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# tat06 - T HE U NIVERSITY OF S YDNEY P URE M ATHEMATICS...

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T HE U NIVERSITY OF S YDNEY P URE M ATHEMATICS 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 0 1 0 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 0 } by u 0 =0 , u 1 =1 , u 2 =2 and for all n 0 , u n +3 = u n +2 + 1 4 u n +1 - 1 4 u n . a ) Check that parenleftBigg 4 3 0 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 0 0 1 2 0 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 0 1 0 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 = λ i v i , then we know from lectures that the solution to the recurrence relation is of the form u n = r 1 λ n 1 v 1 + r 2 λ n 2 v 2 + r 3 λ 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

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Linear Mathematics Tutorial 6 — Solutions Page 2 Now, u 0 = parenleftBig u 2 u 1 u 0 parenrightBig = parenleftBig 2 1 0 parenrightBig , so putting n =0 we must have parenleftBig 2 1 0 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 0 parenrightBig = parenleftBigg 4 3 0 1 3 2 1 1 2 3 1 1 3 parenrightBigg parenleftBig 2 1 0 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 n parenleftBigg 1 4 1 2 1 parenrightBigg .
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tat06 - T HE U NIVERSITY OF S YDNEY P URE M ATHEMATICS...

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