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 (F RANCE )
borde43@wanadoo.fr
19 RUE L OUISE M ICHEL
92300 LEVALLOIS-PERRET (F RANCE )
benoit7848c@orange.fr
Received 24 November, 2008; accepted 27 March, 2009
Communicated by L. Tóth
A BSTRACT. 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. I NTRODUCTION
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)
det Rn = M (n) :=
n
X
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
317-08
2
O LIVIER B ORDELLÈS AND B ENOIT C LOITRE
(for any ε > 0). These estimations for | det Rn | remain unproven,
h
i but Vaughan [6] showed that
log n
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 .
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
1+
h n i
i
i=2
n−[n/2]
=2
n
[n/2] Y
1+
i=2
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.
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.
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
!
(mod(j, k + 1) − mod(j, k)) = 1.
Proof.
(i) We have
j
X
µ(k) mod (j, k)
k=1
k
=
j
X
µ(k)
k=1
k
j
j
X
j
µ(k) X
j
j−k
=j
−
µ(k)
k
k
k
k=1
k=1
and we conclude with Lemma 1.1.
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 62, 9 pp.
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A M ATRIX I NEQUALITY
FOR
M ÖBIUS F UNCTIONS
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
k=1
h=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
h=1
k=1
h=1
=j
j
X
µ(k)
k=1
=j
k
j
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
k
and we conclude using (i).
2. A N I NTEGER M ATRIX R ELATED TO
THE
M ÖBIUS F UNCTION
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. 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
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 62, 9 pp.
k
.
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4
O LIVIER B ORDELLÈS AND B ENOIT C LOITRE
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
lij =

1,




P


 j
k=1
if 1 6 i = j 6 n − 1;
µ(k)
,
k
Pn


n

k=1




0,
if i = n
µ(k)
,
k
and 1 6 j 6 n − 1;
if (i, j) = (n, n);
otherwise.
The proof of Theorem 2.1 follows from the lemma below.
Lemma 2.2. We have Γn = Ln Un .
Proof. Set Ln Un = (xij ). When i = 1 we immediately obtain x1j = u1j = γ1j . We also have
xn1 =
n
X
lnk uk1 = ln1 u11 = 1 = γn1 .
k=1
Moreover, using Corollary 1.2 (ii) we get for i = n and 2 6 j 6 n − 1
xnj =
n
X
lnk ukj = ln1 u1j +
k=1
n
X
lnk ukj
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 = γnj
h
k=1
h=1
and, for (i, j) = (n, n), we have similarly
xnn =
n
X
k=1
lnk ukn = ln1 u1n +
n−1
X
lnk ukn + lnn unn
k=2
!
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
!
n
k
X
X
µ(h)
=
(mod(n, k + 1) − mod(n, k)) − 1 = 0 = γnn .
h
k=1
h=1
Finally, for 2 6 i 6 n − 1 and 1 6 j 6 n, we get
xij =
n
X
lik ukj = lii uij = uij = γij .
k=1
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 62, 9 pp.
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A M ATRIX I NEQUALITY
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
FOR
M ÖBIUS F UNCTIONS
5


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
6 −1 −1
 0 0



1
0
0 0
0
0
0
0
7 −1
1
8
0 0
0
0
0
0
0
1
− 105
− 105
0
0
0
0
0
1
0
2
15
Theorem 2.1 now immediately follows from
det Γn = det Ln det Un = (n − 1)! det Ln = n!
n
X
µ(k)
k=1
k
.
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
RH ⇐⇒ det Γn = Oε (n−1/2+ε n!).
2.2. 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 2.1. 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



,
if (i + 1) | m and i - m;
−µ


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
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
j
X
µi (k)
k=i
k
− δij ,
!
(mod(j, k + 1) − mod(j, k)) = δij ,
where δij is the Kronecker symbol.
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 62, 9 pp.
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6
O LIVIER B ORDELLÈS AND B ENOIT C LOITRE
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
Now suppose that 2 6 i < j. By Lemma 1.1, we have
j
j
X
X
j
k
j
µi (k)
=
µ
−
k
i
k
k=i
k=i
i|k, (i+1)-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+1)]
X
[j/(i + 1)]
[j/i]
=
µ(h)
−
µ(h)
h
h
h=1
h=1
[j/i]
X
= 1 − 1 = 0,
which concludes the proof.
This result gives the inverse of Un .
Corollary 2.5. Set Un−1 = (θij ). Then we have
θij =
j
X
µi (k)
k=i
n
X
θin = n
k=i
(1 6 i 6 j 6 n − 1)
k
µi (k)
k
(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 +
j
X
k=1
θ1k ukj
k=2
j
= mod(j, 2) − 1 +
k
X X
µ(h)
k=2
j
=
X
k
X
k=1
h=1
µ(h)
h
h=1
h
!
(mod(j, k + 1) − mod(j, k))
!
(mod(j, k + 1) − mod(j, k)) − 1 = 0
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A M ATRIX I NEQUALITY
FOR
M ÖBIUS F UNCTIONS
7
and, if 2 6 i < j 6 n − 1, then by Lemma 2.4, we have
!
j
k
X
X
µi (h)
Sij =
(mod(j, k + 1) − mod(j, k)) = δij = 0.
h
k=i
h=1
Now suppose that j = n. By Corollary 1.2, we first have
S1n =
n
X
θ1k ukn
k=1
k
n−1
X
X
µ(h)
= θ11 u1n +
h
k=2
h=1
!
(mod(n, k + 1) − mod(n, k)) + θ1n unn
!
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
!
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)
Sin =
h
k=i
h=i
!
n−1
k
X
X
µi (h)
!
=
k=i
h=i
(mod(n, k + 1) − mod(n, k)) + θin unn
(mod(n, k + 1) − mod(n, k)) + n
h
n
X
µi (k)
k=i
n
k
X
X
µi (h)
=
h
k=i
h=i
k
!
(mod(n, k + 1) − mod(n, k)) = δin = 0
which completes the proof.
For example, we get

U8−1
1


0



0



0


=
0


0



0



0
1
2
1
6
1
6
1
− 30
2
15
1
2
1
6
1
− 12
1
− 12
1
− 12
1
− 12
0
1
3
1
12
1
12
1
− 12
1
− 12
0 0
1
4
1
20
1
20
1
20
0 0
0
1
5
1
30
1
30
0 0
0
0
1
6
1
42
0 0
0
0
0
1
7
0 0
0
0
0
0
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 62, 9 pp.
1
8
− 105
− 105






1 

3 


− 35 


4 .
15 


4 
21 


1 
7 


1 
− 23
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8
O LIVIER B ORDELLÈS AND B ENOIT C LOITRE
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
1
µi (k) ,
6
nσn
k
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
min |aii |
2n−1
but, applied here, such a bound is still very far from the PNT.
σn >
A PPENDIX : A P ROOF OF R EDHEFFER ’ S T HEOREM
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 = 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
sik tkj = sij =
k=2
(
1,
0,
if i | j
otherwise
= rij
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/
A M ATRIX I NEQUALITY
FOR
M ÖBIUS F UNCTIONS
9
R EFERENCES
[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/