19 - Linear Independence

Linearly Independent

Let $V$ be a vector space over $\mathbf{F}$
$v_1,v_2,\cdots ,v_m\in V$ are linearly independent if the only solution to

$${\alpha }_1{v}_1+\cdots +{\alpha }_m{v}_m=0_V\text{ where }{\alpha }_1,\cdots ,{\alpha }_m\in \mathbf{F}$$

is

$${\alpha }_1={\alpha }_2=\cdots ={\alpha }_m=0$$

Otherwise the vectors are linearly dependent

Subsets of Vector Spaces being Linearly Independent

Let $S\subseteq V$ then
$S$ is linearly independent of every finite subset of $S$ is linearly independent

Comparing Coefficients (of Linearly Independent Vectors)

Let $S=\{v_1,\cdots ,v_m\}$ where $S\subseteq V$, where $V$ is a vector space
Suppose

$${\alpha }_1{v}_1+\cdots +{\alpha }_m{v}_m={\beta }_1{v}_1+\cdots +{\beta }_m{v}_m$$

for some ${\alpha }_1,{\beta }_1,\cdots ,{\alpha }_m,{\beta }_m\in \mathbf{F}$

Then ${\alpha }_i={\beta }_i$ for $1\le i\le m$

Proof

Rearrange the equation to

$$({\alpha }_1-{\beta }_1)v_1+({\alpha }_2-{\beta }_2)v_2+\cdots +({\alpha }_m-{\beta }_m)v_m=0_V$$

As the vectors $v_1,\cdots ,v_m$ are linearly independent then

$$\begin{array}{rl}{\alpha }_1-{\beta }_1& =0\\ {\alpha }_2-{\beta }_2& =0\\ ⋮& \\ {\alpha }_m-{\beta }_m& =0\end{array}$$

Hence

$${\alpha }_1={\beta }_1,{\alpha }_2={\beta }_2,\cdots ,{\alpha }_m={\beta }_m$$

Useful Examples of Linearly Independent Sets

1. $S=\{1,i\}$ with $V=\mathbb{C}$, the set of complex numbers
2. $S=\{1,x,x^2,\cdots \}$ with $V=\mathbb{R}[x]$, the vector space of polynomials with real coefficients

Extending a linearly independent set lemma

Let $v_1,v_2,\cdots ,v_m$ be linearly independent elements of a vector space $V$
Let $v_{m+1}\in V$ then

$$v_1,v_2,\cdots ,v_m,v_{m+1}\text{ are linearly independent }⟺v_{m+1}\notin ⟨v_1,\cdots ,v_m⟩$$

Proof

1. Proving $⟸$
   Suppose $v_{m+1}\notin \{v_1,\cdots ,v_m\}$
   Take ${\alpha }_1,\cdots ,{\alpha }_{m+1}\in \mathbf{F}$ such that

$${\alpha }_1{v}_1+\cdots +{\alpha }_m{v}_m+{\alpha }_{m+1}{v}_{m+1}=0_V$$

Then for ${\alpha }_{m+1}\ne 0$

$$v_{m+1}=-\frac{1}{{\alpha }_{m+1}}({\alpha }_1{v}_1+\cdots +{\alpha }_m{v}_m)$$

So $v_{m+1}\in V$ however this is a contradiction so ${\alpha }_{m+1}=0$
Then as $v_1,v_2,\cdots ,v_m$ are linearly independent then the solution to

$${\alpha }_1{v}_1+\cdots +{\alpha }_m{v}_m=0_V$$

Is ${\alpha }_1={\alpha }_2=\cdots ={\alpha }_m=0$
So $v_1,v_2,\cdots ,v_m,v_{m+1}$ are linearly independent

2. Proving $⟹$
   Suppose $v_1,v_2,\cdots ,v_m,v_{m+1}$ are linearly independent then
   Suppose $v_m\in ⟨v_1,\cdots ,v_m⟩$ then there exists ${\alpha }_1,\cdots ,{\alpha }_m$ such that

$${\alpha }_1{v}_1+{\alpha }_2{v}_2+\cdots +{\alpha }_m{v}_m=v_{m+1}$$

Rearranging this to get

$${\alpha }_1{v}_1+{\alpha }_2{v}_2+\cdots +{\alpha }_m{v}_m-v_{m+1}=0$$

However as $v_1,v_2,\cdots ,v_{m+1}$ are linearly independent then there exists no solution to the above equation
Hence $v_m\notin ⟨v_1,\cdots ,v_m⟩$

Test for Independence corollary

Let $A$ be a $m\times n$ matrix

$$\mathrm{RRE}(A)\text{ has a }0\text{ row }⟺\text{ rows of }A\text{ are dependent}$$

Proof

We know that $\mathrm{RRE}(A)=BA$, where $B$ is a product of EROs so $B^{-1}$ exists
Suppose the $i$th row of $BA$ is $0$
Then

$$0=\sum _{s=1}^m{b}_{is}\cdot (s\text{th, row of A})$$

Where $b_{is}$ are the entries along the $i$th row of $B$ which cannot all be zero (as $B$ is invertible)
Hence rows of $A$ are linearly dependent