In this paper, we present the permanents and determinants of some Hessenbergmatrices. Also, some special cases for permanents are given

Koparal, Sibel

Çolak, Cemile D.

Ömür, Neşe

University of Niš: Facta Universitatis (E-Journals) / Универзитет у Нишу

FACTA UNIVERSITATIS (NISˇ)Ser. Math. Inform. Vol. 33, No 2 (2018), 153–162https://doi.org/10.22190/FUMI1802153KON CERTAIN HESSENBERG MATRICES RELATED WITHLINEAR RECURRENCESSibel Koparal, Nes¸e O¨mu¨r and Cemile D. C¸olakAbstract. In this paper, we present various results for permanents and determinantsof some Hessenberg matrices. Also, some special cases for permanents are given.Keywords: Hessenberg matrices, permanents, determinants.1. IntroductionMatrix methods are useful tools deriving some properties of linear recurrences.Some authors obtained many connections between certain sequences and perma-nents of Hessenberg matrices in the literature [1]-[4],[6],[10]-[12].The permanent of an n− square matrix An = [aij ] is defined byperAn =∑σ∈Dnn∏i=1aiσ(i),where the summation extends over all permutations σ of the symmetric group Dn.In [9], Minc defined the super-diagonal matrix and showed that the permanentof the matrix equals the order k-Fibonacci number.In [5], Kılıc¸ derived recurrence relations and generating matrices for the sums ofusual tribonacci numbers and 4n subscripted tribonacci sequences {T4n}, and theirsums. Also, the relationships between these sequences and permanents of certainmatrices are obtained.In [6], Kılıc¸ and Tas¸cı found the relationships between the sums of the Fibonacciand Lucas numbers and 1-factors of bipartite graphs.In [7], Kılıc¸ and Tas¸cı defined the n × n tridiagonal Toeplitz (0,−1, 1)-matrixMn = [mi,j ] with mi,i = −1 for 1 ≤ i ≤ n,mi,i+1 = mi+1,i = 1 for 1 ≤ i ≤ n − 1Received November 16, 2017; accepted February 23, 20182010 Mathematics Subject Classification. Primary 11B37; Secondary 05A15153154 S. Koparal, N. O¨mu¨r and C. D. C¸olakand 0 otherwise, and the n × n tridiagonal Toeplitz (0,−1, 1)-matrix Ln = [li,j ]with li,i = −1 for 2 ≤ i ≤ n, li,i+1 = li+1,i = 1 for 1 ≤ i ≤ n− 1, l1,1 = −12 and 0otherwise. They showed perMn = F−(n+1) and perLn =L−n2 , where Fn and Ln isthe nth Fibonacci and Lucas numbers, respectively.In [8], Li showed new Fibonacci-Hessenberg matrices and gave another proof ofthe well-known results relative to the Pell and Perrin numbers.In [3], Kalman showed that the (n+k)-th term of a sequence is defined recursivelyas a linear combination of the preceding k terms:(1.1) un+k = c0un + c1un+1 + ...