---
# System prepended metadata

title: Lecture 2
tags: [' NCTU OCW', ' 莊重', Linear Algebra]

---

---
tags: Linear Algebra, NCTU OCW, 莊重
---

# Lecture 2

## Vector Space

$\{\text{vectors in plane}\}$
$\{\text{vectors in space}\}$

prerequisites:
- $V$: a set
- $F$: a field (體)
  - $\mathbb{R}$
  - $\mathbb{C}$
  - $\mathbb{Q}$
    - $\mathbb{Z}$ is not a field

elements in such field is called "__scalar__"
- operators on $V$
  - $+$
  - $\cdot$
    - with elements in $F$

e.g.:
$V$ is a vector space (over a field $F$)

further, $V$, $F$, and the operators need to obey the following 8 (10) axioms.

### Axioms

- operator $+$ is closed over $V$
$\forall  (x,y)\in V\times V ,~ x+y\in V$
- operator $\cdot$ is closed.
$\alpha x\in V,~\forall \alpha\in F~\forall x\in V$
1. commutative
$x+y=y+x~\forall x,y\in V$
2. associative
$\left(x+y\right)+z=x+\left(y+z\right)\text{ whenever }x,y,z\in V$
3. unit element of operator $+$
$\exists \vec{0}\in V\text{ s.t. }\vec{0}+x=x~\forall x\in V$
4. "anti element" of operator $+$
$\exists Y\in V\text{ s.t. }y+Y=\vec{0}~\forall y\in V$
5. unit element of operator $\cdot$
$\exists 1\in F\text{ s.t. }1\cdot x=x~\forall x\in V$
6. distributive of $\cdot$
$a\left(bx\right)=\left(ab\right)x~\forall x\in V,~a,b\in F$
7. distributive
$a\left(x+y\right)=ax+ay~\forall x,y\in V\text{ and }a\in F$
8. distributive
$\left(a+b\right)x=ax+bx~\forall x\in V\text{ and }a\in F$

in axiom 4, we call $Y$ is 加法反元素 of $y$.

### Examples

- $V=\{\text{vectors in plane}\}=\{\left(a,b\right)\vert~a,b\in\mathbb{R}\}$
  $F=\mathbb{R}$
  $+$: genuine operator $+$ in the Euclidean space
  $\cdot$: genuine operator $\cdot$ in the Euclidean space
  - observe that $\vec{0}$ is $\left(0,0\right)$ in the 2-D Euclidean space.
  - $V$ __IS__ a vector space over field $F$.
- $V=\{\text{vectors in plane}\}=\{\left(a,b\right)\vert~a,b\in\mathbb{R}\}$
  $F=\mathbb{C}$
  $+$: genuine operator $+$ in the Euclidean space
  $\cdot$: genuine operator $\cdot$ in the Euclidean space
  - $V$ is __NOT__ a vector space over field $F$.
    - operator $\cdot$ is not closed
- $V=\{\text{vectors in plane}\}=\{\left(a,b\right)\vert~a,b\in\mathbb{R}\}$
  $F=\mathbb{Q}$
  $+$: genuine operator $+$ in the Euclidean space
  $\cdot$: genuine operator $\cdot$ in the Euclidean space
  - $V$ __IS__ a vector space over field $F$.
- $V=\{\text{vectors in complex plane}\}=\{\left(a,b\right)\vert~a,b\in\mathbb{C}\}$
  $F=\mathbb{C}$
  $+$: genuine operator $+$ in the Euclidean space
  $\cdot$: genuine operator $\cdot$ in the Euclidean space
  - $V$ __IS__ a vector space over field $F$.
- $V=\{\text{vectors in plane}\}=\{\left(a,b\right)\vert~a,b\in\mathbb{R}\}$
  $F=\mathbb{R}$
  $+$: $x+y\overset{\Delta}{=}\left(x_1+y_1, x_2-y_2\right)$
  $\cdot$: genuine operator $\cdot$ in the Euclidean space
  - $V$ is __NOT__ a vector space over field $F$.
  - operator $+$ is not commutative over $V$
  
