# Powers of Complex Persymmetric Antitridiagonal Matrices with Constant Antidiagonals

Authors: Haibo Wang
Journal: ISRN Computational Mathematics
 (2014)

---

## Abstract

We derive a general expression for thepth power (
p
∈
N
) of any complex persymmetric antitridiagonal Hankel (constant antidiagonals) matrices. Numerical examples are presented, which show that our results generalize the results in the related literature (Rimas 2008, Wu 2010, and Rimas 2009).


---

## Body

## 1. Introduction

Solving some difference, differential, and delay differential equations, we meet the necessity to compute the arbitrary positive integer powers of square matrix. Recently, computing the integer powers of antitridiagonal matrices has been a very popular problem. There have been several papers on computing the positive integer powers of various kinds of square matrices by Rimas et al., and others [1–5]. In 2011, the general expression for the entries of the power of complex persymmetric or skew-persymmetric antitridiagonal matrices with constant antidiagonals is presented by Gutiérrez-Gutiérrez [1]. Rimas [2] gave the general expression of the pth power for this type of symmetric odd order antitridiagonal matrices (antitridia
g
n
(
1,0
,
1
)) in 2008. In [3, 4] a similar problem is solved for antitridiagonal matrices having zeros in main skew diagonal and units in the neighbouring diagonals. In 2010, the general expression for the entries of the power of odd order antitridiagonal matrices with zeros in main skew diagonal and elements 1
,
1
,
…
,
1; -
1
,
-
1
,
…
,
-
1 in neighbouring diagonals is derived by Rimas [5]. In 2013, Rimas [6] gave the eigenvalue decomposition for real odd order skew-persymmetric antitridiagonal matrices with constant antidiagonals (antitridia
g
n
(
a
,
b
,
a
)) and derived the general expression for integer powers of such matrices.In the present paper, we derive a general expression for thepth power (
p
∈
N
) of any complex persymmetric antitridiagonal matrices with constant antidiagonals (antitridia
g
n
(
b
,
a
,
b
)). This novel expression is both an extension of the one obtained by Rimas for the powers of the matrix antitridia
g
n
(
1,0
,
1
) with n
∈
N (see [2] for the odd case and [5] for the even case) and an extension of the one obtained by Honglin Wu for the powers of the matrix antitridia
g
n
(
1,1
,
1
) with n
∈
N (see [3] for the even case).