+ ck−1un+k−1in which the initial terms u0 = ... = uk−2 = 0, uk−1 = 1 and c0, c1, ..., ck−1 areconstants.In [10], considering the generalized Fibonacci-Narayana sequence {Gn(a, c, r)},Ramı´rez derived some relations between this sequence and a permanent of one typeof the upper Hessenberg matrix. For example,pera c c · · · c 01 a 0 0 · · · c. . .. . .. . .. . .. . .1 a 0 0 · · · c. . .. . .. . .. . .1 a 0 01 a 01 a= Gn+r−1(a, c, r),where the generalized Fibonacci-Narayana sequence {Gn(a, c, r)}n∈N is defined asfollows:Gn(a, c, r) = aGn−1(a, c, r) + cGn−r(a, c, r), 2 ≤ r ≤ n,with the initial conditions G0(a, c, r) = 0, Gi(a, c, r) = 1, for i = 1, 2, ..., r − 1.In [12], Trojovsky´ defined tridiagonal matrices Bδn=[bδij]in the form1 if i = j or i = j − 1,(−1)j+δif i = j + 1,0 otherwise,where δ ∈ {0, 1} and showeddetBδn={F(n+4−6δ)/2 if n ≡ 0 (mod 2),F(n+1)/2 if n ≡ 1 (mod 2).2. Some resultsIn this section, we define the sequence {Rn(a, b, c, d)} and determine some relation-ships between the terms of this sequence and permanents of certain upper Hessen-On Certain Hessenberg Matrices Related with Linear Recurrences 155berg matrices. A sequence {Rn(a, b, c, d)} is defined by for 3 ≤ d ≤ n,(2.1) Rn(a, b, c, d) = aRn−1(a, b, c, d) + bRn−2(a, b, c, d) + cRn−d(a, b, c, d),in which R0(a, b, c, d) = 0, R1(a, b, c, d) = R2(a, b, c, d) = ... = Rd−2(a, b, c, d) = 1and Rd−1(a, b, c, d) = a. The sequence {Rn(a, b, c, d)} is a generalization of thetribonacci sequence. When a = b = c = 1 and d = 3, Rn(1, 1, 1, 3) = Tn (thenth tribonacci number). If c = 0 and d = 3, the generalized Fibonacci sequence{Un(a, b)} is obtained. If a = b = 1, c = 0 and d = 3, the Fibonacci sequence {Fn}is obtained and if a = c = 1, b = 0 and d = 3, the Narayana sequence is obtained.The generating function R(z) of Rn(a, b, c, d) is given byR(z) =(a− 1 + bz) zd−1 − bz3 − az2 + z(1− z) (1− az − bz2 − czd).Now we give relationships between terms of the sequence {Rn(a, b, c, d)} and thepermanents of certain matrices.For n ≥ 1, define a n × n matrix Hn(a, b, c, d, k, t) = [hi,j ] with hi+1,i = 1 for1 ≤ i ≤ n− 2, hi,i = a for 1 ≤ i ≤ n− 1, hi,i+1 = b for 1 ≤ i ≤ n− 1, h1,i = c for3 ≤ i ≤ d, hi,d+i−1 = c for 2 ≤ i ≤ n− d+ 1, hn,n−1 = k, hn,n = t, and 0 otherwise,i.e.,(2.2) Hn(a, b, c, d, k, t) =a b c · · · c 0 ... 01 a b 0 · · · c ....... . .. . .. . .. . . · · ·. . . 01 a b 0 · · · c. . .. . .. . .. . .. . .. . .. . .. . . 