Journal of Applied Mathematics and Computational Mechanics 2014, 13(3), 21-28
RELATION BETWEEN DETERMINANTS OF TRIDIAGONAL AND CERTAIN PENTADIAGONAL MATRICES
Jolanta Borowska, Lena Łacińska Institute of Mathematics, Czestochowa University of Technology Częstochowa, Poland
[email protected],
[email protected]
Abstract. In this paper we consider a pentadiagonal matrix which consists of only three non-zero bands. We prove that the determinant of such a matrix can be represented by a product of two determinants of corresponding tridiagonal matrices. It is shown that such an approach gives greatly shorter time of computer calculations. Keywords: pentadiagonal matrix, tridiagonal matrix, determinant, recurrence equations
Introduction This paper constitutes generalization of results presented in [1]. Let us consider a pentadiagonal matrix of the form a1 0 c1 0 a 0 2 0 b a3 3 b4 0 O An =
c2 0 a4 O O
c3 0 O O bn − 3
c4 O O O O O 0 an − 3 0 cn − 3 bn − 2 0 an − 2 0 bn −1 0 an −1 bn 0
cn − 2 0 an
(1)
It means that An = [ aij ]nxn where aij = 0 for i − j > 2 and for i − j = 1 . Determinant Wn of matrix An can be represented by a particular solution of fourth order linear recurrence equation (see [2]) Wn+ 4 = an+ 4Wn+3 − an+3bn+ 4 cn+ 2Wn+1 + bn+3bn+4 cn+1cn+ 2Wn
(2)
22
J. Borowska, L. Łacińska
with initial conditions of the form W1 = a1 W = a a 2 1 2 W a = 3W2 − a2 b3c1 3 W4 = a4W3 + b4 c2 (b3c1 − a1a3 )
(3)
Now, let us introduce two auxiliary tridiagonal matrices of order k a1 c1 b a c 3 3 3 b5 a5 c5 O O O ∗ Ak = O O b2 k −5
O a2 k −5 b2 k −3
c2 k −5 a2 k −3 b2 k −1
c2 k −3 a2 k −1
(4)
c2 k − 2 a2 k
(5)
and a2 b 4 ∗∗ Ak =
c2 a4 b6
c4 a6
c6
O O O
O O
O
b2 k − 4
a2 k − 4
c2 k − 4
b2 k − 2
a2 k − 2 b2 k
By Wk* , Wk** we denote the determinants of matrices Ak* , Ak** respectively. Following [2], we can represent determinant Wk* by a particular solution of second order linear recurrence equation Wk∗+ 2 = a2 k + 3Wk∗+1 − b2 k + 3c2 k +1Wk∗
(6)
with initial conditions of the form W1* = a1 * W2 = a1a3 − b3c1
(7)
Relation between determinants of tridiagonal and certain pentadiagonal matrices
23
In the same way we obtain recurrence relation for determinant Wk**
Wk∗+∗2 = a2 k +4Wk∗+∗1 − b2 k +4c2k +2Wk∗∗
(8)
W1** = a2 ** W2 = a2 a4 − b4 c2
(9)
where
The main results In this section we show the relationship between the determinant of matrix An and determinants of matrices Ak* , Ak** . Theorem 1. Let An , Ak* , Ak** be the matrices given by (1), (4), (5) and Wn , Wk* , Wk** be their determinants respectively. Moreover, let us assume that W0** = 1 . Then the following statements hold: 1) If n = 2k − 1 then
Wn = W2k −1 = Wk∗∗−1 ⋅ Wk∗ , k = 1,2,K,
n +1 2
(10)
n 2
(11)
2) If n = 2k then
Wn = W2 k = Wk∗∗ ⋅ Wk∗ , k = 1,2,K,
Proof. We are to prove that every determinant Wn , n ∈ N , given by relationships (10) and (11) satisfies recurrence equation (2) with initial conditions (3). Firstly, we prove that initial conditions (3) can be obtained from (10) and (11). Substituting consecutively k = 1, k = 2 into (10) and (11) we have
W1 = W0** ⋅ W1* W2 = W1** ⋅ W1* W3 = W1** ⋅ W2* W4 = W2** ⋅ W2* Simultaneously from (3), (7) and (9) under the assumption that W0** = 1 we obtain
24
J. Borowska, L. Łacińska
W1 = a1 = 1 ⋅ a1 = W0** ⋅ W1* W2 = a1 ⋅ a2 = W1* ⋅ W1** W3 = a3W2 − a2b3c1 = a1a2 a3 − a2b3c1 = a2 (a1a3 − b3c1 ) = W1** ⋅ W2* W4 = a4W3 + b4 c2 (b3c1 − a1a3 ) = a2 a4 (a1a3 − b3c1 ) − b4 c2 (a1a3 − b3c1 ) =
