Number Theory: An Introduction to Mathematics

5 Application to Hadamard’s Determinant Problem 245

ReplacingAbyDA, we obtainAAt = AtA.ThenAcommutes withAtAand
hence also withJn. Thus the rows and columns ofAall have the same sumsand
AJn=sJn=AtJn. Moreovers^2 = 2 n−1, since

s^2 Jn=sAtJn=AtAJn=( 2 n− 1 )Jn. 

The maximum value of det(AtA)whenn≡3 mod 4 is still a bit of a mystery. We
now consider the remaining case whennis even, but not divisible by 4.

Proposition 20Let A=(αjk)be an n×m matrix with 2 ≤m≤n andαjk=± 1
for all j,k. If n≡2mod4and n> 2 ,then

det(AtA)≤(n− 2 )m−^2 (n− 2 +m)^2 if m is even,

det(AtA)≤(n− 2 )m−^2 (n− 1 +m)(n− 3 +m) if m is odd.

Moreover, equality holds if and only if there is a signed permutation matrix U such that

UtAtAU=

[

L 0

0 M

]

,

where

L=M=(n− 2 )Im/ 2 + 2 Jm/ 2 if m is even,

L=(n− 2 )I(m+ 1 )/ 2 + 2 J(m+ 1 )/ 2 ,M=(n− 2 )I(m− 1 )/ 2 + 2 J(m− 1 )/ 2 if m is odd.

Proof We need only show thatG=AtAsatisfies the hypotheses of Proposition 17
withα=n−2andβ=2. We certainly haveγjj=n.Moreoverallγjkare even,
sincenis even and

γjk=α 1 jα 1 k+···+αnjαnk.

Hence|γjk|≥2ifγjk=0. Finally, ifj,k,,are all different andγj=γk=0, then

∑n

i= 1

(αij+αik)(αij+αi)=n+γjk.

Sincen≡2 mod 4, it follows that alsoγjk≡2 mod 4 and thusγjk=0.

Again there is no guarantee that the upper bound in Proposition 20 is attained.
However the question may be reduced to the existence ofH-matrices ifm=n,n−1.
For supposem≤n−2 and there exists an(n− 2 )×mH-matrixB. If we put

A=

[

B

C

]

,

where

C=

[

er es

er −es

]

,