Linear dependence and independence Linear dependence 1 Definition - - PowerPoint PPT Presentation

LINEAR ALGEBRA

Berkant Ustao˘ glu

CRYPTOLOUNGE.NET

Linear dependence and independence

Linear dependence

1

Definition (linear (in)dependence)

Let { v1, v2, . . . , vk} be a set of vectors. If a1 v1 + a2 v2 + · · · + ak vk = ⇒ a1 = a2 = · · · = ak = 0 then the vectors v1, v2, . . . , vk are called linearly independent otherwise they are linearly dependent.

C4 example

2

    1     x1 +     1     x2 +     1     x3 +     1     x4 =        

Linear dependence

3

Theorem

The standard basis vectors are linearly independent, in

• ther words the columns and rows of I are linearly

independent.

C2 example

4

Are 4 3

• ,

1 1

• and

2 2

• linearly dependent?

4 3

• x1 +

1 1

• x2 +

2 2

• x3 =
• ◮ x1 = x2 = x3 = 0 is solution

◮ x1 = 0, x2 = 2 and x3 = −1 is another solution

C2 example

5

Are 4 3

• and

1 1

• linearly dependent?

4 3

• x1 +

1 1

• x2 =
• 1. x1 = 0 and x2 = 0 is the only solution

CVS example

6

Are 4 3

• and

1 1

• linearly dependent?
• α ⊙

4 3 ⊕

• β ⊙

1 1 = 4 3

• look at

(4α − 4α + 4) + (β − 4β + 4) − 4 (3α − 3α + 3) + (β − 3β + 3) − 3

• =

4 3

• ◮ α = 0 and β = 0 is a solution

◮ α = 7 and β = 0 is a solution

CVS example

7

Is 4 3

• linearly dependent?

α ⊙ 4 3

• =

4 3

• look at

4α − 4α + 4 3α − 3α + 3

• =

4 3

• ◮ α = 0 is a solution

◮ α = 7 is a solution

C2 example

8

Is 4 3

• linearly dependent?

α 4 3

• =

4α 3α

• =
• ◮ α = 0 is the only solution

Functions

9

◮ Are the functions p0(x) = x0, p1(x) = x1 and p2(x) = x2

linearly dependent or independent?

◮ Are the functions 5x0, sin2 x and 3 cos2 x linearly

dependent or independent?

Linear dependence

10

Theorem

Let { v1, v2, . . . , vk} be a collection of vectors. If k = 1 the system of vectors is linearly dependent if and only if

• v1 =

0.

Linear dependence

11

Theorem

Let { v1, v2, . . . , vk} be a collection of vectors. If for some 1 ≤ i ≤ k we have that vi = 0 then the system of vectors is linearly dependent.

Linear dependence

12

Theorem

Let { v1, v2, . . . , vk} be a collection of vectors. If for some 1 ≤ i = j ≤ k we have that vi = vj then the system of vectors is linearly dependent.

Linear dependence

13

Theorem

Let { v1, v2, . . . , vk} be a collection of linearly dependent vectors and k > 1. Then there is an index i such that vi can be written as a linear combination of the remaining vectors.

Linear dependence

14

Theorem

Let { v1, v2, . . . , vk} be a collection of vectors. If a subset of { v1, v2, . . . , vk} is linearly dependent then { v1, v2, . . . , vk} is also linearly dependent.

Linear independence

15

Theorem

Let { v1, v2, . . . , vk} be linearly independent, then any subset of { v1, v2, . . . , vk} is also linearly independent.

Span

16

Theorem

S ⊂ V and u ∈ V, then S ∪ u = S ⇐ ⇒ u ∈ S

Corollary

If u ∈ S, then S \ u = S ⇐ ⇒ u ∈ S

Span

17

Theorem

Let S = { v1, v2, . . . , vk}. Then S is linearly independent if and only if ∀i ∈ [1, . . . , k], S \ vi S

Span

18

Theorem

Let S = { v1, v2, . . . , vk}. Then there is a set B that is linearly independent and S = B

A sufficient condition

19

Theorem

Let { a1, a2, . . . , ak} and

• b1,

b2, . . . , bs

• be two sets of
• vectors. Suppose that for each 1 ≤ i ≤ k we have that

ai is a linear combination of

• b1,

b2, . . . , bs

• . Suppose also

k > s then { a1, a2, . . . , ak} is linearly dependent.

• bs independent

20

• 2

−3 3

• =

2 1

• − 3

1

• +3

1

• 2

4 4

• =

2 1

• + 4

1

• +4

1

• 1

2 2

• =

1

• + 2

1

• +2

1

• 1

3 1

• =

1

• + 3

1

• +

1

• bs dependent

21

8 8 3 3

• =

2 1 1

• − 3

1 1

• +3

2 2 2 2

• 3

3 6 6

• =

1 1

• + 4

1 1

• +

2 2 2 2

• 5

5 7 7

• =

1 1

• + 3

1 1

• +2

2 2 2 2

• 3

3 4 4

• =

1 1

• + 2

1 1

• +

2 2 2 2

k ≤ s

22

• 2

2 −3 1

• =

2 1 1

• − 3

1

• +1

1

• 3
• =

1 1

• + 0

1

• +3

1

• −2
• =

1 1

• − 2

1

• +0

1

k ≤ s

23

2 2 2

• =

2 1 1

• − 0

1 1

• +2

1

• 4

4 4

• =

1 1

• + 2

2 2 3

• −2

1