ENGG 2013 Unit 17 Diagonalization Eigenvector and eigenvalue

ENGG 2013 Unit 17 Diagonalization Eigenvector and eigenvalue Mar, 2011.

EXAMPLE 1 kshum ENGG 2013 2

Q 6 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 Unemployment rate at equilibrium = 0. 2 kshum ENGG 2013 3

Equilibrium Unstable kshum Stable ENGG 2013 4

If stable, how fast does it converge to the equilibrium point? Slow convergence 0. 2 kshum Fast convergence 0. 2 ENGG 2013 5

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 ENGG 2013 6

EXAMPLE 2 kshum ENGG 2013 7

How to count? • Count the number of binary strings of length n with no consecutive ones. kshum ENGG 2013 8

SOLVING RECURRENCE RELATION kshum ENGG 2013 9

• • F 1 = 1 F 2 = 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 ENGG 2013 http: //en. wikipedia. org/wiki/Fibonacci_number Fibonacci numbers 10

A matrix formulation • Define a vector • Initial vector • Find the recurrence relation in matrix form kshum ENGG 2013 11

A general question • Given initial condition and for t 2 Find v(t) for all t. kshum ENGG 2013 12

Matrix power • Need to raise a matrix to a very high power kshum ENGG 2013 13

A trivial special case • Diagonal matrix • The solution is easy to find • Raising a diagonal matrix to the power t is easy. kshum ENGG 2013 14

Decoupled equations • When the equation is diagonal, we have two separate equation, each in one variable kshum ENGG 2013 15

DIAGONALIZATION kshum ENGG 2013 16

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 Dn P– 1. kshum ENGG 2013 17

Example of diagonalizable matrix • Let • A is diagonalizable because we can find a matrix such that kshum ENGG 2013 18

Now we know how fast it converges to 0. 2 • The matrix can be diagonalized kshum ENGG 2013 19

Convergence to equilibrium • The trajectory of the unemployment rate – the initial point is set to 0. 1 kshum ENGG 2013 20

EIGENVECTOR AND EIGENVALUE kshum ENGG 2013 21

How to diagonalize? • How to determine whether a matrix M is diagonalizable? • How to find a matrix P which diagonalizes a matrix M? kshum ENGG 2013 22

From diagonalization to eigenvector • By definition a matrix M is diagonalizable if P– 1 M P = D for some invertible matrix P, and diagonal matrix D. or equivalently, kshum ENGG 2013 23

The columns of P are special • Suppose that kshum ENGG 2013 24

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 Matrix-vector product Scalar product of a vector • This number is called an eigenvalue of A, corresponding to the eigenvector v. kshum ENGG 2013 25

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. k 0 kshum ENGG 2013 26

Geometric meaning • A linear transformation L(x, y) given by: L(x, y) = (x+2 y, 3 x-4 y) x x + 2 y y 3 x – 4 y • If the input is x=1, y=2 for example, the output is x = 5, y = -5. kshum 27

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+2 y, 3 x-4 y) • If input is (2, 1), output is magnified by a factor of 2, i. e. , the eigenvalue is 2. kshum 28

Another invariant direction • • L(x, y) = (x+2 y, 3 x-4 y) 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

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 ENGG 2013 30

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 ENGG 2013 31