#A56 INTEGERS 20 (2020)
A NOTE ON A UNITARY ANALOG TO REDHEFFER’S MATRIX
Olivier Bordell`es Aiguilhe, France [email protected]
Received: 11/2/19, Accepted: 7/7/20, Published: 7/24/20
Abstract We study a unitary analog to Redhe↵er’s matrix. It is first proved that the deter- minant of this matrix is the unitary analogue to that of Redhe↵er’s matrix. We also show that the coe cients of the characteristic polynomial may be expressed as sums of Stirling numbers of the second kind. This implies in particular that 1 is an eigenvalue with algebraic multiplicity greater than that of Redhe↵er’s matrix.
1. Introduction
In 1977, Redhe↵er [7] introduced the matrix R = (r ) ( 0, 1 ) defined by n ij 2 Mn { } 1, if i j or j = 1 rij = | (0, otherwise and has shown that n det Rn = M(n) := µ(k), kX=1 where µ is the M¨obius function and M is the Mertens function. This determinant is clearly related to two of the most famous problems in number theory, namely the Prime Number Theorem (PNT) and the Riemann Hypothesis (RH) since it is well-known that
PNT M(n) = o(n) and RH M(n) = O n1/2+" . () () " ⇣ ⌘ The second estimate remains unproven, but Vaughan [11] showed that 1 is an eigen- value of R with algebraic multiplicity n log n 1, that R has two “domi- n log 2 n nant” eigenvalues such that n1/2, andj thatk the other eigenvalues satisfy ± | ±| ⇣ (log n)2/5. ⌧ INTEGERS: 20 (2020) 2
The purpose of this note is to supply an analogous study to the 0, 1 -matrix { } Rn⇤ = (⇢ij) defined by 1, if i j or j = 1 ⇢ij = k (0, otherwise. Recall that the integer i is said to be a unitary divisor of j, denoted by i j, k whenever i j and gcd i, j = 1. | i For instance, when n = 8, we have 1 1 1 1 1 1 1 1 1 1 0 0 0 1 0 0 01 0 1 0 0 1 0 01 B1 0 0 1 0 0 0 0C R⇤ = B C . 8 B1 0 0 0 1 0 0 0C B C B1 0 0 0 0 1 0 0C B C B1 0 0 0 0 0 1 0C B C B1 0 0 0 0 0 0 1C B C @ A Note that this matrix does not belong to the set of general matrices studied in [2].
This article is organized as follows. In Section 2, we shall use some elementary properties of unitary divisors to determine an LU-decomposition of the matrix Rn⇤ and deduce its determinant. In Section 3, following the ideas of [11], we shall discuss further on the characteristic polynomial of Rn⇤ and the algebraic multiplicity of the eigenvalue 1 of this matrix.
1.1. Notation
In what follows, n > 2 is a fixed integer and the function µ⇤ is the unitary analog of the M¨obius function. We also define
M ⇤(x, n) := µ⇤(k) (x > 0, n N) 2 k6x gcd(Xk,n)=1 and simply write M ⇤(x) := M ⇤(x, 1) which is the unitary analog of the Mertens function. As usual, let 1(n) = 1 and the unitary convolution product of the two arithmetic functions f and g is defined by
(f g)(n) = f(d)g(n/d). d n Xk Finally, from [3, Theorem 2.5] it is known that
!(n) µ⇤(n) = ( 1) , INTEGERS: 20 (2020) 3 where !(n) is the number of distinct prime factors of n, and from [3, Corollary 2.1.2] we have the important convolution identity
1, if n = 1 (µ⇤ 1) (n) = (1) (0, otherwise.
2. The Determinant of Rn⇤
We start with the following basic identities involving unitary divisors which will prove to be useful to determine an LU-type decomposition of the matrix Rn⇤ . Lemma 1. (i) Let i, j be positive integers. Then
1, if i = j µ⇤(d) = 0, otherwise. d j ( iXjk/d k (ii) Let 1 i n be integers. Then n M ⇤ , k = 1. k k n Xik ⇣ ⌘ k Proof.
