Determinants of Some Pentadiagonal Matrices

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In case of Toeplitz pentadiagonal matrices the diagonal vectors are constant vectors, i.e. L =(L,...,L), R =(R,...,R) ∈ Rn+1−2k, l =(l,...,l), r =(r,...,r) ∈ Rn+1−k, d =(d,...,d) ∈ Rn+1, and for the matrix and its determinant the notations An+1,k,2k(L, l, d, r, R) and Dn+1,k,2k(L, l, d, r, R) will be used. Marr and Vineyard [20]) have shown that the product of two 1-tridiagonal Toeplitz matrix is an (α,β) imperfect Toeplitz matrix An+1,1,2 which is related to the corresponding Toeplitz matrix by a two-step recursion. Imperfectness means that the main diagonal is changed from (d,...,d) ∈ Rn+1 to (d − α,...,d − α,d,...,d,d − β,...,d − β) ∈ Rn+1

k n+1−2k k where α, β are given reals.| They{z found} | the{z determinants} | {z of Toeplitz} and imperfect Toeplitz 1, 2- pentadiagonal matrices in terms of Chebyshev polynomials of the second kind. A similar approach was used in [27] to find a formula for the inverse of a 1, 2-pentadiagonal Toeplitz matrix. Imperfect (α,β) k, 2k-pentadiagonal Toeplitz matrices will be denoted by An+1,k,2k(L, l, d, r, R), their determinants (α,β) by Dn+1,k,2k(L, l, d, r, R). A number of papers studied general 1, 2-pentadiagonal matrices and their Toeplitz versions. Al- gorithms and recursive formulas were found for their determinants [24], [15], [11], [17] and inverses [16]. Explicit formulas were found for the determinants of symmetric [14] skew-symmetric [13] and general [17] Toeplitz 1, 2-pentadiagonal matrices. Perhaps the first appearance of k,ℓ-pentadiagonal (Toeplitz Hermitian) matrices (where the distances of the sub- and superdiagonals from the main one are k and ℓ) was in Egerváry and Szász [4] with k + ℓ = n +1, while k-tridiagonal matrices ap- peared first in [12]. In [19] new sub/superdiagonals were added. A graph theoretical approach can be found in [5,7] and some extensions in [21,25]. An exhaustive list of recent references is given in the survey [6]. In spite of the large numbers of papers on pentadiagonal matrices there are only few of them in which determinants are given in terms of the entries. In [8] we developed a method to reduce the determinant of k,ℓ-pentadiagonal matrices to tridiagonal determinants provided that k + ℓ ≥ n +1. If k ≥ (n + 1)/3 then by this method in [9] we determined the determinants of general, Toeplitz and imperfect Toeplitz k, 2k-pentadiagonal matrices. We proved (among others)

Theorem 1. Assuming (n + 1)/3 ≤ k ≤ n/2 the determinant Dn+1,k,2k(L, l, d, r, R) of the general k, 2k-pentadiagonal matrix is

n−2k k−1

(djdj+kdj+2k −djlj+krj+k −LjRjdj+k −ljrjdj+2k +ljRjlj+k +rjLjrj+k) (djdj+k−ljrj). j=0 j n − k Y = Y+1 2 If the condition k + ℓ ≥ n +1 is not satisfied then the method given in [8] is not applicable in general. However, if ℓ = 2k then with some modification it is applicable. In Section 2 we develop this modification and extend Theorem 1 to the case when the restriction (n + 1)/3 ≤ k is dropped. Finally in Section 3 we show how the determinants Dn+1,k,2k(L, l, d, r, R) can be factorized and also discuss the case of Toeplitz determinants. DETERMINANTSOFSOMEPENTADIAGONALMATRICES 3

2. REDUCTION OF GENERAL k, 2k-PENTADIAGONALDETERMINANTSTOTRIDIAGONALONES

Our starting point is the matrix A0 = An+1,k,2k(L, l, d, r, R) where 1 ≤ k < 2k ≤ n. Let n +1= kq + p where 0 ≤ p

