40 CHAPTER 1. REVIEW OF SOME LINEAR ALGEBRA

(−1)n−θ sgnn−1 (i1, · · · , iθ−1, iθ+1, · · · , in) (1.17)

where n = iθ 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 onlyone list and it is just the number 1. Thus one can define sgn1 (1)≡ 1. For the general casewhere n > 1, simply define

sgnn (i1, · · · , in)≡ sign

(∏r<s

(is− ir)

)

This delivers either −1,1, or 0 by definition. What about the other claims? Suppose youswitch ip with iq 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 jp = iq and jq = ip and ifr /∈ {p,q} , jr = ir. See the following illustration

i11

i22 · · · ip

p · · · iqq · · · in

ni11

i22 · · · iq

p · · · ipq · · · in

nj11

j22 · · · jp

p · · · jqq · · · jn

n

Then

sgnn ( j1, · · · , jn)≡ sign

(∏r<s

( js− jr)

)

= sign

(both p,q(ip− iq)

one of p,q

∏p< j<q

(i j− iq) ∏p< j<q

(ip− i j)neither p nor q

∏r<s,r,s/∈{p,q}

(is− ir)

)The last product consists of the product of terms which were in ∏r<s (is− ir) while thetwo products in the middle both introduce q− p− 1 minus signs. Thus their product ispositive. The first factor is of opposite sign to the iq− ip which occured in sgnn (i1, · · · , in) .Therefore, this switch introduced a minus sign and

sgnn ( j1, · · · , jn) =−sgnn (i1, · · · , in)

Now consider the last claim. In computing sgnn (i1, · · · , iθ−1,n, iθ+1, · · · , in) there willbe the product of n−θ negative terms

(iθ+1−n) · · ·(in−n)

and the other terms in the product for computing sgnn (i1, · · · , iθ−1,n, iθ+1, · · · , in) are thosewhich are required to compute sgnn−1 (i1, · · · , iθ−1, iθ+1, · · · , in) multiplied by terms of theform (n− i j) which are nonnegative. It follows that

sgnn (i1, · · · , iθ−1,n, iθ+1, · · · , in) = (−1)n−θ sgnn−1 (i1, · · · , iθ−1, iθ+1, · · · , in)

It is obvious that if there are repeats in the list the function gives 0. ■