Number Theory: An Introduction to Mathematics

5 Application to Hadamard’s Determinant Problem 243

H=

[

LN

Nt M

]

,

whereL,Mare square matrices of ordersr′,s′respectively. By construction all ele-
ments in the first row ofLare positive and all elements in the first row ofNare zero.
Furthermore, by the hypotheses of the proposition, all elements ofMhave absolute
value≥β.
By the induction hypothesis,

δ≤αm−^3 (α+sβ)(α+mβ−β−sβ).

Ifη≤0, it follows immediately that (4) holds with strict inequality. Thus we now
supposeη>0. ThenHis positive definite and hence, by Fischer’s inequality (Propo-
sition 12),η≤detL·detM, with equality only ifN=0. But, by Proposition 14,

detL≤β

∏r′

j= 2

(γjj−γ 12 j/β)≤αr

′− 1

β

and, by Proposition 16,

detM≤αs

′− 1

(α+s′β).

Hence

detG≤αm−^2 (α+sβ)(α+mβ−β−sβ)+αm−^2 β(α+s′β),

Sinces′≤s, it follows that (4) holds and actually with strict inequality ifs′=s.
If equality holds in (4) then, by Proposition 14, we must haveL=αIr′+βJr′,
and by Proposition 16 after normalization we must also haveM=αIs′+βJs′.

5 ApplicationtoHadamard’sDeterminantProblem.................

We have seen that, ifAis ann×mreal matrix with all entries±1, then det(AtA)≤nm,
with strict inequality ifn>2andnis not divisible by 4. The question arises, what
is the maximum value of det(AtA)in such a case? In the present section we use the
results of the previous section to obtain some answers to this question. We consider
first the case wherenis odd.

Proposition 18Let A=(αjk)be an n×m matrix withαjk=± 1 for all j,k. If n is
odd, then

det(AtA)≤(n− 1 )m−^1 (n− 1 +m).

Moreover, equality holds if and only if n≡1mod4and, after changing the signs of
some columns of A,

AtA=(n− 1 )Im+Jm.