246 V Hadamard’s Determinant Problem
andr+s=m,then
AtA=[
(n− 2 )Ir+ 2 Jr 0
0 (n− 2 )Is+ 2 Js]
.
Thus the upper bound in Proposition 20 is attained by takingr=s=m/2whenmis
even andr=(m+ 1 )/2,s=(m− 1 )/2whenmis odd.
Suppose now thatm=nand
AtA=[
L 0
0 L
]
,
whereL=(n− 2 )In/ 2 + 2 Jn/ 2 .IfBis then×(n− 1 )submatrix ofAobtained by
omitting the last column, then
BtB=[
L 0
0 M
]
,
whereM=(n− 2 )In/ 2 − 1 + 2 Jn/ 2 − 1. Thus if the upper bound in Proposition 20 is
attained form=n, then it is also attained form=n−1. Furthermore, since
det(AAt)=det(AtA),it follows from Proposition 20 that thereexists a signed permutation matrixUsuch that
UAAtUt=AtA.ReplacingAbyUA, we obtainAAt=AtA.ThenAcommutes withAtA.If
A=
[
XY
ZW
]
,
is the partition ofAinto square submatrices of ordern/2, it follows thatX,Y,Z,W
all commute withLand hence withJn/ 2. This means that the entries in any row or
any column ofXhave the same sum, which we will denote byx. Similarly the entries
in any row or any column ofY,Z,Whave the same sum, which will be denoted by
y,z,wrespectively. We may assumex,y,w≥0 by replacingAby
[
In/ 2 0
0 ±In/ 2
]
A
[
±In/ 2 0
0 ±In/ 2]
,
We h av e
XtX+ZtZ=YtY+WtW=L, XtY+ZtW= 0 ,and
XXt+YYt=ZZt+WWt=L, XZt+YWt= 0.Postmultiplying byJ, we obtain
x^2 +z^2 =y^2 +w^2 = 2 n− 2 , xy+zw= 0 ,