Multivariable Advanced Calculus

3.5. DETERMINANTS 39= (−1) n+1−r (−1) s−1−r sgn n(i 1 , · · · , r q, · · · , i n+1)= (−1) n+s sgn n(i 1 , · · · , r q, · · · , i n+1)= (−1) 2s−1 (−1) n+1−s sgn n(i 1 , · · · , r q, · · · , i n+1)= − sgn n+1(i 1 , · · · , q, r s)· · · , n + 1, · · · , i n+1 .This proves the existence of the desired function.To see this function is unique, note that you can obtain any ordered list of distinctnumbers from a sequence of switches. If there exist two functions, f and g both satisfying3.19 and 3.20, you could start with f (1, · · · , n) = g (1, · · · , n) and applying the samesequence of switches, eventually arrive at f (i 1 , · · · , i n ) = g (i 1 , · · · , i n ) . If any numbersare repeated, then 3.20 gives both functions are equal to zero for that ordered list. Thisproves the lemma. In what follows sgn will often be used rather than sgn n because the context suppliesthe appropriate n.Definition 3.5.2 Let f be a real valued function which has the set of orderedlists of numbers from {1, · · · , n} as its domain. Define∑(k 1 ,··· ,k n )f (k 1 · · · k n )to be the sum of all the f (k 1 · · · k n ) for all possible choices of ordered lists (k 1 , · · · , k n )of numbers of {1, · · · , n} . For example,∑f (k 1 , k 2 ) = f (1, 2) + f (2, 1) + f (1, 1) + f (2, 2) .(k 1 ,k 2 )Definition 3.5.3 Let (a ij ) = A denote an n × n matrix. The determinant ofA, denoted by det (A) is defined bydet (A) ≡∑(k 1 ,··· ,k n )sgn (k 1 , · · · , k n ) a 1k1 · · · a nknwhere the sum is taken over all ordered lists of numbers from {1, · · · , n}. Note it sufficesto take the sum over only those ordered lists in which there are no repeats because ifthere are, sgn (k 1 , · · · , k n ) = 0 and so that term contributes 0 to the sum.Let A be an n × n matrix, A = (a ij ) and let (r 1 , · · · , r n ) denote an ordered list of nnumbers from {1, · · · , n}. Let A (r 1 , · · · , r n ) denote the matrix whose k th row is the r krow of the matrix, A. Thusdet (A (r 1 , · · · , r n )) =∑sgn (k 1 , · · · , k n ) a r1 k 1· · · a rn k n(3.24)andProposition 3.5.4 Let(k 1,··· ,k n)A (1, · · · , n) = A.(r 1 , · · · , r n )be an ordered list of numbers from {1, · · · , n}. Then=∑(k 1,··· ,k n)sgn (r 1 , · · · , r n ) det (A)sgn (k 1 , · · · , k n ) a r1 k 1· · · a rn k n(3.25)= det (A (r 1 , · · · , r n )) . (3.26)