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

# Lecture 5

## 1.5 Linear Independence and Linear Dependence

### Goals

- Let $V$ be a vector space, $W\subset V$, $W$ is a subspace of $V$
Find the smallest set $S$ s.t. $\text{Span}\left(S\right)=W$
- $\text{L.I.}$ and $\text{L.D.}$
- Properties
- How to check
- Gauss Elimination
- What is __*dimension*__, and what is __*basis*__
- imagine $\text{dimension}\overset{\Delta}{=}\lVert\text{basis}\rVert$, something like that.

### Definition

Suppose $S\subseteq V$ where $V$ is vector space

1. Linear *Dependent*
A $S$ is called $\text{L.D.}$ if there exists __some finite distinct__ vectors $u_i\in S$ and __not all zero__ scalar coefficients $a_i\in F$ s.t.
$\sum{\left(a_iu_i\right)}=0$
2. Linear *Independent*
If $S$ is not $\text{L.D.}$ then it is called $\text{L.I.}$
That is, if for any set of __distinct__ vectors $\{u_i\} \subset S$, we have $\sum\limits_{i=0}^{n}{\left(a_iu_i\right)}=0\implies a_i=0~ \forall i$, then $S$ is $\text{L.I.}$.

#### Example

- $V=\mathbb{R}^2,~ S=\big\{(0,0)\big\}$
$S$ is linear dependent.
- $\vec{0}_V\in S\implies S\text{ is Linear Dependent}$
- $V=\mathbb{R}^2,~ S=\big\{(1,0),(0,1)\big\}$
$S$ is linear independent.
- $V=P_2\left(\mathbb{R}\right)$, $S=\big\{1,x\big\}$
$S$ is linear independent.

### Theorem 1.6

Let $S_1\subseteq S_2\subset V,~ V\text{ vector space}$
then we have the following:
1. $S_2$ linear independent $\implies$ $S_1$ is linear independent
2. $S_1$ linear dependent $\implies$ $S_2$ is linear dependent

### Theorem 1.7

$S\subset V,~ V\text{ vector space},~ S\text{ L.I.},~ v\in V\setminus S$
then:
$$\begin{align}
S\cup\{v\}&\text{ is L.D.} && \iff v\in\text{Span}\left(S\right)\\
S\cup\{v\}&\text{ is L.I.} && \iff v\notin\text{Span}\left(S\right)
\end{align}$$

#### Example

- $V=\mathbb{R}^2,~ S=\{(1,0)\}$
$\text{Span}(S)=\text{the x axis}$
supp. $v=(1,1)$
then $S^{'}\overset{\Delta}{=}S\cup \{v\}$ is linear independent.
supp. $u=(2,0)$
then $S^{"}\overset{\Delta}{=}S\cup \{u\}$ is linear dependent.

__proof__: (theorem 1.6)
$\rightarrow$:
objective: $v\in\text{Span}(S)$
since $S\cup\{v\}$ is linear dependent,
there exists __distinct__ $u_i\in\left(S\cup\{v\}\right),~ i\in\{1,2,\cdots,n\},~ a_i\text{ not all zero}\\
\text{s.t. }\sum\limits_{i=1}^{n}{a_iu_i}=0$
claim: $\exists i\in\{1,2,\cdots,n\}\text{ s.t. }u_i=v$
    proof: by contradiction, or $S$ shall be linear dependet at the beginning.
WLOG let $u_n=v$; furthermore, $a_n\neq 0$, since similar reasons in the above claim.
We ignore the trivial case $v = \vec{0}_V$, since $\vec{0}_V$ is always in span.
Hence, we have $v=u_n=\frac{-1}{a_n}\left(a_1u_1+\cdots+a_{n-1}u_{n-1}\right)$
notice they are __distinct__, thus $\text{RHS}\in\text{Span}(S)$, $\text{Q.E.D.}$
$\leftarrow$
Trivial.
Notice that span implies there's some __finite__ subset of $V$ sums to $v$, from which we could ensure distinct elements, and by moving terms we know the coefficient corresponding to $v$ is never $0$, since $v\in\left(V\setminus S\right)$; by definition, since there's at least one coefficient that could be non-zero, the proposition hence holds.