A MATRIX INEQUALITY FOR MÖBIUS FUNCTIONS
OLIVIER BORDELLÈS
BENOIT CLOITRE
2 allée de la combe
43000 AIGUILHE (France)
EMail: borde43@wanadoo.fr
19 rue Louise Michel
92300 LEVALLOIS-PERRET (France)
EMail: benoit7848c@orange.fr
Möbius Functions
Olivier Bordellès and Benoit Cloitre
vol. 10, iss. 3, art. 62, 2009
Title Page
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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Contents
1
Introduction
1.1 Arithmetic motivation . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Convolution identities for the Möbius function . . . . . . . . . . . .
2
An Integer Matrix Related to the Möbius Function
2.1 The determinant of Γn . . . . . . . . . . . . . . . . .
2.2 A sufficient condition for the PNT and the RH . . . . .
2.2.1 Computation of Un−1 . . . . . . . . . . . . . .
2.2.2 A sufficient condition for the PNT and the RH
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Möbius Functions
Olivier Bordellès and Benoit Cloitre
vol. 10, iss. 3, art. 62, 2009
Title Page
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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 matrix Rn = (rij ) ∈ Mn ({0, 1}) defined by
(
1, if i | j or j = 1;
rij =
0, otherwise
Olivier Bordellès and Benoit Cloitre
vol. 10, iss. 3, art. 62, 2009
Title Page
and has shown that (see appendix)
det Rn = M (n) :=
Möbius Functions
n
X
Contents
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
PNT ⇐⇒ M (n) = o(n) and RH ⇐⇒ M (n) = Oε n1/2+ε
(for any ε > 0). These estimations for | det Rn | remain unproven, but
[6]
h Vaughan
i
log n
showed that 1 is an eigenvalue of Rn with (algebraic) multiplicity n− log 2 −1, that
Rn has two "dominant" eigenvalues λ± such that |λ± | n1/2 , and that the others
eigenvalues satisfy λ (log n)2/5 .
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It should be mentioned that Hadamard’s inequality, which states that
n
Y
| det Rn | 6
kLi k22 ,
2
i=1
where Li is the ith row of Rn and k·k2 is the euclidean norm on Cn , gives
2
(M (n)) 6 n
n Y
i=2
1+
h n i
i
n−[n/2]
=2
n
[n/2] Y
i=2
1+
h n i
i
n−[n/2]
62
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
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.
Möbius Functions
Olivier Bordellès and Benoit Cloitre
vol. 10, iss. 3, art. 62, 2009
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1.2.
Convolution identities for the Möbius function
The function µ, which plays an important role in number theory, satisfies the following well-known convolution identity.
Lemma 1.1. For every real number x > 1 we have
hxi X 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.
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Corollary 1.2. For every integer j > 1 we have
(i)
j
X
µ(k) mod (j, k)
k
k=1
=j
j
X
µ(k)
k=1
k
− 1.
(ii)
j
k
X
X
µ(h)
h
k=1
h=1
Möbius Functions
!
(mod(j, k + 1) − mod(j, k)) = 1.
Olivier Bordellès and Benoit Cloitre
vol. 10, iss. 3, art. 62, 2009
Proof.
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(i) We have
j
X
µ(k) mod (j, k)
k=1
k
=
j
X
µ(k)
k=1
k
j
j
X
µ(k) X
j
j
j−k
=j
−
µ(k)
k
k
k
k=1
k=1
and we conclude with Lemma 1.1.
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(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
! k
j−1
k+1
k
X X µ(h) X
µ(h) X
−
−
(mod(j, m + 1) − mod(j, m))
h
h
m=1
k=1
h=1
h=1
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=j
j
X
µ(k)
k=1
j
=j
k
X µ(k)
k=1
k
−
j−1
X
µ(k + 1) mod (j, k + 1)
k+1
k=1
j
−
X µ(k) mod (j, k)
k=1
and we conclude using (i).
k
Möbius Functions
Olivier Bordellès and Benoit Cloitre
vol. 10, iss. 3, art. 62, 2009
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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 6 j 6 n;



 mod(j, i + 1) − mod(j, i), if 2 6 i 6 n − 1 and 1 6 j 6 n;
γij =

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


3 −1 −1 2
2 −2
0 0


0
4 −1 −1 −1 3 
0 0
Γ8 = 
.
0
0
5 −1 −1 −1
0 0
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.
Möbius Functions
Olivier Bordellès and Benoit Cloitre
vol. 10, iss. 3, art. 62, 2009
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The determinant of Γn
Theorem 2.1. Let n > 2 be an integer and Γn defined as above. Then we have
det Γn = n!
n
X
µ(k)
k=1
k
.
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
Möbius Functions
Olivier Bordellès and Benoit Cloitre
and
lij =

