線性代數 - 向量空間¶

高中學習的向量的性質¶

在高中學習的向量滿足以下性質：

Theorem

\(\vec{x}+\vec{y}=\vec{y}+\vec{x}\)
\(\vec{x}+(\vec{y}+\vec{z})=(\vec{x}+\vec{y})+\vec{z}\)
\(\exists\vec{0},s.t.\vec{x}+\vec{0}=\vec{x},\forall\vec{x}.\)
\(\forall\vec{x},\exists\vec{y},s.t.\vec{x}+\vec{y}=\vec{0}\)
\(\forall\vec{x},1\vec{x}=\vec{x}\)
\(a,b,\in\mathbb{R},(ab)\vec{x}=a(b\vec{x})\)
\(a\in\mathbb{R},a(\vec{x}+\vec{y})=a\vec{x}+a\vec{y}\)
\(a,b,\in\mathbb{R},(a+b)\vec{x}=a\vec{x}+b\vec{x}\)

體¶

Definition

我們說集合\(F\)是一個體，如果\(F\)上有兩種運算\(+,\cdot\)，加法與乘法，滿足：

\(\forall a,b\in F,a+b\in F\)
\(\forall a,b,c\in F,a+(b+c)=(a+b)+c\)
\(\forall a,b\in F,a+b=b+a\)
\(\exists 0\in F,\text{s.t. }\forall a\in F,a+0=a\)
\(\forall a\in F,\exists b\in F,\text{s.t. }a+b=0\)
以及
\(\forall a,b\in F,a\cdot b\in F\)
\(\forall a,b,c\in F,a\cdot (b\cdot c)=(a\cdot b)\cdot c\)
\(\forall a,b\in F,a\cdot b=b\cdot a\)
\(\exists 1\in F,\text{s.t. }\forall a\in F,1\cdot a=a\)
\(\forall a\in F\backslash\{0\},\exists b\in F\backslash{0},\text{s.t. }a+b=0\)
以及分配律

\[ \forall a,b,c\in F,(a+b)\cdot c=a\cdot c+b\cdot c \]

因此可以發現\(\mathbb{Q},\mathbb{R},\mathbb{C}\)都是體，但整數不是，因為\(2\)在整數中沒有乘法反元素，也就是我們找不到整數\(k\)使得\(2\cdot k=1\)。

向量空間¶

Definition

設\(V\)為一集合，且\(F\)為一體。我們說\(V\)是一個over \(F\)的向量空間，如果其上有一加法運算\(+\)，以及\(F\)中與\(V\)之元素有一係數積之運算\(\cdot\)，使得對任意\(x,y\in V\)以及任意\(a\in F\)，我們都有：

\[ x+y\in V,\quad a\cdot x\in V \]

以及
1. \(\forall x,y,z\in V, (x+y)+z=x+(y+z)\)（加法結合律）
2. \(\forall x,y\in V, x+y=y+x\)（加法交換律）
3. \(\exists\mathbf{0}\in V,\text{s.t. }\forall x\in V, x+\mathbf{0}=x\)（加法單位元素）
4. \(\forall x\in V,\exists y\in V,\text{s.t. }x+y=\mathbf{0}\)（加法反元素）
5. \(1\cdot x=x\)（係數積為群作用(1)）
6. \(\forall x\in V, a,b\in F,(ab)\cdot x=a\cdot (b\cdot x)\)（係數積為群作用(2)）
7. \(\forall x,y\in V, a\in F,a\cdot(x+y)=a\cdot x+a\cdot y\)（係數積對\(V\)上加法有分配律）
8. \(\forall a,b\in F, x\in V,(a+b)\cdot x=a\cdot x+b\cdot x\)（係數積對\(F\)上加法有分配律）

Theorem

設\(V\)為向量空間over \(F\)，那麼

\(\forall x,y,z\in V,x+y=x+z\iff y=z\)（加法消去律）
\(\forall x\in V,\exists !y\in V,\text{s.t. }x+y=\mathbf{0},y=(-1)\cdot x\)（加法反元素唯一）
\(\mathbf{0}\in V\)唯一。
\(0\cdot x=\mathbf{0},\forall x\in V\)
\(a\cdot \mathbf{0}=\mathbf{0},\forall a\in F\)