## 2. Derivation of General Expression

In this present paper, we study the entries of positive integer power of ann
×
n complex persymmetric antitridiagonal matrix with constant antidiagonals as follows:
(1)
B
n
=
antitridia
g
n
(
b
,
a
,
b
)
=
(
b
a
b
a
b
b
a
b
⋰
⋰
⋰
b
a
b
a
b
)
,
where a
∈
C,  b
∈
C
∖
(
0
).Consider the followingn
×
n complex Toeplitz tridiagonal matrix:
(2)
A
n
=
tridia
g
n
(
b
,
a
,
b
)
=
(
a
b
b
a
b
b
a
b
⋱
⋱
⋱
b
a
b
b
a
)
.The next trivial result relates the matrixB
n with A
n and with the n
×
n backward identity [1]:
(3)
J
n
=
(
δ
j
,
n
-
k
+
1
)
j
,
k
=
1
n
=
(
1
1
1
⋰
1
1
)
,
where δ is the Kronecker delta.Lemma 1.
Leta
∈
C,  b
∈
C
∖
(
0
), and n
∈
N. Then
(4)
B
n
=
J
n
A
n
,
where B
n
=
antitridia
g
n
(
b
,
a
,
b
) and A
n
=
tridia
g
n
(
b
,
a
,
b
).Proof.
We have(5)
[
J
n
A
n
]
j
,
k
=
∑
h
=
1
n
‍
[
J
n
]
j
,
h
[
A
n
]
h
,
k
=
[
A
n
]
n
+
1
-
j
,
k
=
{
b
,
if
n
+
1
-
(
j
+
k
)
=
-
1
,
a
,
if
n
+
1
-
(
j
+
k
)
=
0
,
b
,
if
n
+
1
-
(
j
+
k
)
=
1
,
0
,
other
.
=
[
B
n
]
j
,
k
.
This completes the proof.We will find theqth power (q
∈
N) of the matrix (1). Theorem 2 relates all positive integer powers of the matrix B
n to A
n and J
n.Theorem 2.
Ifa
∈
C,  b
∈
C
∖
(
0
), and n
∈
N and if B
n
=
antitridia
g
n
(
b
,
a
,
b
), then
(6)
B
n
q
=
{
A
n
q
,
i
f
q
i
s
e
v
e
n
,
J
n
A
n
q
,
i
f
q
i
s
o
d
d
,
where A
n
=
tridia
g
n
(
b
,
a
,
b
).Proof.
We will proceed by induction onq. The case q
=
1 is obvious.
Suppose that the result is true forq
≥
1 and consider case q
+
1.
By the induction hypothesis we have(7)
B
n
q
+
1
=
B
n
B
n
q
=
{
B
n
J
n
A
n
q
,
if
q
+
1
is
even
,
B
n
A
n
q
,
if
q
+
1
is
odd
.
Since B
n
=
J
n
A
n we obtain that
(8)
B
n
q
+
1
=
{
J
n
A
n
J
n
A
n
q
,
if
q
+
1
is
even
,
J
n
A
n
(
A
n
q
)
,
if
q
+
1
is
odd
.
Since A
n is symmetric and J
n
-
1
=
J
n, we have
(9)
B
n
q
=
{
A
n
q
,
if
q
is
even
,
J
n
A
n
q
,
if
q
is
odd
.
This completes the proof.Next, we have to solveA
n
q.We begin this work by reviewing a theorem regarding the Hermitian Toeplitz tridiagonal matrixA
n.Theorem 3.
Leta
∈
C,  b
∈
C
∖
(
0
), and n
∈
N. Then A
n has eigenvalues
(10)
λ
i
=
a
+
2
|
b
|
cos
⁡
⁡
i
π
n
+
1
,
1
≤
j
≤
n
.Proof.
See [7].With the tridiagonal matrixA
n
=
tridia
g
n
(
b
,
a
,
b
), we associate the polynomial sequence {
P
i
}
-
1
≤
i
≤
n characterized by a three-term recurrence relation:
(11)
x
P
i
(
x
)
=
b
P
i
+
1
(
x
)
+
a
P
i
(
x
)
+
b
P
i
-
1
(
x
)
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
i
=
0,1
,
…
,
n
-
1
.With initial conditionsP
-
1
(
x
)
=
0 and P
0
(
x
)
=
1, we can write the relations (11) in matrix form:
(12)
x
Q
n
-
1
(
x
)
=
A
n
Q
n
-
1
(
x
)
+
P
n
(
x
)
E
n
,
where Q
n
-
1
(
x
)
=
[
P
0
(
x
)
,
P
1
(
x
)
,
…
,
P
n
-
1
(
x
)
]
T and E
n
=
[
0,0
,
…
,
0,1
]
T
∈
R
n.Lemma 4.
Fori
≥
0, the degree of the polynomial P
i is i and P
i and P
i
+
1 has no common root.Proof.
See [7].One can show that the characteristic polynomial ofA
n is precisely (
Π
k
=
1
n
c
k
)
P
n
(
x
). Hence the eigenvalues of A
n are exactly the roots of P
n.If(
λ
j
)
0
≤
j
≤
n
-
1 are the roots of the polynomial P
n, then it follows from (12) that each λ
j is an eigenvalue of the matrix A
n and Q
n
-
1
(
x
)
=
[
P
0
(
x
)
,
P
1
(
x
)
,
…
,
P
n
-
1
(
x
)
]
T is a corresponding eigenvector [5, 7, 8]. This observation should be taken into account elsewhere in the paper.The polynomials{
P
i
}
0
≤
i
≤
n verify the well-known Christoffel-Darboux Identity.Lemma 5.
We have(13)
∑
i
=
0
n
-
1
‍
P
i
(
x
)
P
i
(
y
)
=
P
n
(
y
)
P
n
-
1
(
x
)
-
P
n
(
x
)
P
n
-
1
(
y
)
y
-
x
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
f
o
r
x
≠
y
.Proof.
See [7].Tendingy to x in formula (13), we get
(14)
∑
i
=
0
n
-
1
‍
P
i
(
x
)
P
i
(
x
)
=
P
n
′
(
x
)
P
n
-
1
(
x
)
-
P
n
(
x
)
P
n
-
1
′
(
x
)
.Since the matrixA
n has distinct eigenvalues λ
0
,
λ
1
,
…
,
λ
n
-
1, thus, the eigendecomposition of the matrix A
n is
(15)
A
n
=
T
D
T
-
1
,
where D
=
diag
⁡
(
λ
0
,
λ
1
,
…
,
λ
n
-
1
) and T is the transforming matrix formed by the eigenvectors of A
n. Namely, T
=
(
t
i
,
j
=
P
i
-
1
(
λ
j
-
1
)
)
1
≤
i
,
j
≤
n, where {
P
i
}
-
1
≤
i
≤
n are defined as above.Lemma 6.
IfT
-
1
=
(
s
i
,
j
)
1
≤
i
,
j
≤
n, then
(16)
s
i
,
j
=
P
j
-
1
(
λ
i
-
1
)
P
n
′
(
λ
i
-
1
)
P
n
-
1
(
λ
i
-
1
)
.Proof.
By using the relations (13) and (14), we obtain
(17)
∑
k
=
1
n
‍
s
i
,
k
t
k
,
j
=
∑
k
=
1
n
‍
P
k
-
1
(
λ
i
-
1
)
P
k
-
1
(
λ
j
-
1
)
P
n
′
(
λ
i
-
1
)
P
n
-
1
(
λ
i
-
1
)
=
δ
i
,
j
,
where δ
i
,
j
=
1 if i
=
j and δ
i
,
j
=
0 if i
≠
j.
This completes the proof.Forq
∈
N, we have A
n
q
=
T
D
q
T
-
1.We get immediately the following.Theorem 7.
Assume thatq
∈
N and A
n
q
=
(
α
i
,
j
(
q
)
)
1
≤
i
,
j
≤
n. Then
(18)
α
i
,
j
(
q
)
=
∑
k
=
1
n
‍
λ
k
-
1
q
P
i
-
1
(
λ
k
-
1
)
P
j
-
1
(
λ
k
-
1
)
P
n
′
(
λ
k
-
1
)
P
n
-
1
(
λ
k
-
1
)
.
By using the Cauchy Integral Formula, we can give another expressions of the coefficients α
i
,
j
(
q
) as follows:
(19)
α
i
,
j
(
q
)
=
∫
C
‍
z
q
P
i
-
1
(
z
)
P
j
-
1
(
z
)
P
n
(
z
)
P
n
-
1
(
z
)
d
z
,
where C is a closed curve containing the roots of P
n and no roots of P
n
-
1.Corollary 8.
If the matrixA
n is nonsingular with A
-
1
=
(
α
i
,
j
(
-
1
)
)
1
≤
i
,
j
≤
n, then