### Axioms (Revisited)

Def: Given a set V, a field F, and two operations on V,
then V is called a vector space over a field F provided that the 10 axioms mentioned earlier are satisfied.
Elements in $V$ are called __vector__ (__向量__), whereas elements in $F$ are called __scalar__ (__純量__)

### More Examples

- $V=\{\text{vectors in n-dimensional Euclidean space}\}=\{\left(a_1,\cdots,a_n\right)\vert~a_i\in\mathbb{R}~\forall i\in\mathbb{Z},~1\leq i\leq n\}$
  $F=\mathbb{R}$
  $+$: genuine operator $+$ in the n-dimensional Euclidean space
  $\cdot$: genuine operator $\cdot$ in the n-dimensional Euclidean space
  - observe that $\vec{0}$ is $\left(0,0\right)$ in the 2-D Euclidean space.
  - $V$ __IS__ a vector space over field $F$.
- Suppose $V=F^n$, with genuine operators, then it's vector space for $F\in\{ \mathbb{R}, \mathbb{C}, \mathbb{Q} \}$, $n\in\mathbb{Z}^+$
- $V=\{\text{n by n square matrices}\}\\
=\Bigg\{
\begin{bmatrix} a_{11} & a_{12} & \dots & a_{1n} \\
a_{21} & a_{22} & \dots & a_{2n} \\
\vdots & \vdots & \ddots \\
a_{n1} & a_{n2} & \dots & a_{nn}
\end{bmatrix} = \left[ a_{ij} \right] \Bigg\vert ~ a_{ij}\in F,~ i,j\in\mathbb{Z},~ 1\leq i,j\leq n
\Bigg\}$
  $F=F$
  $+$: $A=\left[a_{ij}\right], B=\left[b_{ij}\right],\\
  A+B=\left[a_{ij}+b_{ij}\right]$
  $\cdot$: $A=\left[a_{ij}\right], \alpha A=\left[\alpha a_{ij}\right]$
  - such $V$ __IS__ a vector space over $F$
- $V=\{f\vert ~f:S\to F,~ S\text{ is non empty},~ F\text{ is a field}\}$
  $F=F$
  $+$: $f+g=h,~ h(s)=f(s)+g(s)$
  $\cdot$: $\alpha\in F,~ f\in V, \left(\alpha f\right)\left(s\right)=\alpha\left(f\left(s\right)\right)$
  - such $V$ __IS__ a vector space over $F$.
    - thanks to $F$ being a field
  - notation: $V=\mathcal{F}(S,F)$
    - $V$ is set of function of $S$ to $F$
- $V=\mathcal{F}\left(S,F^n\right)$
  $F=F$
  $+$: $f+g=h,~ h(s)=f(s)+g(s)$
  $\cdot$: $\alpha\in F,~ f\in V, \left(\alpha f\right)\left(s\right)=\alpha\left(f\left(s\right)\right)$
  - such $V$ __IS__ a vector space over $F$.
- $V=\mathcal{F}\left(S,{\left(F^{'}\right)}^n\right)$
  $F'$ is another field not necessarily equal to $F$
  $+$: $f+g=h,~ h(s)=f(s)+g(s)$
  $\cdot$: $\alpha\in F,~ f\in V, \left(\alpha f\right)\left(s\right)=\alpha\left(f\left(s\right)\right)$
  - such $V$ is __NOT NECESSARILY__ a vector space over $F$.
- $V=\{\left(a_1,a_2\right)\vert~ a_1,a_2\in\mathbb{R}\}$
  $F=\mathbb{R}$
  $+$: $a+b\overset{\Delta}{=}\left(a_1+b_1,0\right)$
  $\cdot$: $\alpha a\overset{\Delta}{=}\left(\alpha a_1,0\right)$
  - such $V$ is __NOT__ a vector space
    - there's no anti element of operator $+$ for some elements in $V$, e.g. (1,1) have no anti element of operator $+$.
    - there's no unit element of operator $\cdot$.
- $V=\{\left(a_1,a_2\right)\vert~ a_1,a_2\in\mathbb{R}\}$
  $F=\mathbb{R}$
  $+$: genuine operations
  $\cdot$: $\alpha a\overset{\Delta}{=}
  \begin{cases}
    \left(0,0\right),& \text{if } \alpha=0\\
    \left(\alpha a_1,\frac{a_2}{\alpha}\right), & \text{if } \alpha\neq0
\end{cases}$
  - such $V$ is __NOT__ a vector space
    - axiom $8$ does __NOT__ hold
  
