---
tags: Linear Algebra
---

## Vector Space
It's **my lecture notes**, [**1.2 Vector Space**](https://www.youtube.com/watch?v=FjN_sxRBmQg&list=PLj6E8qlqmkFtjxknKFtdxc1_SxNBXgpbo&index=2), [Linear Algebra 1, NYCU OCW](https://ocw.nycu.edu.tw/?course_page=all-course%2Fcollege-of-science%2Fam%2F%E7%B7%9A%E6%80%A7%E4%BB%A3%E6%95%B8%E4%B8%80-linear-algebra-i-%E6%87%89%E7%94%A8%E6%95%B8%E5%AD%B8%E7%B3%BB-%E8%8E%8A%E9%87%8D%E8%80%81%E5%B8%AB)

## Facebook 嘅討論
- [2023-11-28 02:50](https://www.facebook.com/groups/978651839991898/posts/1053216632535418/)
- [2023-11-29 04:09](https://www.facebook.com/groups/978651839991898/posts/1053738029149945/)
- [2023-11-29 23:29](https://www.facebook.com/groups/978651839991898/posts/1054121652444916/)

## Definition of Vector Spaces
A vector space $V$ over a field $F$ consists of a set on which two operations,
**addition** and **scalar multiplication**, are satisfied with the following axioms

(VS-1) $x + y \in V$ whenever $x,\ y \in V$

(VS 0) $\alpha x \in V$ whenever $\alpha \in F\ and\ x \in V$

(VS 1) $x + y = y + x\;\; \forall\ x,\ y \in V$ (commutative)

(VS 2) $(x + y) + z = x + (y + z)\;\; \forall\ x,\ y,\ z \in V$ (associative)

(VS 3) $\exists\ \mathbf{0} \in V$ such that $x + \mathbf{0} = x,\;\: x \in V$

(VS 4) For each $x \in V, \; \exists\:y\in V$ such that $x + y = \mathbf{0}$

(VS 5) For each $x \in V$, $\mathbf{1} \cdot x=x\;, \mathbf{1} \in F$

(VS 6) $(ab)x = a(bx)\;\; \forall\; a,\ b \in F\ and\ x \in V$ (distributive)

(VS 7) $a(x + y) = ax + ay\;\; \forall\; a,\ b \in F\ and\ x \in V$ (distributive)

(VS 8) $(a + b)x = ax + bx\;\; \forall\; a,\ b \in F\ and\ x \in V$ (distributive)

### Examples 01
Can a vector space be constructed using the following definition?

$S = \{(a_1, a_2): a_1, a_2 \in R\},\ F = R$

Addition: $(a_1, a_2) + (b_1, b_2) = (a_1 + b_1,\ a_2 - b_2)$

Scalar Multiplication: $c(a_1, a_2) = (ca_1, ca_2)$

##### Proof:

Let $x = (a_1,\ a_2),\; y = (b_1,\ b_2)$

$x + y = (a_1,\ a_2) + (b_1,\ b_2) = (a_1 + b_1,\ a_2 - b_2)$
$y + x = (b_1,\ b_2) + (a_1,\ a_2) = (b_1 + a_1,\ b_2 - a_2)$

$\implies$ $(a_1,\ a_2) + (b_1,\ b_2) \neq (b_1,\ b_2) + (a_1,\ a_2)$
$\implies$ $x + y \neq y + x$

The definition can't satisfy the conditions (VS 1) $x + y = y + x\;\; \forall\ x,\ y \in V$, so it can't be constructed.

## Definition of Field
A field $F$ is a set on which two operations $+$ and $\cdot$ are defined so that, for each pair of elements $x, y \in F$ , there are unique elements $x + y$ and $x \cdot y \in F$ for which the following axioms hold for all elements $a, b, c \in F$.
1. a + b = b + a
2. (a + b) + c = a + (b + c)
3. There exist distinct element $\mathbf{0} \in F$ and $1 \in F$ such that $\mathbf{0} + a = a$ and $\mathbf{1} \cdot a = a$
4. For each element $a \in F$ and each nonzero element $b \in F$ , there exist elements $c \in F$ and $d \in F$ such that $a + c = \mathbf{0}$ and $b \cdot d = \mathbf{1}$
5. $a \cdot (b + c) = a \cdot b + a \cdot c$

### Example 01
$\mathbb{C}$, $\mathbb{R}$ and $\mathbb{Q}$ are the common fields.

### Example 02
Is $V$ a vector space over the complex number $\mathbb{C}$?

$$
V = \{\begin{bmatrix}a_{1} \\a_{2} \\\vdots \\a_{n}\end{bmatrix}:
a_i \in R,\; i = 1, 2, \cdots, ..., n \},\: F = \mathbb{C}
$$

##### Proof:

let $z = i,\ z \in \mathbb{C} = F$,

$v_1 = \begin{bmatrix}a_{1} \\a_{2} \\\vdots \\a_{n}\end{bmatrix} \in V, a_i \in R$

$z \cdot v_1 \implies a_i \in \mathbb{C}$

It doesn't satisfy the axiom (VS 0) $\alpha x \in V$ whenever $\alpha \in F\ and\ x \in V$, so it can't be constructed.

## Theorem 1.1
If $x,\ y,\ z \in V$, a Vector Space $V$ such that $x + z = y + z\,$ then $x = y$.

#### Corollary 1
The element $\mathbf{0}$ is **unique** in the axioms (VS 3) of the definition of the vector space.

(VS 3) $\exists\ \mathbf{0} \in V$ such that $x + \mathbf{0} = x,\;\: x \in V$

#### Corollary 2
The vector $y$ is **unique** in the axioms (VS 4) of the definition of the vector space.

(VS 4) For each $x \in V, \; \exists\:y\in V$ such that $x + y = \mathbf{0}$

## Theorem 1.2
$V$ is a vector space, then
1. $0 \cdot x = \mathbf{0}$, $0 \in F$, $x \in V$, $\mathbf{0} \in V$
2. $(-a)x = -(ax) = a(-x)$, $a \in F$, $x \in V$
3. $a \cdot \mathbf{0} = \mathbf{0}$, $a \in F$，$\mathbf{0} \in V$

## Exercises in [Linear Algebra](https://www.amazon.com/Linear-Algebra-4th-Stephen-Friedberg/dp/0130084514)