Linearly Dependent, Linearly Independent.

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Definition: Linearly Dependent

Let $V$ be a vector space over a field $F$. A subset $S \subseteq V$ is called linearly dependent if there exist a finite number of distinct vectors $u_1, u_2, …, u_n \in S$ and $a_1, a_2,…, a_n \in F$, not all zero such that

$$a_1v_1 + a_2v_2 + … + a_nv_n = 0$$

Example: $\R^2$

Prove that $S= \set{(1,0), (0,1), (2,-3)}$ is linearly dependent.

Proof 
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We just need to find one not all zero coefficient solution.

$-2(1,0) + 3(0,1) +1(2,-3) = (0,0)$

$\therefore S$ is linearly dependent.

Definition: Linearly Independent

Let $V$ be a vector space over $F$, a subset of $S\subseteq V$.

$S$ is called linearly independent if it is NOT linearly dependent.

Equivalently, $S$ is called linearly independent if for every finite subset of vector $v_1, v_2, …, v_n \in S$ the only solution of the equation (in $a_i$) $$a_1v_1 + a_2v_2 + … + a_nv_n = 0$$ is $$a_1=a_2=…=a_n=0$$

Note

1. $V/F, \set{\emptyset}$ is linearly independent.

Proof 
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Assume that $\emptyset$ is linear dependent.

$v_1, v_2, …, v_n \in \emptyset$ and $a_1,a_2, …, a_n \in F$ such that at least one of the $a_i$s is non-zero and $a_1v_1 + a_2v_2 + … + a_nv_n = 0$.

However, since there is no $v \in \emptyset$, there doesn’t exist a $a$ that is non-zero.

Therefore, the empty set is linearly independent.

2. $V/F, \set{v}\ (v \neq 0)$ is linearly independent.

3. if $0 \in S, S$ is linearly dependent. (In particular, $\set{0}$ is linearly dependent.)

Proof 
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Assume that it is linearly dependent:

$\exists a \in F, a\neq 0.$ s.t. $$av=0$$

$\because a\in F$ and $a\neq 0, \therefore a^{-1} \in F$.

$$a^{-1}(av)= a^{-1}0 \implies v = 0$$

This is a contradiction to $v\neq 0$.

Example: $\R^2$

$S=\set{(1,0),(0,1)}$ is linearly independent.

Proof 
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$a(1,0) + b(0,1) = (0,0)$

$\implies (a,b) = (0,0)$

$a= 0, b=0$

$\therefore S$ is linearly independent.

Example: $\R^2$

$S=\set{(1,1),(3,-3)}$ is linearly independent.

Proof 
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$a(1,1) + b(2,-3) = (0,0)$

$(a+2b,a-3b) = (0,0)$

$$ \begin{align*} a+2b=0 \br a-3b=0 \end{align*} $$

$\implies (a,b) = (0,0)$

$a= 0, b=0$

$T$ is linearly independent.

Theorem 1.6

$V/F, S_1, S_2 \subseteq V$.

If $S_1$ is linearly dependent and $S_2\supseteq S_1$, then $S_2$ is also linearly dependent.

If $S_1$ is linearly independent and $S_2 \subseteq S_1$, then $S_2$ is also linearly independent.

Theorem 1.7

$V/F, S \subseteq V, v \in V,$ $S$ is linearly independent.

$S \cup \set{v}$ is linearly dependent $\iff v \in \spa{S}$.

Proof 
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$\impliedby$

$v\in \spa{S}$.

$\therefore \exists v_1, v_2, …, v_n \in S$ and $a_1, a_2, …, a_n \in F$ s.t. $$v=a_1v_1 + a_2v_2 + … + a_nv_n$$

$a_1v_1+a_2v_2 + … a_nv_n + (-1)v = 0$

$S\cup \set{v}$ is linearly dependent.

$\implies$

$S \cup \set{v}$ is linearly dependent.

$\therefore$ for $v_1, v_2, …, v_n, v \in S\cup \set{v}, \exists a_1, a_2, …, a_n \in F$ s.t.

$$a_1v_1+a_2v_2 + … + a_nv_n + av = 0$$ at least one of $a_1, a_2, … ,a_n, a$ is non-zero.

Claim $a \neq 0$.

If $a = 0$, then we have $a_1v_1 + a_2v_2 + … + a_nv_n = 0$.

Since $S$ is linearly independent $\set{v_1, v_2, …, v_n} \subseteq S$ is also linearly independent.

From theorem we get $a_1=a_2=a_3= … = a_n = 0= a$

This is a contradiction to the condition $a_1, a_2, …, a_n, a$ not all zero

$\therefore a \neq 0$

$a_1v_1+a_2v_2 + … + a_nv_n + av = 0$ $av = -a_1v_1 -a_2v_2 - … -a_nv_n$ $v = -\frac{a_1}{a}v_1 -\frac{a_2}{a}v_2 - … -\frac{a_n}{a}a_nv_n$

$\therefore v \in \spa{S}$.

Example

$V=\R, F=\R$

$S=\set{1}$

$x\in V = \R$

$\forall r \in \R, S=\set{r}$ is a generating set.

Example

$V=\C, F=\C$

$S = {z} \subset V = \C, z \neq z$

$w \in V = \C$

$w = \frac w z \cdot z$

Example

$V=\C, F=\R$

$s = {1, z}, z\neq 0$

$w\in V = \C$

NOTTRUE! $w = \frac w z \cdot z$