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

2. EXAMPLE 1 kshumENGG20132

3. 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 kshumENGG20133  Unemployment rate at equilibrium = 0.2

4. Equilibrium kshumENGG20134 Unstable Stable

5. If stable, how fast does it converge to the equilibrium point? kshumENGG20135 0.2 Fast convergenceSlow convergence

6. 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. kshumENGG20136

7. EXAMPLE 2 kshumENGG20137

8. How to count? Count the number of binary strings of length n with no consecutive ones. kshumENGG20138

9. SOLVING RECURRENCE RELATION kshumENGG20139

10. Fibonacci numbers F 1 = 1 F 2 = 1 For n > 2, F n = F n-1 +F n-2. The Fibonacci numbers are – 1,1,3,5,8,13,21,34,55,89,144 kshumENGG201310 http://en.wikipedia.org/wiki/Fibonacci_number

11. A matrix formulation Define a vector Initial vector Find the recurrence relation in matrix form kshumENGG201311

12. A general question Given initial condition and for t  2 Find v(t) for all t. kshumENGG201312

13. Matrix power Need to raise a matrix to a very high power kshumENGG201313

14. A trivial special case Diagonal matrix The solution is easy to find Raising a diagonal matrix to the power t is easy. kshumENGG201314

15. Decoupled equations When the equation is diagonal, we have two separate equation, each in one variable kshumENGG201315

16. DIAGONALIZATION kshumENGG201316

17. 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. – M n = (P D P –1 ) (P D P –1 ) (P D P –1 ) … (P D P –1 ) = P D n P –1. kshumENGG201317

18. Example of diagonalizable matrix Let A is diagonalizable because we can find a matrix such that kshumENGG201318

19. Now we know how fast it converges to 0.2 The matrix can be diagonalized kshumENGG201319

20. Convergence to equilibrium The trajectory of the unemployment rate – the initial point is set to 0.1 kshumENGG201320

21. EIGENVECTOR AND EIGENVALUE kshumENGG201321

22. How to diagonalize? How to determine whether a matrix M is diagonalizable? How to find a matrix P which diagonalizes a matrix M? kshumENGG201322

23. 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, kshumENGG201323

24. The columns of P are special Suppose that kshumENGG201324

25. 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. kshumENGG201325 Matrix-vector productScalar product of a vector

26. 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. kshumENGG201326 k  0

27. 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. kshum27 x  x + 2y y  3x – 4y

28. 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. kshum28

29. 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. kshum29

30. 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. kshumENGG201330

31. 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. kshumENGG201331