Skip to main content

Section 1.8 Linear Independence

We have seen how vector equations relate to a system of equations and how to frame different questions in terms of whether it is possible to find a linear combination from a set S that equals a target vector. This was the same as asking if our target vector was in the span of S. When we looked at vector equations, we also looked at whether there was a unique linear combination or whether there are many ways to write a target vector as a linear combination.
In this section we will introduce the idea of linear independence and how that relates to the uniqueness of these linear combinations. Let’s consider the homogeneous vector equation:
c1vβ†’1+c2vβ†’2+…+cnvβ†’n=0β†’
Notice that this always has a solution. What is it?
Is it possible to have another solution to this equation? Let’s generate a couple of examples.

Activity 1.8.1.

(a)

Give a set of two vectors from R3, v→1 and v→2, such that c1v→1+c2v→2=0→ has more than one solution. Justify your answer.

(b)

Give a set of two vectors from R3, v→1 and v→2, such that c1v→1+c2v→2=0→ has only one solution. How do you know there is only one solution?

Definition 1.8.1.

A set of vectors S is linearly independent if the only linear combination of elements of S that equals the zero vector is the trivial linear combination. In other words, S being a linear independent set implies that if c1v1β†’+c2v2β†’+...+ckvkβ†’=0β†’ where viβ†’βˆˆS, then all ci=0.
A set of vectors S is linearly dependent if the set is not linearly independent. More specifically, there exists a solution to c1v1β†’+c2v2β†’+...+ckvkβ†’=0β†’ where viβ†’βˆˆS and at least one of the cjβ‰ 0.

Activity 1.8.2.

(c)

Choose a vector vβ†’ so that the set {[230],[βˆ’1βˆ’12],vβ†’} is linearly independent, where vβ†’βˆˆR3.

(d)

Is your choice of vβ†’ in Span({[230],[βˆ’1βˆ’12]})? Show why or why not.