(a2 a4 − b4c2 )(a1a3 − b3c1 ) = W2** ⋅ W2* So, it was shown that initial conditions (3) can be obtained from (10) and (11). Secondly we are to show that every determinant Wn , n > 4 , given by relationships (10) and (11) satisfies recurrence equation (2). Substituting n = 2k – 1 into (2) we have (12)
W2 k +3 = a2 k +3W2 k + 2 − a2 k +2b2 k +3c2 k +1W2 k + b2 k + 2b2 k +3c2 k +1c2 kW2 k −1
At the same time from (10) and (11) we get W2 k + 3 = Wk*+*1Wk*+ 2 W2 k + 2 = Wk*+*1Wk*+1
(13)
W2 k = Wk**Wk* W2 k −1 = Wk*−*1Wk*
Hence we have to show that equation
Wk*+*1Wk*+ 2 = a2 k +3Wk*+*1Wk*+1 − a2k + 2b2k +3c2 k +1Wk**Wk* + b2k + 2b2 k +3c2 k +1c2 kWk*−*1Wk* (14) holds for every k ∈ N . We start with the left-hand side of equation (14). Bearing in mind (8) we obtain
(
)(
)
Wk*+*1Wk*+ 2 = a2 k + 2Wk** − b2 k + 2 c2 kWk∗∗−1 a2 k + 3Wk∗+1 − b2 k + 3c2 k +1Wk∗ = = a2 k + 2 a2 k + 3Wk∗∗Wk∗+1 − a2 k + 2b2 k + 3c2 k +1Wk∗∗Wk∗ − a2 k + 3b2 k + 2c2 kWk∗∗−1Wk∗+1 +
(
)
+ b2 k + 2b2 k + 3c2 k c2 k +1Wk∗∗−1Wk∗ = a2 k + 3Wk*+1 a2 k + 2Wk** − b2 k + 2 c2 kWk*−*1 + − a2 k + 2b2 k + 3c2 k +1Wk**Wk*
+ b2 k + 2b2 k + 3c2 k c2 k +1Wk*−*1Wk*
=
= a2 k + 3Wk*+1Wk*+*1 − a2 k + 2b2 k + 3c2 k +1Wk**Wk* + b2 k + 2b2 k + 3c2 k c2 k +1Wk*−*1Wk*
Let us observe that at the end of the above transformations we have obtained the right-hand side of equation (14). Now, let us substitute n = 2k into equation (2) W2 k + 4 = a2 k + 4W2 k + 3 − a2 k + 3b2 k + 4 c2 k + 2W2 k +1 + b2 k + 3b2 k + 4c2 k + 2 c2 k +1W2 k
25
Relation between determinants of tridiagonal and certain pentadiagonal matrices
From (10) and (11) we get
W2 k + 4 = Wk*+*2Wk*+ 2
(15)
W2 k +1 = Wk**Wk*+1
Moreover, from (13) we take the formulae for W2k + 3 and W2 k . Hence we must prove that the equation
Wk*+*2Wk*+ 2 = a2 k + 4Wk*+*1Wk*+2 − a2 k +3b2k + 4c2 k + 2Wk**Wk*+1 + b2k +3b2 k +4 c2 k +2 c2 k +1Wk**Wk* (16) holds for every k ∈ N . We start with the left-hand side of the equation (16). Bearing in mind (6) and (8) we have
(
)(
)
Wk*+* 2Wk*+ 2 = a2 k + 4Wk*+*1 − b2 k + 4 c2 k + 2Wk** a2 k + 3Wk*+1 − b2 k + 3c2 k +1Wk* = = a2 k + 3 a2 k + 4Wk*+*1Wk*+1 − a2 k + 4b2 k + 3c2 k +1Wk*+*1Wk* − a2 k + 3b2 k + 4 c2 k + 2Wk**Wk*+1 +
(
)
+ b2 k + 3b2 k + 4 c2 k +1c2 k + 2Wk**Wk* = a2 k + 4 a2 k + 3Wk*+1 − b2 k + 3c2 k +1Wk* Wk*+*1 + − a2 k + 3b2 k + 4 c2 k + 2Wk**Wk*+1 + b2 k + 3b2 k + 4 c2 k +1c2 k + 2Wk**Wk* = = a2 k + 4Wk*+ 2Wk*+*1 − a2 k + 3b2 k + 4 c2 k + 2Wk**Wk*+1 + b2 k + 3b2 k + 4 c2 k +1c2 k + 2Wk**Wk*
Hence we have obtained the right-hand side of equation (16), which ends the proof. Example Let us consider a special form of pentadiagonal matrix (1) in which elements n n on diagonals are defined by sequences of the form (ak )k =1 = k , (bk )k = 3 = 2k , (ck )kn=−12 = 2k − 3. Moreover, we assume that n = 10 6 , i.e. the matrix has the order 10 6. The value of the determinant of the considered matrix will be obtained in two ways. Firstly we apply fourth order linear recurrence equation (2) with initial conditions (3), which in this case have the forms Wn + 4 = ( n + 4)Wn + 3 − ( n + 3)( 2n + 8)( 2n + 1)Wn +1 + ( 2n + 6)( 2n + 8)( 2n − 1)(2n + 1)Wn W1 = 1, W2 = 2, W3 = 18, W4 = 0
where n = 1,2,...,106 − 4 . Let us observe that the above recurrence equation has functional coefficients, hence it is impossible to solve this equation using known analytical methods [3]. The proper algorithm for computing the determinant of this matrix will be implemented in the Maple system [4]. We apply the following steps:
26
J. Borowska, L. Łacińska
a := evalf ([seq(n, n = 1..106 )]) : b := evalf ([0,0, seq(2n, n = 3..106 )]) : c := evalf ([seq(2n − 3, n = 1..106 )]) : W [1] := a[1] : W [ 2] := a[1]⋅ a[2] : W [3] := a[3]⋅ W [2] − a[2]⋅ b[3]⋅ c[1] : W [ 4] := a[4] ⋅ W [3] + b[4] ⋅ c[2] ⋅ (b[3] ⋅ c[1] − a[1] ⋅ a[3]) : for n from 1 to 999996 do W [n + 4] := a[n + 4] ⋅ W [n + 3] − a[n + 3] ⋅ b[n + 4] ⋅ c[n + 2]⋅ W [n + 1] + b[n + 3] ⋅ b[n + 4] ⋅ c[n + 1] ⋅ c[n + 2] ⋅ W [n] : end do : print (evalf (W [1000000])) Finally we get − 6.43082825 ×105866732 as the value of determinant of the matrix under consideration. Secondly, we apply Theorem 1, from which we have ∗∗ ∗ W1000000 = W500000 ⋅ W500000
where determinants W * , W ** will be obtained from formulae (6), (7) and (8), (9). These formulae in this case have the following forms:
Wk∗+ 2 = (2k + 3)Wk∗+1 − (4k + 6 )(4k − 1)Wk∗ W1* = 1, W2* = 9 Wk∗∗+2 = (2k + 4 )Wk∗∗+1 − (4k + 8)(4k + 1)Wk∗∗ W1** = 2, W2** = 0 Let us denote F = W * , G = W ** and apply the following syntax:
a := evalf ([seq(n, n = 1..106 )]) : b := evalf ([0,0, seq(2n, n = 3..106 )]) : c := evalf ([seq(2n − 3, n = 1..106 )]) : F [1] := a[1] : F [ 2] := a[1] ⋅ a[3] − b[3] ⋅ c[1] :
Relation between determinants of tridiagonal and certain pentadiagonal matrices
27
for k from 1 to 499998 do F [k + 2] := a[2k + 3] ⋅ F [k + 1] − b[ 2k + 3] ⋅ c[2k + 1]⋅ F [k ] : end do : G[1] := a[2] : G[ 2] := a[2]⋅ a[4] − b[4]⋅ c[2] : for k from 1 to 499998 do G[k + 2] := a[2k + 4] ⋅ G[k + 1] − b[2k + 4] ⋅ c[2k + 2]⋅ G[k ] : end do : print (evalf ( F [500000] ⋅ G[500000]))
Finally we get − 6.430935419 ×105866732 as the value of determinant of the matrix under consideration. The presented results were obtained with the Maple default precision (Digits = 10). It has to be emphasized that the running time of the first algorithm was 40 s, whilst of the second was 17 s.
Conclusions It was shown that the determinant of the pentadiagonal matrix, which consists of only three non-zero bands, can be represented by a product of two determinants of corresponding tridiagonal matrices. This means that this determinant is obtained as a product of particular solutions of two second order homogeneous linear recurrence equations. In the presented example we dealt with a 10 6 × 106 pentadiagonal matrix. In order to obtain the determinant of the considered matrix we used two approaches. Firstly we had a formulated algorithm based on proper fourth order homogeneous linear recurrence equation. In second algorithm we used the theorem proved in this paper. Both algorithms were implemented in the Maple system. It turned out that time of calculations was shorter in the second approach.
References [1] Borowska J., Łacińska L., Rychlewska J., On determinant of certain pentadiagonal matrix, Journal of Applied Mathematics and Computational Mechanics 2013, 12(3), 21-26. [2] Borowska J., Łacińska L., Rychlewska J., A system of linear recurrence equations for determinant of pentadiagonal matrix, Journal of Applied Mathematics and Computational Mechanics 2014, 2(13), 5-12. [3] Elaydi S., An Introduction to Difference Equations, Springer, 2005. [4] Adams P., Smith K., Vyborny R., Introduction to Mathematics with Maple, World Scientific, 2004.
28
J. Borowska, L. Łacińska
