A construction for generalized hadamard matrices GH(4q, EA(q))

Journal of Statistical North-Holland

Planning

and Inference

103

11 (1985) 103-l 10

A CONSTRUCTION FOR GENERALIZED MATRICES GH(4q, EA(q)) Jeremy

HADAMARD

E. DAWSON

Australian National University, Canberra, Australia Received

19 January

Recommended

1984

by J. Seberry

and N.M.

Singhi

Abstract: We give a construction for a generalized Hadamard matrix GH(4q, EA(q)) as a 4 x 4 matrix of q x q blocks, for q an odd prime power other than 3 or 5. Each block is a GH(q, EA(q)) and certain matrix

combinations

of 4 blocks

exists for every prime

power

form

GH(Zq,EA(q))

matrices.

Hence

a GH(4q,EA(q))

q.

AMS Subject Classification: Primary

62K10,

Key words and phrases: Orthogonal

62K15;

Secondary

OSB15.

arrays.

For q an odd prime power, Street (1979) and Jungnickel (1979) have given a construction for a GH(2q, EA(q)), i.e. a generalized Hadamard matrix of order 2q over EA(q), the elementary abelian group of order q (which is also the additive group of GF(q)). Their constructions are cases of the following theorem.

Theorem 1. Let q be an odd prime power. Let the matrix H be formed by four blocks Hik, i, k = 1,2, where each Hik is a q x q matrix whose rows and columns are indexed by the elements of GF(q). Let (HikIm,

=

2

aikz

+ PikXZ

+

YikX

2

where a11

a12

-a21

-

a22 (1)

Plll321

=

Y11-

Yl2

=

P12/322 Y21

-

’

Y22 (2)

PllP12

P21P22

all-a21 4 X(P11

’

y11-~12

PllP21

&l/322-P12P21

pll=

P12P21

P22)

Pll =

0

1985, Elsevier

(3) P22

’

(4)

-1,

and the numerators and denominators a GH(2q, EA(q)) matrix. 0378-3758/85/$3.30

P12P21

Science

of (I), (2) and (3) are non-zero. Then H is

Publishers

B.V. (North-Holland)

J.E. Dawson

104

/ Generalized

Hadamard

matrices

Proof.

Each Hik is a generalized Hadamard matrix (with the group written additively), since the differences between corresponding elements of rows x and y are {(oikZ2+/3ikXZ+ YikX2)-(((Y;kz2+P;ky~+YikY2) = {fiik(X-_Y)Z+ Now,

: ZEGF(~)}

yik(X2-_Y2) : ZEGF(q))

= GF(q).

the differences between corresponding elements of rows 1,x and 2, y of H are ,;,

Uolk-%k)z2+(Plk

are working here with multi-sets)

(we

(alk-a2k)

k=l,2

blkX--P2ky

(

“I

=

~-P~~Y)z+Y~Ax~-Y~~Y~:zEGF(~)}

z+

2(alk

2

+Ck:zEGF(q)

- a2k) >

where c, = yrkx2 - y2ky2 -

1

(PlkX-b2kY)2

4(% -

a2k)

>

{(a,k-a2k)Z2+Ck:ZEGF(q)}.

=,;,

We shall show that Cr = C2 (for any x, y); (1) and (4) imply that of all - a21 and or2 - (Yap,one is a square and the other is not. Now {(ark - 02k)z2 : z E GF(q)} runs through each square (resp. non-square) twice and zero once if ark - o2k is a square (resp. non-square), and so ,_j,

{(alk-a2k)z2:

zEGF(q))

= GF(q)UGF(d.

Adding the constant Ct (=C,) throughout makes no difference, and so the set of differences between corresponding elements of the two rows runs through GF(q) twice. To show that Cr = C2 (regardless of the choice of x and y), we look separately at the terms in x2, xy and y2. Thus we require Pf2

El

0)

= Y12-

Y114&l

4@12

- a211

PllP21

=

or1 - a21

- a221

P12P22 al2

- a22

(ii) ’

P:l -Y21-

’

4&t -

Pi2 = -Y22

(iii)

Wl2

a21)

- a221

’

Now (ii) is equivalent to (1); (3) gives 1 Yll

-Y12

=

Wll

- a211

(

p,

PL

=

_

Pll

P12P21 P22

>

P?2 (by

4@lI

- a21)

-

Wl2

-

a221

(l)),

J.E. Dawson

which implies 4

/ Generalized

Hadamard

105

matrices

(i). Also (2) and (3) give a11 - a21 Y21 - Y22 Pll

which similarly

P21

implies

Pll~22-P12/32l

P21 P22

PllP12P2lP22

(iii). Thus

We use this construction

Ct = C2, and the proof

is complete.

