242 V Hadamard’s Determinant Problem
with equality only ifγjk=γ 1 jγ 1 k/βfor 2≤j<k≤m. Henceη≤αm−^1 β, with
equality only ifγjj=α+βfor 2≤ j≤mandγjk=βfor 1≤j <k≤m.
Consequently
detG≤αm−^1 (α+mβ−β)+αm−^1 β=αm−^1 (α+mβ),with equality only ifG=αIm+βJm.
A square matrix will be called asigned permutation matrixif each row and column
contains only one nonzero entry and this entry is 1 or−1.
Proposition 17Let G=(γjk)be an m×m positive definite real symmetric matrix
such thatγjj≤α+βfor all j and eitherγjk= 0 or|γjk|≥βfor all j,k, where
α,β > 0.
Suppose in addition thatγik=γjk= 0 impliesγij= 0 .Then
detG≤αm−^2 (α+mβ/ 2 )^2 if m is even,
detG≤αm−^2 (α+(m+ 1 )β/ 2 )(α+(m− 1 )β/ 2 ) if m is odd.(3)
Moreover, equality holds if and only if there is a signed permutation matrix U such that
UtGU=[
L 0
0 M
]
,
where
L=M=αIm/ 2 +βJm/ 2 if m is even,
L=αI(m+ 1 )/ 2 +βJ(m+ 1 )/ 2 ,M=αI(m− 1 )/ 2 +βJ(m− 1 )/ 2 if m is odd.Proof We are going to establish the inequality
detG≤αm−^2 (α+sβ)(α+mβ−sβ), (4)wheresis the maximum number of zero elements in any row ofG. Since, as a function
of the real variables, the quadratic on the right of (4) attains its maximum value for
s=m/2, and has the same value fors=(m+ 1 )/2asfors=(m− 1 )/2, this will
imply (3). It will also imply that if equality holds in (3), thens=m/2ifmis even and
s=(m+ 1 )/2or(m− 1 )/2ifmis odd.
Form=2 it is easily verified that (4) holds. We assumem>2 and use induction.
By performing the same signed permutation on rows and columns, we may suppose
that the second row ofGhas the maximum numbersof zero elements, and that all
nonzero elements of the first row are positive and precede the zero elements. All the
hypotheses of the proposition remain satisfied by the matrixGafter this operation.
Lets′be the number of zero elements in the first row and putr′=m−s′.Asin
the proof of Proposition 16, we have
detG=(γ 11 −β)δ+η,whereδis the determinant of the matrix obtained fromGby omitting the first row and
column andηis the determinant of the matrixHobtained fromGby replacingγ 11 by
β. We partitionHin the form