1,





Pj

k=1
if 1 6 i = j 6 n − 1;
µ(k)
,
k
Pn


n

k=1




0,
µ(k)
,
k
if i = n
and 1 6 j 6 n − 1;
otherwise.
Lemma 2.2. We have Γn = Ln Un .
Moreover, using Corollary 1.2 (ii) we get for i = n and 2 6 j 6 n − 1
lnk ukj = ln1 u1j +
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k=1
k=1
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Page 8 of 17
Proof. Set Ln Un = (xij ). When i = 1 we immediately obtain x1j = u1j = γ1j . We
also have
n
X
xn1 =
lnk uk1 = ln1 u11 = 1 = γn1 .
xnj =
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if (i, j) = (n, n);
The proof of Theorem 2.1 follows from the lemma below.
n
X
vol. 10, iss. 3, art. 62, 2009
n
X
k=2
lnk ukj
!
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 = γnj
h
k=1
h=1
and, for (i, j) = (n, n), we have similarly
xnn =
n
X
lnk ukn = ln1 u1n +
k=1
= mod(n, 2) − 1 +
n−1
X
Möbius Functions
Olivier Bordellès and Benoit Cloitre
lnk ukn + lnn unn
n−1
X
k=2
k
X
k=2
h=1
µ(h)
h
!
(mod(n, k + 1) − mod(n, k))
n
X
µ(k)
k
k=1
!
n
k
X
X
µ(h)
=
(mod(n, k + 1) − mod(n, k)) − 1 = 0 = γnn .
h
k=1
h=1
+n
Finally, for 2 6 i 6 n − 1 and 1 6 j 6 n, we get
xij =
n
X
k=1
lik ukj = lii uij = uij = γij .
vol. 10, iss. 3, art. 62, 2009
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Example 2.2.

1 0
0 1

0 0

0 0
Γ8 = 
0 0
0 0

0 0
1 12
0
0
1
0
0
0
0
0
0
0
1
0
0
0
1
6
1
6
0
0
0
0
1
0
0
1
− 30
0
0
0
0
0
1
0
2
15


0
0
1 −1 0 −1 0 −1 0 −1
0
0  0 2 −1 1
1
0
0
2


0
0  0 0
3 −1 −1 2
2 −2


0
0  0 0
0
4 −1 −1 −1 3 

.
0
0  0 0
0
0
5 −1 −1 −1


0
0  0 0
0
0
0
6 −1 −1

1
0  0 0
0
0
0
0
7 −1
1
8
0 0
0
0
0
0
0
1
− 105
− 105
Möbius Functions
Olivier Bordellès and Benoit Cloitre
vol. 10, iss. 3, art. 62, 2009
Theorem 2.1 now immediately follows from
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det Γn = det Ln det Un = (n − 1)! det Ln = n!
n
X
µ(k)
k=1
k
.
Contents
We easily deduce the following criteria for the PNT and the RH.
Corollary 2.3. For any real number ε > 0 we have
PNT ⇐⇒ det Γn = o(n!) and
2.2.
RH ⇐⇒ det Γn = Oε (n
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−1/2+ε
n!).
A sufficient condition for the PNT and the RH
2.2.1. Computation of Un−1
The inverse of Un uses a Möbius-type function denoted by µi which 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 function µi by µi (1) = 1 and, for any integer m > 2, by
 m
µ i ,
if i | m and (i + 1) - m;






m
−µ i+1
,
if (i + 1) | m and i - m;
µi (m) :=

m

µ mi − µ i+1
, if i(i + 1) | m;





