5 Linear independence - Auburn University

5

5.1

Linear independence

Introduction

Let x1 , x2 , and x3 be three vectors in Rn . There is always one way to get a

linear combination of these vectors to equal zero, namely,

0x1 + 0x2 + 0x3 = 0.

But suppose that there¡¯s another way. For instance,

2x1 + 5x2 + 4x3 = 0.

In this case, we say that the vectors are ¡°linearly dependent.¡± The reason for

the terminology is that we can solve for one of the vectors in terms of the others,

say,

5

x1 = ? x2 ? 2x3 .

2

So x1 ¡°depends¡± on the other two vectors.

5.2

Definition and examples

Linear dependence/independence.

We say that vectors x1 , x2 , . . . , xs in Rn are linearly dependent if there are scalars ¦Á1 , ¦Á2 , . . . , ¦Ás not all zero such that

¦Á1 x1 + ¦Á2 x2 + ¡¤ ¡¤ ¡¤ + ¦Ás xs = 0.

We say that the vectors are linearly independent if they are

not linearly dependent, that is, if

¦Á1 x1 + ¦Á2 x2 + ¡¤ ¡¤ ¡¤ + ¦Ás xs = 0

implies

¦Ái = 0 for all i.

In other words, the vectors x1 , x2 , . . . , xs are linearly dependent if there is a

way to get a linear combination of them to equal 0 without making all of the

scalar factors 0, and, on the other hand, they are linearly independent if the

only way to get a linear combination of them to equal 0 is by making all of the

scalar factors 0.

We say that the set {x1 , x2 , . . . , x2 } is linearly dependent if the vectors x1 , x2 ,

. . . , xs are linearly dependent (and similarly for linearly independent).

1

5 LINEAR INDEPENDENCE

2

5.2.1 Example Determine whether the following vectors in R2 are linearly

dependent or linearly independent:

 

 

 

?1

5

1

x1 =

, x2 =

, x3 =

.

3

6

4

Solution

Suppose we have a linear combination of the vectors equal to 0:

¦Á1



¦Á1 x1 + ¦Á2 x2 + ¦Á3 x3 = 0

   

 



1

5

?1

0

=

+ ¦Á3

+ ¦Á2

4

6

3

0

  



0

?¦Á1 + 5¦Á2 + ¦Á3

=

.

0

3¦Á1 + 6¦Á2 + 4¦Á3

Equating components we get a system with augmented matrix









?1 5 1 0

?1 5 1 0 3

¡«

.

3 6 4 0

0

21 7 0

Since ¦Á3 is free, we can choose it to be anything. In particular, we can choose

it to be nonzero. Therefore, the vectors are linearly dependent.

5.2.2 Example Determine whether the following vectors in R3 are linearly

dependent or linearly independent:

? ?

? ?

? ?

1

?2

1

x1 = ?2? , x2 = ? 1 ? , x3 = ?0? .

3

0

1

Solution

Suppose we have a linear combination of the vectors equal to 0:

¦Á1 x1 + ¦Á2 x2 + ¦Á3 x3 = 0

? ?

? ?

? ? ? ?

1

?2

1

0

¦Á1 ?2? + ¦Á2 ? 1 ? + ¦Á3 ?0? = ?0?

3

0

1

0

? ? ?

?

¦Á1 ? 2¦Á2 + ¦Á3

0

? 2¦Á1 + ¦Á2 ? = ?0? .

3¦Á1 + ¦Á3

0

Equating components we get a system with augmented matrix

?

1

? 2

3

?2 1

1 0

0 1

?

0 ?2

0 ?

0

?3

¡«

¡«

?

1 ?2 1 0

? 0 5 ?2 0 ??6

5

0 6 ?2 0

?

?

1 ?2

1

0

? 0

5

?2 0 ? .

0

0

2

0

?

5 LINEAR INDEPENDENCE

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Since there is a pivot in every column except for the augmented column, there

is a unique solution, namely, ¦Á1 = 0, ¦Á2 = 0, and ¦Á3 = 0.

The computation shows that the only way to get a linear combination of the

vectors to equal 0 is by making all of the scalar factors 0. Therefore, the vectors

are linearly independent.

The solutions to these last two examples show that the question of whether

some given vectors are linearly independent can be answered just by looking

at a row-reduced form of the matrix obtained by writing the vectors side by

side. The following theorem uses a new term: A matrix has full rank if a

row-reduced form of the matrix has a pivot in every column.

Theorem. Let x1 , x2 , . . . , xs be vectors in Rn and let A be the

matrix formed by writing these vectors side by side:





A = x1 x2 ¡¤ ¡¤ ¡¤ xs .

The vectors x1 , x2 , . . . , xs are linearly independent if and only

if A has full rank.

5.2.3 Example Use the last theorem to determine whether the vectors

[1, 3, ?1, 0]T , [4, 9, ?2, 1]T , and [2, 3, 0, 1]T are linearly independent.

Solution We have

?

?

1

4 2 ?3

? 3

9 3 ?

?

?

? ?1 ?2 0 ?

0

1 1

1

¡«

¡«

?

1 4

2

? 0 ?3 ?3 ?2

1

?

?