Proof

若\(x+y=x+z\)，則\(\exists u\in V,\text{s.t. }x+u=\mathbf{0}\)，因此

\[ x+y+u=x+u+y=0+y=y=z \]

另一方向顯然。

若\(y,z\in V\)都使得\(x+y=\mathbf{0}=x+z\)，則根據加法消去律得到\(y=z\)。因此加法反元素唯一，並且因為\(x+(-1)x=(1-1)x=0x=\mathbf{0}\)，可以知道\(0x=\mathbf{0}\)

若\(\mathbf{0},\mathbf{0}'\)都是加法單位元素，那麼

\[ \mathbf{0}+\mathbf{0}'=\mathbf{0}=\mathbf{0}' \]

所以加法單位元素唯一。

\[ a\cdot\mathbf{0}=a(\mathbf{0}+\mathbf{0})=a\mathbf{0}+a\mathbf{0} \]

利用加法消去律得到\(a\cdot\mathbf{0}=(a-a)\mathbf{0}=\mathbf{0}\)

向量空間的範例¶

Example

\[ \mathbb{R}^n:=\{(a_1,a_2,\dots,a_n)\mid a_i\in\mathbb{R}\} \]

對所有\(\mathbf{x}=(x_1,\dots,x_n),\mathbf{y}=(y_1,\dots y_n)\in\mathbb{R}^n, a\in\mathbb{R}\)，定義加法：

\[ \mathbf{x}+\mathbf{y}:=(x_1+y_1,\dots,x_n+y_n)\in\mathbb{R}^n \]

以及係數積：

\[ a\cdot\mathbf{x}:=(ax_1,\dots ax_n)\in\mathbb{R}^n \]

則\(\mathbb{R}^n\)是一個向量空間over \(\mathbb{R}\)。

如果係數積改為\(\mathbb{Q}\)，則\(\mathbb{R}^n\)依然是向量空間over \(\mathbb{Q}\)，但若改為\(\mathbb{C}\)，則係數積將不會滿足封閉性，也就是說\(\exists z\in\mathbb{C},\text{s.t. }z\cdot\mathbf{x}\not\in\mathbb{R}^n\)。此時\(\mathbb{R}^n\)就不能成為一向量空間。

Example

所有函數的向量空間

設\(S\)為一非空集合以及\(F\)為一個體。定義集合

\[ \mathcal{F}(S,F):=\{f\mid f:S\to F\text{ is a function.}\} \]

定義加法：

\[ \forall f,g\in\mathcal{F}(S,F), (f+g)(s):=f(s)+g(s),\forall s\in S \]

以及係數積：

\[ \forall a\in F, f\in\mathcal{F}(S,F), (a\cdot f)(s):=a\cdot f(s),\forall s\in S \]

則\(\mathcal{F}(S,F)\)是一向量空間over \(F\)。

如果說\(T\)為任意集合，那麼\(\mathcal{F}(S,T)\)並不一定是一個向量空間，因為\(T\)上不一定有加法或乘法，就算有，也不一定滿足體的要求。

向量空間的子空間¶

Definition

設\(V\)是一個向量空間over \(F\)。我們說\(W\subset V\)是一個\(V\)的子空間，如果其繼乘了\(V\)上的加法和係數積，並且滿足：

\(\forall x,y\in W,x+y\in W\)
\(\forall x\in W,a\in F,a\cdot x\in W\)
\(\exists\mathbf{0}_W\in W,\text{s.t. }\forall x\in W,x+\mathbf{0}_W=x\)
\(\forall x\in W,\exists y\in W,\text{s.t. }x+y=\mathbf{0}_W\)

Theorem

設\(V\)是一個向量空間且\(W\)為\(V\)的子集合，則：

若\(W\)是\(V\)的子空間，則\(W\)中的零向量就是\(V\)的零向量。
\(W\)是一個子空間若且唯若\(\forall x,y\in W,a\in F,ax+y\in W\)且\(\mathbf{0}\in W\)。
若\(W\)是\(V\)的子空間，則我們不需要檢驗加法反元素是否存在\(W\)中

Proof

設\(W\)中之零向量為\(\mathbf{0}_W\)，\(V\)中之零向量為\(0_V\)，則

\[ \forall x\in W,0x=\mathbf{0}_V\in W \]

