Number Theory: An Introduction to Mathematics

4 Some Matrix Theory 241

Lemma 15If C=αIm+βJmfor some realα,β,then

detC=αm−^1 (α+mβ).

Moreover, ifdetC= 0 ,thenC−^1 =γIm+δJm,whereδ=−βα−^1 (α+mβ)−^1 and
γ=α−^1.

Proof Subtract the first row ofCfrom each of the remaining rows, and then add to the
first column of the resulting matrix each of the remaining columns. These operations
do not alter the determinant and replaceCby an upper triangular matrix with main
diagonal entriesα+mβ(once) andα(m−1 times). Hence detC=αm−^1 (α+mβ).
If detC=0andifγ,δare defined as in the statement of the lemma, then from
Jm^2 =mJmit follows directly that

(αIm+βJm)(γIm+δJm)=Im. 

Proposition 16Let G=(γjk)be an m×m positive definite real symmetric matrix
such that|γjk|≥βfor all j,k andγjj≤α+βfor all j , whereα,β > 0 .Then

detG≤αm−^1 (α+mβ). (2)

Moreover, equality holds if and only if there exists a diagonal matrix D, with main
diagonal elements± 1 , such that

DGD=αIm+βJm.

Proof The result is trivial ifm=1 and is easily verified ifm=2. We assumem> 2
and use induction onm. By replacingGbyDGD,whereDis a diagonal matrix
whose main diagonal elements have absolute value 1, we may suppose thatγ 1 k≥ 0
for 2≤k≤m. Since the determinant is a linear function of its rows, 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
byβ. By the induction hypothesis,

δ≤αm−^2 (α+mβ−β).

Ifη≤0, it follows that

detG≤αm−^1 (α+mβ−β)<αm−^1 (α+mβ).

Thus we now supposeη>0. ThenHis positive definite, since the submatrix
obtained by omitting the first row and column is positive definite. By Proposition 14,

η≤β

∏m

j= 2

(γjj−γ 12 j/β),