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向量空間
This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.
\(\newcommand{\trans}{^\top} \newcommand{\adj}{^{\rm adj}} \newcommand{\cof}{^{\rm cof}} \newcommand{\inp}[2]{\left\langle#1,#2\right\rangle} \newcommand{\dunion}{\mathbin{\dot\cup}} \newcommand{\bzero}{\mathbf{0}} \newcommand{\bone}{\mathbf{1}} \newcommand{\ba}{\mathbf{a}} \newcommand{\bb}{\mathbf{b}} \newcommand{\bc}{\mathbf{c}} \newcommand{\bd}{\mathbf{d}} \newcommand{\be}{\mathbf{e}} \newcommand{\bh}{\mathbf{h}} \newcommand{\bp}{\mathbf{p}} \newcommand{\bq}{\mathbf{q}} \newcommand{\br}{\mathbf{r}} \newcommand{\bx}{\mathbf{x}} \newcommand{\by}{\mathbf{y}} \newcommand{\bz}{\mathbf{z}} \newcommand{\bu}{\mathbf{u}} \newcommand{\bv}{\mathbf{v}} \newcommand{\bw}{\mathbf{w}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\nul}{\operatorname{null}} \newcommand{\rank}{\operatorname{rank}} %\newcommand{\ker}{\operatorname{ker}} \newcommand{\range}{\operatorname{range}} \newcommand{\Col}{\operatorname{Col}} \newcommand{\Row}{\operatorname{Row}} \newcommand{\spec}{\operatorname{spec}} \newcommand{\vspan}{\operatorname{span}} \newcommand{\Vol}{\operatorname{Vol}} \newcommand{\sgn}{\operatorname{sgn}} \newcommand{\idmap}{\operatorname{id}} \newcommand{\am}{\operatorname{am}} \newcommand{\gm}{\operatorname{gm}} \newcommand{\mult}{\operatorname{mult}} \newcommand{\iner}{\operatorname{iner}}\)
Main idea
The vectors in \(\mathbb{R}^n\) enjoy some nice properties, such as \({\bf u} + {\bf v} = {\bf v} + {\bf u}\).
Although one can always expand a vector \({\bf x}\) in \(\mathbb{R}^n\) as \((x_1,\ldots, x_n)\), this seemingly natural setting is not always required.
For example, linear combinations \(c_1{\bf u}_1 + \cdots + c_d{\bf u}_d\) are defined without really examing the entries.
Similarly, we may say
apple
andbanana
are vectors.We can make a linear combination like
1 apple + 2 banana
and say
{ apple, banana }
is a linearly independent setsince intuitively there is no way to replace
apples
withbananas
such as1 apple + 2 banana = 0
.Definition (Vector space)
A vector space over \(\mathbb{R}\) consists of three parts:
The addition has to have the following basic properties:
The scalar multiplication has to cooperate with the addition in a good mannar:
6. For any \(k\in\mathbb{R}\) and \({\bf v}\in V\), \(k\cdot {\bf v}\in V\). (closed under scalar multiplication)
7. For any \(k,\ell\in\mathbb{R}\) and \({\bf v}\in V\), \((k + \ell)\cdot {\bf v} = k\cdot{\bf v} + \ell\cdot{\bf v}\). (distributive over scalar addition)
8. For any \(k\in\mathbb{R}\) and \({\bf u}, {\bf v}\in V\), \(k\cdot({\bf u} + {\bf v}) = k\cdot {\bf u} + k\cdot{\bf v}\). (distributive over vector addition)
9. For any \(k,\ell\in\mathbb{R}\) and \({\bf v}\in V\), \((rs)\cdot {\bf v} = r\cdot(s\cdot{\bf v})\). (associative over field multiplication and scalar multiplication)
10. For \(1\in\mathbb{R}\) and \({\bf v}\in V\), \(1\cdot {\bf v} = {\bf v}\). (\(1\) is the identity operation)
Obviously, one may check \(\mathbb{R}^n\) along with the classical vector addition \(+\) and the scalar multiplication \(\cdot\) is a vector space.
In addition, the following can be viewed as vector spaces with the standard vector addition and scalar multiplication:
There are still many instances of vector spaces.
