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2.C Linear independence

The next goal is to measure the size of a subspace by finding its dimension. The dimension of a subspace is the minimal number of vectors that span it. Above we found some finite spanning sets, but you might ask: How do I know whether or not I’ve found a minimal spanning set? To be able to answer that, we will first introduce the terminology of linear independence.

Definition 2.19:

Let $\vec{v_{1}},\vec{v_{2}},\cdots,\vec{v_{n}}\in V$ be vectors in a vector space. If one of them, say $\vec{v_{i}}$ , is a linear combination (see Definition 2.3) of the others, then we say the whole sequence is linearly dependent. Otherwise, we call the sequence linearly independent.

The following theorem is the standard test for linear independence.

Theorem 2.20.

Let $V$ be a vector space over a field $F$ . A sequence of vectors $\vec{v_{1}},\cdots,\vec{v_{r}}$ is linearly independent if and only if, for any scalars $\alpha_{1},\cdots,\alpha_{r}\in F$ , we have that

\alpha_{1}\vec{v_{1}}+\alpha_{2}\vec{v_{2}}+\cdots+\alpha_{r}\vec{v_{r}}=\vec{0}

implies that $\alpha_{1}=\alpha_{2}=\cdots=\alpha_{r}=0$ .

Some sources define that phrase “linearly independent” by the condition in Theorem 2.20, rather than the definition we gave. Since they are logically equivalent, if makes no difference which one you use, and throughout this module we will find it convenient check linear independence using Theorem 2.20 without always stating the theorem number.

Example 2.21.

i.

Is the sequence of vectors $(1,0,1),(2,1,0),(0,-1,1)\in{\mathbb{R}}^{3}$ linearly independent?
Solution: We will use Theorem 2.20. So assume $a,b,c\in{\mathbb{R}}$ are such that

$a(1,0,1)+b(2,1,0)+c(0,-1,1)=(0,0,0).$

Then we obtain the following three equations for the scalars $a, b, c$ :
1. $a+2b=0$
2. $b-c=0$
3. $a+c=0$
Solving these equations immediately proves that $a=b=c=0$ . In other words, we have proved that

$a\vec{v_{1}}+b\vec{v_{2}}+c\vec{v_{3}}=\vec{0}$

implies $a=b=c=0$ . Therefore, by Theorem 2.20, this sequence of vectors is linearly independent.
ii.

Prove that the sequence $(1,2,-1,1),(1,2,1,3),(0,0,-1,-1)\in{\mathbb{R}}^{4}$ is linearly dependent.
Solution: Assume $a,b,c\in{\mathbb{R}}$ are scalars such that

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

Then we obtain four equations:
1. $a+b=0$
2. $2a+2b=0$
3. $-a+b-c=0$
4. $a+3b-c=0$
In fact, the solution set to this system of equations is

$S=\{(a,-a,-2a)\;|\;a\in{\mathbb{R}}\}=\operatorname{span}_{{\mathbb{R}}}\{(1,-% 1,-2)\}.$

In particular, this proves $(1,2,-1,1)-(1,2,1,3)-2(0,0,-1,-1)=(0,0,0,0)$ . Therefore we can write one of the vectors as a linear combination of the others, so these three vectors are linearly dependent.

Exercise 2.22:

Prove that the following sequences are linearly independent.

i.

$(2,0,-1),(3,1,1),(1,0,5)$ in the vector space ${\mathbb{R}}^{3}$ .
ii.

The vectors $\vec{e_{1}}=(1,0,\cdots,0)$ , $\vec{e_{2}}=(0,1,0,\cdots,0)$ , $\cdots$ , $\vec{e_{n}}=(0,\cdots,0,1)$ in the vector space ${\mathbb{R}}^{n}$ . This is called the standard basis of ${\mathbb{R}}^{n}$ .
iii.

The polynomials $1,x,x^{2},\cdots,x^{n}$ in the vector space of polynomials $\mathcal{P}_{n}({\mathbb{R}})$ .

[End of Exercise]

Exercise 2.23:

i.

Give an example of a pair of vectors in ${\mathbb{R}}^{2}$ which is linearly independent, and all 4 of their coordinates are rational but not integers.
ii.

Give an example of a sequence of three vectors in ${\mathbb{R}}^{3}$ which is linearly independent, and such that none of their coordinates are rational numbers.
iii.

Can you find a sequence of 100 vectors in ${\mathbb{R}}^{2}$ which is linearly independent?

[End of Exercise]