ON ASYMPTOTIC PROPERTIES OF THE RANK OF A SPECIAL RANDOM ADJACENCY MATRIX

Consider the matrix ∆ n = (( I( X i + X j > 0) )) i,j =1 , 2 ,...,n where { X i } are i.i.d. and their distribution is continuous and symmetric around 0. We show that the rank r n of this matrix is equal in distribution to 2 P n − 1 i =1 I( ξ i = 1 , ξ i +1 = 0) + I( ξ n = 1) where ξ i i.i.d. ∼ Ber(1 , 1 / 2) . As a consequence √ n ( r n /n − 1 / 2) is asymptotically normal with mean zero and variance 1 / 4. We also show that n − 1 r n converges to 1 / 2 almost surely.

The motivation for studying this matrix arises from the study of random network models, known as threshold models. Suppose that there is a collection of n nodes. Each node is assigned a fitness value and links are drawn among nodes when the total fitness crosses a threshold. This gives rise to a good-get-richer mechanism, in which sites with larger fitness are more likely to become hubs (i.e., to be connected). The scale free random network generated in this way is often used as a model in social networking, friendship networks, peer-to-peer (P2P) networks and networks of computer programs. Many features, such as power-law degree distributions, clustering, and short path lengths etc., of this random network has been studied in the physics literature extensively (see, for example, Caldarelli et. al. (2002), Söderberg (2002), Masuda et. al. (2005)).
Suppose that the fitness value of the sites are represented by the random variables {X i } and we connect two points when their accumulated fitness is above the threshold 0. The matrix ∆ n then represents the adjacency matrix of the above random graph.
Note that ∆ n is a random matrix with zero and one entries. There are a few results known for the rank of random matrices with zero and one entries. See for example , Costello, Tao and Vu (2006). In particular, the latter shows that the matrix with i.i.d. entries which are Bernoulli with probability 1/2 each is almost surely nonsingular as n → ∞.
Suppose that the distribution of X 1 is continuous and symmetric around 0. Then the entries of the matrix ∆ n are identically distributed, being Bernoulli with probability 1/2 but now there is strong dependency among them. We show that the rank r n of this matrix is asymptotically half of the dimension. Indeed, the rank of ∆ n , can be approximated in distribution by the sum of a 1-dependent stationary sequence. As a consequence it is asymptotically normal with asymptotic mean n/2 and variance n/4. Further, r n /n converges to 1/2 almost surely. In what follows, X D = Y means the two random variables X and Y have the same probability distributions.
Since any row column permutation leaves the eigenvalues unchanged, the eigenvalues of ∆ n are the same as the eigenvalues of the matrix (( we have the following equality in distribution:. eigenvalues of (( I(S σ(i∧j) = 1) )) D = eigenvalues of (( I(S i∧j = 1) )).
To complete the proof, we will need the following Lemma We will first finish the proof of the theorem assuming the lemma and then prove the lemma.
From the discussion above, and from Lemma 1, we have is a stationary 1-dependent sequence of random variables with mean 1/4. Part (A) now follows immediately.
From equation (1) above, the asymptotic distribution of √ n rank(∆ n )/n − 1/2 is the same as that of n −1/2 2 n−1 i=1 I(ξ i = 1, ξ i+1 = 0) − n/2 . But by the central limit theorem for m-dependent stationary sequence (see for example Brockwell and Davis, 1990, page 213), the latter is asymptotically normal with mean zero and variance σ 2 . This variance is calculated as This proves Part (B) of the theorem.
To prove Part (C), first observe that By using the fact that the summands are all bounded, it is easy to check that the moment convergence in the central limit theorem for m-dependent sequences hold. That is Combining all the above, we conclude that Claim: Suppose that 1 ≤ k ≤ n − 1. Let 1 ≤ m 1 < m 2 < · · · < m k ≤ n be such that

This implies that
Then the matrix A may be reduced to the following matrix B = ((b ij )) by appropriate row column transformations.
Proof of the claim: By definition of the matrix A, its m 1 th row (resp. column) has all ones starting from the m 1 th column (resp. row). Thus we may visualise A as follows: . . . a n,n−1 a n,n We move the first (m 1 − 1) columns of A to the extreme east end and then move the m 1 th row to the extreme south to get the following matrix A 1 which has the same rank as that of the original matrix A. 1 a n,m1+1 . . . a n,n−1 a n,n 0 . . . 0 Now leave the first column, the last row, the first (m 1 − 1) zero rows and the last (m 1 − 1) zero columns intact and consider the remaining (n − m 1 ) × (n − m 1 ) submatrix. This is a function of ξ m1+1 , ξ m1+2 , . . . , ξ n and write it in the way that A was written and repeat the procedure.
Note that now when we move the rows and columns, we move extreme south and east of only the submatrix but the remaining part of the matrix does not alter. It is easy to see that in (k − 1) more steps we obtain B. This proves the claim.
We return to the proof of the Lemma. Note that b ij = 1 iff b n−j+1,n−i+1 = 1. In other words, B is anti-symmetric (symmetric about its anti-diagonal). Indeed, B is the n×n anti-symmetric 0-1 matrix containing minimum number of ones such that its i-th column contains n − m i + i ones from bottom for 1 ≤ i ≤ k.
Let u i denote the number of 1's in the i-th column of B. Then, u 1 = n − m 1 + 1, u 2 = n − m 2 + 2, . . . , u k = n − m k + k. Since m 1 < m 2 < · · · < m k we have u 1 ≥ u 2 ≥ · · · ≥ u k . Also from the anti-symmetry of B we get for k < r ≤ n, u r = #{i : 1 ≤ i ≤ k, u i ≥ r}. This immediately implies that u r ≥ u r+1 for all r > k. Since u k = n − m k + k ≥ k and u k+1 = #{i : 1 ≤ i ≤ k, u i ≥ k + 1} ≤ k, it follows that u k ≥ u k+1 . In fact u k = u k+1 is not possible since then both are equal to k but then by definition of u k+1 < k. Thus {u i } n i=1 is a nonincreasing sequence.
. . > . . . = u k . We will consider the following two cases separately: Case I: m k < n. In this case u k = (n − m k ) + k ≥ k + 1. If we write out the whole u-sequence, it looks like this The first row enumerates the column positions and the second row provides count of the number of one's in that column, that is, the corresponding u i .
The rank of the matrix B can be easily computed by looking at the u-sequence. Each distinct non zero block contributes one to the rank. In this case, rank(B) = 2d and since ξ n = 0, we have also d = n−1 i=1 I(ξ i = 1, ξ i+1 = 0). And therefore, This completes the proof of the Lemma.