因此\(\mathbf{0}_V\in W\)，又\(\mathbf{0}_W+\mathbf{0}_V=\mathbf{0}_W=\mathbf{0}_V\)，可以知道\(W\)中的零向量就是\(V\)中的零向量，為了方便都簡寫為\(\mathbf{0}\)。

如果\(W\)是\(V\)的子空間，則\(\forall a\in F,x\in W\)我們都有\(ax\in W\)，因此任取\(y\in W\)，\(ax+y\in W\)。\(\mathbf{0}\in W\)根據定義成立。

如果\(\forall x,y\in W,a\in F,ax+y\in W\)且\(\mathbf{0}\in W\)，那麼：

\[ 1x+y=x+y\in W, ax+\mathbf{0}=ax\in W \]

並且\(-1x\in W\)，所以\(\forall x\in W,\exists! y=-x\in W,\text{s.t. }x+y=\mathbf{0}\)

若\(W\)是\(V\)的子空間，則因為\(\forall x\in W,-x\in W\)且\(x-x=\mathbf{0}\)，我們可以知道加法反元素必定落在\(W\)中。

Thm:

Theorem

設\(V\)為一向量空間over \(F\)。
任意\(V\)的子空間的交集還是\(V\)的子空間。

Proof

設\(\{W_{\alpha}\}_{\alpha\in\Lambda}\)為一些\(V\)之子空間。令

\[ W:=\bigcap_{\alpha\in\Lambda}W_{\alpha} \]

若\(x,y\in W,a\in F\)，那麼因為所有\(W_\alpha\)都是\(V\)的子空間，我們有

\[ \forall\alpha\in\Lambda,ax+y\in W_\alpha,\mathbf{0}\in W_\alpha \]

因此\(ax+y\in W,\mathbf{0}\in W\)

所以\(W\)為\(V\)的子空間。

子空間的和¶

Definition

設\(W_1,W_2\)都是\(V\)這個向量空間的子空間。定義其和為

\[ W_1+W_2:=\{w_1+w_2\mid w_1\in W_1,w_2\in W_2\} \]

Theorem

\(W_1+W_2\)為\(V\)之子空間

Proof

設\(a\in F,x,y\in W_1+W_2\)，那麼

\[ \exists w_1,w_1'\in W_1,w_2,w_2'\in W_2,\text{s.t. }\begin{cases}x&=w_1+w_2\\y&=w_1'+w_2'\end{cases} \]

因此

\[ ax+y=aw_1+aw_2+w_1'+w_2'=(aw_1+w_1')+(aw_2+w_2')\in W_1+W_2 \]

因為\(\mathbf{0}\in W_1\cap W_2\)，所以

\[ \mathbf{0}+\mathbf{0}=\mathbf{0}\in W_1+W_2 \]

故\(W_1+W_2\)為\(V\)的子空間。

子空間的平移¶

Definition

設\(v\in V,W\)為\(V\)之子空間，定義

\[ v+W:=\{v+w\mid w\in W\} \]

以及

\[ W+v:=\{w+v\mid w\in W\} \]

顯而易見的，這兩個集合是一樣的，我們稱為\(W\)關於\(v\)的Coset。

Theorem

給定\(v\in V\)，\(w\in W\)，\(v+W=W\)若且唯若\(v\in W\)。

也就是說\(v+W\)是\(V\)之子空間若且唯若\(v\in W\)

Definition

定義

\[ V/W:=\{v+W\mid v\in V\} \]

並且定義

\[ (v_1+W)+(v_2+W):=(v_1+v_2)+W \]

以及

\[ \forall c\in F, c(v+W):=cv+W \]

那麼\(V/W\)是一向量空間over \(F\)，稱為\(V\)關於\(W\)的商空間。

Example

設\(V=\mathbb{R}^3,F=\mathbb{R}\)。若定義

\[ W:=\{(x_1,x_2,0)\mid x_1,x_2\in\mathbb{R}\} \]

那麼\((a,b,c)+W,a,b,c\in\mathbb{R}\)其實就是所有\(x_3=c\)之平面集合。
並且可以感覺到所有\(V/W\)裡面的向量大概就只有向上或向下移動，因為只有第三個座標可以自由選取，也就是維度為\(1\)。

在之後介紹完基底與線性變換將會再回來講這些事情。

直和¶

Definition

設\(V\)為一向量空間over \(F\)。
給定子空間\(W_1,W_2,W_3\)，我們說\(W_3\)是\(W_1\)跟\(W_2\)的直和的意思是

\[ W_1\cap W_2=\{0\},W_1+W_2=W_3 \]