The 2kth superdiagonal remains unchanged, however all its elements R0,...,Rk−1 multiplied one by one by l0/d0,..., −lk−1/dk−1, respectively move down by k units and added to rk,...,r2k−1 thus these elements change to (1) (3) rj = rj − Rj−klj−k/dj−k , if j = k, . . . , 2k − 1. (l) Please note that the position of entries −lj /dj (j = 0,...,k − 1) in the matrix B1 is the same as the position of lj (j = 0,...,k − 1) in the matrix An+1,k,ℓ(L, l, d, r, R), the position of the entries (r) (L) (R) we want to eliminate. The matrices B1 , B1 , B1 in the following multiplications also have similar structures. (l) (r) (ii) Multiply B1 A0 from the right by B1 ∈Mn+1 with entries 1 if i = j, (4) b(1,r) := −r /d if i =0,...,k − 1; j = i + k, ij  i i  0 otherwise (l) The effect of this, is multiplication of the columns 0,...,k −1 of B1 A0 by −r0/d0,..., −rk−1/dk−1 and addition of these products to the columns k, . . . , 2k − 1, respectively. The elements r0,...,rk−1 in these columns disappear the diagonal elements remain unchanged. The 2kth subdiagonal remains unchanged, however its elements L0,...,Lk−1 multiplied one by one by −r0/d0,..., −rk−1/dk−1 respectively move to the right by k units and added to the elements l2k,...,l3k−1, hence these elements change to (1) (5) lj = lj − Lj−krj−k/dj−k , if j =2k, . . . , 3k − 1.

(l) (r) (L) (iii) Next multiply B1 A0B1 from the left by B1 ∈Mn+1 with entries 1 if i = j, (6) b(1,L) := −L /d if i =2k, . . . , 3k − 1; j = i − 2k, ij  j j  0 otherwise  4 LÁSZLÓ LOSONCZI

(l) (r) The effect of this on the matrix B1 A0B1 is multiplication of its rows 0,...,n − 2k by −L0/d0,..., −Ln−2k/dn−2k and adding these products to the rows 2k, . . . , 3k − 1, respectively. The elements L0,...,Lk−1 of the 2kth subdiagonal disappear, the elements d2k,...,d3k−1 of the main diagonal change to

(1) (7) dj = dj − Lj−2kRj−2k/dj−2k , for j =2k, . . . , 3k − 1.

(L) (l) (r) (R) (iv) Finally multiply B1 B1 A0B1 from the right by B1 ∈Mn+1 with entries 1 if i = j, (8) b(1,R) := −R /d if i =0,...,k − 1; j = i +2k, ij  i i  0 otherwise (L) (l) (r) The effect of this on the matrixB1 B1 A0B1 is multiplication of its columns 0,...,k − 1 by −R0/d0,..., −Rk−1/dk−1 and adding these products to the columns 2k, . . . , 3k −1 respectively. The elements R0,...,Rk−1 vanish and the main diagonal does not change. (L) (l) (r) (R) With this the first process ended. The matrices B1 , B1 , B1 , B1 are the same as in [8,9] however the transformation rules for the entries are different. In those papers during the first (and subse- quent) processes the diagonal vectors l, r just shortened while the diagonal vectors L, R transformed. Here just the opposite happened: the diagonals L, R shortened and l, r transformed. The results of the first process are summarized in Theorem 2. Suppose that 1 ≤ k ≤ 2k ≤ n,k < (n + 1)/3. Then

(L) (l) (r) (R) ∗ (9) A1 := B1 B1 A0B1 B1 = E1 A1 where E1 ∈Mk is a diagonal matrix with diagonal elements d0M,...,dk−1 and ∗ L(1) l(1) d(1) r(1) R(1) A1 = An+1−k,k,2k( , , , , ). isa k, 2k-pentadiagonal matrix with main and other diagonal vectors (called ﬁrst iterated diagonals) (1) n+1−3k (1) n+1−3k L =(Lk,...,Ln−2k) ∈ R R =(Rk,...,Rn−2k) ∈ R l(1) (1) (1) Rn+1−2k r(1) (1) (1) Rn+1−2k =(lk ,...,ln−k) ∈ =(rk ,...,rn−k) ∈ d(1) (1) (1) Rn+1−k =(dk ,...,dn ) ∈ where

