A Matrix Inequality for Möbius Functions

Volume 10 (2009), Issue 3, Article 62, 9 pp.

A MATRIX INEQUALITY FOR MÖBIUS FUNCTIONS

OLIVIER BORDELLÈS AND BENOIT CLOITRE 2 ALLÉEDELACOMBE 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 BORDELLÈSAND 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 INEQUALITYFOR 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 BORDELLÈSAND 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 INEQUALITYFOR 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 . The inverse of Un uses a Möbius-type function denoted by µi which we deﬁne below.

Deﬁnition 2.1. Set µ1 = µ the well-known Möbius function and, for any integer i > 2, we deﬁne the Möbius function µi by µi(1) = 1 and, for any integer m > 2, by  m µ , if i | m and (i + 1) - m;  i     m −µ , if (i + 1) | m and i - m;  i + 1 µi(m) :=   m m µ − µ , if i(i + 1) | m;  i i + 1    0, otherwise. The following result completes and generalizes Lemma 1.1 and Corollary 1.2.

Lemma 2.4. For all integers i, j > 2 we have j X j µ (k) = δ , i k ij k=i j j X µi(k) mod (j, k) X µi(k) = j − δ , k k ij k=i k=i j k ! X X µi(h) (mod(j, k + 1) − mod(j, k)) = δ , h ij k=i h=i

where δij is the Kronecker symbol.

J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 62, 9 pp. http://jipam.vu.edu.au/ 6 OLIVIER BORDELLÈSAND BENOIT CLOITRE

Proof. We only prove the ﬁrst identity, the proof of the two others being strictly identical to the identities of Corollary 1.2. Without loss of generality, one can suppose that 2 6 i 6 j. If i = j then we have j X j j j µ (k) = µ (j) = µ = 1. i k j j j k=i Now suppose that 2 6 i < j. By Lemma 1.1, we have j j j X j X k j X k j µ (k) = µ − µ i k i k i + 1 k k=i k=i k=i i|k, (i+1)-k (i+1)|k, i-k j X k k j + µ − µ i i + 1 k k=i i(i+1)|k j j X k j X k j = µ − µ i k i + 1 k k=i k=i i|k (i+1)|k [j/i] [j/(i+1)] X [j/i] X [j/(i + 1)] = µ(h) − µ(h) h h h=1 h=1 = 1 − 1 = 0, which concludes the proof.

This result gives the inverse of Un. −1 Corollary 2.5. Set Un = (θij). Then we have j X µi(k) θ = (1 i j n − 1) ij k 6 6 6 k=i n X µi(k) θ = n (1 i n). in k 6 6 k=i −1 Proof. Since Un is upper triangular, it sufﬁces to show that, for all integers 1 6 i 6 j 6 n, we have j X θikukj = δij. k=i In what follows, we set Sij as the sum on the left-hand side We easily check that Sjj = 1 for every integer 1 6 j 6 n. Now suppose that 1 6 i < j 6 n − 1. By Corollary 1.2, we ﬁrst have j j X X S1j = θ1kukj = θ11u1j + θ1kukj 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 h k=1 h=1

J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 62, 9 pp. http://jipam.vu.edu.au/ AMATRIX INEQUALITYFOR MÖBIUS FUNCTIONS 7

and, if 2 6 i < j 6 n − 1, then by Lemma 2.4, we have j k ! X X µi(h) S = (mod(j, k + 1) − mod(j, k)) = δ = 0. ij h ij k=i h=1 Now suppose that j = n. By Corollary 1.2, we ﬁrst have

