# Explicit Expression for Arbitrary Positive Integer Powers of Special Sparse Matrices

Authors: Mohammad Beiranvand; Mojtaba Ghasemi Kamalvand
Journal: Journal of Mathematics
 (2022)

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## Abstract

Sparse matrices appear frequently in mathematical models. In this paper, we firstly present a general expression for the entries of ther th r∈ℕ power of a certain n-square sparse matrix, in terms of the Chebyshev polynomials of the second kind. Secondly, we present a method for integer positive powers of the skew matrix corresponding to these sparse matrices. This method will be inspiring to calculate the positive integer powers of the similar matrices. Finally, we present some examples to illustrate our results. Also, we give maple 18 procedures in order to verify our calculations.


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## Body

## 1. Introduction

In order to solve some difference equations, differential equations, and delay differential equations, we meet the necessity to compute the arbitrary positive integer powers of the square matrix [1–5]. Many papers discuss this topic [6–19]. In this paper, we obtain the entries of positive integer powers of a n-square certain sparse matrix of the order n=4p+3p∈ℕ of the following form:(1)A=00200−1−10−1−101010110−1−10−1−100⋱00200.We compute ther th power r∈ℕ of the matrix and the skew matrix corresponding to it, by using the expression Ar=TJrT−1 [14], where J is Jordan’s form of the matrix A, and T is the transforming matrix A. Although these types of matrices may have applications, we investigate these matrices purely from a mathematical point of view. In this article, interesting formulas are presented that will be attractive for math enthusiasts. Now, we are beginning with the following definition.Definition 1.
(see [15]). The Chebyshev polynomial Unx of the second kind is a polynomial of degree n in x defined by(2)Unx=sinn+1θsinθ,whenx=cosθ,n=0,1,2,….
With the definition above, we find thatUnx satisfies the recurrence relation(3)Unx=2xUn−1x−Un−2x,U0x=1,U1x=2x,n=2,3,….


## 2. Eigenvalues and Jordan’s Form of MatrixA

In this section, we calculate the eigenvalues of matrixA which are given in (1). The following lemma helps us to calculate the eigenvalues of tridiagonal matrices.Lemma 1.
(see [16]). Let Hn,n=1,2,… be a sequence of tridiagonal matrices of the following form:(4)Hn=h11h12h21h22h23h32h33h34h43h44h450h54⋱0hn−2n−1hn−1n−2hn−1n−1hn−1nhnn−1hnn.Then, the successive determinants ofHn are given by the following recursive formula:(5)H1=h11,H2=h11h22−h12h21,Hn=hnn|Hn−1|−hn−1nhnn−1Hn−2.Let {H∗n,n=1,2,… be a sequence of the tridiagonal matrix of the following form:(6)H∗n=h11−h12−h21h22−h23−h32h33h34h43h44h450h54⋱0−1p2hn−2n−1−1p2hn−1n−2hn−1n−1−1q2hn−1n−1q2hnn−1hnn,where (p and q = 1, 2, 3, …). Since the matrices Hn and H∗n have the same recursive formula, it can be written as follows:(7)Hn=H∗n.Theorem 1.
The eigenvalues ofA, as given in (4), are as follows:(8)λk=−2cosk−1πn−1,k=1,2,…,n.Proof.
Let(9)Dnx=x020x−1−1x−1−1x101x11x−1−1x−1−1x0⋱020x,Δnx=x11x11x101x11x11x11⋱0x11x.
If we calculate the determinant given in (9) by using the Laplace expansion, then we have(10)Dnx=x2−4Δn−2x.
On the other hand, from Lemma1, we get(11)Δnx=xΔn−1x−Δn−2x,n≥3,Δ1x=x,Δ2x=x2−1.
Solving difference equation (11), we obtain(12)Δnx=Unx2,where Unx is the n th degree Chebyshev polynomial of the second kind. From (10) and (12) follows.(13)Dnx=x2−4Un−2x2,Dn−x=A−xI=x2−4Un−2−x2.
Hence, the eigenvalues ofA are(14)λk=−2cosk−1πn−1,k=1,2,…,n.
Since all the eigenvaluesλk for k=1,2,…,n are simple, each eigenvalue λk corresponds single Jordan cell J1λk in the matrix J. Taking this into account, we write down Jordan’s form of the matrix A.(15)J=diagλ1,λ2,λ3,…,λn.
And applying the relationλk=−λn−k+1,k=1,2,…,n−1/2, we can write(16)J=diagλ1,λ2,λ3,…,0,…,−λ3,−λ2,−λ1.