(1) lj − Lj−krj−k/dj−k (j = k, . . . , 2k − 1), lj = lj (j =2k,...,n − k) (10) (1) rj − Rj−klj−k/dj−k (j = k, . . . , 2k − 1) rj = rj (j =2k,...,n − k) and

dj − lj−krj−k/dj−k (j = k, . . . , 2k − 1), (1) (11) d = d − L − R − /d − (j =2k, . . . , 3k − 1), j  j j 2k j 2k j 2k  dj (j =3k,...,n), Proof. From the description of the four matrix multiplication (9) follows while (3), (5) show the correctness of (10) and (2), (7) show that (11) is valid. DETERMINANTSOFSOMEPENTADIAGONALMATRICES 5

For the convenience of later calculations all letters li,di,ri referring to the original matrix will be (0) labeled with superscripts but occasionally we omit this label. We omit from this labeling Li, Ri since these numbers did not change during our process. Define for s = 1, 2,...,q − 1 the matrices of Mn+1 (for s = 1 these coincide with the matrices (1), (4), (6), (8)) which we use in further calculations by 1 if i = j, (l) (s,l) (s−1) (s−1) Bs = b = −l /d if i = sk, . . . , (s + 1)k − 1; j = i − k, ij  j j  0 otherwise,  1 if i = j, (r) (s,r) (s−1) (s−1) Bs = b = −r /d if i =(s − 1)k,...,sk − 1; j = i + k, ij  i i  0 otherwise, .  1 if i = j, (L) (s,L) (s−1) (s−1) Bs = b = −L /d if i =(s + 1)k, . . . , (s + 2)k − 1; j = i − 2k, ij  j j  0 otherwise,  1 if i = j, (R) (s,R) (s−1) (s−1) Bs = b = −R /d if i =(s − 1)k,...,sk − 1; j = i +2k, ij  i i  0 otherwise. We define the first and second matrices for s = q too, but in this case the index sets are restricted to i = qk,...,n; j = i − k and i = (q − 1)k,...,n − k; j = i + k respectively, if p = 0 then both sets are empty and the first and second matrices are degenerate to unit matrices. Similarly for the third and fourth matrices with s = q − 1 the index domains are i = qk,...,n; j = i − 2k and i =(q − 2)k,...,n − 2k; j = i +2k respectively and if p =0 then both sets are empty and the third and fourth matrices are degenerate to unit matrices. (s−1) (s−1) (s−1) In the above definitions the quantities lj ,rj ,dj can be obtained inductively by continuing the iterations of (10), (11) for s =1, 2,..., (q − 2, q − 1, q) as follows (s−1) (s−1) (s−1) (s) l − L − r /d (j = sk, . . . , (s + 1)k − 1), l = j j k j−k j−k j l (j =(s + 1)k,...,n − k) j (12) (s−1) (s−1) (s−1) (s) r − R − l /d (j = sk, . . . , (s + 1)k − 1), r = j j k j−k j−k j r (j =(s + 1)k,...,n − k) j and (s−1) (s−1) (s−1) (s−1) dj − lj−k rj−k /dj−k (j = sk, . . . , (s + 1)k − 1), (13) d(s) = (s−1) (s−1) j  dj − Lj−2kRj−2k/dj−2k (j =(s + 1)k, . . . , (s + 2)k − 1),  dj (j =(s + 2)k,...,n), These definitions are valid if − We put the last three values − − of in parenthesis  s < q 2. q 2, q 1, q s since for them the definitions should be modified. For these values of s the index groups in (12), (13) may run out of their ranges j = sk,...,n − k and j = sk,...,n therefore cannot be defined, or the index groups may be restricted. For example (13) is valid for s ≤ q − 2, for s = q − 1 the second index group should be restricted to j = qk,...,n if p > 0 and empty if p = 0 with empty third index group. For s = q the diagonal 6 LÁSZLÓ LOSONCZI

(q) elements dj can be deﬁned only if p > 0, in this case for s = q in (13) the ﬁrst index group should be restricted to j = qk,...,n with empty second and third index groups.

