Determinants of Legendre symbol matrices Robin Chapman Department of Mathematics University of Exeter Exeter, EX4 4QE, UK [email protected] 3 October 2003 Abstract We study and evaluate determinants of various matrices built up from the Legendre symbol defined modulo a prime p. MSC class: 11C20 1 Introduction While performing the research for [3] the author was led to study Hankel matrices Ap and Bp defined for each odd prime p by a1 a2 a3 ··· a(p−1)/2 a a a ··· a 2 3 4 (p+1)/2 a a a ··· a Ap = 3 4 5 (p+3)/2 . .. . a(p−1)/2 a(p+1)/2 a(p+3)/2 ··· ap−2 and b1 b2 b3 ··· b(p−1)/2 b b b ··· b 2 3 4 (p+1)/2 b b b ··· b Bp = 3 4 5 (p+3)/2 . .. . b(p−1)/2 b(p+1)/2 b(p+3)/2 ··· bp−2 1 where the entries are given by 1 i 1 if j is a quadratic residue modulo p, a = 1 + = j 2 p 0 if j is a quadratic nonresidue modulo p, and 1 i 0 if j is a quadratic residue modulo p, b = 1 − = j 2 p 1 if j is a quadratic nonresidue modulo p, j p being the Legendre symbol. He observed that for p up to 300 (the limits of his computing power) det Ap = det Bp = −1 whenever p ≡ 3 (mod 4) and p 6= 3. There seemed to be no obvious pattern for the values when p ≡ 1 (mod 4). Here we prove that det Ap = det Bp = −1 whenever p ≡ 3 (mod 4) and p 6= 3, and also identify the values of det Ap and det Bp = −1 when p ≡ 1 (mod 4). We also consider various related determinants. The author wishes to thank the many mathematicians who have com- mented on this problem to him, notably Roland Bacher, Neil Sloane and especially Benne de Weger who simplified his original arguments greatly, and also Bill Hart for pointing out various errors in earlier versions. 2 2 Quadratic residue codes over C We require some preliminaries on quadratic residue codes over C. These will be defined by analogy with the quadratic residue codes over finite fields used in the theory of error-correcting codes. Fix an odd prime p and let A = C[T ]/ hT p − 1i. We may identify this p Pp−1 j quotient ring with the vector space C with the polynomial j=0 ajT cor- responding to the vector (a0, . , ap−1). Let ζ = exp(2πi/p) be a primitive p-th root of unity and let µ = {ζj : 0 ≤ j < p} denote the group of all p-th roots of unity in C. Define µ+ as the set of j j − j j ζ ∈ µ with p = 1 and µ as the set of ζ ∈ µ with p = −1. Thus µ is the disjoint union of µ+, µ− and {1}. Note that + k2 1 µ = {ζ : 1 ≤ k ≤ 2 (p − 1)} and + k2 1 {1} ∪ µ = {ζ : 0 ≤ k ≤ 2 (p − 1)}, Let p−1 X j τ = ζj. p j=1 3 This Gauss sum satisfies τ 2 = (−1)(p−1)/2p, indeed √ p if p ≡ 1 (mod 4), τ = √ i p if p ≡ 3 (mod 4), a result due to Gauss, [1, Chapter 5, Section 4, Theorem 7]. Moreover p−1 X j k ζjk = τ p p j=1 for all integers k. If f ∈ A, it is meaningful to write f(η) for η ∈ µ. Define Q = f ∈ A : f(η) = 0 for all η ∈ µ− . p Pp−1 j Equivalently Q is the set of (a0, . , ap−1) ∈ C with j=0 ajη = 0 for all η ∈ µ−. We regard Q as a quadratic residue code over C. Clearly Q is an 1 ideal of A, with dimension 2 (p + 1) over C, and is generated by Y G(T ) = (T − η). η∈µ− Let us write (p−3)/2 (p−1)/2 X j G(T ) = T + gjT . j=0 Considered as a vector subspace of Cp, Q is generated by the rows of the 1 2 (p + 1)/2 by p matrix g0 g1 g2 ··· 1 0 0 ··· 0 0 g g ··· g 1 0 ··· 0 0 1 (p−3)/2 0 0 g ··· g g 1 ··· 0 Γ = 0 (p−5)/2 (p−3)/2 . . .. .. . 0 0 0 ··· g0 g1 g2 ··· 1 4 For future reference consider some examples of polynomials in Q. Let p−1 p−1 X j X F = τ + T j and U = T j. p j=1 j=0 For η = ζk ∈ µ−, p−1 X j k F (η) = τ + ζjk = τ + τ = 0 p p j=1 and ηp − 1 U(η) = = 0 η − 1 so F , U ∈ Q. 1 We now prove a lemma concerning the evaluation of size 2 (p + 1) deter- minants by means of evaluating polynomials at roots of unity. 