記為\(W_1\oplus W_2=W_3\)。

Example

設\(V=\mathbb{R}^3,F=\mathbb{R}\)。定義

\[ W_1:=\{(x_1,x_2,0)\mid x_1,x_2\in\mathbb{R}\} \]

以及

\[ W_2:=\{(0,0,x_3)\mid x_3\in\mathbb{R}\} \]

那麼\(W_1\cap W_2=\{0\}\)，並且\(W_1+W_2=V\)，這樣一來，\(V\)就被拆解為比較簡單的兩個子空間的和，並且這兩個子空間的交集是零空間，不會有向量被重複算到。

線性相依與線性獨立¶

Definition

設\(V\)是一個向量空間over \(F\)，\(S\subset V\)為一\(V\)之子集合。
我們說\(S\)是線性相依的，如果存在有限個不全為\(0\)之係數\(a_1,a_2,\dots,a_n\in F\)，以及\(n\)個\(S\)中的相異向量

\[ v_1,\dots,v_n\in S,\text{s.t. }\sum_{i=1}^na_iv_i=0 \]

我們說\(S\)是線性獨立如果他不是線性相依的，也就是說，任選\(n\)個\(S\)中的相異向量\(v_1,\dots,v_n\)，如果有一組係數\(a_1,\dots,a_n\in F\)使得

\[ \sum_{i=1}^{n}a_iv_i=0 \]

那麼只有可能是\(a_1=a_2=\dots=a_n=0\)。

我們說\(a_1=a_2=\dots=a_n=0\)是以下等式之顯然解

\[ \sum_{i=1}^{n}a_iv_i=0 \]

如果\(a_i\)不全為\(0\)，並且使得以上等式成立，則稱\(a_i\)為此等式的非顯然解。

Definition

設\(V\)是一個向量空間over \(F\)。如果\(v\in V\)且存在\(a_1,\dots,a_n\in F\)，以及\(v_1,\dots,v_n\in V\)，使得

\[ v=\sum_{i=1}^{n}a_iv_i \]

那麼\(v\)就是\(v_i\)的一個線性組合。

若給定\(V\)之子集合\(S\)，定義\(S\)的線性生成空間，記作\(\text{span}(S)\)，為

\[ \text{span}(S):=\left\lbrace v\in V\ \Bigg{|}\ \exists a_k\in F,u_k\in S,1\le k\le n,s.t.\ v=\sum_{k=1}^na_ku_k\right\rbrace \]

其實就是所有\(S\)中可能組合出來的向量，當然線性組合必須是有限個向量相加。
如果\(S\)是空集合，我們就定義\(\text{span}(S)=\{0\}\)

\(\varnothing\) 是線性獨立的，因為沒有向量可以使得其有非顯然解，故其非線性相依，所以是線性獨立的。

Theorem

設\(V\)是一個向量空間over \(F\)，\(S\subset V\)為一\(V\)之子集合，則\(\text{span}(S)\)為\(V\)之子空間。

Proof

如果\(S\)是空集合，則根據定義\(0\in\text{span}(S)\)。
如果\(S\)非空，任取\(s\in S\)，則\(0s\in \text{span}(S)\)，所以\(0\in\text{span}(S)\)。

設\(v_1,v_2\in\text{span}(S),c\in F\)，那麼就有\(a_1,\dots,a_n,b_1,\dots,b_m\in F\)以及向量

\[ s_1,\dots,s_n,s_{n+1},\dots,s_{n+m},\text{s.t. }v_1=\sum_{i=1}^{n}a_is_i,v_2=\sum_{i=1}^{m}b_is_{n+i} \]

所以

\[ cv_1+v_2=\sum_{i=1}^{n}ca_is_i+\sum_{i=1}^{m}b_is_{n+i} \]

若定義

\[ r_i=\begin{cases}ca_i,&\text{if }i\le n\\b_{i-n},&\text{if }i>n\end{cases},i\in\{1,\dots,n+m\} \]

那麼

\[ cv_1+v_2=\sum_{i=1}^{n+m}r_is_i\in\text{span}(S) \]