01 a b0 k t.Then we give the following Theorem.Theorem 2.1. Let Hn(a, b, c, d, k, t) be the matrix defined in (2.2). Then, forn > 1 and d ≥ 3,(2.3) perHn(a, b, c, d, k, t) = kRn+d−2(a, b, c, d)− (ka− t)Rn+d−3(a, b, c, d),where the real numbers k and t.Proof. (Induction on n) If n = 1, then we haveperH1(a, b, c, d, k, t) = t = kRd−1(a, b, c, d)− (ka− t)Rd−2(a, b, c, d).156 S. Koparal, N. O¨mu¨r and C. D. C¸olakSuppose that the equation holds for n− 1. Then we show that the equation holdsfor n. Expanding the perHn with respect to the last column d times, we writeperHn(a, b, c, d, k, t)= aperHn−1(a, b, c, d, k, t) + bperHn−2(a, b, c, d, k, t) + cperHn−d(a, b, c, d, k, t).By our assumption, we haveperHn(a, b, c, d, k, t) = a (kRn+d−3(a, b, c, d)− (ka− t)Rn+d−4(a, b, c, d))+b (kRn+d−4(a, b, c, d)− (ka− t)Rn+d−5(a, b, c, d))+c (kRn−2(a, b, c, d)− (ka− t)Rn−3(a, b, c, d))= kRn+d−2(a, b, c, d)− (ka− t)Rn+d−3(a, b, c, d).Thus, the proof is complete.When t = a and k = 1 in (2.3), we have perHn(a, b, c, d, 1, a) = Rn+d−2(a, b, c, d).For n ≥ 1, define a n × n matrix En(a, b, c, d, k, t) = [ei,j] with ei+1,i = −1 for1 ≤ i ≤ n− 2, ei,i = a for 1 ≤ i ≤ n− 1, ei,i+1 = b for 1 ≤ i ≤ n− 1, e1,i = c for3 ≤ i ≤ d, ei,d+i−1 = c for 2 ≤ i ≤ n−d+1, en,n−1 = −k, en,n = t, and 0 otherwise,i.e.,En(a, b, c, d, k, t) =a b c · · · c 0 ... 0−1 a b 0 · · · c ....... . .. . .. . .. . . · · ·. . . 0−1 a b 0 · · · c. . .. . .. . .. . .. . .. . .. . .. . . 0−1 a b0 −k t.It is clearly showed from [2] thatdetEn(a, b, c, d, k, t) = perHn(a, b, c, d, k, t).Now, we take the n× n matrix Hn(a, b, c, 3, k, t) by the following form:Hn(a, b, c, 3, k, t) =a b c 01 a b c. . .. . .. . .. . .1 a b c. . .. . .. . .. . .. . .. . .. . . c1 a b0 k t.On Certain Hessenberg Matrices Related with Linear Recurrences 157Then, we have(2.4) perHn(a, b, c, 3, k, t) = kRn+1(a, b, c, 3)− (ka− t)Rn(a, b, c, 3).For example, from [9], for a = b = c = t = k = 1 in (2.4), we have thatperHn(1, 1, 1, 3, 1, 1) = Tn+1 = perF(n, 3),where Tn is the nth tribonacci number.For n > 1; we define an n× n matrix Wn(a, b, c, 3, k, t) as in the compact form,by the definition of Hn(a, b, c, 3, k, t);(2.5) Wn(a, b, c, 3, k, t) =1 1 ... 110... Hn−1(a, b, c, 3, k, t)0.Now, we have the following theorem:Theorem 2.2. Let Wn(a, b, c, 3, k, t) be the matrix defined in (2.5). Then, forn > 2perWn(a, b, c, 3, k, t) = kn∑i=1Ri(a, b, c, 3)− (ka− t)n∑i=1Ri−1(a, b, c, 3).Proof. (Induction on n) If n = 3, we writeperW3(a, b, c, 3, k, t)= k + t+ at+ bk = k3∑i=1Ri(a, b, c, 3)− (ka− t)3∑i=1Ri−1(a, b, c, 3).Suppose that the equation holds for n. Then we show that the equation holds forn + 1. From the definitions of matrices Hn(a, b, c, 3, k, t) and Wn(a, b, c, 3, k, t),expanding the perWn+1(a, b, c, 3, k, t) with respect to the first column gives usperWn+1(a, b, c, 3, k, t) = perHn(a, b, c, 3, k, t) + perWn(a, b, c, 3, k, t).By our assumption and (2.4), we haveperWn+1(a, b, c, 3, k, t)= kRn+1(a, b, c, 