Number Theory: An Introduction to Mathematics

226 V Hadamard’s Determinant Problem

Let

A=

[

αβ

γδ

]

,

where at least one ofα,β,γ,δis nonzero. By multiplyingAon the left, or on the right,

or both byWwe may suppose thatα=0. Now, by multiplyingAon the right or left

byRα, we may suppose thatα=1. Next, by multiplyingAon the right byU−β,we

may further suppose thatβ=0. Finally, by multiplyingAon the left byV−γ,wemay

also suppose thatγ=0.

The preceding argument is valid even ifFis a division ring. In what follows we

will use the commutativity of multiplication inF.

We are now going to show thatif d :E → F is a map such that d(ST)=

d(S)d(T)for all S,T∈E,then either d(S)= 0 for every S∈Eor d(S)= 1 for

every S∈E.

Ifd(T)= 0forsomeT ∈ E,thend(I)= d(T)d(T−^1 )= 0andd(S) =

d(I)d(S)=0foreveryS∈E. Thus we now supposed(S)=0foreveryS∈E.

Then, in the same way,d(I)=1andd(S−^1 )=d(S)−^1 for everyS∈E.

It is easily verified that

UλUμ=Uλ+μ, VλVμ=Vλ+μ,

W−^1 =−W, W−^1 VμW=U−μ.

It follows that

d(Vμ)=d(U−μ)=d(Uμ)−^1.

Also, for anyρ=0,

R−ρ^1 UλRρ=Uλρ 2.

Henced(Uλρ 2 )=d(Uλ)andd(Uλ(ρ (^2) − 1 ))=1.
If the fieldFcontains more than three elements, thenρ^2 − 1 =0 for some nonzero
ρ∈F.Sinceλ(ρ^2 − 1 )runs through the nonzero elements ofFat the same time asλ,
it follows thatd(Uλ)=1foreveryλ∈F. Hence alsod(Vμ)=1foreveryμ∈F
andd(S)=1forallS∈E.
IfFcontains 2 elements, thend(S)=1foreveryS∈Eis the only possibility. If
Fcontains 3 elements, thend(S)=±1foreveryS∈E. Henced(S−^1 )=d(S)and
d(S^2 )=1. SinceU 2 =U 12 andU 1 =U 2 −^1 , this impliesd(Uλ)=1foreveryλ∈F,
and the rest follows as before.
The preceding discussion is easily extended to higher dimensions. Put
Uij(λ)=In+λEij,
for anyi,j∈{ 1 ,...,n}withi=j,whereEijis then×nmatrix with all entries 0
except the(i,j)-th, which is 1, and letSLn(F)denote the set of allA∈Mnwhich
are products of finitely many matricesUij(λ).ThenSLn(F)is a group under matrix
multiplication.
IfA∈MnandA=O, then there existS,T ∈SLn(F)and a positive integer
r≤nsuch that