(L) (l) (r) (R) In the second process we calculate the product A2 := B2 B2 A1B2 B2 then we continue similarly s − 2(s < q − 1) times to get j=s j=s (L) (l) (r) (R) As := Bs+1−jBs+1−j A0 Bj Bj . j=1 ! j=1 ! Y Y For As we have the decomposition ∗ As = Esk As where Esk ∈Msk is a diagonal matrix with diagonalM elements

(1) (1) (s−1) (s−1) d0,...,dk−1,dk ,...,d2k−1,...,d(s−1)k,...,dsk−1 and ∗ L(s) l(s) d(s) r(s) R(s) As = An+1−sk,k,2k( , , , , ) with (s) n+1−(s+2)k (s) n+1−(s+2)k L =(Lsk,...,Ln−2k) ∈ R R =(Rsk,...,Rn−2k) ∈ R l(s) (s−1) (s−1) Rn+1−(s+1)k r(s) (s−1) (s−1) Rn+1−(s+1)k =(lsk ,...,ln−k ) ∈ =(rsk ,...,rn−k ) ∈ d(s) (s−1) (s−1) Rn+1−sk =(dsk ,...,dn ) ∈ and the elements of these iterated diagonals are given by (12), (13). L(s) R(s) ∗ For s = q − 2 the dimension of the vectors , is p. This means that if p = 0 then Aq−2 is ∗ tridiagonal, and if p > 0 then Aq−1 is tridiagonal. Therefore the cases p = 0 and p > 0 should be treated differently. Theorem 3. Let 1 ≤ k < 2k ≤ n, n +1= kq (i.e. p =0) q > 3. Then

(l) (r) Aq−1 := Bq−2Aq−2Bq−2 = Eqk ∈Mn+1 is a diagonal matrix with diagonal elements

(1) (1) (q−1) (q−1) d0,...,dk−1,dk ,...,d2k−1,...,d(q−1)k,...,dqk−1 deﬁned by (13). (l) (r) If p> 0 then multiplying Aq−1 from the left by Bq and the from the right by Bq we get a diagonal matrix. Theorem 4. Let 1 ≤ k < 2k ≤ n, n +1= kq + p, 0 ≤ p

(l) (r) Aq := Bq Aq−1Bq = Eqk+p ∈Mn+1 is a diagonal matrix with diagonal elements

(1) (1) (q−1) (q−1) (q) (q) d0,...,dk−1,dk ,...,d2k−1,...,d(q−1)k,...,dqk−1 ,dqk ,...,dqk+p−1 deﬁned by (13). For the determinant we have DETERMINANTSOFSOMEPENTADIAGONALMATRICES 7

Theorem 5. Let 1 ≤ k < 2k ≤ n, n +1= kq + p, 0 ≤ p < k, and suppose that either q =3,p> 0, or q > 3. Then the determinant of An+1,k,2k(L, l, d, r, R) is

k−1 p−1 L l d r R (0) (1) ··· (q−1) (q) Dn+1,k,2k( , , , , )= dj dj+k dj+(q−1)k dj+qk j=0 j=0 (14) Y Y p−1 k−1 (0) (1) ··· (q) (0) (1) ··· (q−1) = dj dj+k dj+qk dj dj+k dj+(q−1)k j=0 j=p Y Y

−1 (s) where dj (s =1,...,q; j = sk,...,n) are deﬁned by (11), (13) and := 1. j=0 Q To express (14) in terms of the entries of A0 is quite complicated if q is large. Next we do this calculation if q =3. For j =0,...,k − 1 we have

l r d(0)d(1) = d d − j j = d d − l r j j+k j j+k d j j+k j j j and

(1) (1) Lj rj Rj lj lj+k − rj+k − (2) (1) lj+krj+k LjRj dj dj d = d − = dj+2k − − j+2k j+2k (1) lj rj d dj dj k − j+k + dj

d d − L R (l d − L r )(r d − R l ) = j j+2k j j − j+k j j j j+k j j j dj dj(dj+kdj − ljrj)

(d d − L R )(d d − l r ) − (l d − L r )(r d − R l ) = j j+2k j j j+k j j j j+k j j j j+k j j j dj(dj+kdj − ljrj)

d d d − d l r − d L R − d l r + R l l + L r r = j j+k j+2k j+2k j j j+k j j j j+k j+k j j j+k j j j+k , djdj+k − ljrj

(0) (1) (2) dj dj+kdj+2k = djdj+kdj+2k − dj+2kljrj − dj+kLjRj − djlj+krj+k + Rjljlj+k + Ljrjrj+k.