And the nice thing about them is all the things we have learnt so far also apply to them, including span, linearly independence, dimension, ans so on.
We usually consider \(\mathbb{R}\) as the scalar, but any other number system with a field structure also works.
A field is, roughly speaking, a set of elements equipped with customized \(+\), \(-\), \(\times\), and \(\div\).
Common examples are \(\mathbb{R}\), \(\mathbb{C}\), and \(\mathbb{Q}\).
In contrast, \(\mathbb{Z}\) is not a field since it does not allow division.
Side stories
Experiments
Exercise 1
執行以下程式碼。
令 \(S = \{ p_1, p_2, p_3 \}\)。
藉由
seed = 0
得到 \[𝚙_𝟷=−5𝑥^2+3𝑥+1\] \[𝚙_𝟸=15𝑥^2−8𝑥−3\] \[𝚙_𝟹=45𝑥^2−23𝑥−9\] \[𝚋=65𝑥^2−34𝑥−13 \]第一題全
Exercise 1(a)
將 \(b\) 寫成 \(S\) 的線性組合。
\(Ans:\) \[\left[\begin{array}{ccc|c} p_1 & p_2 & p_3 &b \end{array}\right].\] \[\left[\begin{array}{ccc|c} 1 & -3 & -9 & -13\\ 3 & -8 & -23 & -34\\ -5 & 15 & 45 & 65 \end{array}\right].\] \[\left[\begin{array}{ccc|c} 1 & 0 & 3 & 2\\ 0 & 1 & 4 & 5\\ 0 & 0 & 0 & 0 \end{array}\right].\]
\[b = - p_1 + p_2 + p_3 \]
Exercise 1(b)
判斷 \(S\) 是否線性獨立?
\(Ans:\) \[\begin{bmatrix} p_1 & p_2 & p_3 \end{bmatrix} \]\begin{bmatrix} 1 & -3 & -9\\ 3 & -8 & -23\\ -5 & 15 & 45 \end{bmatrix} \[\begin{bmatrix} 1 & 0 & 3\\ 0 & 1 & 4\\ 0 & 0 & 0 \end{bmatrix}\] \[-3p_1-4p_2+p_3=0\] 係數非皆為零,因此 \(S\) 非線性獨立
Exercise 1©
判斷 \(\operatorname{span}(S)\) 是否等於所有二次以下的多項式?
\(Ans:\)
\(p_3=3p_1+4p_2\)
基底為 \(p_1\) 和 \(p_2\),兩個元素的基底無法形成三維空間,因此 \(S\) 無法生出所有二次以下的多項式。
Exercises
Exercise 2
令
\[\begin{aligned} p_1 &= (x + 1)(x + 2), \\ p_2 &= (x + 1)(x + 3) \end{aligned} \]
且 \(S = \{p_1, p_2\}\)。
Exercise 2(a)
寫出一個 \(S\) 的線性組合。
\(Ans:\) \[\begin{aligned} p_1 &= (x + 1)(x + 2)=x^2+3x+2 \\ p_2 &= (x + 1)(x + 3)=x^2+4x+3 \end{aligned} \] \(p_1 + p_2 = 2x^2+7x+5\)
Exercise 2(b)
判斷 \(S\) 是否線性獨立。
\(Ans:\)
把 \(p_1\) 和 \(p_2\) 的係數寫成矩陣可以得到\[ \begin{bmatrix} 1 & 3 & 2 \\ 1 & 4 & 3 \\ \end{bmatrix} \ \rightarrow\ \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ \end{bmatrix} \] \(p_1, p_2\) 無法互相寫成彼此的線性組合,故 \(S\) 為線性獨立。
Exercise 2©
判斷 \(\operatorname{span}(S)\) 是否等於所有二次以下的多項式。
\(Ans:\)
基底為 \(p_1\) 和 \(p_2\),兩個元素的基底無法形成三維空間,因此 \(S\) 無法生出所有二次以下的多項式。
Exercise 3
令
\[\begin{aligned} A_1 &= \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}, \\ A_2 &= \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}, \\ A_3 &= \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \end{aligned} \]
