More on the Wilson W tk ( v ) matrices

For integers 0 6 t 6 k 6 v − t, let X be a v-set, and let Wtk(v) be a ( v t ) × ( v k ) inclusion matrix where rows and columns are indexed by t-subsets and k-subsets of X, respectively, and for row T and column K, Wtk(v)(T,K) = 1 if T ⊆ K and zero otherwise. Since Wtk(v) is a full rank matrix, by reordering the columns of Wtk(v) we can write Wtk(v) = (S|N), where N denotes a set of independent columns of Wtk(v). In this paper, first by classifying t-subsets and k-subsets, we present a new decomposition of Wtk(v). Then by employing this decomposition, the Leibniz Triangle, and a known right inverse of Wtk(v), we construct the inverse of N and consequently special basis for the null space (known as the standard basis) of Wtk(v).


Introduction
Integers t, k, and v with 0 t k v − t are considered.Let X be a linearly ordered v-set, and let For the sake of brevity, we will denote a set {a 1 , . . ., a i } by the string "a 1 . . .a i ", and assuming that a 1 < a 2 < • • • < a i .The elements of X k and X t are called blocks and t-subsets, respectively.
The inclusion matrix W tk (v) (known as Wilson matrix) is defined to be a v t by v k (0, 1)-matrix whose rows and columns are indexed by (and referred to) the members of the electronic journal of combinatorics 21(2) (2014), #P2.53 X t and X k , respectively, and where For the sake of convenience, sometimes we use W tk or just a bare W for W tk (v).Let S = x 1 x 2 . . .x n be a finite set, and let F be an arbitrary ring.An F -collection of the elements of S is a function f : S → F, with the vector representation f (x 1 ), • • • , f (x n ) T , for i, 1 i n, f (x i ) is defined to be the value of x i in f .It is well known that W tk is a full rank matrix over Q [8].As a linear operator, W tk acts on a Z -collection of blocks, and algebraically counts the number of times that any member of X t appears in the blocks of the collection.In the set of our notations, for any matrix M , the free Z-module generated by rows and columns of matrix M will be denoted by row Z (M ) and col Z (M ), respectively, and null Z (M ) will be the free Z -module orthogonal to row Z (M ).
Let 1 be the all 1 vector, and let λ be a nonnegative integer.We call the following equation the fundamental equation of design theory: Every integral solution of equation ( 1) is called a signed t-(v, k, λ) design.For more on this, see [4,8].
Since W is a full rank matrix, by reordering the columns of W we can write W as W = (S|N ), where N denotes a set of independent columns of W .Therefore, there is a matrix C such that N −1 (S|N ) = (C|I).Let S be a matrix defined by stacking an identity matrix above the matrix −C,i.e., S := I −C .Since W S = 0 and W is full rank, the columns of S form a basis for null Z (W ).Now, we would like to give a rather comprehensive view on the problem addressed in this paper: We start with the halving conjecture.In 1987 A. Hartman [9] stated the following conjecture which is now known as the halving conjecture: ).Up to our knowledge, the conjecture has been settled for t = 2 utilizing a recursive construction [2], and some infinite classes have been constructed too [10].
Since every (1, −1)-vector in null Z (W ) is a linear combination of the columns of S, therefore the null space of the W should be studied more carefully.For this, we have to know the components, row structure and column structure of S.
• In what follows, an explicit formula for the entries of N −1 and consequently a closed formula for the entries of S are presented.
• For the row structure of S, there are two conjectures on the table : -The elements of every row of S have the same sign.
In [1] these two conjectures have been settled for t = 2 and k = 3.
• In [1] the columns of S 23 (v) have been classified into five classes and by utilizing these classes the correctness of the halving conjecture has been established.
Example 2. For A = 3458, we obtain: For B = 1478, we obtain: In [6,14], a decomposition of W tk (v) is presented: Now, we propose a new ordering of blocks and t-subsets and consequently a new decomposition.
Definition 3.For given t, k, and X, let If we order every B i and T j in reverse lexicographic ordering, then B 0 , B 1 , . . ., B k and T 0 , T 1 , . . ., T t are orders on X k and X t , respectively.This ordering is called R-ordering.
Here the rows and the columns of W tk (t + k) are indexed by T t and B k elements, respectively.In passing we note that the matrix Example 4. The above decomposition of W 23 (7) is: . Table 1.The decompostion of W 23 (7).
(In tables throughout this paper, unless otherwise indicated, blanks are zeros.) Remark 5. To obtain the inverse of N , first we construct the inverse of W tk (t + k).In the next section, we introduce a right inverse of W .

