This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 18.06 Problem Set 7 Solutions Problem 1: Do problem 11 from section 8.3. Solution A Markov matrix must conserve probability. Hence the columns must sum to 1: A = . 7 . 1 . 2 . 1 . 6 . 3 . 2 . 3 . 5 (1) The steady state vector x satisfies Ax = x . In otherwords, x is an eigenvector of A with eigenvalue 1. Therefore x solves: ( A I ) x =  . 3 . 1 . 2 . 1 . 4 . 3 . 2 . 3 . 5 x 1 x 2 x 3 = 0 (2)  . 11 . 11 . 1 . 4 . 3 . 11 . 11 x 1 x 2 x 3 = 0 (3) (4) Take x 3 = 1, then x 2 = 1 and . 1 x 1 = . 4 . 3 = . 1 so x 1 = 1 and x = (111) T . In general, a symmetric Markov matrix A T = A has a steady solution x = (111) T . This follows from: A is Markov columns of A sum to 1. A T = A rows of A sum to 1. The vector x = (111) T sums the row elements. Hence, Ax = x . Problem 2: Do problem 12 from section 8.3. Solution B = ( A I ) x = . 2 . 3 . 2 . 3 (5) 1 The eigenvalues satisfy the characteristic equation: ( . 2 )( . 3 ) ( . 2)( . 3) = 0 (6) ( + . 5) = 0 (7) The eigenvalues are = 0 and = . 5. In general, when A is Markov, A I will have a = 0 eigenvalue. Specifically, this follows from the (more general) fact that if is an eigenvalue of any matrix A , then  c is an eigenvalue of A cI . Markov matrices have an eigenvalue of 1, hence A I must have an eigenvalue 1 1 = 0. The eigenvectors of B are, x 1 = ( . 3 ,. 2) T corresponding to = 0, and x 2 = (1 , 1) T corresponding to = . 5. A general solution to the ODE du dt = ( A I ) u has the form: u = c 1 e t x 1 + c 2 e . 5 t x 2 (8) Here c 1 and c 2 are integration constants (determined by initial values). As t , u becomes: u c 1 x 1 (9) Problem 3: Do problem 16 section 8.3. Solution A = . 4 . 2 . 3 . 2 . 4 . 3 . 4 . 4 . 4 (10) Since A is a Markov matrix, we know = 1 is an eigenvalue. In addition, detA = 0, so = 0 must also be an eigenvalue (ie det ( A I ) = 0 ). The third eigenvalue = . 2 can be found by i ) inspection, ii ) using MATLAB , iii ) by direct computation. Using MATLAB, we can diagonalize A = S S 1 , where the columns of S are eigenvectors of A: S = . 5145 . 7071 . 4082 . 5145 . 7071 . 4082 . 686 . 8165 (11) = 1 0 0 . 2 0 0 0 0 (12) 2 Now A k = S k S 1 , so: A k u = ( S S 1 ) k u (13) = S k S 1 u (14) (15) Since S 1 = . 5831 . 5831 . 5831 . 7071 . 7071 . 4899 . 4899 . 7348 (16) then...
View
Full
Document
This note was uploaded on 05/06/2010 for the course 18 18.06 taught by Professor Strang during the Fall '08 term at MIT.
 Fall '08
 Strang

Click to edit the document details