故\(\text{span}(S)\)為\(V\)之子空間。

Theorem

設\(V\)是一個向量空間over \(F\)，\(W\)為其子空間，\(S\subset V\)為一\(V\)之子集合，並且\(S\subset W\)，則\(W\supset\text{span}(S)\)。
也就是說\(\text{span}(S)\)是包含\(S\)的最小子空間。

Proof

設\(v\in\text{span}(S)\)，那麼\(v\)就可以寫成\(s\)中有限多個向量的線性組合

\[ v=\sum_{i=1}^{n}a_is_i,a_i\in F,s_i\in S \]

又因為\(s_i\in S\subset W\)，得到\(s_i\in W\)，所以根據\(W\)對於加法與係數積的封閉性，我們就有\(v\in W\)，這就說明了\(W\supset\text{span}(S)\)。

Theorem

設\(V\)是一個向量空間over \(F\)，\(S\subset V\)為一\(V\)之子集合，若\(S\)是線性相依的，則包含\(S\)的任意\(V\)的子集合都是線性相依的。
反過來說就是任意線性獨立集合之子集合也是線性獨立的。

Proof

設\(S'\supset S\)，因為\(S\)是線性相依的，存在一些\(s_1,\dots,s_n\in S\)有非顯然解。既然\(S'\supset S\)，\(s_1,\dots,s_n\in S'\)，所以\(S'\)當然也有非顯然解，所以也是線性相依的。

Definition

設\(V\)是一個向量空間over \(F\)，\(S\subset V\)為一\(V\)之子集合，我們說\(S\)是\(V\)之生成集，如果

\[ \text{span}(S)=V \]

Theorem

設\(V\)是一個向量空間over \(F\)，\(S\subset V\)為一\(V\)之線性獨立的子集合。令\(v\in V\)，那麼\(S\cup\{v\}\)是線性相依若且唯若\(v\in\text{span}(S)\)。

Proof

設\(S\cup \{v\}\)線性相依。如果\(v=0\)，那顯然\(v\in\text{span}(S)\)。
如果\(v\neq 0\)，我們就知道\(\exists a_1,\dots,a_n\in F,v_1,\dots,v_n\in S\cup\{v\}\)使得

\[ \sum_{i=1}^{n}a_iv_i=0 \]

且\(a_i\)不全為\(0\)，不失一般性假設\(a_1\neq0\)，如果沒有\(i\)使得\(v=v_i\)，那麼因為\(S\)是線性獨立，\(a_i\)必須全為\(0\)，出現矛盾，所以必定有\(i\)使得\(v=v_i\)。這就代表

\[ -a_iv=\sum_{j\neq i}a_jv_j \]

所以\(a_iv\in \text{span}(S)\)，又因為\(\text{span}(S)\)是子空間，所以\(v\in\text{span}(S)\)。

設\(v\in\text{span}(S)\)，如果\(v=0\)，那麼\(1v=0\)，所以\(S\cup\{v\}\)是線性相依。
如果\(v\neq0\)，將\(v\)寫作\(S\)中之線性組合：

\[ v=\sum_{i=1}^{n}a_iv_i \]

將會發現

\[ 0=-v+\sum_{i=1}^{n}a_iv_i \]

也就是說\(S\cup\{v\}\)中可以找到有限個向量有非顯然解，所以\(S\cup\{v\}\)是線性相依。

向量空間的基底¶

Definition

設\(V\)是一個向量空間over \(F\)，\(\beta\subset V\)為一\(V\)之線性獨立的子集合。

\(\beta\)是一個基底如果

Defintion:

Let \(V\) be a vector space.
A basis \(\beta\) is a linearly independent subset of \(V\) that generates \(V\).

\(\beta=\{u_1,u_2,\dots,u_k\}\) is a basis
if and only if
\(\forall v\in V\) can be uniquely written as the linear combination of \(u_1,u_2,\dots,u_k\).

Proof

(\(\implies\))

Suppose \(v=\displaystyle\sum_{1\le k\le n}a_ku_k\) and \(v=\displaystyle\sum_{1\le k\le n}b_ku_k\)

Then \(\mathit0=\displaystyle\sum_{1\le k\le n}(a_k-b_k)u_k\)
And since \(\beta\) is linearly independent, we know that \(a_k=b_k,1\le k\le n\).

Therefore if \(\beta=\{u_1,u_2,\dots,u_k\}\) is a basis then \(\forall v\in V\) can be uniquely written as the linear combination of \(u_1,u_2,\dots,u_k\).

