tat06 - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear...

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Unformatted text preview: THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear Mathematics 2010 Tutorial 6 Solutions 1. Suppose that A and P are defined as follows: A = parenleftbigg 1 1 4 1 4 1 0 0 1 parenrightbigg and P = parenleftBigg 1 1 4 1 4 1 1 2 1 2 1 1 1 parenrightBigg . Define a sequence of numbers { u n | n } by u = 0 , u 1 = 1 , u 2 = 2 and for all n , u n +3 = u n +2 + 1 4 u n +1- 1 4 u n . a ) Check that parenleftBigg 4 3 1 3 2 1 1 2 3 1 1 3 parenrightBigg is the inverse of P by calculating its product with P . b ) Evaluate P 1 AP . c ) Hence, find the eigenvalues of A and a basis of R 3 which consists of eigenvectors for A . d ) Express the recurrence relation for u n given above in matrix form. e ) Find a formula for u n using parts (c) and (d). f ) Find lim n u n . Solution a ) Direct calculation shows that PP 1 = parenleftBig 1 0 0 0 1 0 0 0 1 parenrightBig . b ) Direct calculation shows that P 1 AP = parenleftBigg 1 0 1 2 0 0 1 2 parenrightBigg . c ) By part (b), the eigenvalues of A are 1 , 1 2 and- 1 2 , and braceleftBigparenleftBig 1 1 1 parenrightBig , parenleftBigg 1 4 1 2 1 parenrightBigg , parenleftBigg 1 4 1 2 1 parenrightBigg bracerightBig is a basis of R 3 consisting of eigenvectors for A . d ) Write u n = parenleftBig u n +2 u n +1 u n parenrightBig . Then u n +1 = parenleftBig u n +3 u n +2 u n +1 parenrightBig = parenleftBig u n +3 u n +2 u n +1 parenrightBig = parenleftbigg u n +2 + 1 4 u n +1 1 4 u n u n +2 u n +1 parenrightbigg = parenleftbigg 1 1 4 1 4 1 0 0 1 parenrightbigg parenleftBig u n +2 u n +1 u n parenrightBig . That is, u n +1 = A u n . e ) If { v 1 , v 2 , v 3 } is a basis of R 3 consisting of eigenvectors for A , with A v i = i v i , then we know from lectures that the solution to the recurrence relation is of the form u n = r 1 n 1 v 1 + r 2 n 2 v 2 + r 3 n 3 v 3 , for some r 1 , r 2 , r 3 R . Hence, by part (c), we have that u n = r 1 (1) n parenleftBig 1 1 1 parenrightBig + r 2 ( 1 2 ) n parenleftBigg 1 4 1 2 1 parenrightBigg + r 3 (- 1 2 ) n parenleftBigg 1 4 1 2 1 parenrightBigg . Math 2061: Tutorial 6 Solutions S.B and A.M 8/2/2010 Linear Mathematics Tutorial 6 Solutions Page 2 Now, u = parenleftBig u 2 u 1 u parenrightBig = parenleftBig 2 1 parenrightBig , so putting n = 0 we must have parenleftBig 2 1 parenrightBig = r 1 parenleftBig 1 1 1 parenrightBig + r 2 parenleftBigg 1 4 1 2 1 parenrightBigg + r 3 parenleftBigg 1 4 1 2 1 parenrightBigg = parenleftBigg 1 1 4 1 4 1 1 2 1 2 1 1 1 parenrightBigg parenleftBig r 1 r 2 r 3 parenrightBig = P parenleftBig r 1 r 2 r 3 parenrightBig . Hence, parenleftBig r 1 r 2 r 3 parenrightBig = P 1 parenleftBig 2 1 parenrightBig = parenleftBigg 4 3 1 3 2 1 1 2 3 1 1 3 parenrightBigg parenleftBig 2 1 parenrightBig = parenleftBigg 8 3 3 1 3 parenrightBigg , and so u n = 8 3 parenleftBig 1 1 1 parenrightBig- 3 2 n parenleftBigg 1 4 1 2 1 parenrightBigg + ( 1) n 3 2...
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This note was uploaded on 02/06/2012 for the course MATH 2061 taught by Professor Notsure during the Three '09 term at University of Sydney.

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tat06 - THE UNIVERSITY OF SYDNEY PURE MATHEMATICS Linear...

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