且 \(S = \{A_1, A_2, A_3\}\)。
Exercise 3(a)
寫出一個 \(S\) 的線性組合。
\(Ans:\)
\(A_1 + A_2 + A_3 = \begin{bmatrix} 4 & 4 \\ 4 & 4 \end{bmatrix}\)
Exercise 3(b)
判斷 \(S\) 是否線性獨立。
\(Ans:\)
由 \(A_1 + A_2 + A_3 = \begin{bmatrix} 4 & 4 \\ 4 & 4 \end{bmatrix} = 4A_1\)
可得出 \(A_3 = 3A_1 - A_2\)。
因為 \(A_3\) 是 \(A_1\) 和 \(A_2\) 的線性組合,所以 \(S\) 不是線性獨立。
Exercise 3©
判斷 \(\operatorname{span}(S)\) 是否等於所有 \(2\times 2\) 的實係數矩陣。
\(Ans:\)
令 \[V = \{A_1, A_2\},\]
因為 \(A_3\) 是 \(A_1\) 和 \(A_2\) 的線性組合,所以 \(\operatorname{span}(V) = \operatorname{span}(S)\)。
而把 \(A_1\) , \(A_2\)的係數重新寫成 \(4\times 2\) 矩陣可得
\[ \begin{bmatrix} 1 & 1 \\ 1 & 2 \\ 1 & 2 \\ 1 & 1 \end{bmatrix}\]
經過列運算可得
\[R = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0 \end{bmatrix}, \]
代表 \(A_1\) 和 \(A_2\) 是 \(V\) 的基底且 \(V\) 線性獨立。
因此 \(V\) 是 \(\operatorname{span}(S)\) 的基底,且 \(\operatorname{span}(S)\) 的維度為 \(2\)。
然而 \(2\times 2\) 實係數矩陣的維度為 \(4\),
所以 \(\operatorname{span}(S)\) 無法做出所有的 \(2\times 2\) 實係數矩陣。
Exercise 4
以下考慮一種奇怪的加法 \(\oplus\) 和乘法 \(\otimes\)。
考慮 \(\mathbb{R}_+\) 為所有正實數的集合、\(+\) 和 \(\times\) 分別是實數的正常加法和乘法。
若 \(a,b\in\mathbb{R}_+\) 定義 \(a \oplus b = a \times b\)﹐
若 \(k\in\mathbb{R}\) 而 \(a\in\mathbb{R}_+\) 定義 \(k\otimes a = a^k\)。
驗證 \((\mathbb{R}_+, \oplus, \otimes)\) 配合在一起是一個向量空間。
Let \(a, b, c \in \mathbb{R}_{+}, k, l \in \mathbb{R}\)
(1) \(a\oplus b = ab \in \mathbb{R}_{+}\)
(2) \(a\oplus b = ab =ba = b\oplus a\)
(3) \((a\oplus b) \oplus c = ab \oplus c = abc = a\oplus bc = a \oplus (b \oplus c)\)
(4) \(1\in \mathbb{R}_{+}\), then we have \(a\oplus 1 = a.\)
(5) \(\frac{1}{a} \in \mathbb{R}_{+}\), then we have \(a\oplus \frac{1}{a} = 1.\)
(6) \(k \otimes a = a^{k} \in \mathbb{R}_{+}\)
(7) \((k+l)\otimes a = a^{k+l} = a^{k}a^{l} = (k \otimes a) \oplus (l \otimes a)\)
(8) \(k\otimes (a\oplus b) = k\times ab = (ab)^{k} = a^{k}b^{k} = k\otimes a \oplus k\otimes b\)
(9) \((kl)\otimes a = a^{kl} = (a^{l})^{k} = k\otimes (l\otimes a)\)
(10) For \(1 \in \mathbb{R}, 1\otimes a =a\)
By (1) to (10), we can say that \(\mathbb{R}_{+}\) is a vector space with \(\oplus\) and \(\otimes.\)
Exercise 5
若 \(V\) 為一向量空間。
證明以下性質。
Exercise 5(a)
對於任何 \({\bf v}\in V\)﹐都有 \(0\cdot {\bf v} = {\bf 0}\)。
(這個式子的口頭敘述是:
實數的 \(0\) 乘上任何一個向量
都是零向量。)
Consider the equality \(0\bv = (0 + 0)\bv = 0\bv + 0\bv\).
