Volume 10 (2009), Issue 3, Article 62, 9 pp. A MATRIX INEQUALITY FOR MÖBIUS FUNCTIONS OLIVIER BORDELLÈS AND BENOIT CLOITRE 2 ALLÉE DE LA COMBE 43000 AIGUILHE (FRANCE) [email protected] 19 RUE LOUISE MICHEL 92300 LEVALLOIS-PERRET (FRANCE) [email protected] Received 24 November, 2008; accepted 27 March, 2009 Communicated by L. Tóth ABSTRACT. The aim of this note is the study of an integer matrix whose determinant is related to the Möbius function. We derive a number-theoretic inequality involving sums of a certain class of Möbius functions and obtain a sufficient condition for the Riemann hypothesis depending on an integer triangular matrix. We also provide an alternative proof of Redheffer’s theorem based upon a LU decomposition of the Redheffer’s matrix. Key words and phrases: Determinants, Dirichlet convolution, Möbius functions, Singular values. 2000 Mathematics Subject Classification. 15A15, 11A25, 15A18, 11C20. 1. INTRODUCTION In what follows, [t] is the integer part of t and, for integers i, j > 1, we set mod(j, i) := j − i[j/i]. 1.1. Arithmetic motivation. In 1977, Redheffer [5] introduced the matrix Rn = (rij) ∈ Mn({0, 1}) defined by ( 1, if i | j or j = 1; rij = 0, otherwise and has shown that (see appendix) n X det Rn = M(n) := µ(k), k=1 where µ is the Möbius function and M is the Mertens function. This determinant is clearly related to two of the most famous problems in number theory, the Prime Number Theorem (PNT) and the Riemann Hypothesis (RH). Indeed, it is well-known that 1/2+ε PNT ⇐⇒ M(n) = o(n) and RH ⇐⇒ M(n) = Oε n 317-08 2 OLIVIER BORDELLÈS AND BENOIT CLOITRE (for any ε > 0). These estimations for | det Rn| remain unproven, but Vaughan [6] showed that h log n i 1 is an eigenvalue of Rn with (algebraic) multiplicity n− log 2 −1, that Rn has two "dominant" 1/2 2/5 eigenvalues λ± such that |λ±| n , and that the others eigenvalues satisfy λ (log n) . It should be mentioned that Hadamard’s inequality, which states that n 2 Y 2 | det Rn| 6 kLik2, i=1 n where Li is the ith row of Rn and k·k2 is the euclidean norm on C , gives n [n/2] Y hni Y hni n + [n/2] (M(n))2 n 1 + = 2n−[n/2]n 1 + 2n−[n/2] , 6 i i 6 n i=2 i=2 which is very far from the trivial bound |M(n)| 6 n so that it seems likely that general matrix analysis tools cannot be used to provide an elementary proof of the PNT. In this work we study an integer matrix whose determinant is also related to the Möbius function. This will provide a new criteria for the PNT and the RH (see Corollary 2.3 below). In an attempt to go further, we will prove an inequality for a class of Möbius functions and deduce a sufficient condition for the PNT and the RH in terms of the smallest singular value of a triangular matrix. 1.2. Convolution identities for the Möbius function. The function µ, which plays an impor- tant role in number theory, satisfies the following well-known convolution identity. Lemma 1.1. For every real number x > 1 we have X hxi X x µ(k) = M = 1. k d k6x d6x One can find a proof for example in [1]. The following corollary will be useful. Corollary 1.2. For every integer j > 1 we have (i) j j X µ(k) mod (j, k) X µ(k) = j − 1. k k k=1 k=1 (ii) j k ! X X µ(h) (mod(j, k + 1) − mod(j, k)) = 1. h k=1 h=1 Proof. (i) We have j j j j X µ(k) mod (j, k) X µ(k) j X µ(k) X j = j − k = j − µ(k) k k k k k k=1 k=1 k=1 k=1 and we conclude with Lemma 1.1. J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 62, 9 pp. http://jipam.vu.edu.au/ AMATRIX INEQUALITY FOR MÖBIUS FUNCTIONS 3 (ii) Using Abel summation we get j k ! X X µ(h) (mod(j, k + 1) − mod(j, k)) h k=1 h=1 j ! j X µ(h) X = (mod(j, k + 1) − mod(j, k)) h h=1 k=1 j−1 k+1 k ! k X X µ(h) X µ(h) X − − (mod(j, m + 1) − mod(j, m)) h h k=1 h=1 h=1 m=1 j j−1 X µ(k) X µ(k + 1) mod (j, k + 1) = j − k k + 1 k=1 k=1 j j X µ(k) X µ(k) mod (j, k) = j − k k k=1 k=1 and we conclude using (i). 