(20)
A
n
-
1
=
T
D
-
1
T
-
1
,
α
i
,
j
(
-
1
)
=
∑
k
=
1
n
‍
1
λ
k
-
1
P
i
-
1
(
λ
k
-
1
)
P
j
-
1
(
λ
k
-
1
)
P
n
′
(
λ
k
-
1
)
P
n
-
1
(
λ
k
-
1
)
.
By using the Cauchy Integral Formula, we can give other expressions of the coefficients α
i
,
j
(
-
1
):
(21)
α
i
,
j
(
-
1
)
=
∫
C
‍
z
-
1
P
i
-
1
(
z
)
P
j
-
1
(
z
)
P
n
(
z
)
P
n
-
1
(
z
)
d
z
,
where C is a closed curve containing the roots of P
n and no roots of P
n
-
1.Theorem 9.
Assume thatq
∈
N and B
n
q
=
(
β
i
,
j
(
q
)
)
1
≤
i
,
j
≤
n. Then
(22)
β
i
,
j
(
q
)
=
{
∑
k
=
1
n
‍
λ
k
-
1
q
P
i
-
1
(
λ
k
-
1
)
P
j
-
1
(
λ
k
-
1
)
P
n
′
(
λ
k
-
1
)
P
n
-
1
(
λ
k
-
1
)
,
i
f
q
i
s
e
v
e
n
,
∑
k
=
1
n
‍
λ
k
-
1
q
P
n
-
i
(
λ
k
-
1
)
P
j
-
1
(
λ
k
-
1
)
P
n
′
(
λ
k
-
1
)
P
n
-
1
(
λ
k
-
1
)
,
i
f
q
i
s
o
d
d
.
By using the Cauchy Integral Formula, we can give other expressions of the coefficients β
i
,
j
(
q
):
(23)
β
i
,
j
(
q
)
=
{
∫
C
‍
z
q
P
i
-
1
(
z
)
P
j
-
1
(
z
)
P
n
(
z
)
P
n
-
1
(
z
)
d
z
,
i
f
q
i
s
e
v
e
n
,
∫
C
‍
z
q
P
n
-
i
(
z
)
P
j
-
1
(
z
)
P
n
(
z
)
P
n
-
1
(
z
)
d
z
,
i
f
q
i
s
o
d
d
,
where C is a closed curve containing the roots of P
n and no roots of P
n
-
1.Proof.
From Theorem2 we get
(24)
[
B
n
q
]
i
,
j
=
{
[
A
n
q
]
i
,
j
,
if
q
is
even
,
[
J
n
A
n
q
]
i
,
j
=
∑
n
=
1
n
‍
[
J
n
]
i
,
h
[
A
n
q
]
h
,
j
,
if
q
is
odd
.
=
{
[
A
n
q
]
i
,
j
,
if
q
is
even
,
[
A
n
q
]
n
+
1
-
i
,
j
,
if
q
is
odd
.
Namely,
(25)
β
i
,
j
(
q
)
=
{
α
i
,
j
(
q
)
,
if
q
is
even
,
α
n
+
1
-
i
,
j
(
q
)
,
if
q
is
odd
.
From Theorem 7 it follows that
(26)
β
i
,
j
(
q
)
=
{
∑
k
=
1
n
‍
λ
k
-
1
q
P
i
-
1
(
λ
k
-
1
)
P
j
-
1
(
λ
k
-
1
)
P
n
′
(
λ
k
-
1
)
P
n
-
1
(
λ
k
-
1
)
,
if
q
is
even
,
∑
k
=
1
n
‍
λ
k
-
1
q
P
n
-
i
(
λ
k
-
1
)
P
j
-
1
(
λ
k
-
1
)
P
n
′
(
λ
k
-
1
)
P
n
-
1
(
λ
k
-
1
)
,
if
q
is
odd
.
By using the Cauchy Integral Formula, we can give other expressions of the coefficients β
i
,
j
(
q
):
(27)
β
i
,
j
(
q
)
=
{
∫
C
‍
z
q
P
i
-
1
(
z
)
P
j
-
1
(
z
)
P
n
(
z
)
P
n
-
1
(
z
)
d
z
,
if
q
is
even
,
∫
C
‍
z
q
P
n
-
i
(
z
)
P
j
-
1
(
z
)
P
n
(
z
)
P
n
-
1
(
z
)
d
z
,
if
q
is
odd
.
This completes the proof.Corollary 10.
Assume thatq
∈
N and B
n
-
1
=
(
β
i
,
j
(
-
1
)
)
1
≤
i
,
j
≤
n. Then
(28)
β
i
,
j
(
-
1
)
=
∑
k
=
1
n
‍
1
λ
k
-
1
P
i
-
1
(
λ
k
-
1
)
P
j
-
1
(
λ
k
-
1
)
P
n
′
(
λ
k
-
1
)
P
n
-
1
(
λ
k
-
1
)
.
By using the Cauchy Integral Formula, we can give other expressions of the coefficients β
i
,
j
(
-
1
):
(29)
β
i
,
j
(
-
1
)
=
∫
C
‍
z
(
-
1
)
P
i
-
1
(
z
)
P
j
-
1
(
z
)
P
n
(
z
)
P
n
-
1
(
z
)
d
z
,
where C is a closed curve containing the roots of P
n and no roots of P
n
-
1.


