Selçuk J. Appl. Math. Selçuk Journal of Special Issue. pp. 81-86, 2010 Applied Mathematics
On Computing th ( = 2 ∈ ) Powers for One Type Even Order Antipentadiagonal Matrix
Hümeyra Kıyak, ˙Irem Gürses, Durmu¸s Bozkurt
Selçuk University, Science Faculty, Department of Mathematics, 42003, Kampus, Konya, Türkiye
e-mail: db ozkurt@ selcuk.edu.tr
Presented in 2National Workshop of Konya Ere˘gli Kemal Akman College, 13-14 May 2010.
Abstract. In this paper, we derive the general expression of the th ( = 2 ∈ N) power for one type of even order antipentadiagonal matrix.
Key words: Antipentadiagonal, Matrix power,Chebyshev Polynomial. 2000 Mathematics Subject Classification: 15A60, 33C45.
1. Introduction
Anti-pentadiagonal matrices find application in many fields, such as solution of boundary value problems, numerical analysis and high order harmonic spec-tral filtering theory [1]. In this paper, we derive the general expression of the th power for one type of symmetric anti-pentadiagonal matrices of even order.
2. Derivation of Formulation Let us define a matrix as following :
= ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ −1 0 0 1 0 0 0 −1 0 0 0 −1 1 0 0 0 1 . .. . .. . .. . .. . .. 1 0 0 0 1 0 0 0 −1 0 0 1 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦
It is known that, the th power of a matrix is obtained by the expression = −1[2] where is Jordan’s form and is the transforming matrix. The
matrices and are derived by the eigenvalues and eigenvectors of the matrix . The eigenvalues of a matrix are derived by the characteristic equation of the matrix which is defined by
() = | − | = 0
here = − and is th order identity matrix. Let us denote
∆() = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ −1 1 1 −1 −1 . .. ... ... −1 −1 1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ and () = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ −1 0 0 1 0 0 0 −1 0 0 0 −1 1 0 0 0 1 . .. . .. . .. . .. . .. 1 0 0 0 1 0 0 0 −1 0 0 1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ − − ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ 0 0 0 0 0 0 0 0 . .. 0 0 . .. 0 . .. ... ... 0 0 0 0 0 0 0 0 0 0 0 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯
= ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ −1 0 0 1 0 0 0 . .. −1 0 0 0 −1 1 0 0 0 1 −1 0 0 −1 1 0 0 1 . .. . .. . .. . .. . .. 1 0 0 0 1 . .. 0 0 0 −1 0 0 1 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯
We can write the following difference equation,
(1) ∆() = ∆−1() + ∆−2()
where ∆0() = 1 ∆1() = ∆2() = 2+ 1. Solution of the difference
equation (1) is: 1= +√2+ 4 2 2= −√2+ 4 2 ⇒ ∆= 1(1)+ 2(2)
Writing ∆0 = 1 and ∆1 = on the equation, we obtain 1 = 22−−1 and
2= 2−−11. Substituting 1 2 1 2on ∆we obtain
∆ = Ã +√2+ 4 2√2+ 4 ! Ã +√2+ 4 2 ! − Ã −√2+ 4 2√2+ 4 ! Ã −√2+ 4 2 ! = 1 2+1 ¡ +√2+ 4¢+1−¡ −√2+ 4¢+1 ¡√ 2+ 4¢
By definition of second kind Chebyshev polynomials [4] and rewriting¡ 2
¢ on the the following equality instead of ;
() = 1 2 ¡ +√2− 1¢+1−¡ −√2− 1¢+1 ¡√ 2− 1¢
Then, ( 2) = 1 2 µ 2 + q¡ 2 ¢2 − 1 ¶+1 − µ 2 − q¡ 2 ¢2 − 1 ¶+1 µq¡ 2 ¢2 − 1 ¶ = 1 2 ¡ 2 ¢+1n¡ +√2+ 4¢+1−¡ −√2+ 4¢+1o ¡ 2 ¢ ¡√ 2+ 4¢ = 1 2 µ 2 ¶¡ +√2+ 4¢+1−¡ −√2+ 4¢+1 ¡√ 2+ 4¢ = () 2+1 ¡ +√2+ 4¢+1−¡ −√2+ 4¢+1 ¡√ 2+ 4¢ = () 1 2+1 ¡ +√2+ 4¢+1−¡ −√2+ 4¢+1 ¡√ 2+ 4¢ | {z } ∆ ( 2) = () ∆() (2) ∆() = (−)( 2) and () =¡∆ 2() ¢2 From (2): (3) () = µ (−)2 2( 2) ¶2
All roots of the polynomial () are included in the interval [−1 1] and
ob-tained by the relation
= 2 cos
2
+ 2 = 1 2
So we can write the form of the matrix .