0

to form our GH(4q,EA(q)).

2. Let H consist of 16 q x q blocks Hik, i, k= 1,2,3,4, lowing 12 submatrices are GH(2q, EA(q)) matrices:

Lemma

HI,

H12

H13

H14

H31

H32

H33

H34

ff21

H22

H23

H24

H41

H42

H43

H44

ff1l

H13

ffl2

H14

Hz1

H23

H22

H24

H31

H33

H32

H34

ff41

H43

H42

&I

H1,

ff14

H12

H13

H21

H24

ff22

H23

H41

H44

H42

ff43

H31

H34

H32

H33

Then H is a GH(4q, EA(q))

such that the fol-

matrix.

Proof. Consider, for example, rows 2,x and 3,~. The differences between corresponding elements of them consist of each element of GF(q) occurring twice in 2; 2; and twice in 2; $1, i.e. four times in all. The same happens with any pair of rows. 0 We need to determine the 48 coefficients oik, Pik, yik, i, k = 1,2,3,4, such that conditions (l)-(4) of Theorem 1, with indices changed appropriately, are satisfied for each of these 12 submatrices. For q = 3 this is not possible, as the four ait must be distinct. and q=9.

We have a method

to construct

a, p and y otherwise,

except for q = 5

3. Let A, B, C,D be non-zero elements of GF(q) such that Am2+ Bp2+ Cm2+ De2 = 0 and x(ABCD) = - 1. Let

Lemma

where a=a’=O, c=

b is arbitrary non-zero,

bAW-BC) (A2+B2)D

d

’

=

bA(AC+BO (A2+B2)C

’

106

J.E.

Dawson / Generalized Hadamard

A2+B2 b’ = ~ 4b ’

matrices

c, = b, A(AD + BC) (A2+B2)D ’

d, = b,A(AC-BD) (A2+B2)C

For i, k = 1,2,3,4, let (Hik)xz = aikz2 + PikXZ + yikX2, taking the values of aik, Yik from the matrices above. Then each of the 12 matrices of Lemma GH(2q, Wq)).

* Pik

and 2 is a

PrOOf. We show that (Yik, Pik and yik Satisfy equations (l)-(4) in respect of each of the 12 matrices of Lemma 2. Equation (1) for the 12 submatrices becomes

a-b d-c AB=Dc’ (occurring

four times each),

al-d’ AD= (four

times each).

a-c b-d AC=BD’ and equation

b’- c’ BC’

(2) becomes

a’- b’ c’-d’ AB =F’

(AC-BD) 4ABCD

a’- c’ AC=

of a =a’=O, equation

By substitution

b d’ AB AD

a-d c-b AD=cB

’

c b’ ACAB

(AD-BC) 4ABCD

d’- b’ DB

(3) gives

’

d c’ --= ADAC

(AB-CD) 4ABCD

in each case equation (4) is x(ABCD) = -1. The reader are satisfied (this requires frequent use of the identity A2B2C2 + A2B2D2 + A2C2D2 + B2C2D2 = 0, as implied by the hypothesis). The result now follows from Theorem 1. q (four times may verify

It remains some results

each) and that these

to show that there exist A, B, C, D as in Lemma on finite fields.

3. We first assemble

Lemma

4. (i) For q an odd prime power, 427, any non-zero element a of GF(q) is the sum of two non-zero squares. (ii) Further, if q = 1 (mod 4) and the fourth powers (including zero) form a subfield of GF(q), then q=9. Proof.

(i) Since XF&C4jX(X) = 0

and

c X(X)X(Y) =X;OX z-1 x+y=(l (

>

=-x(-l),

a simple counting argument shows that (x(x), X(Y)) = (1,l) for at least t - 1 pairs (x, y), where q = 4t + 1. is of order pm where m ) n. In this case, then, (ii) Let q=p”; a subfield 4=(p”-l)/(p”-l)= 1 +pm+...+p(“-l)m, whence n=2 and pm=3. 0 Theorem 5. If q is an odd prime power other than 3, 5 or 9, then there is a generalized Hadamard matrix of order 4q over EA(q).

