Transcription

1 Math 6023 Topics: Design and Graph Theory Lecture Notes 2: Another class of square designs arises from Hadamard matrices. DEFINITION AND BASIC PROPERTIES An n x n matrix H = h ij is an Hadamard matrix of order n if the entries of H are either +1 or -1 and such that HH t = ni, where H t is the transpose of H and I is the order n identity matrix. Put another way, a (+1,-1)-matrix is Hadamard if the inner product of two distinct rows is 0 and the inner product of a row with itself is n. A few examples [check the definition for each] of Hadamard matrices are; These matrices were first considered as Hadamard determinants. They were so named because the determinant of an Hadamard matrix satisfies equality in Hadamard's determinant theorem, which states that if X = x ij is a matrix of order n where x ij 1 for all i and j, then det X n n/2 It is apparent that if the rows and columns of an Hadamard matrix are permuted, the matrix remains Hadamard. It is also true that if any row or column is multiplied by -1, the Hadamard property is retained.[prove this] Thus, it is always possible to arrange to have the first row and first column of an Hadamard matrix contain only +1 entries. An Hadamard matrix in this form is said to be normalized. Theorem - The order of an Hadamard matrix is 1,2 or 4n, n an integer. Proof: [1] is an Hadamard matrix of order 1 and the first example above is an Hadamard matrix of order 2. Suppose now that H is an Hadamard matrix of order h 3. Normalize H and rearrange the first three rows to look like: ÀÄÄ ÚÄÄÙ ÀÄÄ ÚÄÄÙ ÀÄÄ ÚÄÄÙ ÀÄÄ ÚÄÄÙ x y z w Where x,y,z,w are the numbers of columns of each type. Then since the order is h, x + y + z + w = h and taking the inner products of rows 1 and 2, 1 and 3, and, 2 and 3 we get

2 x + y - z - w = 0 x - y + z - w = 0 x - y - z + w = 0. Solving this system of equations gives, x = y = z = w = h/4. Thus, the integer h must be divisible by 4. Corollary - If H is a normalized Hadamard matrix of order 4n, then every row (column) except the first has 2n minus ones and 2n plus ones, further n minus ones in any row (column) overlap with n minus ones in each other row (column). Proof: This is a direct result of the above proof since any two rows other than the first can take the place of the second and third rows in the proof. The same argument can be applied to the columns. Hadamard matrices are known for many of the possible orders, the smallest order for which the existence of an Hadamard matrix is in doubt is currently 668. (The previous smallest order of 428 was constructed in June 2004 by Kharaghani & Tayfeh-Rezaie.) CONSTRUCTION METHODS While there are a great many construction methods for Hadamard matrices, we will only consider two. The first, one of the simplest, is the direct product construction. Construction - Given Hadamard matrices H 1 of order n and of order m the direct product of these two matrices, represented by: H [h11 2 h12 H 2 h1n H 2 h 21 h 22 h 2n ] h n1 h n2 h nn where H = h ij, is an Hadamard matrix of order nm. Proof: [Left as an exercise]. Example: Let H = [ ], H * = then the direct product H H* is [ ]

3 [ H* H * *] H * H = [ ] Homework: Starting with the Hadamard matrix of order 2, repeatedly use the direct product construction to construct an Hadamard matrix of order 32. Hadamard matrices constructed this way are also called Sylvester matrices. A second construction is due to Paley(1933). Construction: Let q be a prime power 3 mod 4, and let F = GF(q), and Q be the set of nonzero squares in F. Define the matrix P = p ij = 1 if i - j Q and -1 otherwise. Border P with a row and column of 1's to get an Hadamard matrix. Example: Let q = 11, then Q = {1,4,9,5,3} and the Hadamard matrix constructed this way is H = [ ]. Hadamard 2-Designs Hadamard matrices of order 4t can be used to create symmetric 2-designs, which are called Hadamard 2-Designs. The construction actually forms the incidence matrix of the 2-design. The Hadamard designs have parameters v = 4t - 1, k = 2t - 1 and = t - 1, or v = 4t - 1, k = 2t, and = t. The construction, as we shall see, is reversible, so that 2-designs with these parameters can be used to construct Hadamard matrices.

4 Let H be an Hadamard matrix of order 4t. First normalize the matrix H (so that the first row and column are just +1's), then remove the first row and column. The 4t-1 x 4t-1 matrix which remains, say A, has 2t -1's in each row and column and 2t-1 +1's in each row and column, so the row and column sums are always -1 for A. The inner product of two distinct rows of A will be -1 and the product of a row with itself will be 4t-1. These statements are summarized by the matrix equations, AJ = JA T = -J and AA T = 4tI - J where I is the identity matrix and J is the all one matrix of the appropriate order. Now construct the matrix B = ½(A + J). B is a (0,1)-matrix, whose row and column sums are 2t-1, i.e., BJ = JB = (2t-1)J. Furthermore, the matrix equation, BB T = ti + (t-1)j is satisfied [verify]. Comparing this to (1.11), we see that B is the incidence matrix of a symmetric (since B is a square matrix) 2-design with v = 4t-1, k = 2t-1 and = t-1. Similarly, if C = ½(J - A), C will be the incidence matrix of a (4t-1,2t,t)-design. For example, let H be the 8 8 Hadamard matrix above. In normalized form we have, H = A = B = So, labeling the rows of B with {1,2,...,7} we have the (7,3,1)-design whose 7 blocks are: {1,4,5} {2,4,6} {3,4,7} {1,2,3} {1,6,7} {2,5,7} {3,5,6} The matrix C is B with the 0's and 1's interchanged (a design obtained by interchanging the 0's and 1's of an incidence matrix is called the complementary design of the original) and its blocks form the (7,4,2)-design: {2,3,6,7} {1,3,5,7} {1,2,5,6} {4,5,6,7} {2,3,4,5} {1,3,4,6} {1,2,4,7}. Exercise: Prove that an Hadamard design (i.e. a symmetric 2-design with either of these sets of parameters) can be used to construct an Hadamard matrix. Example: The point-hyperplane design of PG(n,2) is a Hadamard 2-design, so in particular the square 2-(7,3,1) design (projective plane of order 2) is a Hadamard 2-design. (1.32) Definition. Let D = (X, B) be a t-(v,k, ) design and p a point. The derived design D p has point set X - {p} and as block set all the blocks of D which contain p with p removed. It is a (t- 1)- (v-1, k-1, ) design. Note that derived designs with respect to different points may not be

