A MATRIX INEQUALITY FOR MÖBIUS FUNCTIONS

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Möbius Functions Olivier Bordellès and Benoit Cloitre

vol. 10, iss. 3, art. 62, 2009

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A MATRIX INEQUALITY FOR MÖBIUS FUNCTIONS

OLIVIER BORDELLÈS BENOIT CLOITRE

2 allée de la combe 19 rue Louise Michel

43000 AIGUILHE (France) 92300 LEVALLOIS-PERRET (France)

EMail:borde43@wanadoo.fr EMail:benoit7848c@orange.fr

Received: 24 November, 2008

Accepted: 27 March, 2009

Communicated by: L. Tóth

2000 AMS Sub. Class.: 15A15, 11A25, 15A18, 11C20.

Key words: Determinants, Dirichlet convolution, Möbius functions, Singular values.

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.

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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 matrixR_n = (r_ij)∈ M_n({0,1})defined by r_ij =

(1, ifi|j or j = 1;

0, otherwise and has shown that (see appendix)

detR_n=M(n) :=

n

X

k=1

µ(k),

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

PNT⇐⇒M(n) = o(n) and RH⇐⇒M(n) = O_ε n^1/2+ε

(for anyε > 0). These estimations for|detRn|remain unproven, but Vaughan [6]

showed that1is an eigenvalue ofR_nwith (algebraic) multiplicityn−h

logn log 2

i−1, that R_n has two "dominant" eigenvalues λ± such that |λ±| n^1/2, and that the others eigenvalues satisfyλ(logn)^2/5.

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It should be mentioned that Hadamard’s inequality, which states that

|detR_n|² 6

n

Y

i=1

kL_ik²₂,

whereL_i is theith row ofR_nandk·k₂is the euclidean norm onCⁿ, gives

(M(n))² 6n

n

Y

i=2

1 +hn

i i

= 2^n−[n/2]n

[n/2]

Y

i=2

1 +hn

i i

62^n−[n/2]

n+ [n/2]

n

, 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 Corollary2.3below). 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 functionµ, which plays an important role in number theory, satisfies the following well-known convolution identity.

Lemma 1.1. For every real numberx>1we have X

k6x

µ(k) hx

k i

=X

d6x

M x

d

= 1.

One can find a proof for example in [1]. The following corollary will be useful.

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Corollary 1.2. For every integerj >1we have (i)

j

X

k=1

µ(k) mod (j, k)

k =j

j

X

k=1

µ(k) k −1.

(ii)

j

X

k=1 k

X

h=1

µ(h) h

!

(mod(j, k+ 1)−mod(j, k)) = 1.

Proof.

(i) We have

j

X

k=1

µ(k) mod (j, k)

k =

j

X

k=1

µ(k) k

j−k

j k

=j

j

X

k=1

µ(k)

k −

j

X

k=1

µ(k) j

k

and we conclude with Lemma1.1.

(ii) Using Abel summation we get

j

X

k=1 k

X

h=1

µ(h) h

!

(mod(j, k+ 1)−mod(j, k))

=

j

X

h=1

µ(h) h

! _j X

k=1

−

j−1

X

k=1 k+1

X

h=1

µ(h)

h −

k

X

h=1

µ(h) h

! _k X

m=1

(mod(j, m+ 1)−mod(j, m))

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=j

j

X

k=1

µ(k)

k −

j−1

X

k=1

µ(k+ 1) mod (j, k+ 1) k+ 1

=j

j

X

k=1

µ(k)

k −

j

X

k=1

µ(k) mod (j, k) k

and we conclude using (i).

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2. An Integer Matrix Related to the Möbius Function

We now consider the matrixΓ_n= (γ_ij)defined by

γ_ij =











mod(j,2)−1, if i= 1 and 26j 6n;

mod(j, i+ 1)−mod(j, i), if 26i6n−1 and 16j 6n;

1, if (i, j)∈ {(1,1),(n,1)};

0, otherwise.

The matrixΓ_nis almost upper triangular except the entryγ_n1 = 1which is nonzero.

Note that it is easy to check that|γ_ij| 6ifor every1 6i, j 6 nand thatγ_ij =−1 if[j/2]< i < j.

Example 2.1.

Γ₈ =







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

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 ofΓ_n

Theorem 2.1. Letn >2be an integer andΓ_ndefined as above. Then we have det Γ_n=n!

n

X

k=1

µ(k) k .

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A possible proof of Theorem2.1uses a LU decomposition of the matrixΓ_n. Let L_n= (l_ij)andU_n = (u_ij)be the matrices defined by

u_ij =











0, if (i, j) = (n,1);

1, if (i, j) = (n, n);

γ_ij, otherwise and

l_ij =











1, if 16i=j 6n−1;

Pj k=1

µ(k)

k , if i=n and 16j 6n−1;

nPn k=1

µ(k)

k , if (i, j) = (n, n);

0, otherwise.

The proof of Theorem2.1follows from the lemma below.

