1.6 Linear Independence(1)

Linearly Independent

A set of vectors $\begin{Bmatrix}\bold{v_1}, \cdots, \bold{v_p}\end{Bmatrix}$ in $\mathbb{R}^n$ is said to be linearly independent if the vector equation $x_1\bold{v_1} + \cdots + x_p\bold{v_p} = 0$ has only trivial solutions

 

 

Linearly Dependent

The set $\begin{Bmatrix}\bold{v_1}, \cdots, \bold{v_p}\end{Bmatrix}$ in $\mathbb{R}^n$ is said to be linearly dependent if there exist weights $c_1, \cdots, c_p$, not all zero, such that $c_1\bold{v_1} + \cdots + c_p\bold{v_p} = 0$

not all zero는 all zero가 아닌 solution 이 하나라도 존재한다는 뜻이다

여기서 유의할 것은, not all zero는 모두 all zero가 아니라는 뜻이 아니라, zero가 아닌 weight가 하나라도 존재하면 된다는 뜻이기 때문에

$c_1\bold{v_1} + \cdots + c_p\bold{v_p} = 0$ 에서 임의의 벡터 $\bold{v_j}$ 에 대해 $\bold{v_j} = a_1\bold{v_1} + \cdots + a_p\bold{v_p}$ 꼴로 항상 표현이 가능한 것은 아니다

(왜냐하면, $\bold{v_j}$의 weight가 0이라면, $\bold{v_j}$를 이항한 후 나머지 vector 들을 0으로 나눌 수가 없기 때문이다)

 

 

Example1.

$\bold{v_1} = \begin{bmatrix} 1\\2\\3 \end{bmatrix} \bold{v_2} = \begin{bmatrix} 4\\5\\6 \end{bmatrix} \bold{v_3} = \begin{bmatrix} 2\\1\\0 \end{bmatrix}$

Solution

$\begin{bmatrix} 1 & 4 & 2 & 0\\ 2 & 5 & 1 & 0\\ 3 & 6 & 0 & 0\\ \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & -2 & 0\\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 0\\ \end{bmatrix}$ → linearly dependent!

(pivot 포지션이 두개이므로 결국 하나의 free variable이 생긴다. 이는 non trivial 이므로 linearly independent 하다)

so,

$\begin{bmatrix} 1 & 0 & -2 & 0\\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 0\\ \end{bmatrix} \sim \begin{aligned} x_1 -2x_3 = 0\\ x_2 + x_3 = 0\\ 0 =0 \end{aligned} \\if,\ x_3=1,\ then\ x_1=2,\ x_2=-1\\\therefore 2\bold{v_1}-\bold{v_2}+\bold{v_3} = 0$

so, the vector set has lienar dependency in the form of $2\bold{v_1}-\bold{v_2}+\bold{v_3} = 0$

 

 

Linear Independence of Matrix Columns

$A = \begin{bmatrix}\bold{a_1} \cdots \bold{a_n}\end{bmatrix}$

$A\bold{x} = 0\\ x_1\bold{a_1} + x_2\bold{a_2} + \cdots + x_n\bold{a_n}$

The columns of a matrix $A$ are linearly independent if and only if the equations $A\bold{x} = 0$ has only the trivial solution

Example2. Determine if the columns of the following matrix are linearly independent

$A = \begin{bmatrix} 0 & 1 & 4\\ 1 & 2 & -1\\ 5 & 8 & 0\\ \end{bmatrix}$

solution

$A\bold{x} = 0 \sim \begin{bmatrix} 0 & 1 & 4 & 0\\ 1 & 2 & -1 & 0\\ 5 & 8 & 0 & 0\\ \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & 1 & 0\\ \end{bmatrix}$

so, $x_1 = x_2 = x_3 = 0$ → trivial!

so, linearly independent!

 

 

Sets of One Vector

If a set contains only one vector, $\bold{v}$, then the set is linearly independent, only when $v \not = 0$

 

 

Sets of Two Vectors

• A set $\begin{Bmatrix}\bold{v_1},\bold{v_2}\end{Bmatrix}$ is linearly dependent if at least one of the vectors is a multiple of the other
• The set is linearly independent, if and only if, neither of the vectors is multiple of the other