0,
otherwise.
The following result completes and generalizes Lemma 1.1 and Corollary 1.2.
Lemma 2.5. For all integers i, j > 2 we have
j
X
j
µi (k)
= δij ,
k
k=i
j
X
µi (k) mod (j, k)
k
k=i
j
k
X X
µi (h)
k=i
h=i
h
=j
Olivier Bordellès and Benoit Cloitre
vol. 10, iss. 3, art. 62, 2009
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j
X
µi (k)
k=i
Möbius Functions
k
− δij ,
!
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(mod(j, k + 1) − mod(j, k)) = δij ,
where δij is the Kronecker symbol.
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
µi (k)
= µj (j)
=µ
= 1.
k
j
j
k=i
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Now suppose that 2 6 i < j. By Lemma 1.1, we have
j
X
j
µi (k)
=
k
k=i
j
X
k=i
i|k, (i+1)-k
k
j
µ
−
i
k
j
X
µ
k=i
(i+1)|k, i-k
k
i+1
j
k
j
X
k
k
j
+
µ
−µ
i
i+1
k
k=i
i(i+1)|k
j
j
X
X
k
j
k
j
=
µ
µ
−
i
k
i+1
k
k=i
k=i
i|k
(i+1)|k
[j/i]
X
Möbius Functions
Olivier Bordellès and Benoit Cloitre
vol. 10, iss. 3, art. 62, 2009
Title Page
[j/(i+1)]
X
[j/i]
[j/(i + 1)]
µ(h)
=
−
µ(h)
h
h
h=1
h=1
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= 1 − 1 = 0,
J
I
which concludes the proof.
Page 12 of 17
This result gives the inverse of Un .
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Corollary 2.6. Set Un−1 = (θij ). Then we have
θij =
Contents
j
X
µi (k)
k=i
n
X
θin = n
k=i
k
µi (k)
k
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(1 6 i 6 j 6 n − 1)
(1 6 i 6 n).
Proof. Since Un−1 is upper triangular, it suffices to show that, for all integers 1 6 i 6
j 6 n, we have
j
X
θik ukj = 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
S1j =
j
X
θ1k ukj = θ11 u1j +
k=1
j
X
Olivier Bordellès and Benoit Cloitre
vol. 10, iss. 3, art. 62, 2009
θ1k ukj
k=2
j
= mod(j, 2) − 1 +
k
X X
µ(h)
k=2
j
k
X
X
µ(h)
=
h
k=1
h=1
Möbius Functions
h=1
h
!
Title Page
(mod(j, k + 1) − mod(j, k))
!
(mod(j, k + 1) − mod(j, k)) − 1 = 0
and, if 2 6 i < j 6 n − 1, then by Lemma 2.5, we have
!
j
k
X
X
µi (h)
(mod(j, k + 1) − mod(j, k)) = δij = 0.
Sij =
h
k=i
h=1
Now suppose that j = n. By Corollary 1.2, we first have
!
n
n−1
k
X
X
X
µ(h)
θ1k ukn = θ11 u1n +
(mod(n, k + 1) − mod(n, k)) + θ1n unn
S1n =
h
k=1
k=2
h=1
!
n−1
k
n
X
X
X
µ(h)
µ(k)
= mod(n, 2) − 1 +
(mod(n, k + 1) − mod(n, k)) + n
h
k
k=2
h=1
k=1
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n
k
X
X
µ(h)
=
h
k=1
h=1
!
(mod(n, k + 1) − mod(n, k)) − 1 = 0
and, if 2 6 i 6 n − 1, we have
!
n−1
k
X
X
µi (h)
Sin =
(mod(n, k + 1) − mod(n, k)) + θin unn
h
k=i
h=i
!
n
n−1
k
X
X
X
µi (h)
µi (k)
(mod(n, k + 1) − mod(n, k)) + n
=
h
k
k=i
k=i
h=i
!
n
k
X
X
µi (h)
=
(mod(n, k + 1) − mod(n, k)) = δin = 0
h
k=i
h=i
which completes the proof.
For example, we get

U8−1
1

0

0


0

=
0


0

0

0
1
2
1
2
0 0
1
12
1
4
1
− 30
1
− 12
1
12
1
20
2
15
1
− 12
1
− 12
1
20
1
− 105
1
− 12
1
− 12
1
20
0 0
0
1
5
0 0
0
0
1
30
1
6
0 0
0
0
0
1
30
1
42
1
7
4
15
4
21
1
7
0 0
0
0
0
0
1
0
1
6
1
6
1
3
1
6
1
− 12
8
− 105

− 23 

1 
3 

− 35 

Möbius Functions
Olivier Bordellès and Benoit Cloitre
vol. 10, iss. 3, art. 62, 2009
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
.







Page 14 of 17
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2.2.2. A sufficient condition for the PNT and the RH
Corollary 2.7. 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
σn ε n−1+ε ,
where ε > 0 is any real number, is sufficient to prove the PNT. Similarly, any estimate
of the form
σn ε n−1/2−ε
where ε > 0 is any real number, is sufficient to prove the RH.
Proof. The result follows at once from the well-known inequalities
kUn−1 k2
> max |θij | > |θin |
16i,j6n
(see [3]), where k·k2 is the spectral norm, and the fact that σn = kUn−1 k−1
2 .
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
σn >
min |aii |
2n−1
but, applied here, such a bound is still very far from the PNT.
Möbius Functions
Olivier Bordellès and Benoit Cloitre
vol. 10, iss. 3, art. 62, 2009
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Appendix : A Proof of Redheffer’s Theorem
Let Sn = (sij ) and Tn = (tij ) be the matrices defined by

M (n/i),

(


1, if i | j;
sij =
and tij = 1,

0, otherwise


0,
if j = 1;
if i = j > 2;
otherwise.
Möbius Functions
Olivier Bordellès and Benoit Cloitre
Proposition 2.8. We have Rn = Sn Tn . In particular, det Rn = M (n).
Proof. Set Sn Tn = (xij ). If j = 1, by Lemma 1.1, we have
n
X
X n
X
n/i
xi1 =
sik tk1 =
M
=
M
= 1 = ri1 .
k
d
k=1
k6n
d6n/i
i|k
If j > 2, then t1j = 0 and thus
xij =
n
X
k=2
sik tkj = sij =
(
1,
0,
if i | j
otherwise
vol. 10, iss. 3, art. 62, 2009
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= rij
which is the desired result. The second assertion follows at once from
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det Rn = det Sn det Tn = det Tn = M (n).
The proof is complete.
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
Press, 1985.
AND
C.R. JOHNSON, Matrix Analysis, Cambridge University
Möbius Functions
Olivier Bordellès and Benoit Cloitre
[4] F. LEMEIRE, Bounds for condition number of triangular and trapezoid matrices,
BIT, 15 (1975), 58–64.
vol. 10, iss. 3, art. 62, 2009
[5] R.M. REDHEFFER, Eine explizit lösbare Optimierungsaufgabe, Internat.
Schiftenreihe Numer. Math., 36 (1977), 213–216.
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[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.
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