2. AN INTEGER MATRIX RELATED TO THE MÖBIUS FUNCTION We now consider the matrix Γn = (γij) defined by mod(j, 2) − 1, if i = 1 and 2 j n; 6 6 mod(j, i + 1) − mod(j, i), if 2 i n − 1 and 1 j n; γ = 6 6 6 6 ij 1, if (i, j) ∈ {(1, 1), (n, 1)}; 0, otherwise. The matrix Γn is almost upper triangular except the entry γn1 = 1 which is nonzero. Note that it is easy to check that |γij| 6 i for every 1 6 i, j 6 n and that γij = −1 if [j/2] < i < j. Example 2.1. 1 −1 0 −1 0 −1 0 −1 0 2 −1 1 1 0 0 2 0 0 3 −1 −1 2 2 −2 0 0 0 4 −1 −1 −1 3 Γ8 = . 0 0 0 0 5 −1 −1 −1 0 0 0 0 0 6 −1 −1 0 0 0 0 0 0 7 −1 1 0 0 0 0 0 0 0 2.1. The determinant of Γn. Theorem 2.1. Let n > 2 be an integer and Γn defined as above. Then we have n X µ(k) det Γ = n! . n k k=1 J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 62, 9 pp. http://jipam.vu.edu.au/ 4 OLIVIER BORDELLÈS AND BENOIT CLOITRE A possible proof of Theorem 2.1 uses a LU decomposition of the matrix Γn. Let Ln = (lij) and Un = (uij) be the matrices defined by 0, if (i, j) = (n, 1); uij = 1, if (i, j) = (n, n); γij, otherwise and 1, if 1 i = j n − 1; 6 6 Pj µ(k) k=1 k , if i = n and 1 6 j 6 n − 1; lij = Pn µ(k) n k=1 k , if (i, j) = (n, n); 0, otherwise. The proof of Theorem 2.1 follows from the lemma below. Lemma 2.2. We have Γn = LnUn. Proof. Set LnUn = (xij). When i = 1 we immediately obtain x1j = u1j = γ1j. We also have n X xn1 = lnkuk1 = ln1u11 = 1 = γn1. k=1 Moreover, using Corollary 1.2 (ii) we get for i = n and 2 6 j 6 n − 1 n n X X xnj = lnkukj = ln1u1j + lnkukj k=1 k=2 j k ! X X µ(h) = mod(j, 2) − 1 + (mod(j, k + 1) − mod(j, k)) h k=2 h=1 j k ! X X µ(h) = (mod(j, k + 1) − mod(j, k)) − 1 = 0 = γ h nj k=1 h=1 and, for (i, j) = (n, n), we have similarly n n−1 X X xnn = lnkukn = ln1u1n + lnkukn + lnnunn k=1 k=2 n−1 k ! n X X µ(h) X µ(k) = mod(n, 2) − 1 + (mod(n, k + 1) − mod(n, k)) + n h k k=2 h=1 k=1 n k ! X X µ(h) = (mod(n, k + 1) − mod(n, k)) − 1 = 0 = γ . h nn k=1 h=1 Finally, for 2 6 i 6 n − 1 and 1 6 j 6 n, we get n X xij = likukj = liiuij = uij = γij. k=1 J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 62, 9 pp. http://jipam.vu.edu.au/ AMATRIX INEQUALITY FOR MÖBIUS FUNCTIONS 5 Example 2.2. 1 0 0 0 0 0 0 0 1 −1 0 −1 0 −1 0 −1 0 1 0 0 0 0 0 0 0 2 −1 1 1 0 0 2 0 0 1 0 0 0 0 0 0 0 3 −1 −1 2 2 −2 0 0 0 1 0 0 0 0 0 0 0 4 −1 −1 −1 3 Γ8 = . 0 0 0 0 1 0 0 0 0 0 0 0 5 −1 −1 −1 0 0 0 0 0 1 0 0 0 0 0 0 0 6 −1 −1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 7 −1 1 1 1 1 2 1 8 1 2 6 6 − 30 15 − 105 − 105 0 0 0 0 0 0 0 1 Theorem 2.1 now immediately follows from n X µ(k) det Γ = det L det U = (n − 1)! det L = n! . n n n n k k=1 We easily deduce the following criteria for the PNT and the RH. Corollary 2.3. For any real number ε > 0 we have −1/2+ε PNT ⇐⇒ det Γn = o(n!) and RH ⇐⇒ det Γn = Oε(n n!). 2.2. A sufficient condition for the PNT and the RH. −1 2.2.1. Computation of Un .