Number Theory: An Introduction to Mathematics

240 V Hadamard’s Determinant Problem

such that

TtG 1 T=diag[γ 1 ,...,γp], TtH 1 T=diag[δ 1 ,...,δp].

SinceG 3 −^1 is positive definite,ut(G 1 −H 1 )u≥0foranyu∈Rp. Henceγi≥δi> 0
fori= 1 ,...,pand detG 1 ≥detH 1. Moreover detG 1 =detH 1 only ifγi=δifor
i= 1 ,...,p.
Hence if detG 1 =detH 1 ,thenG 1 =H 1 ,i.e.G 2 G− 31 Gt 2 =0. ThuswtG− 31 w= 0
for any vectorw = Gt 2 v.SincewtG− 31 w = 0 impliesw = 0, it follows that
G 2 =0.

From Proposition 12 we obtain by induction

Proposition 13If G=(γjk)is an m×m positive definite real symmetric matrix, then

detG≤γ 11 γ 22 ···γmm,

with equality if and only if G is a diagonal matrix.

By applying Proposition 13 to the matrixG=AtA, we obtain again Proposition 3.

Proposition 13 may be sharpened in the following way:

Proposition 14If G=(γjk)is an m×m positive definite real symmetric matrix, then

detG≤γ 11

∏m

j= 2

(γjj−γ 12 j/γ 11 ),

with equality if and only ifγjk=γ 1 jγ 1 k/γ 11 for 2 ≤j<k≤m.

Proof If

T=

[

1 g

0 Im− 1

]

,

whereg=(−γ 12 /γ 11 ,...,−γ 1 m/γ 11 ),then

TtGT=

[

γ 11 0

0 H

]

,

whereH=(ηjk)is an(m− 1 )×(m− 1 )positive definite real symmetric matrix with

entries

ηjk=γjk−γ 1 jγ 1 k/γ 11 ( 2 ≤j≤k≤m).

Since detG=γ 11 detH, the result now follows from Proposition 13. 

Some further inequalities for the determinants of positive definite matrices will

now be derived, which will be applied to Hadamard’s determinant problem in the next

section. We again denote byJmthem×mmatrix whose entries are all 1.