(i) If i , j, then the sum is equal to 0 since (d j and i j/d) implies i j. k k k If i j, then k (d j and i j/d) if and only if d j/i k k k so that using (1) we get
1, if j/i = 1 µ⇤(d) = µ⇤(d) = 0, otherwise. d j d j/i ( iXjk/d Xk k (ii) Using the identity above, we get j n 1 = µ⇤ = µ⇤(d) = M ⇤ , k . k k j n k j k n d n/k k n ✓ ◆ ⇣ ⌘ X Xi kk Xik gcd(Xd,k)=1 Xik k k k The proof is complete. INTEGERS: 20 (2020) 4
Let S = (s ) and T = (t ) be the (n n)-matrices defined by n ij n ij ⇥
M ⇤(n/i, i), if j = 1 1, if i j sij = k and tij = 81, if i = j 2 (0, otherwise <>0, otherwise.
For instance :> 1 1 1 1 1 1 1 1 4 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 00 0 1 0 0 1 0 01 0 0 0 1 0 0 0 0 0 1 B0 0 0 1 0 0 0 0C B 1 0 0 1 0 0 0 0 C S = B C and T = B C . 8 B0 0 0 0 1 0 0 0C 8 B 1 0 0 0 1 0 0 0 C B C B C B0 0 0 0 0 1 0 0C B 1 0 0 0 0 1 0 0 C B C B C B0 0 0 0 0 0 1 0C B 1 0 0 0 0 0 1 0 C B C B C B0 0 0 0 0 0 0 1C B 1 0 0 0 0 0 0 1 C B C B C @ A @ A We are now in a position to prove the first result concerning the matrix Rn⇤ .
Theorem 2. Let n 2 be an integer. Then R⇤ = S T . In particular n n n n
det Rn⇤ = M ⇤(n) = µ⇤(k). kX=1
Proof. Set SnTn = (xij). If j = 1, using Lemma 1 (ii) we get
n n x = s t = M ⇤ , k = 1 = ⇢ . i1 ik k1 k i1 k=1 k6n X Xi k ⇣ ⌘ k
If j > 2, then t1j = 0 and thus
n 1, if i j xij = siktkj = sij = k = ⇢ij, (0, otherwise kX=2 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.
Corollary 3. The Riemann hypothesis is true if and only if, for each " > 0
1/2+" det Rn⇤ = O n . ⇣ ⌘ INTEGERS: 20 (2020) 5
3. The Characteristic Polynomial of Rn⇤ 3.1. The “Trivial” Eigenvalue 1
log n Let ` = log 2 . It is proved in [11] that 1 is an eigenvalue of the Redhe↵er’s matrix Rn of algebraicj k multiplicity equal to n ` 1. We will show in this section that the algebraic multiplicity mn of the eigenvalue 1 of Rn⇤ may be somewhat larger. To this end, we first note that the method developed in [2, 11] to determine the characteristic polynomial of Redhe↵er type matrices can readily be adapted to the matrix Rn⇤ , which yields
` n n 2 n k 1 det ( I R⇤ ) = ( 1) (n 1) ( 1) S⇤(n) ( 1) , n n k kX=2 where
Sk⇤(x) = Dk⇤(m) mX6x and
Dk⇤(m) = 1.
m=d1 dk i=j gcdX(d···,d )=1 6 ) i j dj >2
Note that the arithmetic function Dk⇤ is the unitary analogue to the strict divisor function Dk, which can be found in the coe cients of the characteristic polynomial of Rn. Hence, using [10, (14)] and [1, (4)] successively, we get for any m, k Z 1 2 > k k k j k k j k !(m) !(m) D⇤(m) = ( 1) ⌧ ⇤(m) = ( 1) j = k! , k j j j k j=0 j=0 X ✓ ◆ X ✓ ◆ ⇢ where n is the Stirling number of the second kind. In particular, for any m, k k 2 such that !(m) < k, we have D (m) = 0. We are now in a position to prove Z>1 k⇤ the following result.