? 0 2

2 ?3

0 1

1

3

?

?

1

4

2

? 0 ?3 ?3 ?

?.

?

? 0

0

0 ?

0

0

0

?

Since the matrix does not have full rank, the vectors are not linearly independent.

5.3

Facts about linear dependence/independence

The next theorem says that if a vector is written as a linear combination of

linearly independent vectors, then the scaling factors are uniquely determined.

5 LINEAR INDEPENDENCE

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Theorem. Let x1 , x2 , . . . , xs be linearly independent vectors in

Rn . Let x ¡Ê Rn and suppose that

x = ¦Á1 x1 + ¦Á2 x2 + ¡¤ ¡¤ ¡¤ + ¦Ás xs

and also

x = ¦Â1 x1 + ¦Â2 x2 + ¡¤ ¡¤ ¡¤ + ¦Âs xs ,

with ¦Ái , ¦Âi ¡Ê R. Then ¦Ái = ¦Âi for each i.

Proof. We have

¦Á1 x1 + ¦Á2 x2 + ¡¤ ¡¤ ¡¤ + ¦Ás xs = x = ¦Â1 x1 + ¦Â2 x2 + ¡¤ ¡¤ ¡¤ + ¦Âs xs ,

so moving all terms to the left-hand side and collecting like terms, we get

(¦Á1 ? ¦Â1 )x1 + (¦Á2 ? ¦Â2 )x2 + ¡¤ ¡¤ ¡¤ + (¦Ás ? ¦Âs )xs = 0.

Since the vectors are linearly independent, the only way to get a linear combination of them to equal 0 is by making all of the scalar factors 0. We conclude

that ¦Ái ? ¦Âi = 0 for all i, that is, ¦Ái = ¦Âi for all i.

Theorem. The vectors x1 , x2 , . . . , xs (s ¡Ý 2) in Rn are linearly

dependent if and only if one of the vectors can be written as a

linear combination of the others.

Due to the ¡°if and only if¡± phrase, the theorem is saying two things: (1) if the

vectors are linearly dependent, then one can be written as a linear combination

of the others, and (2) if one vector can be written as a linear combination of

the others, then the vectors are linearly dependent.

We omit the proof, but the following example illustrates the main ideas:

5.3.1

Example

(a) Let x1 , x2 , x3 , x4 be vectors in Rn and suppose that

3x1 ? 5x2 + 0x3 + 7x4 = 0,

which shows that the vectors are linear dependent. Actually write one as

a linear combination of the others.

(b) Let x1 , x2 , x3 , x4 , x5 be vectors in Rn and suppose that one of the vectors

can be written as a linear combination of the others, say,

x4 = 3x1 + 0x2 + (?2)x3 + 0x5 .

Show that the vectors are linearly dependent.

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Solution

(a) We can solve the equation for any of the vectors with nonzero

scalar factor, say x1 :

3x1 = 5x2 + 0x3 ? 7x4

5

7

x1 = x2 + 0x3 ? x4 .

3

3

It is the last step of dividing both sides by the scalar factor 3 that uses

the fact that it is nonzero.

(We could have solved for either x2 or x4 as well, but not x3 .)

(b) Moving all terms to the left, we get

?3x1 + 0x2 + 2x3 + 1x4 + 0x5 = 0.

This is a linear combination of the vectors equaling 0 with not all scalar

factors equal to 0 (the vector x4 has scalar factor 1). Therefore, the vectors

are linearly dependent.

A linearly dependent list of vectors has a redundancy in the sense that one

of the vectors can be removed without affecting the span. The next example

illustrates this.

5.3.2 Example Let x1 , x2 , and x3 be vectors in Rn and put S = Span{x1 ,

x2 , x3 }. Show that if the vectors x1 , x2 , and x3 are linearly dependent, then S

is the span of two of these vectors.

Solution Assume that the vectors x1 , x2 , and x3 are linearly dependent. By

the previous theorem, one of the vectors is a linear combination of the others.

By renumbering the vectors, if necessary, we may assume that this vector is x3 ,

so that x3 = ¦Á1 x1 + ¦Á2 x2 for some scalars ¦Á1 and ¦Á2 .

We claim that S = Span{x1 , x2 }. To show that these two sets are equal we

show that each is a subset of the other. Let x be a vector in S. Then x =

¦Â1 x1 + ¦Â2 x2 + ¦Â3 x3 for some scalars ¦Â1 , ¦Â2 , and ¦Â3 . Therefore,

x = ¦Â1 x1 + ¦Â2 x2 + ¦Â3 x3

= ¦Â1 x1 + ¦Â2 x2 + ¦Â3 (¦Á1 x1 + ¦Á2 x2 )

= (¦Â1 + ¦Â3 ¦Á1 )x1 + (¦Â2 + ¦Â3 ¦Á2 )x2

and this shows that x is in Span{x1 , x2 }. Thus, S ? Span{x1 , x2 }.

For the other inclusion, we note that x1 and x2 are in S (see Exercise 4¨C2) and

S is a subspace of Rn (see Section 4.4) so every linear combination of these

vectors must also lie in S by the closure properties of subspace. Therefore,

Span{x1 , x2 } ? S.

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