(\(\Longleftarrow\))

If \(\forall v\in V\) can be uniquely written as the linear combination of \(u_1,u_2\dots u_k\), then \(\beta=\{u_1,u_2,\dots,u_k\}\) is a basis.

\(\forall v\in V\) can be uniquely written as the linear combination of \(u_1,u_2\dots u_k\) immediately implies that \(\textrm{span}(\beta)=V\)

\(\forall v\in V,\exists !\{a_k\in F\mid 1\le k\le n\}, s.t.\)
\(v=\displaystyle\sum_{1\le k\le n}a_ku_k\)

\(\therefore\mathit0=\displaystyle\sum_{1\le k\le n}a_ku_k\)

Since this expression is unique, \(\implies a_k=0,1\le k\le n\).
\(\therefore\beta\) is linearly independent.

Therefore \(\beta\) is a basis of \(V\).

If a vector space \(V\) is generated by a finite set \(S\), then some subsets of \(S\) is a basis of \(V\).

Proof

Case 1: \(S=\varnothing\) or \(S=\{0\}\)
\(\therefore\textrm{span}(S)=\{0\}=V\)
\(\therefore\varnothing\) is indeed a subset of \(V\).

Case 2: \(S\not=\varnothing\)

Let \(S=\{u_1,u_2,\dots,u_n\}\)

If \(|S|=1\) and \(\textrm{span}(S)=V\), then \(S\) is a basis.

Otherwise, take \(\beta=\{u_1\}\) and keep adding vectors \(u_2,u_3,\dots,u_k\in S\) into \(\beta\)
Since \(S\) is finite, this process must end with \(\beta=\{u_2,u_3,\dots,u_k\},k\le n\) is a linearly independent subset of \(S\)

(i)\(\beta=S\)
Then \(\textrm{span}(\beta)=\textrm{span}(S)=V\)
and \(\beta\) is linearly independent
\(\therefore\beta\) is a basis of \(V\)

(ii)
Adding any veotors \(v\in S-\{\beta\}\) into \(\beta\) would make \(\beta\) a linearly dependent subset.

Then we have to prove when this happens \(\textrm{span}(\beta)=V\)

\(\because\textrm{span}(S)=V\iff\textrm{span}(S)\subseteq V\land\textrm{span}(S)\supseteq V\)
\(\therefore\) We can prove that \(\textrm{span}(S)\subseteq\textrm{span}(\beta)\) since \(\textrm{span}(\beta)\) is always true.
Moreover, by thm, we can prove that \(\textrm{span}(\beta)\supseteq S\)

\(\because\beta\cup\{v\}\) is linearly dependent and \(\beta\) is linearly independent, therefore
\(v=\displaystyle\sum_{1\le i\le k}a_iu_i,a_i\in F\) and \(a_i\) are not all zeros.
\(\implies v\in\textrm{span}(\beta)\)
\(\therefore v\in S\implies v\in\textrm{span}(\beta)\)
\(\implies S\subseteq\textrm{span}(\beta) \\\implies\textrm{span}(S)\subseteq\textrm{span}(\beta) \\\implies V\subseteq\textrm{span}(\beta) \\\implies V=\textrm{span}(\beta)\)
Therefore, \(\beta\) is a basis of \(V\) and \(\beta\subseteq S\).

The replacement theorem¶

Let \(V\) be a vector space.
Let \(G\) be a finite generating set of \(V\) and \(|G|=n\).
If \(L\) is a linear independent subset of \(V\) and \(|L|=m\)
Then \(m\le n\) and \(\exists H\subseteq G,|H|=n-m\) such that \(\textrm{span}(L\cup H)=V\).

Proof

Using Induction on \(m\) to prove.

\(m=0:\)
\(L=\varnothing\) is linearly independent.
\(\therefore\) take \(H=G\), then
\(0\le n\) and \(|H|=n-0\) and \(\textrm{span}(L\cup H)=\textrm{span}(G)=V\)

Therefore for \(m=0\) the thm is true.

