136 lecture slides 3 post

Lecture 3

Content: Linear Independence

Math 136 Dan Wolczuk Faculty of Mathematics University of Waterloo

Last class we saw that we could simplify a vector equation fora set spanned by a set of vectors by removing vectors whichcould be written as a linear combination of the other vectors.We now generalize our method.

Theorem 1.1.2

Let v 1, . . . ,v k Rn. There exists a vector v i such thatv i Span{v 1, . . . ,v i1,v i+1, . . . ,v k} if and only if

Span{v 1, . . . ,v k} = Span{v 1, . . . ,v i1,v i+1, . . . ,v k}

Proof:


Consider a set of vectors {v 1, . . . ,v k} in Rn. We want toform an equation that will tell us if one of the vectors in theset is a linear combination of the others.

If there exists such a vector, then that means there existsc1, . . . , ck R such that

civ i = c1v 1 + + ci1v i1 + ci+1v i+1 + + ckv k

where ci 6= 0.We can rearrange this as

0 = c1

v 1 + + ckv k


On the other hand, if the equation

0 = c1

v 1 + + ckv khas a solution where one of the coefficients, say cj, isnon-zero, then we can rewrite it as

v j = c1cj

v 1 cj1cj

v j1 cj+1cj

v j+1 ckcj

v k


We have proven that a spanning set can be simplified if and

only if the equation0 = c1

v 1 + + ckv k has a solutionwhere at least one of the coefficients is non-zero.

Definition

A set of vectors {v 1, . . . ,v k} in Rn is said to be linearlydependent if there exist coefficients c1, . . . , ck not all zerosuch that

0 = c1v 1 + + ckv k

A set of vectors {v 1, . . . ,v k} is said to be linearlyindependent if the only solution to

0 = c1

v 1 + + ckv kis c1 = c2 = = ck = 0 (called the trivial solution).


Theorem 1.1.3

A set of vectors is linearly dependent if and only if one of thevectors can be written as a linear combination of the othervectors.


Example

Determine which sets are linearly independent and which arelinearly dependent.

(a)

{[12

],

[11

]}Solution:


Example


(b)

13

1

, 21

1

,262

Solution:


Example


(c)

131

,01

2

,00

0

Solution:


We can learn things from the examples. From (c), we can seethat we have the following theorem.

Theorem 1.1.4

If a set of vectors {v 1, . . . ,v k} contains the zero vector thenit is linearly dependent.

Proof:


Also, from (a), we get the following result

Theorem

If v 1,v 2 Rn are two vectors such that neither v 1 nor v 2is a scalar multiple of the other, then {v 1,v 2} is linearlyindependent.


Remark

Observe that for determining whether a set {v 1, . . . ,v k} inRn is linearly dependent or linearly independent requiresdetermining solutions of the vector equation

c1v 1 + + ckv k = 0 . However, this equation actually

represents n equations (one for each entry of the vectors) in kunknowns c1, . . . , ck. In the next chapter, we will look at howto efficiently solve such systems of equations.


What we have derived above is that the simplest spanning setfor a given set is one that is linearly independent. Hence, wemake the following definition.

Definition

If a subset S of Rn can be written as a span of vectorsv 1, . . . ,v k where {v 1, . . . ,v k} is linearly independent, thenwe say that {v 1, . . . ,v k} is a basis for S.We define a basis for the set {

0 } to be the empty set.


136 lecture slides 3 post

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