1 Lemma 1 Let M be a 2 (p + 1)-by-p matrix all of whose rows lie in Q. Suppose that the j-th row of M corresponds to the polynomial fj ∈ A. Then det Φ(M) det M 0 = det Φ(Γ) 0 1 where M is the matrix consisting of the last 2 (p + 1) columns of M and 1 1 Φ(M) is the 2 (p + 1)-by- 2 (p + 1) matrix with entries k2 1 1 Φ(M)j,k = fj(ζ ) (1 ≤ j ≤ 2 (p + 1), 0 ≤ k ≤ 2 (p − 1)) and Φ(Γ) is defined similarly, with Γ replacing M. 5 Proof By hypothesis, M = AΓ for some matrix A. Hence M 0 = AΓ0 where 0 1 0 Γ is the matrix consisting of the last 2 (p + 1) columns of Γ. But Γ is lower triangular with all diagonal entries equal to 1, and so det(M 0) = det(A), 2 p−1 t 2 (p−1)/2 t Let vη = (1, η, η , . , η ) and wη = (1, η, η , . , η ) for η ∈ µ. Since the rows of Γ are cyclic shifts of the first row, it follows that Γvη = G(η)wη. + For η ∈ {1} ∪ µ , G(η) 6= 0. Let V be the matrix with columns vζk2 for 1 k2 0 ≤ k ≤ 2 (p + 1). The columns of ΓV are then the vectors G(ζ )wζk2 for 1 0 ≤ k ≤ 2 (p + 1) which are linearly independent (by the nonsingularity of a Vandermonde matrix). Now MV = Φ(M) and ΓV = Φ(Γ). As M = AΓ then Φ(M) = AΦ(Γ). As Φ(Γ) is nonsingular we conclude that det Φ(M) det M 0 = det(A) = . det Φ(Γ) The matrix Φ(Γ) is a Vandermonde matrix multiplied by a diagonal ma- trix, and so has determinant Y Y 2 2 G(η) · (ζk − ζj ). η∈{1}∪µ+ 0≤j<k≤(p−1)/2 6 3 The main theorem We now state and prove the main theorem. It is convenient to introduce two new classes of Hankel matrix, each containing an indeterminate x. Let ∗ 1 1 Cp(x) and Cp (x) be the Hankel matrices of sizes 2 (p − 1) and 2 (p + 1) i+j−1 ∗ with (i, j)-entry p . It is plain that det Cp(x) and det Cp (x) are linear 1 1 polynomials in x. Also Ap = 2 Cp(1) and Bp = − 2 Cp(−1). Another special case is Cp(0) which is the Hankel matrix with entries the Legendre symbols 1 2 p−2 p , p ,..., p . ∗ We evaluate det Cp(x) and det Cp (x) by introducing a matrix Mp(x, y) of 1 size 2 (p + 1) containing a further parameter y. Its entries are ( j+k−1 x + p if 1 ≤ j + k − 1 < p, Mp(x, y)j,k = x + y if j + k − 1 = p. The only entry in Mp(x, y) containing y that in the lower right corner. As ∗ Mp(x, 0) = Cp (x) and deleting the last row and last column of Mp(x, y) yields Cp(x) then ∗ det Mp(x, y) = det Cp (x) + y det Cp(x). Since Mp(x, y) is a polynomial with integer coefficients, to evaluate det Cp(x) ∗ and det Cp (x) it suffices to calculate Mp(x, y) for some irrational value of y, and it turns out that the Gauss sum τ is a particularly convenient choice. 7 Theorem 2 Let p ≥ 5 be prime. If p ≡ 3 (mod 4) then (p−1)/2 det Mp(x, y) = 2 (1 − xy). If p ≡ 1 (mod 4) then √ (p−1)/4 (p−1)/2 √ −hp det Mp(x, p) = (−1) 2 (1 + x p)εp where εp > 1 and hp are respectively the fundamental unit and class number √ of the field Q( p). Proof Define a new matrix Np(x, y) as follows: Np(x, y) has (p + 1)/2 rows and p columns, and ( j−i p + x if j 6= i, (Np(x, y))i,j = x + y if j = i. 0 The key observation here is that Mp(x, y) is essentially the submatrix Np(x, y) formed by taking the last (p + 1)/2 columns of Np(x, y). In detail, if p ≡ 1 0 (mod 4) then Mp(x, y) is the left-right reflection of Np(x, y) and if p ≡ 3 0 (mod 4) then Mp(x, y) is the left-right reflection of −Np(−x, −y). Thus (p−1)/4 0 (−1) det Np(x, y) if p ≡ 1 (mod 4), det Mp(x, y) = (p+1)/4 0 (−1) det Np(−x, −y) if p ≡ 3 (mod 4).