### Theorems

#### Theorem 1

*elimination works*

Supp. $V$ is a V.S. over a field $F$
$x,y,z\in V$, then we have:
$$x+z=y+z\iff x=y$$

__proof__: ($\rightarrow$)

$\begin{align}x
=& x+\vec{0} && \text{axiom 3}\\
=& x+(z+v) && \text{axiom 4, anti }+\text{ of }z\\
=&(x+z)+v && \text{associative of operator }+\\
=&(y+z)+v && \text{prerequisite}\\
=&y+(z+v) && \text{associative of operator }+\\
=&y+\vec{0} \\
=&y && \text{Q.E.D.}\\
\end{align}$

##### Corollary 1

$\vec{0}$ is *unique* in $V$

##### Corollary 2

$\exists !~x^{'}\in V\text{ s.t. }x+x^{'}=\vec{0}$

__remark__:
by Corollary 2, $\forall x\in V$ we could denote its anti element of operator $+$ by $-x$

__proof__: (corollary 1)

supp. there exists $\vec{0}$ and $\vec{0}^{'}$ s.t.
$\left( x+\vec{0}=x\land x+\vec{0}^{'}=x\right), ~\forall x\in V$
by theorem 1, elimination works, we have:
$\vec{0}=\vec{0}^{'}\\ \text{Q.E.D.}$

__proof__: (corollary 2)

use theorem 1 and commutative law of operator $+$

#### Theorem 2

Let $V$ be a vector space over a set $F$,
then we have the followings:
1. $0x=\vec{0}, ~\forall x\in V$
2. $-\left(ax\right)=\left(-a\right)x=a\left(-x\right), ~\forall x\in V ~\forall a\in F$
3. $a\vec{0}=\vec{0}, ~\forall a\in F$

__proof__: (theorem 2.1)
$\begin{align}0x=&(0+0)x\\
=&(0x)+(0x) &&\text{axiom 8, distributive}\\
=&0x &&\text{prerequisite}\\
=&\vec{0}+0x &&\text{zero vector}
\end{align}$
apply threorem 1.1 (elimination works), $\text{Q.E.D.}$

__proof__: (theorem 2.2)
first we want $-(ax)=(-a)x$
equivalent condition is $(ax)+\left(\left(-a\right)x\right)=\vec{0}$
since additive anti element exists and unique, we know $((-a)\vec{x}) = -(a\vec{x})$.
$$\begin{align}ax+(-a)x=&\left(a+\left(-a\right)\right)x && \text{axiom 8, distro}\\
=& 0x &&F\text{ is a field}\\
=& \vec{0} &&\text{theorem 2.1}\\
& &&\text{Q.E.D}
\end{align}$$
secondly, we want $(-a)x=a(-x)$
from first part, we have
$$-x=(-1)x$$
now:
$$\begin{align}a(-x)=&a\left(\left(-1\right)x\right)&& \text{first part}\\
=&\left(a\left(-1\right)\right)x &&\text{axiom 6, associative}\\
=&\left(-a\right)x &&F\text{ is a field}\\
& && \text{Q.E.D.}
\end{align}$$

__proof__: (theorem 2.3)
apply axiom 7

# My Questions:

- prove or find counter example:
  the unit element of operator $\cdot$, as defined in axiom $5$, must also be the unit element of multiplication in field $F$.
  - the proposition **doesn't** hold:
    - consider $V=\{0\}$ and $F=\mathbb{R}$
      then any $r\in\mathbb{R}$ is the unit element.
    - $V=\mathbb{R}$, $F=\mathbb{R}$, $+$ is genuine $+$, and $\cdot$ is $\mathbb{R}$ multiplication divided by $a\in\mathbb{R}$, where $a\neq 0\in\mathbb{R}$