[PPT] - Eigenvectors and Eigenvalues Repeated application 1 1 0 0 A PowerPoint Presentation

SLIDE 1

Eigenvectors and Eigenvalues

SLIDE 2

Repeated application

1

A =   1 −1 2   A2n+1 =   1 −1 22n+1   A2n =   1 1 22n  

SLIDE 3

General LA

2

A =   −9 14 4 3 −2 −18 22 9   An =?

SLIDE 4

General LA

3

  −9 14 4 3 −2 −18 22 9   =   2 2 3 1 1 2 3 4     2 −3 1     3 −1 −2 2 −2 −1 −3 2 2  

SLIDE 5

General LA

4

  −9 14 4 3 −2 −18 22 9     2 1 2   =2   2 1 2     −9 14 4 3 −2 −18 22 9     2 3   = − 3   2 3     −9 14 4 3 −2 −18 22 9     3 1 4   =1   3 1 4  

SLIDE 6

Division of polynomials

5

Theorem

Let a(x) and b(x) be polynomials such that b(x) = 0(x), then there are quotient q(x) and remainder r(x) polynomials such that a(x) = q(x)b(x) + r(x) satisfying deg r(x) < deg b(x).

SLIDE 7

Corollaries

6

Corollary

Let a(x) and b(x) = x − λ be polynomials, then a(x) = q(x)(x − λ) + a(λ).

Corollary

a(x) has root λ if and only if x − λ divides a(x).

SLIDE 8

Fundamental Theorem of Algebra

7

Theorem

Every non-constant polynomial with complex coefficients has at least one complex root.

SLIDE 9

Eigenvalues and eigenvectors

8

Definition (eigenvalues and eigenvectors)

Let A be a square matrix. A non-zero vector u is an eigenvector for A if A u = λ u for some λ. The value λ is called eigenvalue for the eigenvector u.

SLIDE 10

Eigenvalues and eigenvectors

9

A u = λ u is A u = λI u (A − λI) u = Want linearly dependent columns!

SLIDE 11

Example

10

from   −9 14 4 3 −2 −18 22 9     2 1 2   = 2   2 1 2   to   −11 14 4 3 −2 −2 −18 22 7     2 1 2   =    

SLIDE 12

Example

11

from   −9 14 4 3 −2 −18 22 9     2 3   = −3   2 3   to   −6 14 4 3 3 −2 −18 22 12     2 3   =    

SLIDE 13

Example

12

from   −9 14 4 3 −2 −18 22 9     3 1 4   =   3 1 4   to   −10 14 4 3 −1 −2 −18 22 8     3 1 4   =    

SLIDE 14

Goal

13

Given A =      a11 a12 · · · a1n a21 a22 a2n . . . ... . . . an1 an2 · · · ann      are the columns of A linearly dependent.

SLIDE 15

Computing Eigenvalues and Eigenvectors

14

1. compute the determinant of A − λI
2. compute the roots λ1, . . . , λn of the resulting

polynomial, the n (possibly repeated) roots are the eigenvalues

3. for each eigenvalue i find the corresponding

eigenvectors by computing (A − λiI) x =

SLIDE 16

Example

15

A =   −9 14 4 3 −2 −18 22 9   det(A − λI) = λ3 − 7λ + 6 = (λ − 2) · (λ − 1) · (λ + 3) so λ0 = 2 λ1 = 1 λ2 = −3

SLIDE 17

A − λ0I

16

=   −9 14 4 3 −2 −18 22 9   −2   1 1 1   =   −11 14 4 3 −2 −2 −18 22 7   solve   −11 14 4 3 −2 −2 −18 22 7   x =     ⇒   α   2 1 2   | α ∈ C   

SLIDE 18

A − λ1I

17

  −9 14 4 3 −2 −18 22 9   −1   1 1 1   =   −10 14 4 3 −1 −2 −18 22 8   solve   −10 14 4 3 −1 −2 −18 22 8   x =     ⇒   α   3 1 4   | α ∈ C   

SLIDE 19

A − λ2I

18

  −9 14 4 3 −2 −18 22 9   −(−3)   1 1 1   =   −6 14 4 3 3 −2 −18 22 12   solve   −6 14 4 3 3 −2 −18 22 12   x =     ⇒   α   2 3   | α ∈ C   

SLIDE 20

Example

19

A = 4 −5 2 −3

det(A − λI) = λ2 − λ − 2 = (λ − 2) · (λ + 1)

◮ characteristic polynomial:

λ2 − λ − 2

◮ characteristic equation:

λ2 − λ − 2 = 0 λ0 = 2 λ1 = −1

SLIDE 21

Example

20

A − λ0I = 4 −5 2 −3

− 2

1 1

=

2 −5 2 −5

solve

2 −5 2 −5

x =
⇒
α

5 2

| α ∈ C

SLIDE 22

Example

21

A − λ1I = 4 −5 2 −3

− (−1)

1 1

=

5 −5 2 −2

solve

5 −5 2 −2

x =
⇒
α

1 1

| α ∈ C

SLIDE 23

Example double root

22

A = 2 2

det(A − λI) = λ2 − 4λ + 4 = (λ − 2)2

so λ0 = 2 λ1 = 2

SLIDE 24

Example double root

23

A − λ0I = 2 2

− 2

1 1

=
solve
x =
⇒
α0

1

+ α1

1

| α0, α1 ∈ C

SLIDE 25

Example double root again

24

A = 2 1 2

det(A − λI) = λ2 − 4λ + 4 = (λ − 2)2

so λ0 = 2 λ1 = 2

SLIDE 26

Example

25

A − λ0I = 2 1 2

− 2

1 1

=

1

solve

1

x =
⇒
α

1

| α ∈ C

SLIDE 27

Zero is an eigenvalue

26

A = 4 −4 2 −2

det(A − λI) = λ2 − 2λ = (λ − 2) · λ

◮ characteristic polynomial:

λ2 − 2λ

◮ characteristic equation:

λ2 − 2λ = 0 λ0 = 2 λ1 = 0

SLIDE 28

Example

27

A − λ0I = 4 −4 2 −2

− 2

1 1

=

2 −4 2 −4

solve

2 −4 2 −4

x =
⇒
α

2 1

| α ∈ C

SLIDE 29

Example

28

A − λ1I = 4 −4 2 −2

− (0)

1 1

=

4 −4 2 −2

solve

4 −4 2 −2

x =
⇒
α

1 1

| α ∈ C

SLIDE 30

Characteristic polynomial

29

Definition

Let T ∈ Mn×n. The characteristic polynomial of T is det (T − λI) . The characteristic equation of T is det (T − λI) = 0. The characteristic polynomial of a linear transformation φ is the characteristic polynomial of any matrix representation RB→B(φ)

SLIDE 31

Existence

30

Theorem

Let φ be any linear transformation on a non-trivial vector

space. Then φ has at least one eigenvalue.

SLIDE 32

Eigenspace

31

Definition

The eigenspace of a transformation φ associated with the eigenvalue λ is Vλ =

ζ | φ(

ζ ) = λ ζ

.

Theorem

An eigenspace is a subspace.

SLIDE 33

Algebraic vs geometric multiplicity

32

Theorem

Let a linear transformation φ has a characteristic polynomials p(λ) = (x − λ1)m1(x − λ2)m2 . . . (x − λk)mk Then eigenvalue λk has algebraic multiplicity mk and geometric multiplicity dim Vλk

SLIDE 34

Linear independence

33

Theorem

A set of eigenvectors, corresponding to distinct eigenvalues is linearly independent