The following Lemma will be essential in the definition of the determinant.
Lemma 22.0.1There exists a function,sgn_{n}whichmaps each ordered list of numbers from
{1,⋅⋅⋅,n}
toone of the three numbers, 0,1, or −1 which also has the following properties.
sgnn(1,⋅⋅⋅,n) = 1 (22.1)
(22.1)
sgn (i1,⋅⋅⋅,p,⋅⋅⋅,q,⋅⋅⋅,in) = − sgn(i1,⋅⋅⋅,q,⋅⋅⋅,p,⋅⋅⋅,in) (22.2)
n n
(22.2)
In words, the second property states that if two of the numbers are switched, the value of the function ismultiplied by −1. Also, in the case where n > 1 and
{i1,⋅⋅⋅,in }
=
{1,⋅⋅⋅,n }
so that every number from
{1,⋅⋅⋅,n}
appears in the ordered list,
(i1,⋅⋅⋅,in)
,
sgn(i1,⋅⋅⋅,iθ−1,n,iθ+1,⋅⋅⋅,in) ≡
n
n−θ
(− 1) sgn−n1(i1,⋅⋅⋅,iθ−1,iθ+1,⋅⋅⋅,in) (22.3)
(22.3)
where n = i_{θ}in the ordered list,
(i1,⋅⋅⋅,in)
.
Proof: Define sign
(x)
= 1 if x > 0,−1 if x < 0 and 0 if x = 0. If n = 1, there is only one list and it is
just the number 1. Thus one can define sgn_{1}
(1)
≡ 1. For the general case where n > 1, simply
define
( )
∏
sgnn(i1,⋅⋅⋅,in) ≡ sign (is − ir)
r<s
This delivers either −1,1, or 0 by definition. What about the other claims? Suppose you switch i_{p} with i_{q}
where p < q so two numbers in the ordered list
(i1,⋅⋅⋅,in)
are switched. Denote the new ordered list of
numbers as
(j1,⋅⋅⋅,jn)
. Thus j_{p} = i_{q} and j_{q} = i_{p} and if r
∕∈
{p,q}
, j_{r} = i_{r}. See the following
illustration
PICT
Then
( ∏ )
sgn(j1,⋅⋅⋅,jn) ≡ sign (js − jr)
n r<s
( )
◜--------one◞of◟ p,q------◝ neither p nor q
= sign ||(bioth− p,iq) ∏ (i − i) ∏ (i − i) ∏ (i − i )||
( p qp<j<q j q p<j<q p j r<s,r,s∕∈{p,q} s r )
The last product consists of the product of terms which were in the un-switched product ∏_{r<s}
(is − ir)
so
produces no change in sign, while the two products in the middle both introduce q − p − 1 minus
signs. Thus their product produces no change in sign. The first factor is of opposite sign to
the i_{q}− i_{p} which occured in sgn_{n}
(i1,⋅⋅⋅,in)
. Therefore, this switch introduced a minus sign
and
sgn (j1,⋅⋅⋅,jn) = − sgn (i1,⋅⋅⋅,in)
n n
Now consider the last claim. In computing sgn_{n}
(i1,⋅⋅⋅,iθ−1,n,iθ+1,⋅⋅⋅,in)
there will be the product
of n − θ negative terms
(i − n )⋅⋅⋅(i − n)
θ+1 n
and the other terms in the product for computing sgn_{n}
(i1,⋅⋅⋅,iθ−1,n,iθ+1,⋅⋅⋅,in)
are those which are
required to compute sgn_{n−1}
(i1,⋅⋅⋅,iθ−1,iθ+1,⋅⋅⋅,in )
multiplied by terms of the form
(n− ij)
which are
nonnegative. It follows that
sgn (i1,⋅⋅⋅,iθ−1,n,iθ+1,⋅⋅⋅,in) = (− 1)n−θsgn (i1,⋅⋅⋅,iθ−1,iθ+1,⋅⋅⋅,in)
n n− 1
It is obvious that if there are repeats in the list the function gives 0. ■
Lemma 22.0.2Every ordered list of distinct numbers from
{1,2,⋅⋅⋅,n}
can be obtained from everyother such ordered list by a finite number of switches. Also,sgn_{n}is unique.
Proof:This is obvious if n = 1 or 2. Suppose then that it is true for sets of n− 1 elements. Take two
ordered lists of numbers, P_{1},P_{2}. Make one switch in both to place n at the end. Call the result P_{1}^{n} and
P_{2}^{n}. Then using induction, there are finitely many switches in P_{1}^{n} so that it will coincide with P_{2}^{n}. Now
switch the n in what results to where it was in P_{2}.
To see sgn_{n} is unique, if there exist two functions, f and g both satisfying 22.1 and 22.2, you could
start with f
(1,⋅⋅⋅,n )
= g
(1,⋅⋅⋅,n)
= 1 and applying the same sequence of switches, eventually arrive at
f
(i1,⋅⋅⋅,in)
= g
(i1,⋅⋅⋅,in)
. If any numbers are repeated, then 22.2 gives both functions are equal to zero
for that ordered list. ■
Definition 22.0.3When you have an ordered list of distinct numbers from
{1,2,⋅⋅⋅,n}
,say
(i1,⋅⋅⋅,in),
this ordered list is called apermutation. The symbol for all such permutations is S_{n}. The numbersgn_{n}
(i,⋅⋅⋅,i )
1 n
is called the sign of the permutation.
A permutation can also be considered as a function from the set
{1,2,⋅⋅⋅,n} to {1,2,⋅⋅⋅,n}
as follows. Let f
(k)
= i_{k}. Permutations are of fundamental importance in certain areas of math. For
example, it was by considering permutations that Galois was able to give a criterion for solution of
polynomial equations by radicals, but this is a different direction than what is being attempted
here.
In what follows sgn will often be used rather than sgn_{n} because the context supplies the appropriate
n.