Suppose \(m=k\) the thm is true
That is, if \(L_0=\{u_1,u_2\dots u_k\}\) is a linear independent subset of \(V\) and \(|L_0|=k\), then \(k\le n\) and \(\exists H_0=\{v_1,v_2,\dots v_{n-k}\}\subseteq G,s.t.\textrm{span}(L_0\cup H_0)=V\)
W.L.O.G let \(u_1\not=\mathit0\), then \(\exists a_2,\dots a_k,b_1,\dots b_{n-k},s.t.\)
\(\displaystyle\sum_{2\le i\le k}a_iu_i+\displaystyle\sum_{1\le i\le n-k}b_iv_i=u_1\)
then we know that \(\displaystyle\sum_{1\le i\le n-k}b_iv_i\not=\mathit0\) since \(L_0\) is linearly independent.

Then check \(m=k+1\)
Let \(L=L_0\cup\{u_{k+1}\}\) be linearly independent
Suppose \(k+1>n\)
\(\because n\ge k\therefore n=k\)
\(\therefore\textrm{span}(L_0)=V\) since \(H_0\) becomes \(\varnothing\)
\(\because u_{k+1}\in V=\textrm{span}(L_0)\)
\(\implies u_{k+1}=\displaystyle\sum_{1\le i\le k}c_iu_i,c_i\not=0\) for some \(i\)
\(\implies L\) is linearly dependent.
This is a contradiction.
Therefore \(k+1\le n\)

Since
\(u_{k+1}\in\textrm{span}(L_0\cup H_0) \\\implies u_{k+1}=\displaystyle\sum_{1\le i\le k}a_iu_i+\displaystyle\sum_{1\le i\le n-k}b_iv_i\)
And \(L\) is linearly independent, therefore \(\displaystyle\sum_{1\le i\le n-k}b_iv_i\not=\mathit0\)

W.L.O.G let \({b_{n-k}}\not=0\) and \(v_{n-k}\not=\mathit0\)
then \(v_{n-k}\in\textrm{span}(\{u_1,\dots,u_k,u_{k+1},v_1,\dots,v_{n-k-1}\})\)
Therefore \(v_{n-k}\) can be replaced by \(u_{k+1}\).
so choose \(H=H_0-\{v_{n-k}\},L=L_0\cup\{u_{k+1}\}\), then \(|H|=n-k-1\)
then \(\textrm{span}(L\cup H)=V\).

Therefore for \(m=k+1\), the theorem is true.

Therefore by Induction, we know that the Replacement theorem is true.

Meaning of the replacement theorem

A finite generating set of a vector space would make all linearly independent subsets of this vector space unable to contain more than some finite number of vectors.

We can always find some veotor in this fintie generating set and adding them into the linearly independent subsets of this vector space, which will also become a generating subset, and can possibly create a basis.

Moreover, this can imply that a basis must have a finite nubmer if the generating set of the vector space is finite.

So you probably can see the finite-dimensional stuff is coming soon.

Corollary:

Let \(V\) be a vector space
If \(V\) has a finite basis, then all basis of \(V\) are finite, and contain same number of vectors.

\(pf:\)
Let \(\beta\) be a basis of \(V\) and \(|\beta|=n<\infty\).

Let \(\gamma\) be any other basis of \(V\) and \(|\gamma|=m\)

Suppose \(m>n\), then \(m\ge n+1\).
We can find a subset \(S\) of \(\gamma\) such that \(|S|=n+1\).

Since \(\beta\) is a finite generating set of \(V\) and \(S\) is linearly independent, it follows the replacement theorem that \(n\ge n+1\).
This is a contradiction.

Now suppose \(m<n<\infty\)

Since \(\gamma\) is a finite generating set of \(V\), by the replacement theorem, we know that \(m\ge n\) is also a contradiction.

Therefore \(m=n\), this also implies that all basis are finite.

Definition:
If a vector space \(V\) contains a finite basis, then \(V\) is called finite-dimentional.
Let the finite basis contains \(n\) vectors, then we write \(\dim{V}=n\), by Corollary, we know that this \(n\) is unique for any vector space \(V\) that contains a finite basis.

Corollary:

Let \(V\) be a vector space and \(\dim{V}=n<\infty\)

(a) Any finite generating set contains at least \(n\) vectors, and a finite generating set is a basis if it contains exactly \(n\) vectors.
(b) Any linearly independent subset of \(V\) that contains \(n\) vectors is a basis of \(V\)
(c) If \(L\subseteq V\) and \(L\) is linearly independent, then there is a basis \(\beta\) that satisfies \(L\subseteq\beta\)
i.e. \(L\) can be extended to a basis by adding some vectors in to it.

