31
ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011.

ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

  • View
    225

  • Download
    1

Embed Size (px)

Citation preview

Page 1: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

ENGG2013 Unit 17

Diagonalization Eigenvector and eigenvalue

Mar, 2011.

Page 2: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

EXAMPLE 1

kshum ENGG2013 2

Page 3: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

Q6 in midterm• u(t): unemployment rate in the t-th month.• e(t)= 1-u(t)• The unemployment rate in the next month is

given by a matrix multiplication

• Equilibrium: Solve

kshum ENGG2013 3

Unemployment rate at equilibrium = 0.2

Page 4: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

Equilibrium

kshum ENGG2013 4

Unstable Stable

Page 5: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

If stable, how fast does it converge to the equilibrium point?

kshum ENGG2013 5

0.2 0.2

Fast convergenceSlow convergence

Page 6: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

Question

• Suppose that the initial unemployment rate at the first month is x(1), (for example x(1)=0.25), and suppose that the unemployment evolves by matrix multiplication

Find an analytic expression for x(t), for all t.

kshum ENGG2013 6

Page 7: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

EXAMPLE 2

kshum ENGG2013 7

Page 8: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

How to count?

• Count the number of binary strings of length n with no consecutive ones.

kshum ENGG2013 8

Page 9: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

SOLVING RECURRENCE RELATION

kshum ENGG2013 9

Page 10: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

Fibonacci numbers

• F1 = 1

• F2 = 1

• For n > 2, Fn = Fn-1+Fn-2.

• The Fibonacci numbers are– 1,1,3,5,8,13,21,34,55,89,144

kshum ENGG2013 10

http

://e

n.w

ikip

edia

.org

/wik

i/Fib

onac

ci_n

umb

er

Page 11: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

A matrix formulation

• Define a vector

• Initial vector

• Find the recurrence relation in matrix form

kshum ENGG2013 11

Page 12: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

A general question

• Given initial condition

and for t 2

Find v(t) for all t.

kshum ENGG2013 12

Page 13: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

Matrix power

• Need to raise a matrix to a very high power

kshum ENGG2013 13

Page 14: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

A trivial special case

• Diagonal matrix

• The solution is easy to find

• Raising a diagonal matrix to the power t is easy.

kshum ENGG2013 14

Page 15: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

Decoupled equations

• When the equation is diagonal, we have two separate equation, each in one variable

kshum ENGG2013 15

Page 16: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

DIAGONALIZATION

kshum ENGG2013 16

Page 17: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

Problem reduction

• A square matrix M is called diagonalizable if we can find an invertible matrix, say P, such that the product P–1 M P is a diagonal matrix.

• A diagonalizable matrix can be raised to a high power easily. – Suppose that P–1 M P = D, D diagonal. – M = P D P–1.– Mn = (P D P–1) (P D P–1) (P D P–1) … (P D P–1) = P Dn P–1.

kshum ENGG2013 17

Page 18: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

Example of diagonalizable matrix

• Let

• A is diagonalizable because we can find a matrix

such that

kshum ENGG2013 18

Page 19: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

Now we know how fast it converges to 0.2

• The matrix can be diagonalized

kshum ENGG2013 19

Page 20: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

Convergence to equilibrium

• The trajectory of the unemployment rate– the initial point is set to 0.1

kshum ENGG2013 20

1 2 3 4 5 6 7 8 9 100.1

0.11

0.12

0.13

0.14

0.15

0.16

0.17

0.18

0.19

0.2

month (t)

Une

mpl

oym

ent

rate

Page 21: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

EIGENVECTOR AND EIGENVALUE

kshum ENGG2013 21

Page 22: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

How to diagonalize?

• How to determine whether a matrix M is diagonalizable?

• How to find a matrix P which diagonalizes a matrix M?

kshum ENGG2013 22

Page 23: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

From diagonalization to eigenvector

• By definition a matrix M is diagonalizable ifP–1 M P = D

for some invertible matrix P, and diagonal matrix D.

or equivalently,

kshum ENGG2013 23

Page 24: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

The columns of P are special• Suppose that

kshum ENGG2013 24

Page 25: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

Definition

• Given a square matrix A, a non-zero vector v is called an eigenvector of A, if we an find a real number (which may be zero), such that

• This number is called an eigenvalue of A, corresponding to the eigenvector v.

kshum ENGG2013 25

Matrix-vector product Scalar product of a vector

Page 26: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

Important notes

• If v is an eigenvector of A with eigenvalue , then any non-zero scalar multiple of v also satisfies the definition of eigenvector.

kshum ENGG2013 26

k 0

Page 27: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

Geometric meaning• A linear transformation L(x,y) given by: L(x,y) = (x+2y, 3x-4y)

• If the input is x=1, y=2 for example, the output is x = 5, y = -5.

kshum 27

x x + 2yy 3x – 4y

Page 28: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

Invariant direction• An Eigenvector points at a direction which is invariant under the linear

transformation induced by the matrix.• The eigenvalue is interpreted as the magnification factor.• L(x,y) = (x+2y, 3x-4y)• If input is (2,1), output is magnified by a factor of 2, i.e., the eigenvalue is 2.

kshum 28

Page 29: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

Another invariant direction• L(x,y) = (x+2y, 3x-4y)• If input is (-1/3,1), output is (5/3,-5). The length is increased by a factor of 5, and

the direction is reversed. The corresponding eigenvalue is -5.

kshum 29

Page 30: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

Eigenvalue and eigenvector of

First eigenvalue = 2, with eigenvector

where k is any nonzero real number.

Second eigenvalue = -5, with eigenvector

where k is any nonzero real number.

kshum ENGG2013 30

Page 31: ENGG2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011

Summary

• Motivation: want to solve recurrence relations.

• Formulation using matrix multiplication• Need to raise a matrix to an arbitrary power• Raising a matrix to some power can be easily

done if the matrix is diagonalizable.• Diagonalization can be done by eigenvalue

and eigenvector.

kshum ENGG2013 31