3)− (ka− t)Rn(a, b, c, 3)+kn∑i=1Ri(a, b, c, 3)− (ka− t)n∑i=1Ri−1(a, b, c, 3)= kn+1∑i=1Ri(a, b, c, 3)− (ka− t)n+1∑i=1Ri−1(a, b, c, 3).Thus the proof is obtained.158 S. Koparal, N. O¨mu¨r and C. D. C¸olakWhen a = b = c = 1 in (2.1), the sequence {Rn} , special case of the sequence{Rn(a, b, c, d)} , is defined by the recurrence(2.6) Rn = Rn−1 +Rn−2 +Rn−d, 3 ≤ d ≤ nin which R0 = 0, R1 = R2 = ... = Rd−2 = Rd−1 = 1 and especially from (1.1), thesequence {Sn} is defined by(2.7) Sn = Sn−1 + Sn−2 + Sn−d, 3 ≤ d ≤ nin which S0 = S1 = S2 = ... = Sd−3 = 0 and Sd−2 = Sd−1 = 1. For d = 3 in (2.6)and (2.7), the sequences {Rn} and {Sn} coincide as the tribonacci sequence.For n > 1, define an n× n matrix Zn = [zi,j ] with zi+1,i = 1 for 1 ≤ i ≤ n− 1,zi,i = 1 for 1 ≤ i ≤ n, z1,2 = 0, zi,i+1 = 1 for 2 ≤ i ≤ n− 1, z1,i = 1 for 3 ≤ i ≤ d,zi,d+i−1 = 1 for 2 ≤ i ≤ n− d+ 1 and 0 otherwise, i.e.,(2.8) Zn =1 0 1 · · · 1 0 ... 01 1 1 0 · · · 1 ....... . .. . .. . .. . . · · ·. . . 01 1 1 0 · · · 1. . .. . .. . .. . .. . .. . .. . .. . . 01 1 10 1 1.Then we give the following Theorem.Theorem 2.3. Let Zn be the matrix defined in (2.8). Then, for n ≥ 5 and d ≥ 3,perZn = Rn+d−3 +Rn+d−4 −Rn+d−5 + Sn−5.Proof. We prove this by induction on n. For n = 5, we writeperZ5 = 2Rd+1 + 1 = Rd+2 +Rd+1 −Rd + S0.The claim is true for n = 5. Assume that the claim is true for n− 1. Thus we showthat the claim is true for n. Expanding the perZn according to the last column dtimes, we haveperZn = perZn−1 + perZn−2 + perZn−d.By our assumption, we haveperZn = Rn+d−4 +Rn+d−5 −Rn+d−6 + Sn−6+Rn+d−5 +Rn+d−6 −Rn+d−7 + Sn−7+Rn−3 +Rn−4 −Rn−5 + Sn−d−5.On Certain Hessenberg Matrices Related with Linear Recurrences 159From the sequences {Rn} and {Sn}, we writeperZn = Rn+d−3 +Rn+d−4 −Rn+d−5 + Sn−5.So the proof is complete.For n ≥ 1, define the n× n matrix Vδn = [vi,j ] with vi+1,i = 1 for 1 ≤ i ≤ n− 2,vi,i = 1 for 1 ≤ i ≤ n− 1, v1,i+1 = vi,i+1 = (−1)i−δfor 1 ≤ i ≤ n and 0 otherwise,where δ ∈ {0, 1} . i.e.,(2.9) Vδn =1 (−1)1−δ (−1)2−δ · · · (−1)n−2−δ (−1)n−1−δ1 1 (−1)2−δ 0 · · · ...0. . .. . .. . .. . . · · ·. . .1 1 0 · · ·. . .. . . (−1)n−2−δ . . .. . . 1 (−1)n−1−δ0 1 1.Theorem 2.4. Let Vδnbe the matrix defined in (2.9). Then, for n ≥ 1,detVδn=(−1)δ(F(n+5+3(−1)δ)/2 − 2− (−1)δ)if n ≡ 0 (mod 2),(−1)δ(F(n+5)/2 − 2 + (−1)δ)if n ≡ 1 (mod 2),where Fn is the nth Fibonacci number.Proof. For n = 1, detV01= 1 = (F3 − 1) , detV11= 1 = − (F3 − 3) and for n = 2,detV12 = 0 = − (F2 − 1) , detV02 = 2 = (F5 − 3) .We show that the claim is true for n − 1. Using expansion on the first columnof detV 1n , we get as followsdetV1n= detB0n−1− detV0n−1.From (1.2) and the induction