## 3. General Expression for the Entries ofAr

In this section, we firstly find the transforming matrixT and its inverse for expression A=TJT−1 by using Chebyshev polynomial second kind. Secondly, we derive a general expression for the entries of matrix Ar for r∈ℕ. From relation, A=TJT−1 follows(17)AT=TJ,where T=T1,T2,…,Tn and Ti=T1i,T2i,…,Tnit, for, i=1,2,…,n; then, we can write(18)AT1,T2,T3,…,Tn=T1,T2,T3,…,Tndiagλ1,λ2,λ3,…,λn.Now, we have(19)ATi=λiTi,i=1,2,3,…,n.From (19) follows.(20)2Tni−T3i−T2i−T4i−T3i+T5iT4i+T6iT5i−T7i−T6i−T8i−T7i+T9iT8i+T10iT9i−T11i−T10i−T12i⋮−Tn−2i2T1i=λiT1iλiT2iλiT3iλiT4iλiT5iλiT6iλiT7iλiT8iλiT9iλiT10iλiT11i⋮λiTn−1iλiTni,i=1,2,3,…,n.Solving the set of systems (20), we find the eigenvectors of the matrix A.(21)T1=α0000000000⋮−α,Tn=α0000000000⋮α,Ti=0c0βU0λi2c1βU1λi2c2βU2λi2c3βU3λi2c4βU4λi2c5βU5λi2c6βU6λi2c7βU7λi2c8βU8λi2c9βU9λi2⋮0,i=2,3,…,n−1,where Ukx is the k th degree Chebyshev polynomial of the second kind, α and β are arbitrary numbers, and(22)ci=−1,if,i∈X=4p−3|p=1,2,…,1,if,i∈ℕ∪0−X.Taking into account (21) and (22) and choosing α=β=1 arbitrarily, we write down the transforming matrix T:(23)T=1000…10U0λ22U0λ32U0λ42…00−U1λ22−U1λ32−U1λ42…00U2λ22U2λ32U2λ42…00U3λ22U3λ32U3λ42…00U4λ22U4λ32U4λ42…00−U5λ22−U5λ32−U5λ42…00U6λ22U6λ32U6λ42…00U7λ22U7λ32U7λ42…00U8λ22U8λ32U8λ42…00−U9λ22−U9λ32−U9λ42…0⋮⋮⋮⋮…⋮−1000…1.Denotingj th column of the inverse matrix T−1 by τjT−1=τ1,τ2,τ3,…,τn, here,(24)τj=τ1j,τ2j,τ3j,…,τnjt,j=1,2,…,n.Then, we can write(25)τ1,τ2,τ3,…,τn−1,τnA=Jτ1,τ2,τ3,…,τn−1,τn.From (25), follows.(26)2τn,−τ3,−τ2−τ4,−τ3+τ5,τ4+τ6,…,τn−2,2τ1=Jτ1,Jτ2,…,Jτn−1,Jτn.Letτi2=mi for i=1,2,3,…,n, then from solving the set of systems (26), we can conclude τ11=−τ1n=τn1=τnn=ρ which ρ is an arbitrary number, τi1=τin=0 for i=2,3,…,n−1. Also, we can conclude τi1=τin=0 for i=2,3,…n−1, τ1j=τnj=0 for j=2,3,…n−1, and other elements of the matrix T−1 are equal mi× (Chebyshev polynomials second kind in eigenvalues matrix A). So we can write the columns matrix T−1 as follows:(27)τ1=ρ000⋮0ρ,τn=−ρ000⋮0ρ,τj=dj−20m2Uj−2λ22m2Uj−2λ32m3Uj−2λ42⋮mn−1Uj−2λn−120,j=2,3,…n−1,where(28)dj=−1if,j∈X=4p−3|p=1,2,…,1if,j∈ℕ∪0−X.Taking into account (27) and (28), we can write down the matrix T−1:(29)T−1=ρ000000m2U0λ22−m2U1λ22m2U2λ22m2U3λ22m2U4λ220m3U0λ32−m3U1λ32m3U2λ32m3U3λ32m3U4λ320m4U0λ42−m4U1λ42m4U2λ42m4U3λ42m4U4λ42⋮⋮⋮⋮⋮⋮0mn−1U0λn−12−mn−1U1λn−12mn−1U2λn−12mn−1U3λn−12mn−1U4λn−12ρ00000,00000…−ρ−m2U5λ22m2U6λ22m2U7λ22m2U8λ22−m2U9λ22…0−m3U5λ32m3U6λ32m3U7λ32m3U8λ32−m3U9λ32…0−m4U5λ42m4U6λ42m4U7λ42m4U8λ42−m4U9λ42…0⋮⋮⋮⋮⋮…⋮−mn−1U5λn−12mn−1U6λn−12mn−1U7λn−12mn−1U8λn−12−mn−1U9λn−12…000000…ρ.FromT−1T=I, we have(30)ρ+ρ=1⇒ρ=12.And, fori=2,3,…,n−1,(31)mi∑j=0n−3Uj2λi2=1⇒mi∑j=0n−3Uj2−cosi−1πn−1=1⇒mi∑j=0n−3Uj2cosn−iπn−1=1⇒mi∑j=0n−3sinj+1n−i/n−1πsinn−i/n−1π2=1⇒mi∑j=1n−2sinji−1/n−1πsini−1/n−1π2=1⇒misin2i−1/n−1π∑j=1n−2sin2ji−1πn−1=1⇒mi2sin2i−1/n−1π∑j=1n−21−cos2ji−1πn−1=1⇒mi2sin2i−1/n−1πn−2−∑j=1n−2cos2ji−1πn−1=1⇒mi2sin2i−1/n−1πn−32+sini−1/n−1π2sini−1/n−1π=1∑j=1Ncosjθ=−12+sinN+1/2θ2sinθ/2⇒min−12sin2i−1/n−1π=1⇒mi=2sin2i−1/n−1πn−1.=21−cos2i−1/n−1πn−1=21−−λi/22n−1.Therefore,(32)mi=4−λi22n−1,i=2,3,…n−1.From (30) and (32) follows.(33)T−1=12n−1n−10000000000…−n−104−λ22U0λ22−4−λ22U1λ22+++−4−λ22U5λ224−λ22U6λ22++−…004−λ32U0λ32−4−λ32U1λ32+++−4−λ32U5λ324−λ32U6λ32++−…004−λ42U0λ42−4−λ42U1λ42+++−4−λ42U5λ424−λ42U6λ42++−…0⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮⋮…⋮04−λn−12U0λn−12−4−λn−12U1λn−12+++−4−λn−12U5λn−124−λn−12U6λn−12++−…0n−10000000000…n−1.By combining (15), (23), (32) and using the equality Ar=TJrT−1, we compute the r th powers of the matrix A. Therefore, i,j th entry of matrix Ar can be given as follows:(34)Arij=TJrT−1ij=12n−1×n−11jλ1r+λnrifi=1,j=1,2,…,n−1n−1in−λ1r+λnrifj=1,i=2,3,…n∑k=2n−1ci−2dj−2λkr4−λk2Ui−2λk2Uj−2λk2ifi,j=2,3,…,n−1n−11i−λ1r+λnrifj=n,i=1,2,…n−1n−1jnλ1r+λnrifi=n,j=2,3,…,n.For any real numberx, x = the largest integer that is less than or equal to x.In the continuation of this article, we calculate the positive integer powers of the skew matrices corresponding to matrix (1) as follows:(35)B=200−100−10−1010−1101−101−10−10−1⋰000002.Let(36)J=101⋰101.And partition matricesAr and JAr are as follows:(1)
Ifr is an even number,(37)Ar=2rAr2r,JAr=2rJAr2r.(2)
Ifr is an odd number,(38)Ar=2rAr2r,JAr=2rJAr2r.With simple calculations, we have(39)B=AJ.Theorem 2.
If matricesA,B and J have the form (1), (34) and (35), respectively, then r th power r∈ℕ of matrix Bn can compute as follows.
Fork=1,2,3,….State 1.
Ifn=4m+1, m=1,2,…, then(40)Br=Ar,forr=2k,JAr,forr=2k+1.State 2.
Ifn=4m+3, m=1,2,…, then(41)Br=2r−Ar2r,forr=4k−2,JAr,forr=4k−1,Ar,forr=4k,2r−JAr2r,forr=4k+1.Proof.
We prove this theorem by induction onk.State 3.
The base case ofk=1 is true, because, with simple calculations, we have(42)B2=A2,B3=JA3.
We suppose that the result is true fork>1 and consider case k+1.
By the induction hypothesis, we have(43)B2k=A2k,B2k+1=JA2k+1.
We alsoshow that the case k+1 is true(44)B2k+2=B2kB2=A2kA2=A2k+2,B2k+3=B2k+1B2=JA2k+1A2=JA2k+3.State 4.
The casek=1 is true, and after the necessary calculations, we have(45)B2=22−A222,B3=JA3,B4=A4,B5=25−JA525.
We suppose that the result is true fork>1 and show that case k+1 is true.
By the induction hypothesis, we have(46)B4k−2=24k−2−A4k−224k−2,B4k−1=JA4k−1,B4k=A4k,B4k+1=24k+1−JA4k+124k+1.
We also show that casek+1 is true(47)B4k+2=B4k−2B4=24k−2−A4k−224k−222−A2222=24k−2−A4k−224k−224A424=24k+2−A4k+224k+2,B4k+3=B4k−1B4=JA4k−1A4=JA4k+3,B4k+4=B4kB4=A4kA4=A4k+4B4k+5=B4k+1B4=24k+1−JA4k+124k+122−A2222=24k+1−A4k+124k+124A424=24k+5−A4k+524k+5.
Thus, the formulas also hold fork+1, and the induction arguments are completed.
We can computei,j th entry of matrix Br in (35) by using formula (34) and Theorem 2.


