MATH535_HW6 - Serge Ballif 1 Define the linear...

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Serge Ballif MATH 535 Homework 6 October 12, 2007 1. Define the linear transformation U on n by v 7→ Fv , where nF is the Vandermonde matrix for the values ω j n := exp ((2 πij ) /n ) , i = 0 , . . . , n - 1 . a) Show that the sum S j = n X k =1 ( ω j n ) k = ( n if n | j 0 otherwise by multiplying the expression by ω i n and simplifying. b) Conclude that U is unitary. c) Show that, for u = ( z 1 , . . . , z n ) t n , U - 1 u = Uv , where v = ( z n , . . . , z 1 ) t . Remark: This mapping has revolutionized mathematical modelling and is the basis for the fastest known method for multiplication of extremely large integers. (a) If n | j , then ω i n = e 2 πij/n = 1, so each of the summands of S j is 1. In this case S j = n . If n - j , then we can examine S j j n = n - 1 k =0 ( ω j n ) k , a geometric series with sum S j ω j n = ( ω j n ) n - 1 ω j n - 1 = 1 - 1 ω j n - 1 = 0. Thus, S j = 0 in the case n - j . (b) We know that the columns of a vandermonde matrix with distinct values is a basis for the space V . U is unitary iff the column vectors form an orthonormal basis of V . To see that the basis is indeed orthonormal we note that the inner product of the i -th and j -th columns is δ ij as determined by the formula in part (a). (c) If we take n = 3, j = 1, we get U = 1 3 1 1 1 1 ω 3 ω 2 3 1 ω 2 3 ω 4 3 . Then for u = 1 2 3 we get a counterexample to the desired statement. Hence, the statement is false.
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