dim(U)=s,dim(W)=n−s∴dim(U+W)=dim(U)+dim(W)∴U+W是直和\begin{aligned} &\because dim(\pmb{U}) = s, \space dim(\pmb{W}) = n-s \\ &\therefore dim(\pmb{U}+\pmb{W}) = dim(\pmb{U}) +dim(\pmb{W}) \\ &\therefore \pmb{U}+\pmb{W} 是直和 \end{aligned} ∵dim(UUU)=s, dim(WWW)=n−s∴dim(UUU+WWW)=dim(UUU)+dim(WWW)∴UUU+WWW是直和

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双语矩阵论课程笔记（2）—— 【chapter 1】 Vector Spaces (Linear Spaces)

文章目录

1. Definitions and Examples

1.1 Number filed（数域）

1.2 Algebraic systems（代数系统）

1.3 Linear space / Vector space（线性空间 / 向量空间）

1.3.1 Definition

1.3.2 Remark on Linear space

1.3.3 Verify a linear space

2. Linear Dependence and Independence（线性相关和线性无关）

2.1 Basic concept

2.1.1 Linear combinations of vectors（线性组合）

2.1.2 linearly represented（线性表示）

2.1.3 linear dependence and Independence（线性相关和线性无关）

2.2 Theorem and corollary

3. Basis and Dimension（基和维数）

3.1 Dimension（维数）

3.2 Basis and coordinate（基和坐标）

3.2.1 Definition

3.2.2 Remark on basis

3.3.3 Transition Matrix（过渡矩阵）

3.3.4 Examples

4. Linear subspace（线性子空间）

4.1 Basic concept

4.1.1 linear subspace（线性子空间）

4.1.2 direct sum（直和）

4.2 Sum of subspaces

4.3 Theorems

4.3.1 Theorem 1

4.3.2 Theorem 2

4.3.3 Theorem 3

4.3.4 Theorem 4 (Dimension Formula 维数公式)

4.3.5 Theorem 5 (Equivalent to direct sum 直和等价于)

4.3.6 Theorem 6 (Direct sum decomposition 直和分解)

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