This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: summationdisplay summationdisplay summationdisplay bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg 18.06 PSET 8 SOLUTIONS APRIL 15, 2010 Problem 1. ( 6.3, #14) The matrix in this question is skewsymmetric ( A T = A ): 0 c b u = cu 2 bu 3 1 d u = c 0 a u or u 2 = au 3 cu 1 dt b a 0 u = bu 1 au 2 3 (a) The derivative of bardbl u ( t ) bardbl 2 = u 1 2 + u 2 2 + u 3 3 is 2 u 1 u 1 + 2 u 2 u 2 + 2 u 3 u 3 . Substitute u 1 , u 2 , u 3 to get zero . Then bardbl u ( t ) bardbl 2 stays equal to bardbl u (0) bardbl 2 . (b) When A is skewsymmetric, Q = e At is orthogonal. Prove Q T = e At from the series for Q = e At . Then Q T Q = I . Solution. (4 points) (a) 2 u 1 u 1 + 2 u 2 u 2 + 2 u 3 u = 2 u 1 ( cu 2 bu 3 ) + 2 u 2 ( au 3 cu 1 ) + 2 u 3 ( bu 1 au 2 ) = 0 . 3 (b) The important points are that ( A n ) T = ( A T ) n = ( A ) n , and that we can take transpose termwise in a sum: parenleftBigg parenrightBigg T Q T = A n t n = ( A n ) T t n = ( A ) n t n = e At . n ! n ! n ! n =0 n =0 n =0 Then, Q T Q At At 0 = e e = e = I because A and A commute (but I dont think the problem intended for you to have to actually check this!). Problem 2. ( 6.3, #24) Write A = 1 3 as S S 1 . Multiply Se t S 1 to find the matrix exponential 0 e At . Check e At and the derivative of e At when t = 0. Solution. (4 points) 2 2 = 1 0 and S = 1 1 . 3 1 Then, 3 t t t e e Se t S 1 = e 2 3 t 2 0 e This is the identity matrix when t = 0, as it should be. The derivative matrix is e t 3 / 2 e 3 t 1 / 2 e t 3 e 3 t which is equal to A when t = 0, as it should be. 1 bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg bracketleftbigg bracketrightbigg Problem 3. ( 6.3, #28) Centering y = y in Example 3 will produce Y n +1 2 Y n + Y n 1 = ( t ) 2 Y n . This can be written as a onestep difference equation for U = ( Y, Z ): Y n +1 = Y n + tZ n bracketleftbigg 1 0 bracketrightbiggbracketleftbigg Y n +1 bracketrightbigg bracketleftbigg 1 t bracketrightbiggbracketleftbigg Y n bracketrightbigg = . Z n +1 = Z n tY n +1 t 1 Z n +1 1 Z n Invert the matrix on the left side to write this as U n +1 = A U n . Show that det A = 1. Choose the large time step t = 1 and find the eigenvalues 1 and 2 = 1 of A : A = 1 1 1 has  1  =  2  = 1. Show that A 6 is exactly I ....
View
Full
Document
This note was uploaded on 08/22/2011 for the course MATH 1806 taught by Professor Strang during the Fall '10 term at MIT.
 Fall '10
 Strang
 Linear Algebra, Algebra, Derivative

Click to edit the document details