## 3. Numerical Examples

Consider then order antitridiagonal matrix B
n of the following type:
(30)
B
n
=
(
1
a
1
a
1
1
a
1
⋰
⋰
⋰
1
a
1
a
1
)
.
Assume that
(31)
A
n
=
(
a
1
1
a
1
1
a
1
⋱
⋱
⋱
1
a
1
1
a
)
,
J
n
=
(
1
1
⋰
1
1
)
,
where A
n and J
n are n
×
n matrix. The polynomial sequence {
P
i
}
-
1
≤
i
≤
n verifies
(32)
x
P
i
(
x
)
=
P
i
+
1
(
x
)
+
a
P
i
(
x
)
+
P
i
-
1
(
x
)
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
i
=
0,1
,
…
,
n
-
1
,
with initial conditions P
-
1
(
x
)
=
0 and P
0
(
x
)
=
1.By simple calculation we can show that(33)
P
i
(
x
)
=
U
i
(
x
-
a
2
)
,
i
=
0
,
…
,
n
,
where U
i are the Chebyshev polynomials [8] of the second kind which satisfies the three-term recurrence relations:
(34)
2
x
U
i
(
x
)
=
U
i
+
1
(
x
)
+
U
i
-
1
(
x
)
,
with initial conditions U
0
(
x
)
=
1 and U
1
(
x
)
=
2
x.EachU
n satisfies
(35)
U
n
(
x
)
=
sin
⁡
(
(
n
+
1
)
arccos
⁡
x
)
sin
(
arccos
⁡
x
)
,
and thus the roots of U
n
(
x
) are z
k
=
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
), k
=
1
,
…
,
n. Then, the eigenvalues of A are
(36)
λ
k
=
a
+
2
cos
⁡
⁡
k
π
n
+
1
,
k
=
1
,
…
,
n
.We get by Theorem7 the following.Assume thatq
∈
N and A
n
q
=
(
α
i
,
j
(
q
)
)
1
≤
i
,
j
≤
n. Then
(37)
α
i
,
j
(
q
)
=
2
∑
k
=
1
n
‍
(
a
+
2
cos
⁡
⁡
k
π
n
+
1
)
q
h
h
h
h
h
h
×
U
i
-
1
(
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
)
U
j
-
1
(
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
)
U
n
′
(
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
)
U
n
-
1
(
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
)
.If the matrixA
n is nonsingular and A
-
1
=
(
α
i
,
j
(
-
1
)
)
1
≤
i
,
j
≤
n, then
(38)
a
i
,
j
(
-
1
)
=
2
∑
k
=
1
n
‍
1
a
+
2
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
h
h
h
h
h
h
×
U
i
-
1
(
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
)
U
j
-
1
(
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
)
U
n
′
(
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
)
U
n
-
1
(
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
)
.We can obtain the following:(39)
β
i
,
j
(
q
)
=
{
2
∑
k
=
1
n
‍
(
a
+
2
cos
⁡
⁡
k
π
n
+
1
)
q
×
U
i
-
1
(
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
)
U
j
-
1
(
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
)
U
n
′
(
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
)
U
n
-
1
(
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
)
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
if
q
is
even
,
2
∑
k
=
1
n
‍
(
a
+
2
cos
⁡
⁡
k
π
n
+
1
)
q
×
U
n
-
i
(
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
)
U
j
-
1
(
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
)
U
n
′
(
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
)
U
n
-
1
(
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
)
,
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
h
if
q
is
odd
,
β
i
,
j
(
-
1
)
=
2
∑
k
=
1
n
‍
(
a
+
2
cos
⁡
⁡
k
π
n
+
1
)
-
1
h
h
h
h
h
h
×
U
i
-
1
(
