Lesson 12: Linear Independence

Lesson 12: Sections 14.1–2Linear Independence and Rank

Math 20

October 17, 2007

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When is a line not a line?

Today we’ll assume all lines and planes, etc. go through the origin.So in their equations, a = 0.

QuestionWhen does the equation

x = tv

not determine a line?

Answer.When v = 0.

When is a plane not a plane?

Question

What is the “plane” generated by

12−1

and

24−2

?

Answer.It’s actually a line!

QuestionWhat is the set spanned by

121

,

211

,

−110

?

Answer.A plane.

From geometry to algebra

DefinitionLet {a1, a2, . . . , an} be a set of vectors in Rm. We say they arelinearly dependent if there exist constants c1, c2, . . . , ck ∈ R, notall zero, such that

c1a1 + c2a2 + · · ·+ ckan = 0.

If the equation only holds when all c1 = c2 = · · · = cn = 0, thenthe vectors are said to be linearly independent.

Huh?

This definition includes the cases we’ve already seen:

I If v = 0 then 1 · v = 0, so the set {0} is linearly dependent.

I If b = 2a then1 · a− 2b = 0,

so the two are linearly dependent. In fact, two vectors arelinearly dependent if and only if one is a multiple of the other.

I If a3 = a1 − a2 then

1 · a1 − 1 · a2 − 1 · a3 = 0.

Three or more vectors are linearly dependent if and only if oneis a linear combination of the rest.

Example

Decide if a =

(12

)and b =

(34

)are linearly independent.

Answer.Yes, these vectors are linearly independent. If they were linearlydependent, one would be a multiple t of the other. Checking thefirst components, t = 3, but checking the second, t = 2. It can’tbe both.

Example

Decide if 1

21

,

211

,

−110

is linearly independent.

SolutionNo. The redundancy is

a3 = a1 − a2,

or−1 · a1 + 1 · a2 + a3 = 0.

Example

Determine whether the set1

21

,

101

,

141

,

123


SolutionWe seek solutions to the SLE1 1 1 1

2 0 4 21 1 1 3

c1

c2

c3

c4

=

000

Solution (Continued)

We can find these using the augmented matrix 1 1 1 1 02 0 4 2 01 1 1 3 0

←−−2

+

←−−−−

−1

+

1 1 1 1 00 −2 2 0 00 0 0 2 0

×1/2

1 1 1 1 00 1 −1 0 00 0 0 1 0

←−−1

+

1 1 1 0 00 1 −1 0 00 0 0 1 0

←−−1

+

1 0 2 0 00 1 −1 0 00 0 0 1 0

Solution (Continued) 1 0 2 0 00 1 −1 0 00 0 0 1 0

A solution is found by

c4 = 0

c3 is free

c2 = c3

c1 = −2c3.

Let’s choose c3 = 1. We get c2 = 1, c1 = −2. So

−2

121

+ 1

101

+ 1

141

+ 0

123

=

000

.

Example

Determine whether the set1

21

,

223

,

101


SolutionWe seek solutions to the system of linear equations1 2 1

2 2 01 3 1

c1

c2

c3

=

000

.

We can get this by finding the reduced row echelon form of thematrix.

Solution (Continued)

1 2 1

2 2 0

1 3 1

←−−2

+

←−−−−

−1

+

1 2 1

0 − 2 − 2

0 1 0

×−1/2

1 2 1

0 1 1

0 1 0

←−

−1

+

1 2 1

0 1 1

0 0 − 1

1 0 0

0 1 00 0 1

So we see that the only solution is c1 = c2 = c3 = 0. Thus the setis not linearly dependent.

Answer.Yes, the vectors are linearly independent.

Relation to invertibility

Let A be an n × n matrix. If A has an inverse A−1, the onlysolution to Ac = 0 is the zero solution.

This means that there is no linear dependence relation among thecolumns.

FactA is invertible if and only if the columns of A are linearlyindependent, if and only if rref(A) = I.


Let A be an n × n matrix. If A has an inverse A−1, the onlysolution to Ac = 0 is the zero solution.This means that there is no linear dependence relation among thecolumns.




FactA is invertible if and only if the columns of A are linearlyindependent,

if and only if rref(A) = I.

Lesson 12: Linear Independence

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