The following Lemma will be essential in the definition of the determinant.
Lemma A.1.1There exists a function,sgn_{n}whichmaps each ordered list ofnumbers from
{1,⋅⋅⋅,n}
to one of the three numbers, 0,1, or −1 which also has thefollowing properties.
sgn(1,⋅⋅⋅,n) = 1 (1.1)
n
(1.1)
sgn (i1,⋅⋅⋅,p,⋅⋅⋅,q,⋅⋅⋅,in) = − sgn (i1,⋅⋅⋅,q,⋅⋅⋅,p,⋅⋅⋅,in) (1.2)
n n
(1.2)
In words, the second property states that if two of the numbers are switched, thevalue of the function is multiplied by −1. Also, in the case where n > 1 and
{i1,⋅⋅⋅,in}
=
{1,⋅⋅⋅,n}
so that every number from
{1,⋅⋅⋅,n}
appears in the orderedlist,
(i1,⋅⋅⋅,in)
,
sgnn(i1,⋅⋅⋅,iθ−1,n,iθ+1,⋅⋅⋅,in) ≡
(− 1)n−θ sgn (i1,⋅⋅⋅,iθ−1,iθ+1,⋅⋅⋅,in) (1.3)
n−1
(1.3)
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
( ∏ )
sgn (i1,⋅⋅⋅,in) ≡ sign (is − ir)
n 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
( )
∏
sgnn(j1,⋅⋅⋅,jn) ≡ sign (js − jr)
r<s
( one of p,q )
| both p,q◜∏---------◞◟∏---------◝ ne∏ither p nor q |
= sign |((ip − iq) (ij − iq) (ip − ij) (is − ir)|)
p<j<q p<j<q r<s,r,s∕∈{p,q}
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
(iθ+1 − n)⋅⋅⋅(in − 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−θ sgn(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 A.1.2Every ordered list of distinct numbers from
{1,2,⋅⋅⋅,n}
can beobtained from every other 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 1.1 and
1.2, 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 1.2 gives both functions are equal to zero for that ordered list.
■
Definition A.1.3When 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}. Thenumbersgn_{n}
(i1,⋅⋅⋅,in)
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.