(3) If p> 0 then for j =0,...,p we have to calculate dj+3k too.

l(2) r(2) d(2) d(2) − l(2) r(2) (3) (2) − j+2k j+2k j+3k j+2k j+2k j+2k dj+3k = dj+3k (2) = (2) dj+2k dj+2k 8 LÁSZLÓ LOSONCZI where

(2) (1) Lj+kRj+k Lj+kRj+k djdj+kdj+3k − dj+3kljrj − djLj+kRj+k d = d − = dj k − = j+3k j+3k (1) +3 lj rj d dj k − djdj+k − ljrj j+k + dj

(1) Rj lj Lj+k rj+k − (2) (1) Lj+krj+k dj l = l − = lj+2k − j+2k j+2k (1) lj rj d dj k − j+k + dj

d d l − l l r − d L r + L l R = j j+k j+2k j j+2k j j j+k j+k j+k j j djdj+k − ljrj

(1) Lj rj Rj+k lj+k − (2) (1) Rj+klj+k dj r = r − = rj+2k − j+2k j+2k (1) lj rj d dj k − j+k + dj

d d r − r r l − d R l + R r L = j j+k j+2k j j+2k j j j+k j+k j+k j j djdj+k − ljrj

Denote by n1, n2, n3, n4 the numerators by m1, m2, m3, m4 the denominators of the above fraction (2) (2) (2) (2) forms of dj+3k,dj+2k,lj+2k,rj+2k respectively, then we have that

(3) n1n2 − n3n4 dj+kdj − ljrj n1n2 − n3n4 dj+3k = 2 = . (dj+kdj − ljrj) n2 (dj+kdj − ljrj)n2 Therefore (0) (1) (2) (3) (3) n1n2 − n3n4 dj dj+kdj+2kdj+3k = n2dj+3k = . dj+kdj − ljrj Calculating and factorizing the numerator by Maple software we obtain

n1n2 − n3n4 =(dj+kdj − ljrj)M where

M := djdj+kdj+2kdj+3k −djdj+klj+2krj+2k −djdj+2kLj+kRj+k −djdj+3klj+krj+k −dj+kdj+3kLjRj

−dj+2kdj+3kljrj +djLj+krj+krj+2k +djRj+klj+klj+2k +dj+3kLjrjrj+k +dj+3kRjljlj+k

+LjLj+kRjRj+k −LjRj+klj+2krj −Lj+kRjljrj+2k +ljlj+2krjrj+2k, therefore (0) (1) (2) (3) L l d r R dj dj+kdj+2kdj+3k = M = Mn,k,j( , , , , ). Further let

Nn,k,j(L, l, d, r, R) := n2

= djdj+kdj+2k − dj+2kljrj − dj+kLjRj − djlj+krj+k + Rjljlj+k + Ljrjrj+k then we have DETERMINANTSOFSOMEPENTADIAGONALMATRICES 9

Theorem 6. Let 1 ≤ k < 2k ≤ n, n +1=3k + p, 0 ≤ p

p−1 k−1 L l d r R (0) (1) (2) (3) (0) (1) (2) Dn+1,k,2k( , , , , )= dj dj+kdj+2kdj+3k dj dj+kdj+2k j=0 j=p Y Y (15) p−1 k−1

= Mn,k,j(L, l, d, r, R) Nn,k,j(L, l, d, r, R) j=0 j=p Y Y −1 where := 1. j=0 Q 3. k, 2k-PENTADIAGONAL DETERMINANTS, THE TOEPLITZ CASE

In the imperfect Toeplitz case in the quantities Mn,k,j, Nn,k,j of (15) we have to substitute Lj = L, lj = l,rj = r, Rj = R and

in Mn,k,j for j =0,...,p − 1 substitute dj = d − α,dj+k = d,dj+2k = d,dj+3k = d − β,

in Nn,k,j for j = p,...,k − 1 substitute dj = d − α,dj+k = d,dj+2k = d − β. We obtain Theorem 7. Let 1 ≤ k < 2k ≤ n, n +1=3k + p, 0 ≤ p