Theorem 4. Let n > 1. Then the algebraic multiplicity mn of the eigenvalue 1 of Rn⇤ satisfies m = n k , n n where the sequence (kn) of positive integers is given by
k1 = 0 and kn = max (kn 1, !(n) + 1) (n Z 2) . (2) 2 > In particular
1.3841 log n (n>3) (n>6) log n n 1 m n . log log n 6 n 6 log log n ⌫ ⌫ INTEGERS: 20 (2020) 6
Also, for any n > 3 log n 2 log n m = n + O? . n log log n (log log n)2 ✓ ◆ Proof. Since m = 1 = 1 k , we may suppose n 2. We first show by induction 1 1 > that, for any n Z 2, there exists a sequence (kn) of positive integers such that, 2 > for any m 1, . . . , n , !(m) < k , this sequence being given by (2). Indeed, 2 { } n the assertion is obviously true for n = 2 since k2 = 2, and if we assume it for some n 2, then, for any m 1, . . . , n + 1 , either m 1, . . . , n and then > 2 { } 2 { } !(m) < kn by induction hypothesis, or m = n + 1 and !(m) < 1 + !(n + 1), so that, for any m 1, . . . , n + 1 , we get !(m) < max (k , !(n + 1) + 1) = k . 2 { } n n+1 We now prove that kn is the smallest nonnegative integer satisfying this property, i.e., if there exists hn Z 0 such that, for all m 1, . . . , n , !(m) < hn, then 2 > 2 { } kn hn. Suppose on the contrary that hn < kn = max (kn 1, !(n) + 1). If 6 hn < !(n) + 1, then !(n) < hn < !(n) + 1 giving a contradiction, and hence hn < kn 1 = max (kn 2, !(n 1) + 1). Again, if hn < !(n 1)+1, then !(n 1) < hn < !(n 1) + 1 which is impossible, and hence hn < kn 2. Continuing this way we finally get hn < k1 = 1, resulting in a contradiction. Hence for any m 1, . . . , n , we infer that D⇤(m) = 0 for any k k , and thus 2 { } k > n S⇤(n) = 0 (k k ) and S⇤(n) = 0 (k < k ) , k > n k 6 n completing the proof of the first part of the theorem. For the second part, we first numerically check the inequality for n 3, . . . , 29 and assume n 30, so that 2 { } > kn 4. Next, for any k Z 1, define Nk := p1 pk. It is easy to see that kn is > 2 > · · · the unique positive integer such that Nk 1 n < Nk (see also [8, p. 380]), so n 6 n that, from [8, Theorem 11], we derive
1.3841 log Nk 1 1.3841 log n n kn = 1 + ! (Nkn 1) 6 1 + 6 1 + . log log Nk 1 log log n n Furthermore, [8, Theorem 10] yields
log Nkn log n kn = ! (Nkn ) > > , log log Nkn log log n which proves the inequality. We proceed similarly for the last estimate: first check it for n 3, . . . , 2 309 , then assume n 2 310 so that k 6, and use [8, 2 { } > n > Theorem 12] to get
log Nk 1 1.4575 log Nk 1 n n kn 6 1 + + 2 log log Nkn 1 (log log Nk 1) n log Nk 1 2 log Nk 1 n n < + 2 log log Nkn 1 (log log Nk 1) n log n 2 log n 6 + , log log n (log log n)2 INTEGERS: 20 (2020) 7 which terminates the proof of Theorem 4.
3.2. The “Dominant” Eigenvalues We first notice that
!(m) !(m) S⇤(x) = 2 = 2 2 x 2 2 b c mX6x ⇢ mX6x x log x 3 ⇣ = + 2x 0 (2) + o x1/2 . ⇣(2) 2 ⇣ ✓ ◆ ⇣ ⌘
Now following the argument leading to [11, (18)], we deduce that Rn⇤ has two “dominant” eigenvalues satisfying the following estimate. ±
Proposition 5. For all n Z 3 2 >
log n 1 ⇣ 0 1/2 2 = pn + + (2) + O n log n . ± ± 2⇣(2) 2 ⇣ ⇣ ⌘
Acknowledgments. The author gratefully acknowledges the anonymous referee for some corrections and remarks that have significantly improved the paper.
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