All the eigenvalues ( = 12 = 2 ∈ N) are double multiple (the
= (1 1 2 2 2
2).
Using the equation −1 we find the matrices and −1 derive the
ex-pression of the th power = 2 ∈ N of the matrix as following: = −1 = 1 + 2() () = (1 + (−1)+ 2 (+)−) − 4 X =1 (−)(4 + 2) − 2 (− 2 ) × × − 2 (− 2 ) here = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 0 = 2 + 4( = 0 1 2 ) − 1 = 4 + 4 1 (−1)+ = −1 (−1)+ = 1 (−1)+ = 1 where = 0 1 2 = ( −1 1 (−1)= (−1) = 1 (−1)= (−1) = −1 = ( − 1 (−1)= (−1)= 1 (−1)= (−1)= −1 = ( 0 2 = 2 ( = 2 4 6 8 ) = 2 ( = 3 5 7 9 ) 3. Numerical Examples
We can compute arbitrary positive integer powers of the matrix taking into account the derived formula. For example, if = 4;
= ⎡ ⎢ ⎢ ⎣ 0 −1 0 0 1 0 0 0 0 0 0 −1 0 0 1 0 ⎤ ⎥ ⎥ ⎦
form of the matrix is
= (1 1 2 2) = (− − ) =1 6(()) = 1 2 ⎡ ⎢ ⎢ ⎣ 1 2 0 0 2 1 0 0 0 0 1 2 0 0 2 1 ⎤ ⎥ ⎥ ⎦
1() = −(1 + (−1)) 2() = 1 − (−1) and −1 = ⎡ ⎢ ⎢ ⎣ 0 1 0 0 −1 0 0 0 0 0 0 1 0 0 −1 0 ⎤ ⎥ ⎥ ⎦ If = 6 : = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 0 0 0 −1 0 0 0 0 1 0 0 0 0 −1 0 0 0 −1 1 0 0 0 1 0 0 0 0 −1 0 0 0 0 1 0 0 0 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ Jordan form of matrix is:
= (1 1 2 2 3 3) = (− − 0 0 ) = 2 cos 4 = √ 2 = 1 8(()) = 1 4 ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 1 0 0 2 1 0 0 1 2 0 0 1 0 2 3 0 0 2 2 0 0 3 2 0 1 0 0 2 1 0 0 1 2 0 0 1 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ = 2 + 4 1() = −(1 + (−1)) 2() = −(1 − (−1))+1 3() = −(1 + (−1))+2 ∈ N If = 4 + 4 1() = (1 + (−1)) 2() = (1 − (−1))+1 3() = (1 + (−1))+2 ∈ N
Since is a singular matrix, we can not compute the negative integer powers of the matrix.
References
1. Jonas Rimas, On computing of arbitrary positive integer powers for one type of sym-metric pentadiagonal matrices of even order, Applied Mathematics and Computation 203 (2008) 582—591.
2. R. A. Horn, Ch. Johnson, "Matrix Analysis", Cambridge University Press, Cam-bridge, 1985.
3. R. P. Agarwal, "Difference Equations and Inequlities", Marcel Dekker, New York, 1992.
4. J. C. Mason, D. C. Handscomb, "Chebyshev Polynomials", CRC Press, Washington, 2003.
