University:
Massachusetts Institute of Technology
Course:
18.06SC | Linear Algebra
Academic year:

2019
Views:

143

Pages:

3
Author:

Karli Fleming

Exercises on Markov matrices; Fourier series Problem 24.1: (6.4 #7. Introduction to Linear Algebra: Strang) � � 1 b a) Find a symmetric matrix that has a negative eigenvalue. b 1 b) How do you know it must have a negative pivot? c) How do you know it can’t have two negative eigenvalues? Solution: a) The eigenvalues of that matrix are 1 ± b. If b > 1 or b < −1 the matrix has a negative eigenvalue. b) The pivots have the same signs as the eigenvalues. If the matrix has a negative eigenvalue, then it must have a negative pivot. c) To obtain one negative eigenvalue, we choose either b > 1 or b < −1 (as stated in part (a)). If we choose b > 1, then λ1 = 1 + b will be positive while λ2 = 1 − b will be negative. Alternatively, if we choose b < −1, then λ1 = 1 + b will be negative while λ2 = 1 − b will be positive. Therefore this matrix cannot have two negative eigenvalues. Problem 24.2: (6.4 #23.) Which of these classes of matrices do A and B belong to: invertible, orthogonal, projection, permutation, diagonalizable, Markov? ⎡ ⎤ ⎡ ⎤ 0 0 1 1 1 1 1 A = ⎣ 0 1 0 ⎦ B = ⎣ 1 1 1 ⎦. 3 1 0 0 1 1 1 Which of these factorizations are possible for A and B: LU, QR, SΛS−1 , or QΛQ T ? Solution: a) For A : A is invertible. det A = −1 �= 0. T AA = I. A is orthogonal. A is not a projection. A2 = I �= A. A has one 1 in each row and column with A is a permutation. 0’s elsewhere. A = A T , so A is symmetric. A is diagonalizable. Each column of A sums to one. A is Markov. A = LU is not possible because A11 = 0. QR is possible because A has independent columns, SΛS−1 is possible because it is diagonalizable, and QΛQ T is possible because it is symmetric. b) For B : B is not invertible. det B = 0. BB T �= I. B is not orthogonal. 2 B is a projection. B = B. B does not have one 1 in each row and each B is not a permutation. column, with 0’s elsewhere. B = B T so B is symmetric. B is diagonalizable. Each column of B sums to one. B is Markov. B = LU is possible but U only contains one nonzero pivot. QR is im possible because B has dependent columns, SΛS−1 is possible because it is diagonalizable, and QΛQ T is possible because it is symmetric. Problem 24.3: (8.3 #11.) Complete A to a Markov matrix and ﬁnd the steady state eigenvector. When A is a symmetric Markov matrix, why is x1 = (1, . . . , 1) its steady state? ⎡ ⎤ .7 .1 .2 A = ⎣ .1 .6 .3 ⎦ . Solution: Matrix A becomes: ⎡ ⎤ .7 .1 .2 A = ⎣ .1 .6 .3 ⎦ , .2 .3 .5 with steady state vector (1,1,1). When A is a symmetric Markov matrix, the elements of each row sum to one. The elements of each row of A − I then sum to zero. Since the steady state vector x is the eigenvector associated with eigenvalue λ = 1, we solve ( A − λI )x = ( A − I )x = 0 to get x = (1, . . . , 1).