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20105ee205A_1_2010ee205A_1_HW9_sol

# 20105ee205A_1_2010ee205A_1_HW9_sol - Chapter 11 Linear...

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Chapter 11 Linear Di ff erential and Di ff erence 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 + + k =0 1 k ! - 1 P = I +( e - 1) P I +1 . 718 P. 2. Suppose x, y IR n and let A = xy T . Further, let α = x T y . Show that e tA = I + g ( t, α ) xy T where g ( t, α ) = 1 α ( e α t - 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 = I + txy T if α = 0 I + 1 α ( e α t - 1) xy T if α = 0 . 3. Let A = I X 0 - I where X IR m × n is arbitrary. Show that e A = eI sinh 1 X 0 1 e I . Answer 11.2 Since A 2 = I , it is clear that A k = A if k is odd I if k is even for k = 0 , 1 , 2 , . . . e A = + k =0 1 k ! A k = + k =0 1 k ! I k odd 1 k ! X 0 + k =0 ( - 1) k k ! I = + k =0 1 k ! I 1 2 + k =0 1 k ! X - + k =0 ( - 1) k k ! X 0 + k =0 ( - 1) k k ! I = eI 1 2 ( e - e - 1 ) X 0 e - 1 I = 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 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 0 I n - I n 0 where I n de- notes the n × n identity matrix. A matrix A IR 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 = λ x for some eigenvector x . Since H is Hamiltonian, ( - λ ) Kx = K ( - H ) x = KK - 1 H T Kx = H T Kx , 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 / λ must also be an eigenvalue of S . Answer 11.4 Since λ is an eigenvalue of S , we can write Sx = λ x for some eigenvector x . Since S is symplectic, it is nonsingu- lar. Thus (1 / λ ) Kx = K ( S - 1 ) x = KK - 1 S T Kx = S T Kx , so 1 / λ is an eigenvalue of S T , and therefore of S .

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