J.E. Dawson

/ Generalized

Hadamard

matrices

107

Proof. We look at cases. If x(-l)

= -1, set A = B = 1, then by Lemma 4(i) we can find C and D such that Am2 + Be2 + C-2 + De2 = 0. If x(ABCD) = 1, replace D by -D, and the hypotheses of Lemma 3 are satisfied. On the other hand, let x(- 1) = 1. Suppose also that there exist non-zero x, y, z such that x(x), x(y) and x(z) are not all equal and x2 +y2 + z2 = 0. Let X(X) #x(y). Then 0 = (x2-y2)(x2+y2+22)

=x4-y4+x2z2-yzzz,

a sum of four squares; let A =xm2, etc. Then &4BCD) =x(x2y2xzyz(-1)) =x(xy) = -1, satisfying Lemma 3. Suppose alternatively that x2 +y2 + z2 = 0 (x, y, z # 0) implies X(X) =x(y) =x(z). Choose x, y such that I = -1 and x2 +y2 #0 (which is possible for q> 5). It follows that x(x2+y2)= -1. Suppose that there exist u and u such that x(u)= x(u)= 1 and x(u2+u2)=-1. Thus we may let s~=-_(u~+u~)/(x~+~~). Then 0=s2x2 +s2y2 + u2 + u2, the sum of four non-zero squares; setting A = (.sx)-’ etc. we get &lBCD) =x(sxsyuo) = -1, again satisfying Lemma 3. Suppose therefore that x(u) =x(u) = 1 implies that x(u2+ u2) = 1 (or 0); then if w2 = -(u2 + u2) #0 our earlier supposition implies that K(W) = 1. Thus, in the case q= 1 (mod S), when -1 is a fourth power, u2 + v2 = -w2, a fourth power. It follows that the fourth powers form a subfield; however, this cannot occur unless q = 9, by Lemma 4(ii). In the case q =p”= 5 (mod 8), when -1 is not a fourth power, we have, for fourth powers u2 and u2, -(u2 + u2) = w2, a fourth power. Then we show by induction that 3k + 1 is a fourth power for kz 0. Suppose 3k+ 1 is a fourth power, as is 1; we have -(l + 1) = -2 and -((3k+l)+l)=-3k-2 are fourth powers, whence -((-3k-2)-2)=3(k+l)+l is also. Since ~23, this implies that -1 is a fourth power, a contradiction. Thus in all cases there exist non-zero A, B, C, D such that K2 + BP2 + Cm2 + DP2 = 0 and x(ABCD) = -1. The result now follows from Lemmas 3 and 2. 0 We note that, for any 4 x 4 matrices a, /I, y forming a solution to (l)-(4) in respect of the 12 matrices in Lemma 2, any of the following changes produces another solution: (a) adding a constant to each entry in any column of (Y, (b) adding a constant to each entry in any row of y, (c) multiplying each entry of any column of /I by a non-zero constant k and multiplying each of the same column of (Y by k2, (d) multiplying each entry of any row of /3 by a non-zero constant k and multiplying each entry of the same row of y by k2, (e) multiplying each entry of a by a non-zero constant k and multiplying each entry of y by l/k, (f) multiplying by -1 each element of any of the sets {&i, p22, &, &}, ~P12?P21yb34pf143}y

{P13pf124yb31pb42}

Or {b149b23,b329fi41}?

performing any permutation on the row indices permutation on their column indices, and (h) transposing /I and interchanging (r and yT, (g)

of a, /I and y, and the same

108

J.E. Dawson / Generalized Hadamard matrices

In seeking solutions, we may therefore assume that all alk, yii =O, and all Pik, pii = 1. If q= 1 (mod 3), a solution can also be found as follows. Let o (21) be a cube root of 1, and let A,& C be non-zero elements of GF(q) such that 1/A-1=02(1/B-1)=c0(l/C-1). Let 0

0

0

0

1

A

C

B

1ACB P=

A-1 ‘=

4A

0

A

w2C

WB

0

B

w2A

WC

1

1

BAC

1

C

1

B

I-

A

1.

’

Then it may be checked that (l)-(3) are satisfied for the 12 matrices of Lemma 2. (Note that 1 + o + o2 = 0, whence l/A + l/B + 1/C= 3 and l/A + o/B + 02/C = 0.) Equation (4) is satisfied in all 12 cases if x(A) =x(BC) = -1. The problem of finding A, B, C of which exactly one, not A, is a square is that of finding x such that exactly one of x+ 1, ox+ 1, 02x+ 1 is a square, or, equivalently, such that not all w’x+ 1 are squares but their product x3 + 1 is. Now x= 2 suffices unless o - o2 (=20+ 1) and 02-o are both squares (when x+ 1 = 3 = -(02 - w)~ is always a square). In this latter case, a counting argument, based on the results XEg(qjX(X) = 0

and

re&(qj X(X+ o’)x(x+u~)

= -1

(o’*o’),

ensures that there exists XE GF(q) such that exactly one of x+ 1,x+ o, x+ o2 (equivalently, one of x + 1, ox + 1, w2x + 1) is a square. The question remains for q = 3, 5 and 9. No solution is possible for q = 3, since a column of (Ymust contain four distinct values. A computer programme was used to look for solutions for q = 5 and 9. It showed that there is no solution for q = 5, and four essentially different (in the sense of the remarks following Theorem 5) solutions for q = 9. These are given in Table 1 (for m # 0, m denotes (I+ fi)m). 6. If q is a prime power, then there is a generalized Hadamard matrix of order 4q over EA(q).

