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Unformatted text preview: Problem Set 2 Solutions Economics 6170 TA: Christopher Handy September 19, 2008 Problem 1 (a) We will show that the vectors x 1 ,...,x n are linearly dependent when n is even and linearly independent when n is odd. Consider a linear combination of the vectors of T : 1 1 1 . . . + 2 1 1 . . . + + n 1 . . . 1 1 + n 1 . . . 1 = n + 1 1 + 2 2 + 3 3 + 4 . . . n 3 + n 2 n 2 + n 1 n 1 + n If n is even, we can find 1 ,..., n not all zero such that this linear combination gives the zero vector: i = ( 1 , if i odd 1 , if i even for all i = 1 ,...,n Thus for n even, the vectors of T are linearly dependent. But if n is odd, these i do not give the zero vector, because then n + 1 6 = 0. In fact, we can show that there are no 1 ,..., n not all zero such that the linear combination above is equal to the zero vector. Suppose there are 1 ,..., n not all zero such that n + 1 1 + 2 . . . n 1 + n = . . . 1 Then, reading from the last equation to the first, since n is odd it must be that n = n 1 = n 2 = = 2 = 1 . Since the 1 ,..., n are not all zero, it must be that each i 6 = 0, so that n + 1 = 2 1 6 = 0. This is a contradiction, so we conclude that when n is odd, the vectors of T are linearly independent. (b) We can prove the vectors of U are linearly independent by contradiction. Suppose they are linearly dependent, so that there are 1 ,..., n not all zero such that 1 e 1 + + n 1 e n 1 + n x = 1 + n x 1 . . . n 1 + n x n 1 n x n = 0 . Since x n 6 = 0, from the last row we must have n = 0. But then the equation above only holds if i = 0 for all i = 1 ,...,n . This contradicts the assumption that the 1 ,..., n are not all zero, so we conclude that the vectors of U are linearly independent....
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This note was uploaded on 10/03/2009 for the course ECON 6170 taught by Professor Mitra during the Fall '08 term at Cornell University (Engineering School).
 Fall '08
 MITRA
 Economics

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