n X S1n = θ1kukn k=1 n−1 k ! X X µ(h) = θ u + (mod(n, k + 1) − mod(n, k)) + θ u 11 1n h 1n nn k=2 h=1 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 h k=1 h=1 and, if 2 6 i 6 n − 1, we have n−1 k ! X X µi(h) S = (mod(n, k + 1) − mod(n, k)) + θ u in h in nn k=i h=i n−1 k ! n X X µi(h) X µi(k) = (mod(n, k + 1) − mod(n, k)) + n h k k=i h=i k=i n k ! X X µi(h) = (mod(n, k + 1) − mod(n, k)) = δ = 0 h in k=i h=i which completes the proof. For example, we get  1 1 1 1 2 1 8  1 2 6 6 − 30 15 − 105 − 105    1 1 1 1 1 1 2  0 2 6 − 12 − 12 − 12 − 12 − 3       1 1 1 1 1 1  0 0 3 12 12 − 12 − 12 3      0 0 0 1 1 1 1 − 3   4 20 20 20 5    U −1 =   . 8 0 0 0 0 1 1 1 4   5 30 30 15      0 0 0 0 0 1 1 4   6 42 21      0 0 0 0 0 0 1 1   7 7      0 0 0 0 0 0 0 1 

J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 62, 9 pp. http://jipam.vu.edu.au/ 8 OLIVIER BORDELLÈSAND BENOIT CLOITRE

2.2.2. A sufﬁcient condition for the PNT and the RH.

Corollary 2.6. For all integers i > 1 and n > 2 we have n X µi(k) 1 6 , k nσn k=i

where σn is the smallest singular value of Un. Thus any estimate of the form −1+ε σn ε n , where ε > 0 is any real number, is sufﬁcient to prove the PNT. Similarly, any estimate of the form −1/2−ε σn ε n where ε > 0 is any real number, is sufﬁcient to prove the RH. Proof. The result follows at once from the well-known inequalities −1 kUn k2 > max |θij| > |θin| 16i,j6n −1 −1 (see [3]), where k·k2 is the spectral norm, and the fact that σn = kUn k2 . Smallest singular values of triangular matrices have been studied by many authors. For example (see [2, 4]), it is known that, if An = (aij) is an invertible upper triangular matrix such that |aii| > |aij| for i < j, then we have min |a | σ ii n > 2n−1 but, applied here, such a bound is still very far from the PNT.

APPENDIX :APROOFOF REDHEFFER’S THEOREM

Let Sn = (sij) and Tn = (tij) be the matrices deﬁned by  ( M(n/i), if j = 1; 1, if i | j;  sij = and tij = 1, if i = j > 2; 0, otherwise  0, otherwise.

Proposition 2.7. We have Rn = SnTn. In particular, det Rn = M(n).

Proof. Set SnTn = (xij). If j = 1, by Lemma 1.1, we have n X X n X n/i x = s t = M = M = 1 = r . i1 ik k1 k d i1 k=1 k6n d6n/i i|k

If j > 2, then t1j = 0 and thus n ( X 1, if i | j xij = siktkj = sij = = rij k=2 0, otherwise which is the desired result. The second assertion follows at once from

det Rn = det Sn det Tn = det Tn = M(n). The proof is complete.

J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 62, 9 pp. http://jipam.vu.edu.au/ AMATRIX INEQUALITYFOR MÖBIUS FUNCTIONS 9

REFERENCES [1] O. BORDELLÈS, Thèmes d’arithmétique, Editions Ellipses, 2006. [2] N.J. HIGHAM, A survey of condition number for triangular matrices, Soc. Ind. Appl. Math., 29 (1987), 575–596.

[3] R.A. HORN AND C.R. JOHNSON, Matrix Analysis, Cambridge University Press, 1985. [4] F. LEMEIRE, Bounds for condition number of triangular and trapezoid matrices, BIT, 15 (1975), 58–64. [5] R.M. REDHEFFER, Eine explizit lösbare Optimierungsaufgabe, Internat. Schiftenreihe Numer. Math., 36 (1977), 213–216. [6] R.C. VAUGHAN, On the eigenvalues of Redheffer’s matrix I, in : Number Theory with an Emphasis on the Markoff Spectrum (Provo, Utah, 1991), 283–296, Lecture Notes in Pure and Appl. Math., 147, Dekker, New-York, 1993.

J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 62, 9 pp. http://jipam.vu.edu.au/