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# Probs6 - Solution to problem 6 Due Tuesday November 8 Rui...

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Solution to problem 6 Due Tuesday November 8 Rui Zhang and T. Witten 11/12/05 properties of determinants A generalized determinant of a set of n arbitrary vectors X 1 , ..., X n in an n -dimensional space is defined as a linear real-valued function f ( X 1 , X 2 , ..., X n ) in each of the n vectors that changes sign when any adjacent pair X i , X i +1 are interchanged. The ordinary determinant is a specific case of a generalized determinant. If we have a basis e 1 , ..., e n , we can express f ( X ... ) in terms of the f ( e ... ) and the set of coefficients x ij in X i = n j =1 x ij e j . Thus the f of the vectors becomes a function on the matrix of x ij . For a given set of X s we define the corresponding matrix as X ¯ . This problem is a review of standard properties of determinants. Feel free to consult standard sources in writing out the answers to these questions. a) Is it possible to make a sequences of adjacent interchanges that return to the original ordering yet that result in a net change of sign? Why not? Solution : Of course it’s not possible, otherwise the function f becomes multivalued, and is not defined properly. Now we need to show that, if there are N adjacent interchanges before the original ordering is returned, N must be an even number. For n different real numbers x 1 , x 2 , ...x n , we consider the polynomial P ( x 1 , x 2 , ...x n ) = i<j ( x i - x j ) . (1) Clearly P is nonzero for any set of distinct x ’s. We show below that it changes sign under all adjacent interchanges. We consider a sequence of N interchanges that restores the original order. If N were odd, we would have P ( x 1 , ..., x n ) = - P ( x 1 , ..., x n ), so that P ( x 1 , ..., x n ) = 0. Since P is manifestly not 0, we conclude that N cannot be odd. If x i and x i +1 are exchanged, the affected factors in P are ( x k - x i )( x k - x i +1 ) with k < i , ( x i - x k )( x i +1 - x k ) with k > i + 1 and ( x i - x i +1 ). Only the ( x i - x i +1 ) factor changes sign; the others are unchanged. Thus P changes sign.

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