Thus, by adding the additive invertse of \(0\bv\) to both side, we have \(0 = 0\bv\).
Exercise 5(b)
對於任何 \({\bf v}\in V\)﹐\(-1\cdot {\bf v} + {\bf v} = {\bf 0}\)。
(這個式子的口頭敘述是:
實數的 \(-1\) 乘上任何一個向量
都是該向量的加法反元素。)
In \(\mathbb{R}\), \(0 = 1 + (-1)\)
Then, we have \(0\bv = [1 + (-1)] \cdot \bv\)
From (a), we know \(0\bv = {\bf 0}= 1 \cdot \bv+ (-1) \cdot\bv\)
Exercise 6
一個體的指的是一個集合 \(F\) 搭配一個加法一個乘法
(想像實數集 \(\mathbb{R}\)、或是有理數集\(\mathbb{Q}\);但整數集 \(\mathbb{Z}\) 則不是一個體。)
使得:
驗證 \(\mathbb{Z}_2 = \{0,1\}\) 搭配
加法 \(a + b = (a + b) \pmod{2}\) 和
乘法 \(a \cdot b = a \cdot b\)
是一個體。
(所以向量空間中的純量也可以代換成 \(\mathbb{Z}_2\)。)
\((a+b) \pmod{2} = (b+a) \pmod{2}\)
\(a \cdot b = b \cdot a\)
\(a \cdot( b \cdot c ) = (a \cdot b) \cdot c\)
\((1+0+0) \pmod{2} = [(1+0) \pmod{2} + 0] \pmod{2} = [1 + (0+0) \pmod{2}] \pmod{2}\)
\((1+1+0) \pmod{2} = [(1+1) \pmod{2} + 0] \pmod{2} = [1 + (1+0) \pmod{2}] \pmod{2}\)
\((1+0+1) \pmod{2} = [(1+0) \pmod{2} + 1] \pmod{2} = [1 + (0+1) \pmod{2}] \pmod{2}\)
\((0+0+0) \pmod{2} = [(0+0) \pmod{2} + 0] \pmod{2} = [0 + (0+0) \pmod{2}] \pmod{2}\)
\((0+1+0) \pmod{2} = [(0+1) \pmod{2} + 0] \pmod{2} = [0 + (1+0) \pmod{2}] \pmod{2}\)
\((0+0+1) \pmod{2} = [(0+0) \pmod{2} + 1] \pmod{2} = [0 + (0+1) \pmod{2}] \pmod{2}\)
\((0+1+1) \pmod{2} = [(0+1) \pmod{2} + 1] \pmod{2} = [0 + (1+1) \pmod{2}] \pmod{2}\)
\((1+1+1) \pmod{2} = [(1+1) \pmod{2} + 1] \pmod{2} = [1 + (1+1) \pmod{2}] \pmod{2}\)
\(0 \cdot [(1 + 0) \pmod{2}]=( 0 \cdot 1 + 0 \cdot 0) \pmod{2}\)
\(0 \cdot [(1 + 1) \pmod{2}]=( 0 \cdot 1 + 0 \cdot 1) \pmod{2}\)
\(1 \cdot [(0 + 1) \pmod{2}]=( 1 \cdot 0 + 1 \cdot 1) \pmod{2}\)
\(1 \cdot [(1 + 1) \pmod{2}]=( 1 \cdot 1 + 1 \cdot 1) \pmod{2}\)
\(1 \cdot [(0 + 0) \pmod{2}]=( 1 \cdot 0 + 1 \cdot 0) \pmod{2}\)
\(0 \cdot [(0 + 0) \pmod{2}]=( 0 \cdot 0 + 0 \cdot 0) \pmod{2}\)
\((0+0) \pmod{2} = 0\)
\((1+0) \pmod{2} = 1\)
\(0 \cdot 1 = 0\)
\(1 \cdot 1 = 1\)
\((0 + 0) \pmod{2} = 0\)
\((1 + 1) \pmod{2} = 0\)
\(1 \cdot 1 = 1\)
故 \(\mathbb{Z}_2 = \{0,1\}\) 搭配
加法 \(a + b = (a + b) \pmod{2}\) 和
乘法 \(a \cdot b = a \cdot b\)
是一個體。
目前分數 6.5