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1

1.7

2012 Pearson Education, Inc.

Linear Equations

in Linear Algebra

LINEAR INDEPENDENCE

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Slide 1.7- 2 2012 Pearson Education, Inc.

LINEAR INDEPENDENCE

Definition:An indexed set of vectors {v1, , vp} in

is said to be linearly independentif the vector

equation

has only the trivial solution. The set {v1, , vp} is

said to be linearly dependentif there exist weights

c1, , cp, not all zero, such that----(1)

n

1 1 2 2

v v ... v 0p p

x x x

1 1 2 2v v ... v 0

p pc c c

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LINEAR INDEPENDENCE

Equation (1) is called a linear dependence relation

among v1, , vp when the weights are not all zero.

An indexed set is linearly dependent if and only if it

is not linearly independent.

Example 1:Let , , and .1

1

v 2

3

2

4

v 5

6

3

2

v 1

0

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a. Determine if the set {v1, v2, v3} is linearlyindependent.

b. If possible, find a linear dependence relation

among v1, v2, and v3.

Solution:We must determine if there is a nontrivial

solution of the following equation.

1 2 3

1 4 2 0

2 5 1 0

3 6 0 0

x x x

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LINEAR INDEPENDENCE

Row operations on the associated augmented matrix

show that

.

x1andx2are basic variables, andx3is free.

Each nonzero value ofx3determines a nontrivialsolution of (1).

Hence, v1, v2, v3are linearly dependent.

1 4 2 0 1 4 2 0

2 5 1 0 0 3 3 03 6 0 0 0 0 0 0

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LINEAR INDEPENDENCE

b. To find a linear dependence relation among v1,

v2, and v3, row reduce the augmented matrix

and write the new system:

Thus, , , andx3is free. Choose any nonzero value forx3say, .

Then and .

1 0 2 0

0 1 1 0

0 0 0 0

1 3

2 3

2 0

0

0 0

x x

x x

1 32x x 2 3x x 3

5x

1 10x

2 5x

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7/20Slide 1.7- 7 2012 Pearson Education, Inc.

LINEAR INDEPENDENCE

Substitute these values into equation (1) and obtain

the equation below.

This is one (out of infinitely many) possible linear

dependence relations among v1, v2, and v3.

1 2 3

10v 5v 5v 0

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LINEAR INDEPENDENCE OF MATRIX COLUMNS

Suppose that we begin with a matrix

instead of a set of vectors.

The matrix equation can be written as

.

Each linear dependence relation among the columns ofA corresponds to a nontrivial solution of .

Thus, the columns of matrixAare linearly independent ifand only if the equation has only the trivial

solution.

1a a

n

A

x 0A

1 1 2 2a a ... a 0n nx x x

x 0A

x 0A

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SETS OF ONE OR TWO VECTORS

A set containing only one vectorsay, vis linearly

independent if and only if vis not the zero vector.

This is because the vector equation has onlythe trivial solution when .

The zero vector is linearly dependent becausehas many nontrivial solutions.

1v 0x v 0

10 0x

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SETS OF ONE OR TWO VECTORS

A set of two vectors {v1, v2} is linearly dependent if

at least one of the vectors is a multiple of the other.

The set is linearly independent if and only if neither

of the vectors is a multiple of the other.

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SETS OF TWO OR MORE VECTORS

Theorem 7: Characterization of Linearly Dependent

Sets

An indexed set of two or more

vectors is linearly dependent if and only if at least oneof the vectors in Sis a linear combination of the

others.

In fact, if Sis linearly dependent and , then

some vj(with ) is a linear combination of the

preceding vectors, v1, , .

1{v ,...,v }

pS

1

v 01j

1v

j

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Proof:If some vjin Sequals a linear combination ofthe other vectors, then vjcan be subtracted from both

sides of the equation, producing a linear dependence

relation with a nonzero weight on vj.

[For instance, if , then

.]

Thus Sis linearly dependent.

Conversely, suppose Sis linearly dependent.

If v1is zero, then it is a (trivial) linear combination of

the other vectors in S.

( 1)

1 2 2 3 3v v vc c

1 2 2 3 3 40 ( 1)v v v 0v ... 0v

pc c

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Otherwise, , and there exist weights c1, , cp, not

all zero, such that

.

Letjbe the largest subscript for which .

If , then , which is impossible because

.

1v 0

1 1 2 2v v ... v 0

p pc c c

0j

c

1j 1 1v 0c

1v 0

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So , and1j

1 1 1v ... v 0v 0v ... 0v 0

j j j j pc c

1 1 1 1v v ... vj j j jc c c

11

1 1v v ... v .

j

j j

j j

cc

c c

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Theorem 7 does notsay that everyvector in a linearly

dependent set is a linear combination of the precedingvectors.

A vector in a linearly dependent set may fail to be alinear combination of the other vectors.

Example 2:Let and . Describe the

set spanned by uand v, and explain why a vector wisin Span {u, v} if and only if {u, v, w} is linearlydependent.

3

u 1

0

1

v 6

0

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Solution:The vectors uand vare linearly

independent because neither vector is a multiple of

the other, and so they span a plane in .

Span {u, v} is thex1x2-plane (with ).

If wis a linear combination of uand v, then {u, v, w}is linearly dependent, by Theorem 7.

Conversely, suppose that {u, v, w} is linearly

dependent. By theorem 7, some vector in {u, v, w} is a linear

combination of the preceding vectors (since ).

That vector must be w, since vis not a multiple of u.

3

3 0x

u 0

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So wis in Span {u, v}. See the figures given below.

Example 2 generalizes to any set {u, v, w} in with

uand vlinearly independent.

The set {u, v, w} will be linearly dependent if and

only if wis in the plane spanned by uand v.

3

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Theorem 8:If a set contains more vectors than thereare entries in each vector, then the set is linearly

dependent. That is, any set {v1, , vp} in is

linearly dependent if .

Proof: Let .

ThenAis , and the equation

corresponds to a system of nequations inp

unknowns. If , there are more variables than equations, so

there must be a free variable.

n

p n

1v v

pA

n p x 0A

p n

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Hence has a nontrivial solution, and thecolumns of A are linearly dependent.

See the figure below for a matrix version of this

theorem.

Theorem 8 says nothing about the case in which the

number of vectors in the set does notexceed the

number of entries in each vector.

x 0A

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Theorem 9:If a set in contains

the zero vector, then the set is linearly dependent.

Proof:By renumbering the vectors, we may suppose

.

Then the equation showsthat Sin linearly dependent.

1{v ,...,v }

pS

n

1v 0

1 21v 0v ... 0v 0p

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