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
)
U
j
-
1
(
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
)
U
n
′
(
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
)
U
n
-
1
(
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
)
)
.Theorem 11.
Consider an odd natural numbern
=
2
m
+
1, m
∈
N. Let B
n
=
antitridia
g
n
(
1,0
,
1
) and λ
k
=
-
2
cos
⁡
(
k
π
/
(
n
+
1
)
) for every 1
≤
k
≤
n. Then
(40)
[
B
n
q
]
i
,
j
=
1
+
(
-
1
)
q
+
i
+
j
2
n
+
2
×
∑
k
=
1
(
n
-
1
)
/
2
‍
β
i
,
j
(
k
)
λ
k
q
(
4
-
λ
k
2
)
U
n
-
i
(
λ
k
2
)
U
n
-
j
(
λ
k
2
)
,
for all q
∈
N and 1
≤
i, j
≤
n, where
(41)
β
i
,
j
(
k
)
=
{
1
,
i
f
i
+
j
i
s
e
v
e
n
,
(
-
1
)
k
-
1
,
i
f
i
+
j
i
s
o
d
d
.
λ
k
(
k
=
1,2
,
…
,
n
) are the eigenvalues of the matrix B
n and U
k
(
x
) is the kth degree Chebyshev polynomial of the second kind.Theorem 12.
Consider an even natural numbern
=
2
m, m
∈
N. Let B
n
=
antitridia
g
n
(
1,0
,
1
) and λ
k
=
-
2
cos
⁡
⁡
(
k
π
/
(
n
+
1
)
) for every 1
≤
k
≤
n. Then
(42)
[
B
n
q
]
i
,
j
=
1
n
+
1
γ
i
,
j
∑
k
=
1
n
/
2
‍
λ
2
k
q
(
4
-
λ
2
k
2
)
U
n
-
i
(
λ
2
k
2
)
U
n
-
j
(
λ
2
k
2
)
for all q
∈
N and 1
≤
i, j
≤
n, where
(43)
γ
i
,
j
=
{
1
,
i
f
i
+
j
i
s
e
v
e
n
,
0
,
i
f
i
+
j
i
s
o
d
d
.
For even order matrix B the following condition is fulfilled: λ
k
≠
0
(
k
=
2,4
,
…
,
n
). This means that even order matrix B
n
=
antitridia
g
n
(
1,0
,
1
) is nonsingular (its determinant is not equal to zero) and derived expression of B
n
q can be applied for computing negative integer powers, as well. Taking q
=
-
1, we get the following expression for elements of the inverse matrix B
n
-
1:
(44)
[
B
n
-
1
]
i
,
j
=
1
n
+
1
γ
i
,
j
∑
k
=
1
n
/
2
‍
λ
2
k
-
1
(
4
-
λ
2
k
2
)
×
U
n
-
i
(
λ
2
k
2
)
U
n
-
j
(
λ
2
k
2
)
,
i
,
j
=
0,1
,
…
,
n
.Theorem 13.
Consider an even natural numbern
=
2
m, m
∈
N. Let B
n
=
antitridia
g
n
(
1,1
,
1
) and λ
k
=
(
-
1
)
k
-
1
(
1
+
2
cos
⁡
(
k
π
/
(
n
+
1
)
)
), (
k
=
1,2
,
…
,
n
). Then
(45)
[
B
n
q
]
i
,
j
=
2
n
+
1
∑
k
=
1
n
‍
λ
k
q
sin
⁡
k
i
π
n
+
1
sin
⁡
k
j
π
n
+
1
=
2
n
+
1
∑
k
=
1
n
‍
[
λ
k
q
+
(
-
1
)
i
+
j
(
λ
k
+
2
(
-
1
)
k
)
q
]
h
h
h
h
h
h
h
h
×
sin
⁡
k
i
π
n
+
1
sin
⁡
k
j
π
n
+
1
.


## 4. Conclusion and Discussion

In this paper, we derive a general expression for thepth power (
p
∈
N
) of any complex persymmetric antitridiagonal Hankel (constant antidiagonals) matrices with constant antidiagonals (antitridia
g
n
(
b
,
a
,
b
)). This novel expression is both an extension of the one obtained by Rimas for the powers of the matrix antitridia
g
n
(
1,0
,
1
) with n
∈
N (see [2] for the odd case and [5] for the even case) and an extension of the one obtained by Honglin Wu for the powers of the matrix antitridia
g
n
(
1,1
,
1
) with n
∈
N (see [3] for the even case). We may safely draw the conclusion that our results generalize the results in the related literature [2, 3, 5].



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*Source: 451270-2014-03-16.xml*