Lemma 2.2. We haveΓ_n =L_nU_n.

Proof. SetL_nU_n = (x_ij). Wheni = 1we immediately obtainx_1j =u_1j =γ_1j.We also have

x_n1 =

n

X

k=1

l_nku_k1 =l_n1u₁₁= 1 =γ_n1.

Moreover, using Corollary1.2(ii) we get fori=nand26j 6n−1 x_nj =

n

X

k=1

l_nku_kj =l_n1u_1j+

n

X

k=2

l_nku_kj

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= mod(j,2)−1 +

j

X

k=2 k

X

h=1

µ(h) h

!

=

j

X

k=1 k

X

h=1

µ(h) h

!

(mod(j, k+ 1)−mod(j, k))−1 = 0 =γ_nj and, for(i, j) = (n, n), we have similarly

x_nn =

n

X

k=1

l_nku_kn =l_n1u_1n+

n−1

X

k=2

l_nku_kn+l_nnu_nn

= mod(n,2)−1 +

n−1

X

k=2 k

X

h=1

µ(h) h

!

(mod(n, k+ 1)−mod(n, k)) +n

n

X

k=1

µ(k) k

=

n

X

k=1 k

X

h=1

µ(h) h

!

(mod(n, k+ 1)−mod(n, k))−1 = 0 =γ_nn. Finally, for26i6n−1and16j 6n, we get

x_ij =

n

X

k=1

l_iku_kj =l_iiu_ij =u_ij =γ_ij.

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Example 2.2.

Γ₈ =







1 0 0 0 0 0 0 0

0 1 0 0 0 0 0 0

0 0 1 0 0 0 0 0

0 0 0 1 0 0 0 0

0 0 0 0 1 0 0 0

0 0 0 0 0 1 0 0

0 0 0 0 0 0 1 0

1 ¹₂ ¹₆ ¹₆ −₃₀¹ ₁₅² −₁₀₅¹ −₁₀₅⁸













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

0 0 0 0 5 −1 −1 −1

0 0 0 0 0 6 −1 −1

0 0 0 0 0 0 7 −1

0 0 0 0 0 0 0 1





 .

Theorem2.1now immediately follows from

det Γ_n = detL_ndetU_n= (n−1)! detL_n=n!

n

X

k=1

µ(k) k .

We easily deduce the following criteria for the PNT and the RH.

Corollary 2.3. For any real numberε >0we have

PNT⇐⇒det Γn=o(n!) and RH⇐⇒det Γn=Oε(n^−1/2+εn!).

2.2. A sufficient condition for the PNT and the RH

2.2.1. Computation ofU_n⁻¹

The inverse ofU_nuses a Möbius-type function denoted byµ_iwhich we define below.

Definition 2.4. Set µ1 = µ the well-known Möbius function and, for any integer

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i>2, we define the Möbius functionµ_i byµ_i(1) = 1and, for any integerm>2, by

µi(m) :=









 µ ^m_i

, if i|m and (i+ 1)-m;

−µ _i+1^m

, if (i+ 1)|m andi-m;

µ ^m_i

−µ _i+1^m

, if i(i+ 1)|m;

0, otherwise.

The following result completes and generalizes Lemma1.1and Corollary1.2.

Lemma 2.5. For all integersi, j >2we have

j

X

k=i

µi(k) j

k

=δij,

j

X

k=i

µ_i(k) mod (j, k)

k =j

j

X

k=i

µ_i(k) k −δ_ij,

j

X

k=i k

X

h=i

µ_i(h) h

!

(mod(j, k+ 1)−mod(j, k)) =δ_ij, whereδij is the Kronecker symbol.

Proof. We only prove the first identity, the proof of the two others being strictly iden- tical to the identities of Corollary1.2. Without loss of generality, one can suppose that26i6j. Ifi=j then we have

j

X

k=i

µ_i(k) j

k

=µ_j(j) j

j

=µ j

j

= 1.

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Now suppose that26i < j. By Lemma1.1, we have

j

X

k=i

µi(k) j

k

=

j

X

i|k,k=i(i+1)-k

µ k

i j k

−

j

X

(i+1)|k, ik=i -k

µ k

i+ 1 j k

+

j

X

i(i+1)|kk=i

µ

k i

−µ k

i+ 1

j k

=

j

X

k=ii|k

µ k

i j k

−

j

X

(i+1)|kk=i

µ k

i+ 1 j k

=

[j/i]

X

h=1

µ(h) [j/i]

h

−

[j/(i+1)]

X

h=1

µ(h)

[j/(i+ 1)]

h

= 1−1 = 0, which concludes the proof.

This result gives the inverse ofUn.

Corollary 2.6. SetU_n⁻¹ = (θ_ij). Then we have θ_ij =

j

X

k=i

µi(k)

k (16i6j 6n−1) θ_in=n

n

X

k=i

µ_i(k)

k (16i6n).

