1 What is a Determinant? 227SAT=diag[1r− 1 ,δ, (^0) n−r]
for some nonzeroδ∈F. The matrixAissingularifr<nandnonsingularifr=n.
HenceA=(αjk)is nonsingular if and only if its transposeAt=(αkj)is nonsingular.
In the nonsingular case we need multiplyAon only one side by a matrix fromSLn(F)
to bring it to the form
Dδ=diag[1n− 1 ,δ].
For ifSAT = Dδ,thenSA= DδT−^1 and this impliesSA= S′Dδ for some
S′∈SLn(F),since
DδUij(λ)=Uij(λδ−^1 )Dδ ifi<j=n,
DδUij(λ)=Uij(δλ)Dδ ifj<i=n,
DδUij(λ)=Uij(λ)Dδ ifi,j=nandi=j.
In the same way as forn=2 it may be shown that, ifd:SLn(F)→Fis a map
such thatd(ST)=d(S)d(T)for allS,T∈SLn(F), then eitherd(S)=0foreveryS
ord(S)=1foreveryS.
Theorem 1There exists a unique map d:Mn→F such that
(i)′d(AB)=d(A)d(B)for all A,B∈Mn,
(ii)′for anyα∈F, if Dα=diag[1n− 1 ,α],thend(Dα)=α.
Proof We consider first uniqueness. Sinced(I)=d(D 1 )=1, we must haved(S)= 1
for everyS∈SLn(F), by what we have just said. Also, if
H=diag[η 1 ,...,ηn− 1 ,0],
thend(H)=0, sinceH=D 0 H. In particular,d(O)=0. IfA∈MnandA=O,
there existS,T∈SLn(F)such that
SAT=diag[1r− 1 ,δ, (^0) n−r],
where 1≤r≤nandδ=0. It follows thatd(A)=0ifr<n,i.e.ifAis singular. On
the other hand ifr=n,i.e.ifAis nonsingular, thenSAT=Dδand henced(A)=δ.
This proves uniqueness.
We consider next existence. For anyA=(αjk)∈Mn,define
detA=
∑
σ∈Sn(sgnσ)α 1 σ 1 α 2 σ 2 ···αnσn,whereσis a permutation of 1, 2 ,...,n,sgnσ =1or−1 according as the permu-
tationσis even or odd, and the summation is over the symmetric groupSnof all
permutations. Several consequences of this definition will now be derived.
(i)if every entry in some row of A is0,thendetA=0.
Proof Every summand vanishes in the expression for detA.