20105ee205A_1_2010ee205A_1_HW9_sol

# 20105ee205A_1_2010ee205A_1_HW9_sol - Chapter 11 Linear...

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Chapter 11 Linear Diferential and Diference Equations 1. Let P IR n × n be a projection. Show that e P I +1 . 718 P . Answer 11.1 e P = + ± k =0 1 k ! P k = I + + ± k =1 1 k ! P = I + ² + ± k =0 1 k ! - 1 ³ P = I +( e - 1) P I +1 . 718 P. 2. Suppose x,y n and let A = xy T . Further, let α = x T y . Show that e tA = I + g ( t,α ) xy T where g ( )= ´ 1 α ( e αt - 1) if α ± =0 t if α = 0. e tA = + ± k =0 t k k ! A k = I + + ± k =1 t k k ! ( xy T ) k = I + + ± k =1 t k k ! x ( y T x ) k - 1 y T 49

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50 CHAPTER 11. LINEAR DIFFERENTIAL AND DIFFERENCE EQUATIONS = I + + ± k =1 t k α k - 1 k ! xy T = ² I + txy T if α =0 I + 1 α ( e αt - 1) xy T if α ± . 3. Let A = ³ IX 0 - I ´ where X IR m × n is arbitrary. Show that e A = ³ eI sinh 1 X 0 1 e I ´ . Answer 11.2 Since A 2 = I , it is clear that A k = ² A if k is odd I if k is even for k , 1 , 2 ,... e A = + ± k =0 1 k ! A k = ³ + k =0 1 k ! I k odd 1 k ! X 0 + k =0 ( - 1) k k ! I ´ = + k =0 1 k ! I 1 2 µ + k =0 1 k ! X - + k =0 ( - 1) k k ! X 0 + k =0 ( - 1) k k ! I = ³ eI 1 2 ( e - e - 1 ) X 0 e - 1 I ´ = ³ eI sinh 1 X 0 1 e I ´ . A clever “trick” to show this is to use the fact that A and e A commute. Since e A must be of the form ³ eI Y 0 e - 1 I ´ , we have Y + e - 1 X = the (1 , 2) -block of Ae A = the (1 , 2) -block of e A A = eX - Y . Thus Y = 1 2 ( e - e - 1 ) X = sinh 1 X . 4. Let K denote the skew-symmetric matrix ³ 0 I n - I n 0 ´ where I n de- notes the n × n identity matrix. A matrix A 2 n × 2 n is said to be Hamiltonian if K - 1 A T K = - A and symplectic if K - 1 A T K = A - 1 .
51 (a) Suppose H is Hamiltonian and let λ be an eigenvalue of H . Show that - λ must also be an eigenvalue of H . Answer 11.3 Since λ is an eigenvalue of H , we can write Hx = λx for some eigenvector x . Since H is Hamiltonian, ( - λ ) Kx = K ( - H ) x = KK - 1 H T = H T , so ( - λ ) is an eigenvalue of H T , and therefore of H . (b) Suppose S is symplectic and let λ be an eigenvalue of S . Show that 1 must also be an eigenvalue of S . Answer 11.4 Since λ is an eigenvalue of S , we can write Sx = λx for some eigenvector x . Since S is symplectic, it is nonsingu- lar. Thus (1 ) = K ( S - 1 ) x = - 1 S T = S T , so 1 is an eigenvalue of S T , and therefore of S .

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## This note was uploaded on 04/19/2011 for the course EE 205A taught by Professor Laub during the Fall '10 term at UCLA.

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20105ee205A_1_2010ee205A_1_HW9_sol - Chapter 11 Linear...

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