Right inverse of W and Leibniz Triangle
Around 1980, Graham, Li, and Li [7] presented a right inverse for W with a closed formula.Later on, Bapat [3] constructed a right inverse for W in a recursive form.The elements of these right inverses are multiples of the entries of Leibniz Triangle (Table 2).Table 2. Leibniz Triangle.
For given 0 r n, the (n, r)-th position of Leibniz Triangle was introduced as the (n, r)-th harmonic coefficient which is defined to be Now we index the rows and the columns of the right inverse of W by the elements of X k and X t , respectively.According to [7] and (4) every entry of this matrix comes from the following relation: where Now back to the inverse of W tk (t + k).We replace v − t by k in (5), and then every element of the inverse of W tk (t + k), denoted by F (B, T ), is defined as where θ = |B ∩ T |.
Let B be an arbitrary block such that where The above formula is easily verified by Maple [13] and exhibits a very interesting relation between Leibniz Triangle and binomial triangle.

The inverse of N
The construction of the inverse of N is based on (6), but first we should partition W into independent and dependent columns.The function which is defined on blocks in [5,11], classifies the blocks into t + 2 classes.Although through that classification independent and dependent columns are separated, the partitioning is not refined enough to be useful for the inverse construction.Here we introduce a new function to partition subsets of X, which is based on R-ordering.
That is to say that there exists an element in T which is not in B. Therefore, N tk (T, B) = 0. Lemma 7.For given t, k, and X, the number of non-starting blocks is Similarly the number of non-starting blocks B such that A ⊆ B and . Now, we have to show that different A's with the same size, produce different t-subsets and different blocks.
and this is a contradiction.Therefore, there exists a bijection from the set of non-starting blocks to all the t-subsets.
Corollary 8.The main diagonal boxes of N tk are square matrices.
Example 9. Table 3 demonstrates the boxing structure of N 23 (6).Now, to show that the columns of N tk are linearly independent, first we define a matrix F tk (v), whose rows and columns are indexed by non-starting blocks and t-subsets, respectively.We note that the non-starting blocks and t-subsets are R-ordered.Then F tk (v) is defined as: where as in (6).Now let M := F tk (v)N tk .Naturally the rows and the columns of M are indexed by non-starting blocks.
For clarity we add the following statements: Lemma 17.Let B and A be two non-starting blocks such that A = B. Then A is a B-changer if and only if M (B, A) = 0.
Proof.For a given block B, let A be a B-changer.By Definition 16 we have Therefore, by (9) it follows that M (B, A) = 0. Now assume that M (B, A) = 0, again by (9) we have k Lemma 13 and (9) we have M (B, A) = 0, which is a contradiction.Therefore, k B > k A and A is a B-changer.
Theorem 18. Suppose that the rows and the columns of matrix N −1 are indexed by non-starting blocks and t-subsets, respectively.For a block B and a t-subset T , we have: Proof.The correctness of the statement of the theorem can be easily established by the elementary row operations.
N In [12] Khosrovshahi and Tayfeh-Rezaie showed that by subtracting 1 from the sum of the columns of the standard basis of W , one obtains a unique signed t-design D. For more on this subject see [15].Here we show that D is also obtained by the sum of the columns of the inverse of N .
Let s i 1 , . . ., s i ( v k ) − ( v t ) be the i-th row of S tk and D = d 1 , . . ., d ( v k ) T .Therefore, Let γ i 1 , . . .γ i ( v t ) be the i-th row of N −1 tk .We have the following identities: Theorem 24.Let η = ( v t ) i=1 Γ i , where Γ i 's are the columns of N −1 tk , then v−t k−t η = D.

Lemma 13 .
Notation.t B := t − (k − k B ).If for two non-starting blocks B and B , k B = k B , then |R tk (B)∩R tk (B )| k B − t B .Proof.Since k B = k B , every element of R tk (B) and R tk (B ) is at most 2k B −k+t = k B +t B by Definition 1. Hence |R tk (B) ∩ R tk (B )| k B − t B .Now, we have
Definition 6.The block B is called a starting block if 0 |R tk (B)| < k − t, and a non-starting block if k − t |R tk (B)| k.Notation.kB :=|R tk (B)|.Now, we omit the columns indexed by the starting blocks from W and we denote the remaining matrix by N tk .If we R-order the t-subsets and non-starting blocks, then: Let B be a non-starting block and T ∈ Definition 10.For given t, k, and X, let B be a non-starting block.For any A ⊆ X such that |A| |B|, A \ B \ R tk (B) denoted by R tk (A, B) is called the root of A with respect to B.
Theorem 15.The columns indexed by non-starting blocks in W are linearly independent.Definition 16.For a given non-starting block B, a block A is called a B-changer, if the following conditions hold: