2. Main ResultsAccording to the following lemmas, Wen-Chyuan Yueh obtains eigenvalue and corresponding eigenvectors for matrix (1), in special cases. In this section, we firstly find the transforming matrices and their invers for the matrix (1). Secondly, we present a general expression for the entries of Am for m∈ℕ.

Lemma 1 (see [<xref ref-type="bibr" rid="B19">19</xref>]).Suppose α=0, β=ac≠0, and ρ=a/c, then the eigenvalues λ1,λ2,⋯,λn of A are given by
(2)λk=b+2accos2kπ2n+1, k=1,2,⋯,n,the corresponding eigenvectors uk=u1k,u2k,⋯,unkT, k=1,2,⋯,n, are given by
(3)ujk=ρj−1sin2kjπ2n+1, j=1,2,⋯,n.

In this paper, we need the following theorem.

Theorem 2.(canonical Jordan’s form [20]). Let A be any square matrix. Then, there exists a nonsingular matrix U which transforms A into a block diagonal matrix J such that
(4)U−1AU=J=diagJn1λ1,Jn2λ2,⋯Jnlλl, n1+n2+⋯+nl=n.which is called the canonical Jordan’s form, λj being the eigenvalues of A and Jk∈Ck×k a Jordan block of the form
(5)J1λ=λ,J2λ=λ10λ,⋯,Jkλ=λ1λ10⋱⋱0λ1.

Since all the eigenvalues λk for k=1,2,⋯,n are distinct (cosθ is a strictly decreasing function of θ on (0,π), and a≠0≠c), columns of the transforming matrix U are the eigenvectors of the matrix (1). Also, all eigenvalues λk correspond to the single Jordan cell J1λk in the matrix J, then we write down Jordan’s form of the matrix A as
(6)J=diagλ1,λ2,⋯,λn.

From (3), we can write the column transforming matrix U as
(7)uk=sin2kπ2n+1,ρ1sin4kπ2n+1,⋯,ρn−1sin2knπ2n+1T,for k=1,2,⋯,n.

Hence,
(8)U=u1,u2,u3,⋯,un.

Let
(9)D=diag1,ρ1,ρ2,ρ3,⋯,ρn−1,U~=u~1,u~2,u~3,⋯,u~n,in which
(10)u~k=sin2kπ2n+1,sin4kπ2n+1,⋯,sin2knπ2n+1T,for k=1,2,⋯,n.

Therefore, we have
(11)U=DU~,(12)U−1=U~−1D−1.

So an explicit expression for U~−1 will suffice.

Theorem 3.Suppose U~ is defined as above, then
(13)U~−1=42n+1U~.

Proof.We show that
(14)u~iTu~j=2n+14,if i=j,0,if i≠j.

From (10), equals sinαsinβ=1/2cosα−β−cosα+β and
(15)∑k=1ncoskθ=sinn+1/2θ2sin1/2−12,follows
(16)u~iTu~j=∑k=1nsin2kiπ2n+1sin2kjπ2n+1=12∑k=1ncos2ki−jπ2n+1−cos2ki+jπ2n+1.

If i=j,
(17)u~iTu~j=n2−12sinn+1/24iπ/2n+12sin4iπ/2n+1/2−12=n2−12sin2iπ2sin2iπ/2n+1−12=n2−120−12=2n+14.

If i≠j,
(18)u~iTu~j=12∑k=1ncos2ki−jπ2n+1−cos2ki+jπ2n+1=12sinn+1/22i−jπ/2n+12sin2i−jπ/2n+1/2−sinn+1/22i+jπ/2n+12sin2i+jπ/2n+1/2=14sini−jπsini−jπ/2n+1−sini+jπsini+jπ/2n+1.

If i and j are even or odd, then i−j and i+j are even; therefore, sini−jπ=sini+jπ=0, so we have
(19)u~iTu~j=0.

If one of i or j is even and the other is odd, then i−j and i+j are odd; therefore, we have
(20)sini−jπ−i−jπ2n+1=sini−jπ2n+1,(21)sini+jπ−i+jπ2n+1=sini+jπ2n+1.

From (18), (20), and (21) follows
(22)u~iTu~j=0.

For derivation of the formula for the entries of Am, from (4), (6), (10), (12), and (13), we can conclude
(23)Am=UJmU−1=UJmU~−1D−1=42n+1Udiadλ1m,λ2m,⋯,λnmU~−1diag1,ac−1/2,ac−2/2,⋯,ac−n−1/2.

By substituting U and U~−1 in the latter equation and doing the necessary computation follows
(24)Ami,j=42n+1aci−j/2∑k=1nλkmsin2ikπ2n+1sin2jkπ2n+1,for i,j=1,2,⋯,n, where λk=b+2accos2kπ/2n+1, k=1,2,⋯,n.

Lemma 4 (see [<xref ref-type="bibr" rid="B19">19</xref>]).Suppose α=−ac≠0 and β=0 and ρ=ac, then the eigenvalues λ1,λ2,⋯,λn of An are given by
(25)λk=b+2accos2k−1π2n+1, k=1,2,⋯n.

The corresponding eigenvectors, vk=v1k,v2k,⋯,vnkT,k=1,2,⋯,n, are given by
(26)vjk=ρj−1cos2k−12j−122n+1, j=1,2,⋯,n.

From (26) for k=1,2,⋯,n, we can write the columns transforming matrix V as
(27)vk=cos2k−1π22n+1,ρ1cos32k−1π22n+1,⋯,ρn−1cos2n−12k−1π22n+1T

Hence,
(28)V=v1,v2,⋯,vn.

Let
(29)D=diag1,ρ1,ρ2,⋯,ρn−1,V~=v~1,v~2,⋯,v~n,in which
(30)v~k=cos2k−1π22n+1,cos32k−1π22n+1,⋯,cos2n−12k−1π22n+1T,for k=1,2,⋯,n.

Therefore, we have
(31)V=DV~,V−1=V~−1D−1.

So an explicit expression for V~−1 will suffice.

Theorem 5.Suppose V~ is defined as above, then
(32)V~−1=42n+1V~.

Proof.We show that
(33)v~iTv~j=2n+14,if i=j,0,if i≠j.

From (30) equals cosαcosβ=1/2cosα−β+cosα+β and
(34)∑k=1ncos2k−1θ=sin2nθ2sinθ,follows
(35)v~iTv~j=∑k=1ncos2k−12i−1π22n+1cos2k−12j−1π22n+1=12∑k=1ncos2k−12i−1π22n+1−2k−12j−1π22n+1+cos2k−12i−1π22n+1+2k−12j−1π22n+1=12∑k=1ncos22k−1i−jπ22n+1+cos22k−1i+j−1π22n+1.

If i=j,
(36)v~iTv~j=n2+12∑k=1ncos22k−12i−1π22n+1=n2+12sin2n2i−1π/2n+12sin2i−1π/2n+1=n2+12sin2n/2n+12i−1π2sin2i−1π/2n+1=n2+12sin2i−1π−2i−1π/2n+12sin2i−1π/2n+1n2+12sin2i−1π/2n+12sin2i−1π/2n+1=2n+14.

If i≠j,
(37)v~iTv~j=12∑k=1ncos2k−1i−jπ2n+1+cos2k−1i+j−1π2n+1=12sin2n2k−1i−jπ/2n+12sin2k−1i−jπ/2n+1+sin2n2k−1i+j−1π/2n+12sin2k−1i+j−1π/2n+1=12sin2ni−jπ/2n+12sini−jπ/2n+1+sin2ni+j−1π/2n+12sini+j−1π/2n+1=12sin1−1/2n+1i−jπ2sini−jπ/2n+1+sin1−1/2n+1i+j−1π2sini+j−1π/2n+1.

If i and j are even or odd, then i−j is even and i+j−1 is odd; therefore,
(38)sin1−12n+1i−jπ=−sini−jπ2n+1,sin1−12n+1i+j−1π=−sini+j−1π2n+1.

So we have
(39)v~iTv~j=0.

If one of i or j is even and the other is odd, then i−j and i+j−1 are even; therefore, we have
(40)sin1−12n+1i−jπ=−sini−jπ2n+1and
(41)sin1−12n+1i+j−1π=−sini+j−1π2n+1.

Therefore,
(42)v~iTv~j=0.

Similar for (24), we can conclude
(43)Ami,j=42n+1aci−j/2∑k=1nλkmcos2k−12i−1π22n+1sin2k−12j−1π22n+1,for i,j=1,2,⋯,n.

Lemma 6 (see [<xref ref-type="bibr" rid="B18">18</xref>]).Suppose α=−β=ac≠0 and ρ=ac, then the eigenvalues λ1,λ2,⋯,λn of An are given by
(44)λk=b+2accos2k−1π2n, k=1,2,⋯,n.

The corresponding eigenvectors are given by
(45)uk=u1k,u2k,⋯,unkT, k=1,2,⋯,n,ujk=ρj−1sin2k−12j−1π4n, j=1,2,⋯,n.

In the case α=−β=−ac≠0, the eigenvalues are given by (44) and the corresponding eigenvectors by
(46)vk=v1k,v2k,⋯,vnkT, k=1,2,⋯,n,vjk=ρj−1cos2k−12j−1π4n, j=1,2,⋯,n.

In these cases which are similar to previous cases, we showed that integer powers of the matrix A for i,j=1,2,⋯,n, respectively,
(47)Ami,j=2naci−j/2∑k=1nλkmsin2k−12i−1π4nsin2k−12j−1π4n,where λk=b+2accos2k−1π/2n, k=1,2,⋯,n, and
(48)Ami,j=2naci−j/2∑k=1nλkmcos2k−12i−1π4ncos2k−12j−1π4n,where λk=b+2accos2k−1π/2n, k=1,2,⋯,n.