Theorem

Proof.

If q is even, then the result is well known (by factoring EA(4) out of a GH(4q, EA(4q)) matrix). If q = 3, the required matrix is given by Seiden (1954). If q = 5, the result is a case of Theorem 1 of Seberry (1980). If q = 9, the result follows from Lemma 2 and Theorem 1, using the matrices a, /I and y given above. The remaining cases form Theorem 5. 0

J.E. Dawson

/ Generalized

Hadamard

matrices

109

Table 1

B

a

Other matrices.

0

0

0

8

8

8

8

0

2

I

3

4

1

6

I

8

5

6

I

0

7

1

6

8

I

3

2

8

3

3

6

0

1

2

5

7

5

4

6

8

2

1

7

0

8

4

2

0

0

0

0

8

8

8

8

4 8

1 7

2 1

5 6

8

5

6

1

0 0

2 7

5 3

1 2

6 7

7

8

1 5

6

6

7 2

1

3

8 8

0

1

0

4

2

8

0

0

0

0

8 5

1

2

8 6

4

3

8 3

2

8

8 8

0

5

0

5

2

6

3

4

6

7

4 7

5

8

3 1

7

2

1 2

3

3

8 8

0

1

0

4

5

8

0 1

0 2

0 7

0 4

8 8

8 1

8 6

8 3

0

4

8

6

0

5

6

1

6

5

3

2

8

7

5

4

0

3

5

2

7

1

2

8

8

2

3

1

0

6

3

I

configurations

be constructed

these

generalized

Corollary 7. For q a prime power, there exist (i) 4q - 1 mutually orthogonal F-squares, F(4q; 4), (ii) an orthogonal array [sq2, sq + 1, q, 21, (iii) a group divisible PBIBD with v = b = 4q2, r = k = 49, 4q groups each of size q, A,=0 and A2=4, and (iv) if q = 1 (mod 3), a BIBD

492-l 4q2, ~ 3

q9

4q2- 1 ---_,49,-3

and generally a BIBD(4q2,(4q2-l)q,4q2-1,49,49-l). Proof. (i) follows from Seberry (1980), Theorem 2. (ii) follows from Shrikhande (1964). (iii) A group G has a natural representation as a transitive permutation group on 1G 1 objects; we replace each entry in the GH(4q, EA(q)) by the q x q permutation matrix representing it to get the incidence matrix of the required design. (iv) For q= 1 (mod 3), take a 49-l 9, -_,4,1 3

J.E.

110

Dawson / Generalized Hadamard

matrices

(which necessarily exists) and replace each variety in it by a group of q varieties. Thus we have a group divisible design with varieties, groups and block size as in the design of (iii), and with A2,= (4q- 1)/3 and A2= 1. We take one copy of this design and (q- I)/3 copies of the design of (iii) to get the BIBD with L = (4q - I)/3 = 1 + 4(q - 1)/3. The second assertion is proved similarly. q Corollary 8. Let q be a prime power. Then there exists a generalized Hadamard matrix GH(2mqk; EA(q)) for m s2k, k? 1.

Take the Kronecker product of copies of GH(q, EA(q)), GH(2q, EA(q)) and GHVq, EA(q)). •I

Proof.

Further corollaries may be deduced from this result along the lines of Corollary 7.

References Jungnickel,

D. (1979). On difference

matrices. Seberry,

matrices,

resolvable

transversal

designs and generalized

Hadamard

Math. Z. 167, 49-60.

J. (1980). A construction

for generalized

Hadamard

matrices.

J. Statist. Plann. Inference

4,

365-368. Seiden,

E. (1954). Construction

Shrikhande,

of orthogonal

S.S. (1964). Generalized

arrays.

Hadamard

Ann.

matrices

Math. Statist. 25, 151-156.

and orthogonal

arrays

of strength

2. Canad.

J. Math. 16, 736-740. Street,

D.J.

(1979). Generalized

toria 8, 131-141.

Hadamard

matrices,

orthogonal

arrays

and F-squares.

Ars Combina-