lecture18

lecture18 - Lecture 18 We begin with a quick review of...

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Unformatted text preview: Lecture 18 We begin with a quick review of permutations (from last lecture). A permutation of order k is a bijective map : { 1 , . . . , k } { 1 , . . . , k } . We denote by S k the set of permutations of order k . The set S k has some nice properties. If S k , then 1 S k . The inverse permutation 1 is defined by 1 ( j ) = i if ( i ) = j . Another nice property is that if , S k , then S k , where ( i ) = ( ( i )). That is, if ( i ) = j and ( j ) = k , then ( i ) = k . Take 1 i < j k , and define i,j ( i ) = j (4.40) i,j ( j ) = i (4.41) i,j ( ) = , = i, j. (4.42) The permutation i,j is a transposition. It is an elementary transposition of j = i + 1. Last time we stated the following theorem. Theorem 4.13. Every permutation can be written as a product = 1 2 r , (4.43) where the i s are elementary transpositions. In the above, we removed the symbol denoting composition of permutations, but the composition is still implied. Last time we also defined the sign of a permutation Definition 4.14. The sign of a permutation is ( 1) = ( 1) r , where r is as in the above theorem. Theorem 4.15. The above definition of sign is well-defined, and ( 1) = ( 1) ( 1) . (4.44) All of the above is discussed in the Multi-linear Algebra Notes. Part of todays homework is to show the following two statements: 1. | S k | = k !. The proof is by induction. 2. ( 1) i,j = 1 . Hint: use induction and i,j = ( j 1 ,j )( i,j 1 )( j 1 ,j ), with i < j ....
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lecture18 - Lecture 18 We begin with a quick review of...

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