Proof

(a)
\(\because\dim{V}=n\)
\(\therefore\exists\beta\) that is a basis containing \(n\) vectors.
Let \(S\) be any finite generating set of \(V\), then by the replacement thm we know that \(|S|\ge n\).
If \(|S|=n\), then by Col 1, \(S\) is a basis of \(V\).

(b)
Let \(S\) be a linearly independent subset of \(V\) and \(|S|=n\) and \(\beta\) is a basis of \(V\)

By the replacement thm, \(\exists H\subseteq\beta,|H|=n-n=0\) such that \(\textrm{span}(S\cup H)=V\)
Since \(H=\varnothing\), this implies that \(S\) is a generating set of \(V\).

Therefore \(S\) is a basis of \(V\).

(c)
Let \(\beta\) be a basis of \(V\)
Since \(L\) is linearly independent, it follows the replacement thm that \(|L|\le n=|\beta|\)
\(\therefore\) we can find a basis \(\beta\) such that \(L\subseteq\beta\)
More precisely, choose \(\beta=L\cup H\) where \(H\) is, by replacement thm, a subset of \(\beta\) and \(H=\beta-\{L\}\)

Theorem

Let \(W\) be a subspace of a vector space \(V\)
If \(\dim{V}=n<\infty\), then \(\dim{W}\le n\).
Moreover if \(\dim{W}=\dim{V}\), then \(W=V\).

That is, every subspace of \(V\) has dimension not greater than the dimension of \(V\).
It's quite obvious since \(W\) is a subset of \(V\), so it must take less (or equal) vectors to generate \(W\).

Proof

The proof is by induction on \(m\).

The zero vector space is a subspace of any vector space, and having a dimension of \(0\). Therefore \(\dim\{\mathit0\}\le n\).
If \(n=0\) then \(V=\{\mathit0\}\).

Let \(m=\dim{W}>0\implies W\not=\{0\}\)
Say \(W\) has a basis \(\{x_1\}\), and \(x_1\not=\mathit0\)
then \(\dim{W}=1\le n\)

Now suppose \(W\) has a basis \(\{x_1,x_2,\dots,x_m\}\),then since \(\{x_1,x_2,\dots,x_m\}\) is linearly independent and is a subset of \(V\), by the replacement thm we know that \(m\le n\).

Suppose \(m=n\), then \(S=\{x_1,x_2,\dots,x_n\}\) is a basis for \(W\), and let \(\beta\) be a basis for \(V\)
By Corollary 2.(b) we know that \(S\) is also a basis for \(V\).
And therefore \(\textrm{span}(S)=\textrm{span}(\beta)=V\implies W=V\)

Therefore :

If \(\dim{V}=n<\infty\), then \(\dim{W}\le n\).
Moreover if \(\dim{W}=\dim{V}\), then \(W=V\).

Definitions:

Let \(S_1,S_2\) be two subsets of a vector space \(V\)
\(S_1+S_2=\{s_1+s_2\mid s_1\in S_1,s_2\in S_2\}\)

The direct sum \(S\) of \(S_1,S_2\), denoted as \(S_1\oplus S_2\), is defined as \(S=S_1+S_2\) where \(S_1\cap S_2=\{\mathit0\}\)

Theorem:
Let \(S_1,S_2\) be two subsets of a vector space \(V\)

(a) \(\text{span}(S_1)+\text{span}(S_2)=\text{span}(S_1\cup S_2)\)
(b) \(\text{span}(S_1\cap S_2)\subseteq\text{span}(S_1)\cap\text{span}(S_2)\)

Let \(W_1,W_2\) be two subspaces of \(V\)
\(W_1+W_2\) is also a subspace of \(V\).

\(\dim{(W_1+W_2)}=\dim{W_1}+\dim{W_2}-(\dim{W_1\cap W_2})\)

Let \(U\) be a subspace of \(V\) and \(U\supseteq W_1\cup W_2\) where \(W_1,W_2\) are subspaces of \(V\).
Then \(U\supseteq W_1+W_2\).
That is, the sum of two subspaces is the minimal subspace of \(V\) containing \(W_1\) and \(W_2\).

Corollary:
\(V=W_1\oplus W_2\) if and only if \(\dim{V}=\dim{W_1}+\dim{W_2}\)