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Filomat 26:4 (2012), 809–826
DOI 10.2298/FIL1204809S
Published by Faculty of Sciences and Mathematics,
University of Niš, Serbia
Available at: http://www.pmf.ni.ac.rs/filomat
An iterative solution to coupled quaternion matrix equations
Caiqin Songa , Guoliang Chena , Xiangyun Zhanga
a Department
of Mathematics in East China Normal University, Shanghai, P.R. China
∑p
T (Xi ),
i=1 1i
T2i (Xi ), · · · , i=1 Tpi (Xi )] = [M1 , M2 , · · · , Mp ], where Tsi , s = 1, 2, · · · , p; is a linear operator from Qmi ×ni
onto Qps ×qs , Ms ∈ Qps ×qs , s = 1, 2, · · · , p. i = 1, 2, · · · , p. by making use of a generalization of the classical
complex conjugate graduate iterative algorithm. Based on the proposed iterative algorithm, the existence
conditions of solution to the above coupled quaternion matrix equations can be determined. When the
considered coupled quaternion matrix equations is consistent, it is proven by using a real inner product
in quaternion space as a tool that a solution can be obtained within finite iterative steps for any initial
quaternion matrices [X1 (0), · · · , Xp (0)] in the absence of round-off errors and the least Frobenius norm
solution can be derived by choosing a special kind of initial quaternion matrices. Furthermore, the optimal
approximation solution to a given quaternion matrix can be derived. Finally, a numerical example is given
to show the efficiency of the presented iterative method.
Abstract.
This note
∑p
∑p studies the iterative solution to the coupled quaternion matrix equations [
i=1
1. Introduction
Throughout this paper, we need the following notations. We denote
the real
field by R, the
⊕ ⊕
⊕ number
complex number field by C and the quaternion field by Q = R
Ri
Rj
Rk, i2 = j2 = k2 = −1,
i j = −ji = k, jk = −k j = i, ki = −ik = j, respectively. For a matrix A ∈ Qm×n , we denote its transpose,
conjugate transpose, trace and Frobenius norm by AT , AH , tr(A) and ∥A∥, respectively. The symbols R(A)
and vec(·) stand for null space and the vec operator, i.e., for A = (a1 , a2 , · · · , an ) ∈ Cm×n , where ai (i = 1, 2 · · · , n)
denotes the ith column of A, vec(A) = (aT1 , aT2 , · · · , aTn )T .
Let LQm×n,p×q denote the set of quaternion linear operators from Qm×n onto Qp×q . Particularly, when
p = m and q = n, LQm×n,p×q is denoted by LQm×n . Let I stand for the identity operator on Qm×n . In the
vector space Qm×n , we define real inner product as: ⟨A, B⟩ = Re[tr(AH B)], for all A, B ∈ Qm×n . Also we have
⟨A, B⟩ = ⟨A, B⟩ and ⟨A, BC⟩ = ⟨BH A, C⟩ = ⟨ACH , B⟩. The adjoint of T ∈ Qm×n is denoted by T∗ . So for all
X, Y ∈ Qm×n , ⟨T(X), Y⟩ = ⟨X, T∗ (Y)⟩. T is called self-adjoint if T∗ = T. Two quaternion matrices X and Y are
said to be orthogonal if ⟨X, Y⟩ = 0.
In the 20th century, quaternion, discovered by Hamilton in 1843, made further appearance in associative
algebra, analysis, topology, and physics. Moreover, quaternion matrices play an important role in computer
science, quantum physics, signal and color image processing, and so on (e.g. [2, 3, 13, 18, 21, 25, 26, 42]). Some
2010 Mathematics Subject Classification. Primary 15A06; Secondary 15A24, 47S10
Keywords. Coupled quaternion matrix equations, quaternion linear operator, real inner product, least Frobenius norm solution,
optimal approximation solution
Received: 28 December 2011; Accepted: 22 April 2012
Communicated by Y. Wei
This project is granted financial support from NSFC (No.11071079, No.10901056), Zhejiang Nature Science Foundation
(No.Y6110043) and the Fundamental Research Funds for the Central Universities
Email address: [email protected] (Guoliang Chen)
C. Song et al. / Filomat 26:4 (2012), 809–826
810
important conclusions which hold on the complex field or real field have been generalized to quaternion
field, such as the Schur’s theorem [30], Cayley-Hamilton’s theorem [6], the Wilandt-Hoffman’s theorem
[37] and Gerschgorin’s theorem [38]. Notice that quaternion algorithm has, up to now, only been proposed
for QR decomposition [14], Jacobi algorithm [4] and the Singular Value Decomposition (SVD) [3]. So in
this paper, we’ll generalize the conjugate gradient iterative algorithm of complex matrix on the quaternion
field.
