20105ee205A_1_2010ee205A_1_HW9_sol

20105ee205A_1_2010ee205A_1_HW9_sol - Chapter 11 Linear...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 .
Background image of page 2
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 .
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

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

Page1 / 9

20105ee205A_1_2010ee205A_1_HW9_sol - Chapter 11 Linear...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online