hypothesis, we havedetV1n={Fn/2 −(F(n+4)/2 − 1)if n ≡ 0 (mod 2),F(n+3)/2 −(F(n+7)/2 − 3)if n ≡ 1 (mod 2),={−(F((n+2)/2 − 1)if n ≡ 0 (mod 2),−(F((n+5)/2 − 3)if n ≡ 1 (mod 2).Similarly, using detV0n = detB1n−1+ detB0n−2+ detV0n−2, the desired result isgiven. We have the proof.160 S. Koparal, N. O¨mu¨r and C. D. C¸olakTheorem 2.5. For n ≥ 1, we haveperV1n ={L(n+2)/2 − 1 if n ≡ 0 (mod 2),L(n−1)/2 − 1 if n ≡ 1 (mod 2),andperV0n ={F(n+2)/2 − 1 if n ≡ 0 (mod 2),F(n+5)/2 − 1 if n ≡ 1 (mod 2),where Fn is the nth Fibonacci number and Ln is the nth Lucas number.Proof. Considering perBδn ={F(n−2+6δ)/2 if n ≡ 0 (mod 2),F(n+1)/2 if n ≡ 1 (mod 2),for δ ∈ {0, 1} andthe equalitiesperV1n= perB0n−1+ perV0n−1and perV0n= perB1n−1− perB0n−2+ perV0n−2,we have the proof from induction on n.3. Some special casesIn this section, we give some special cases of the above theorems:• For b = 1, c = 0 and d = 3, the generalized Fibonacci sequence {Un(a, 1)},perHn (a, 1, 0, 3, k, t)= pera 1 0 · · ·1 a 1 0. . .. . .. . .. . .01 a 1k t= tUn(a, 1) + kUn−1(a, 1),perWn (a, 1, 0, 3, k, t)= per1 1 1 ... 110 Hn−1...=1a(tU(n(a, 1) + (k + t)U(n−1(a, 1) + kUn−2(a, 1) + k(a− 1)− t) ,and then,perWn (a, 1, 0, 3, k, t) = perHn (a, 1, 0, 3, k, t)+perHn−1 (a, 1, 0, 3, k, t)+k(a−1)+t.On Certain Hessenberg Matrices Related with Linear Recurrences 161• For a = 1, b = 2, c = 0 and d = 3 in (2.1), {Jn} is the Jacobsthal sequenceand for n > 2,perHn (1, 2, 0, 3, k, t)= per1 2 01 1 2 0. . .. . .. . .. . . · · ·01 1 2k t= tJn + 2kJn−1.• For a = b = k = 1, c = 0 and d = 3 in (2.1), {Fn} is the Fibonacci sequence,perHn (1, 1, 0, 3, 1, t)= per1 1 01 1 1 0. . .. . .. . .. . . · · ·01 1 11 t= tFn + Fn−1andperWn (1, 1, 0, 3, 1, t) = per1 1 1 ... 110 Hn−1...= t (Fn+1 − 1) + Fn.• For d = 3 in (2.6) and (2.7), {Tn} is the tribonacci sequence andperZn = per1 0 11 1 1 1. . .. . .. . .. . . · · ·11 1 11 1= 2Tn−1 + Tn−3 + Tn−5.REFERENCES1. 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Global Journal of Pure and AppliedMathematics 12 (2016), no. 2, 1885–1892.Sibel KoparalKocaeli UniversityFaculty of Art and ScienceDepartment of MathematicsKocaeli, Turkeysibel.koparal@kocaeli.edu.trNes¸e O¨mu¨rKocaeli UniversityFaculty of Art and ScienceDepartment of MathematicsKocaeli, Turkeyneseomur@kocaeli.edu.trCemile Duygu C¸olakKocaeli UniversityFaculty of Art and ScienceDepartment of MathematicsKocaeli, Turkeycemileduygusener@hotmail.com

On Certain Hessenberg Matrices Related with Linear Recurrences

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On Certain Hessenberg Matrices Related with Linear Recurrences

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