## 4. Numerical Consideration

In this section, we give an example that calculates the powers matrix (1). Calculations given in this section can be verified by the Maple 18 procedure that is given in Appendix.Example 1.
Ifn=7, then let A1 be a 7×7 matrix, given in (1) as in the following:(48)A1=000000200−100000−10−100000−10100000101000001002000000.
2rd, 3rd, and 4rd powers of the matrix A1 are computed as in the following.
From (15), the eigenvalues of the matrix A1 can be written for k=1,2,…,7 as follows:
λk=−2cosk−1π/n−1, namely,λ1=−2,λ2=−3,λ3=−1,λ4=0, λ5=1, λ6=3, and λ7=2.
From (23) and (32), we can write the transforming matrix T and its inverse as follows:(49)T=10000010U0λ22U0λ32U0λ42U0λ52U0λ6200−U1λ22−U1λ32−U1λ42−U1λ52−U1λ6200U2λ22U2λ32U2λ42U2λ52U2λ6200U3λ22U3λ32U3λ42U3λ52U3λ6200U4λ22U4λ32U4λ42U4λ52U4λ620−1000001=100000101111100310−1−30020−10200−310−13001−11−110−1000001,T−1=127−1×7−100000−7−104−λ22U0λ22−4−λ22U1λ224−λ22U2λ224−λ22U3λ224−λ22U4λ22004−λ32U0λ32−4−λ32U1λ324−λ32U2λ324−λ32U3λ324−λ32U4λ32004−λ42U0λ42−4−λ42U1λ424−λ42U2λ424−λ42U3λ424−λ42U4λ42004−λ52U0λ52−4−λ52U1λ524−λ52U2λ524−λ52U3λ524−λ52U4λ52004−λ62U0λ62−4−λ62U1λ624−λ62U2λ624−λ62U3λ624−λ62U4λ6207−1000007−1,=1200000−12011231216−312112001414014−1400130−130130014−140−14−1400112−312163121120120000012.
Then, we get(50)A12=TJ2T−1=400000001010000020−100010201000−1020000010100000004,A13=TJ3T−1=000000800−201000−20−30−1000−30300010302000−102008000000,A14=1600000002030100050−400030603000−40500010302000000016.
Also,i,j th entry of the matrices A12,A13, and A14 can be verified by the formula given in (34).



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