5 isomorphic. A design E is called an extension of D if E has a point p such that E p is isomorphic to D; we call D extendable if it has an extension. Since an extension of a t-(v,k, ) design is a (t+1)-(v+1,k+1, ) design, we have (1.33) Proposition. If a t-(v,k, ) design has an extension, then k + 1 divides b(v+1). Pf. In (1.7) take s = 0 and t = 1 for the extension giving k + 1 divides 1 (v+1) for the (t+1)- design. But 1 = r for the larger design is just b for the smaller design. (1.34) Proposition. The only extendable projective planes are those of orders 2 and 4. Pf. By the previous proposition, if a square 2-(n 2 +n+1, n+1, 1) has an extension then n+2 must divide (n 2 +n+1)(n 2 +n + 2) = n 4 + 2n 3 + 4n 2 + 3n + 2. The division gives a remainder of 12/(n+2), and since this must be an integer we have that n + 2 divides 12, i.e., n = 2,4 or 10. The nonexistence of a projective plane of order 10 establishes the result. Comment: As we shall see, both of these planes are in fact extendable. (1.35) Theorem. (Cameron) If a square 2-(v,k, ) design D is extendable, then one of the following holds: (a) D is a Hadamard 2-design; (b) v = ( + 2)( ), k = ; (c) v = 495, k=39, = 3. Pf: For convenience we will redefine the parameters and consider E a 3-(v,k, ) design which is an extension of the square 2-(v-1, k-1, ) design D. For any point p E, E p will have the same parameters as D and so will be a square design (not necessarily isomorphic to D). Consider any two blocks of E through p, in E p they have points in common, so in E they have +1 points in common. Thus any two blocks which intersect have +1 points in common. Let B be a block of E. If p,q B then for each point r B there are blocks containing p,q and s and meeting B in +1 points, so there are +1 choices for s that will give the same block, hence there are k /( +1) distinct such blocks. As there are 2 = (v-2)/(k-2) = k-1 blocks through p and q, this leaves (k - -1)/( +1) blocks containing p and q and disjoint from B. Thus the incidence structure (block residual) whose points are those outside of B and whose blocks are those disjoint from B, is a 2-(v-k, k, (k- -1)/( +1)) design. The number of blocks in this design can be calculated as: v k v k 1 k 1. k k 1 1 This design may be degenerate, having only 1 block [note, this means that the complement of a block in E is also a block of E]. If this occurs, then v = 2k, k = 2( + 1) and D is a square 2-(4 +3, 2 +1, ) design, i.e. a Hadamard 2-design. Otherwise, we can apply Fisher's inequality and obtain (after some manipulation) that k ( +1)( +2).

6 On the other hand, b = v(v-1)/k = (k 2-3k )(k 2-3k + + 2)/k 2 ; which implies that k divides 2( +1)( +2). If k = 2( +1)( +2) then this same expression shows that divides 3; i.e., = 1 or 3. If =1 then E is a 3-(112,12,1) design, i.e., an extension of a projective plane of order 10, which does not exist. If = 3 we obtain case (c) of the statement. Finally, if k = ( +1)( +2) we obtain case (b). Comment: Under condition (b), when = 1, there is an extendable 2-(21,5,1) design (the projective plane of order 4), when = 2, there are several 2-(56,11,2) designs but Bagchi showed that there is no 3-(57,12,2) design. For higher values of, no square designs are known with these parameters. A design with the parameters of condition (c) is not known. 1.36)Proposition. Any Hadamard 2-design has a unique extension. Pf: From the above proof, any extension of a Hadamard 2-design has the complements of the blocks of the original 2-design as blocks. Thus the original blocks together with some new point on each and the complements of these blocks would give blocks which is the total number for the 3-design. Thus the original design completely determines the extension. Note that we have not proved that this construction gives a 3-design, this is left as an exercise. Example: An example of a 3-(8,4,1) design on the set X = {1,2,...,8} is given by the blocks: {1,2,5,6} {3,4,7,8} {1,3,5,7} {2,4,6,8} {1,4,5,8} {2,3,6,7} {1,2,3,4} {5,6,7,8} {1,2,7,8} {3,4,5,6} {1,3,6,8} {2,4,5,7} {1,4,6,7} {2,3,5,8}. The derived design at the point 8 is {3,4,7} {2,4,6} {1,4,5} {5,6,7} {1,2,7} {1,3,6} {2,3,5} which is easily seen to be a projective plane of order 2. Notice the complements of these blocks are also blocks of the 3-design. This is a particular example of an Hadamard 3-design. Let H be an Hadamard matrix of order > 4 which is standardized. Identify the set S with the columns of H and form the blocks in the following way: For each row of H other than the first form a block consisting of all the columns containing a +1 in this row and also form a block consisting of all the columns which contain a -1. If H is order n, the design formed will be a 3-(n, ½n, ¼n - 1) design, these parameters can also be written as 3-(4 + 4, 2 + 2, ). All Hadamard 3-designs can be obtained from this construction.