(α,β) 4 3 2 2 2 Dn+1,k,2k = d −(α + β)d − (3lr+2LR − αβ)d +(2Lr +2Rl +(α + β)(2lr+LR))d p +L2R2 −2LRlr+l2r2 +(α + β)(Lr2 +Rl2) − αβlr · 3 − 2 − − 2 2 k−p d (α + β)d (2lr+LR αβ)d+Lr +Rl +αβlr In the Toeplitz case we immediately get by α = β =0 from the previous theorem Theorem 8. Let 1 ≤ k < 2k ≤ n, n +1 = 3k + p, 0 ≤ p < k then the determinant of An+1,k,2k(L, l, d, r, R) is

4 2 2 2 2 2 2 2 p Dn+1,k,2k = d −(3lr+2LR)d +(2Lr +2Rl )d+L R −2LRlr+l r k−p · d3 −(2lr+LR)d+Lr2 +Rl2 . Using a suitable rearrangement of k, 2k-pentadiagonal (general or Toeplitz) matrices they can be reduced to the direct sum of 1, 2-pentadiagonal (general or Toeplitz) matrices. In this way we can say more about k, 2k-pentadiagonal determinants. Let n +1= kq + p where 0 ≤ p

Deﬁne the permutation matrix Pσ =(pij) by 1 if j = σ(i), p = ij 0 otherwise. 10 LÁSZLÓ LOSONCZI

T L l d r R Then PσA0Pσ rearranges both the rows and columns of A0 = An+1,k,2k( , , , , ) in the order of the permutation σ. Thus we obtain

p−1 k−1 T (j) (j) (16) PσA0Pσ = Aq+1,1,2 Aq,1,2 j=0 ! j=p ! M M (s) L l d r R where At+1,1,2( (s), (s), (s), (s), (s)) for t = q,s = 0, 1,...,p − 1 and t = q − 1,s = p,p + 1,...,k − 1 are 1, 2-pentadiagonal matrices with diagonal vectors

L(s) =(Ls, Ls+k,...,Ls+(t−2)k), R(s) =(Rs, Rs+k,...,Rs+(t−2)k), l(s) =(ls,ls+k,...,ls+(t−1)k), r(s) =(rs,rs+k,...,rs+(t−1)k), d(s) =(ds,ds+k,...,ds+tk). We remark that σ is the same permutation as the one used by Egerváry and Szász [4], see also N. Bebiano and S. Furtado [1] where a similar decomposition were given. From (16) it follows immediately Theorem 9. Let 1 ≤ k < 2k ≤ n, then we have

p−1 k−1 L l d r R (j) (j) Dn+1,k,2k( , , , , )= Dq+1,1,2 Dq,1,2 j=0 j=p Y Y (s) (s) L l d r R where Dt+1,1,2 denoted the determinant of the matrix At+1,1,2( (s), (s), (s), (s), (s)). For Toeplitz matrices we get (with the notation B⊕j := B ⊕···⊕ B) j

T ⊕p ⊕(k−p) PσAn+1,k,2k(L, l, d, r, R)Pσ = Aq+1,1,2(L, l, d, r,| R) {zAq,1,2}(L, l, d, r, R) (s) L l d r R as in this case At+1,1,2( (s), (s), (s), (s), (s))= At+1,1,2(L, l, d, r, R) for t = q,s =0, 1,...,p − 1 and t = q − 1,s = p,p +1,...,k − 1. For the determinants we get

p k−p Dn+1,k,2k(L, l, d, r, R)= Dq+1,1,2(L, l, d, r, R) Dq,1,2(L, l, d, r, R) thus it is enough to calculate the determinants of 1, 2-pentadiagonal matrices. To do this we could use the iteration formulae (11), (13) (which are now considerably simpler) to find the diagonal elements then by (14) to find the determinants. However it seems easier to apply existing recursion formulae for the determinants. The six term recursion of R.A. Sweet [24] is applicable for the determinants of Toeplitz matrices but its coefficients contain fractions and are more complicated than those of the seven term recursion found by J. Jia, B. Yang, S. Li [17] thus we apply the latter. Let D(n + 1) := Dn+1,1,2(L, l, d, r, R) for n ≥ 2 and let D(−2) = 0,D(−1) = 0,D(0) = 1, D(1) = d, D(2) = d2 − lr. For the determinant D(3) we easily obtain (17) D(3) = d3 − d(LR +2lr)+ Lr2 + Rl2. The recursion of [17] with our notations is D(n)= dD(n − 1)+(LR − lr)D(n − 2)+(Lr2 + Rl2 − 2dLR)D(n − 3) (18) + LR(LR − lr)D(n − 4) + dL2R2D(n − 5) − L3R3D(n − 6) (n =4, 5, 6,... ). DETERMINANTSOFSOMEPENTADIAGONALMATRICES 11

