252 V Hadamard’s Determinant Problem
Twon×nHadamard matricesH 1 ,H 2 are said to beequivalentif one may be
obtained from the other by interchanging two rows or two columns, or by changing
the sign of a row or a column, or by any finite number of such operations. Otherwise
expressed,H 2 =PH 1 Q,wherePandQare signed permutation matrices. Anauto-
morphismof a Hadamard matrixHis an equivalence ofHwith itself:H=PHQ.
SinceP = HQ−^1 H−^1 , the automorphism is uniquely determined by Q. Under
matrix multiplication all admissibleQform a groupG,theautomorphism groupof
the Hadamard matrixH. Evidently−I∈Gand−Icommutes with all elements of
G. The factor groupG/{±I}, obtained by identifyingQand−Q, may be called the
reduced automorphism groupofH.
To illustrate these concepts we will show that all Hadamard matrices of order 12
are equivalent. In fact rather more is true:
Proposition 22Any Hadamard matrix of order 12 may be brought to the form
+++ +++ +++ +++
+++ +++ −−− −−−
+++ −−− +++ −−−
+−+ −+− −+− +−+
++− −−+ −−+ +−+
−++ +−− +−− +−+
+−+ −−+ +−− ++−
+−+ +−− −−+ −++
++− −+− +−− −++
−++ −+− −−+ ++−
++− +−− −+− ++−
−++ −−+ −+− −++(∗)
(where+stands for 1 and−for− 1 ) by changing the signs of some rows and columns,
by permuting the columns, and by permuting the first three rows and the last seven
rows.
Proof LetA=(αjk)be a Hadamard matrix of order 12. By changing the signs of
some columns we may assume that all elements of the first row are+1. Then, by the
orthogonality relations, half the elements of any other row are+1. By permuting the
columns we may assume that all elements in the first half of the second row are+1. It
now follows from the orthogonality relations that in any row after the second the sum
of all elements in each half is zero. Hence,by permuting the columns within each half
we may assume that the third row is the same as the third row of the array (∗) displayed
above. In ther-th row, wherer>3, letρkbe the sum of the entries in thek-th block
of three columns(k= 1 , 2 , 3 , 4 ). The orthogonality relations now imply that
ρ 1 =ρ 4 =−ρ 2 =−ρ 3.In thes-th row, wheres>3ands=r,letσkbe the sum of the entries in thek-th
block of three columns. Then also
σ 1 =σ 4 =−σ 2 =−σ 3.