Distribution of eigenvalues of ensembles of asymmetrically diluted Hopfield matrices

Using Grassmann variables and an analogy with two dimensional electrostatics, we obtain the average eigenvalue distribution ρ(ω) of ensembles of N × N asymmetrically diluted Hopfield matrices in the limit N → ∞. We found that in the limit of strong dilution the distribution is uniform in a circle in the complex plane. Present address: Universidade Federal de Braśılia, Brazil Present address: FaMAF, Universidad Nacional de Córdoba, Argentina 0 Random matrix theory has become an active field of research in mathematics and physics in the last decades. In physics, since the now classical work of Metha[1] on the statistical description of the energy levels of atomic nuclei, random matrices have emerged as an important tool in the study of the localization transition[2, 3], quantum chaos[4], spin glasses[5], neural networks[6, 7], and disordered systems in general. Most of the work deal with ensembles of hermitian or symmetric matrices whose individual properties are well known and can be exploited in more complex situations. In the last years the interest in asymmetric matrices has grown, motivated, besides its intrinsic mathematical value, by problems of dissipative quantum dynamics[8] and the dynamics of neural networks[9]. In a recent paper, Sommers et.al.[10] calculated the average density of eigenvalues ρ(ω) of N ×N random asymmetric matrices in the limit N → ∞, with elements Jij , given by a Gaussian distribution with zero mean and correlations N〈〈J ij〉〉J = 1, N〈〈Jij Jji〉〉J = τ (1) They found that the eigenvalues are uniformly distributed inside an ellipse in the complex plane, whose semi axes depend on the degree of asymmetry of the ensemble τ . Generalizing this result, Lehmann et. al. [11] calculated the joint probability distribution of eigenvalues in Gaussian ensembles of real asymmetric matrices, recovering the elliptic law in the large N limit. In this paper we calculate the average eigenvalue distribution ρ(λ) of an ensemble of asymmetrically diluted Hopfield matrices, whose elements are given by: Jij = cij N p ∑ μ=1 ξ i ξ μ j i, j = 1 . . . N , (2) where {ξ i i = 1 . . .N, μ = 1 . . . p} represents a set of p random patterns. The ξ i are random independent variables that can take the values ±1 with the same probability and the cij are random variables chosen accordingly to the following distribution P (cij) = γδ(cij − 1) + (1− γ)δ(cij). (3) 1 0 ≤ γ ≤ 1 measures the degree of dilution of the matrices. γ = 1 corresponds to symmetric Hopfield matrices whose eigenvalue distribution is known[12]. In order to obtain the distribution ρ(ω) we use an analogy with a two dimensional electrostatic problem introduced in ref.[10]. Let us define the Green function associated to the matrix J G(ω) = 1 N 〈〈Tr 1 Iω − J 〉〉J , (4) where ω = x + iy is a complex variable, I the identity matrix and 〈〈. . .〉〉J denotes an average over the random variables ξ i . If λi, i = 1, . . . , N are the eigenvalues of J , then Tr 1 Iω − J = N ∑ i 1 ω − λi (5) For large N the sum can be approximated by an integral and the Green function becomes G(ω) = ∫ dλ ρ(λ) ω − λ (6) where ρ(λ) is the density of eigenvalues in the plane. The last equation suggests an analogy with a two-dimensional classical electrostatics problem in which ρ(λ) represents the density of charge in the plane. It can be demonstrated[10] that an electrostatic potential Φ exists, satisfying 2ReG = − ∂x , −2ImG = − ∂y (7) and which obeys Poisson’s equation: ∇Φ = −4πρ (8) Thus, in order to determine ρ(ω) we may calculate the potential Φ. Using that det(AB) = detAdetB and detA = detA one can prove that a good definition for Φ can be Φ(ω) = −1/N〈〈 ln det ( (Iω∗ − J )(Iω − J) ) 〉〉J (9) 2 with ω∗ the complex conjugate of ω and J the transpose of J . In what follows we will consider the case N → ∞ and assume that in this limit the average and the ln operations commute [10]. By using a Grassmannian representation[13] of the determinant and adding a matrix ǫ δij , with ǫ positive and infinitesimal in order to avoid zero eigenvalues, we get: exp [−NΦ(ω)] = 〈〈 ∫ ∞