Calculating D(n) for n =4, 5, 6, 7, 8, 9 using (17), (18) with Maple software we obtain d4 −d2(2LR+3lr)+d(2Lr2+2Rl2)+L2R2 −2LRlr+l2r2, d5−d3(3LR+4lr)+d2(3Lr2 +3Rl2)+d(2L2R2−2LRlr+3l2r2)+(L2Rr2 +R2Ll2 −2Llr3 −2Rrl3), d6 −d4(4LR+5lr)+d3(4Lr2 +4Rl2)+d2(4L2R2 +6l2r2)+d(−6Llr3 −6Rl3r) +(−4L2R2lr+L2r4 +6LRl2r2 +R2l4 −l3r3), d7 −d5(5LR+6lr)+d4(5Lr2 +5Rl2)+d3(7L2R2 +4LRlr+10l2r2) −d2(3L2Rr2+3R2Ll2+12Llr3 +12Rrl3)−d(2L3R3 +6L2R2lr−3L2r4 −3R2l4 −15LRl2r2 +4l3r3) +(3L3R2r2 +3R3L2l2 −6L2Rlr3 −6R2Lrl3 +3Ll2r4 +3Rr2l4), d8 −d6(6LR+7lr)+d5(6Lr2 +6Rl2)+d4(11L2R2 +10LRlr+15l2r2) +d3(−8L2Rr2 −8LR2l2 −20Llr3 −20Rl3r) +d2(−6L3R3 −9L2R2lr+6L2r4 +24LRl2r2 +6R2l4 −10l3r3) +d(6L3R2r2 +6L2R3l2 −12L2Rlr3 −12LR2l3r+12Ll2r4 +12Rl4r2) +(R4L4 −6L3R3lr+2L3Rr4 +15L2R2l2r2 −3L2lr5 +2LR3l4 −12LRl3r3 −3R2l5r+l4r4), d9 +(−7LR−8lr)d7 +(7Lr2 +7Rl2)d6 +(16L2R2 +18LRlr+21l2r2)d5 +(−15L2Rr2 −15LR2l2 −30Llr3 −30Rl3r)d4 +(−13L3R3 −16L2R2lr+10L2r4 +30LRl2r2 +10R2l4 −20l3r3)d3 +(12L3R2r2 +12L2R3l2 −12L2Rlr3 −12LR2l3r+30Ll2r4 +30Rl4r2)d2 +(3L4R4 −6L3R3lr+3L3Rr4 +30L2R2l2r2 −12L2lr5 +3LR3l4 −44LRl3r3 −12R2l5r+5l4r4)d +3L4R3r2 +3L3R4l2−16L3R2lr3 +L3r6−16L2R3l3r+18L2Rl2r4+18LR2l4r2−4Ll3r5+R3l6−4Rl5r3. Observing these determinants we see that they are monic polynomials n−2 n j pn(L, l, d, r, R)= d + An,j(L, l, r, R)d j=0 X of degree n in d where the coefﬁcients An,j(L, l, r, R)(j =0,...,n − 2) are polynomials of L, l, r, R of degree n − k which are symmetric in L, R and l,r. Unfortunately for larger n the formulae for these polynomials are too long. With this we have proved Theorem 10. Let 1 ≤ k < 2k ≤ n, n +1 = kq + p, 0 ≤ p

(α,β) (α,β) The imperfect Toeplitz determinants D (n + 1) := Dn+1,k,2k(L, l, d, r, R) can be calculated by the recursion (19) D(α,β)(n)= D(n) − (α + β)D(n − 1) + αβD(n − 2) found by Marr and Vineyard [20] and using the previous theorem. Another possibility is to use a nine term recursion based on (19) and (18).

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FACULTY OF ECONOMICS, UNIVERSITY OF DEBRECEN,HUNGARY Email address: [email protected], [email protected]