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Proof. SinceU_n⁻¹is upper triangular, it suffices to show that, for all integers16i6 j 6n, we have

j

X

k=i

θ_iku_kj =δ_ij.

In what follows, we setS_ij as the sum on the left-hand side

We easily check that S_jj = 1for every integer 1 6 j 6 n. Now suppose that 16i < j 6n−1. By Corollary1.2, we first have

S_1j =

j

X

k=1

θ_1ku_kj =θ₁₁u_1j+

j

X

k=2

θ_1ku_kj

= mod(j,2)−1 +

j

X

k=2 k

X

h=1

µ(h) h

!

=

j

X

k=1 k

X

h=1

µ(h) h

!

(mod(j, k+ 1)−mod(j, k))−1 = 0 and, if26i < j 6n−1, then by Lemma2.5, we have

S_ij =

j

X

k=i k

X

h=1

µi(h) h

!

(mod(j, k+ 1)−mod(j, k)) =δ_ij = 0.

Now suppose thatj =n. By Corollary1.2, we first have S_1n=

n

X

k=1

θ_1ku_kn=θ₁₁u_1n+

n−1

X

k=2 k

X

h=1

µ(h) h

!

(mod(n, k+ 1)−mod(n, k)) +θ_1nu_nn

= mod(n,2)−1 +

n−1

X

k=2 k

X

h=1

µ(h) h

!

n

X

k=1

µ(k) k

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=

n

X

k=1 k

X

h=1

µ(h) h

!

(mod(n, k+ 1)−mod(n, k))−1 = 0 and, if26i6n−1, we have

S_in=

n−1

X

k=i k

X

h=i

µ_i(h) h

!

(mod(n, k+ 1)−mod(n, k)) +θ_inu_nn

=

n−1

X

k=i k

X

h=i

µ_i(h) h

!

n

X

k=i

µ_i(k) k

=

n

X

k=i k

X

h=i

µ_i(h) h

!

(mod(n, k+ 1)−mod(n, k)) = δ_in = 0 which completes the proof.

For example, we get

U₈⁻¹ =







1 ¹₂ ¹₆ ¹₆ −₃₀¹ ₁₅² −₁₀₅¹ −₁₀₅⁸ 0 ¹₂ ¹₆ −₁₂¹ −₁₂¹ −₁₂¹ −₁₂¹ −²₃ 0 0 ¹₃ ₁₂¹ ₁₂¹ −₁₂¹ −₁₂¹ ¹₃ 0 0 0 ¹₄ ₂₀¹ ₂₀¹ ₂₀¹ −³₅ 0 0 0 0 ¹₅ ₃₀¹ ₃₀¹ ₁₅⁴ 0 0 0 0 0 ¹₆ ₄₂¹ ₂₁⁴

0 0 0 0 0 0 ¹₇ ¹₇

0 0 0 0 0 0 0 1





 .

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Corollary 2.7. For all integersi>1andn>2we have

n

X

k=i

µ_i(k) k

6 1

nσ_n,

whereσnis the smallest singular value ofUn. Thus any estimate of the form σ_n_ε n^−1+ε,

whereε >0is any real number, is sufficient to prove the PNT. Similarly, any estimate of the form

σ_n_ε n^−1/2−ε

whereε >0is any real number, is sufficient to prove the RH.

Proof. The result follows at once from the well-known inequalities kU_n⁻¹k₂ > max

16i,j6n|θ_ij|>|θ_in|

(see [3]), wherek·k₂ is the spectral norm, and the fact thatσ_n=kU_n⁻¹k⁻¹₂ .

Smallest singular values of triangular matrices have been studied by many au- thors. For example (see [2,4]), it is known that, ifA_n = (a_ij)is an invertible upper triangular matrix such that|a_ii|>|a_ij|fori < j, then we have

σ_n> min|a_ii| 2ⁿ⁻¹

but, applied here, such a bound is still very far from the PNT.

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Appendix : A Proof of Redheffer’s Theorem

LetS_n= (s_ij)andT_n= (t_ij)be the matrices defined by

s_ij =

(1, if i|j;

0, otherwise

and t_ij =











M(n/i), if j = 1;

1, if i=j >2;

0, otherwise.

Proposition 2.8. We haveR_n =S_nT_n.In particular,detR_n=M(n).

Proof. SetS_nT_n = (x_ij). Ifj = 1,by Lemma1.1, we have x_i1 =

n

X

k=1

s_ikt_k1 =X

k6n i|k

Mn k

= X

d6n/i

M n/i

d

= 1 =r_i1.

Ifj >2, thent_1j = 0and thus x_ij =

n

X

k=2

s_ikt_kj =s_ij =

(1, if i|j 0, otherwise

=r_ij which is the desired result. The second assertion follows at once from

detR_n = detS_ndetT_n = detT_n =M(n).

The proof is complete.

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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 The- ory with an Emphasis on the Markoff Spectrum (Provo, Utah, 1991), 283–296, Lecture Notes in Pure and Appl. Math., 147, Dekker, New-York, 1993.