For the quaternion matrix equation, there are many important results. Jiang and Wei [17] investigate
e = C by using real representation of a quaternion
the solution of the quaternion matrix equation X − AXB
matrix. By making use of complex representation of a quaternion matrix, Huang et al. [16] give the solution
of quaternion matrix equation AXB − CXD = E. Over the real quaternion algebra, Wang [31, 32] considered
bisymmetric and centro-symmetric solution to certain matrix equations and gave the general solution to the
system of quaternion matrix equations A1 X = C1 , A2 X = C2 , A3 X = C3 and A4 X = C4 . In addition, there are
the following matrix equation results over complex field. Iterative methods in [24, 27, 33] were constructed
to obtain the (skew)symmetric solutions of the matrix equations A1 XB1 = C1 , A2 XB2 = C2 , AXB + CYD = E
and AXB + CXT D = E. In [34], the authors presented the necessary and sufficient conditions for the
existence of constant solutions with bi(skew)symmetric constraint to the matrix equations Ai X − YBi = Ci ,
i = 1, 2, · · · , s and Ai XBi − Ci YDi = Ei , i = 1, 2, · · · , s.
Coupled matrix equations have wide applications in several areas, such as stability theory, control theory,
perturbation analysis, and some others fields of pure and applied mathematics. For example, in stability
analysis of linear jump systems with Markovian transitions, the coupled Lyapunov matrix equations
ATi Pi + Pi Ai + Qi +
N
∑
pi j P j = 0, Qi > 0, i ∈ I[1, N]
(1)
j=1
and
Pi = ATi (
N
∑
pij P j ) + Qi , Qi > 0, i ∈ I[1, N]
(2)
j=1
are often encountered [5, 20], where Pi , i ∈ I[1, N] are the matrices to be determined. Due to their wide
applications, coupled matrix equations have attracted considerable attention from many researchers. It was
pointed out that in [7] that the existence of a positive definite solution to the discrete-time Markovian jump
Lyapunov (DMJL) equation (2) is related to the spectral radius of an augmented matrix being less than one.
In addition, the following general coupled Sylvester matrix equations have been investigated
Ai1 X1 Bi1 + Ai2 X2 Bi2 + · · · + Aip Xp Bip = Ei , i ∈ [1, N],
(3)
where Aij ∈ Rri ×n j , Bij ∈ Rmi ×c j , Ei ∈ Rri ×C j , i ∈ [1, N], j ∈ [1, p] are known matrices, and X j ∈ Rn j ×m j , j ∈ [1, p]
are the matrices to be determined. This kind of matrix equations include all the aforementioned matrix
equations as special cases. When p = N, by using the hierarchical identification principle, iterative algorithm
were proposed in [8] for obtaining the unique least-square solution by introducing the block-matrix inner
product. Recently, from an optimization point of view gradient based iterations were constructed in [9]
to solve the general coupled matrix equations (3). The gradient based iterative algorithm [10–12, 41]
and least squares based iterative algorithm [41] for solving (coupled) matrix equations are a novel and
computationally efficient numerical algorithms and were presented based on the hierarchical identification
principle which regards the unknown matrix as the system parameter matrix to be identified. Meanwhile,
Wu et al. [39] have considered the following so-called coupled Sylvester-conjugate matrix by means of the
hierarchical identification principle
p
∑
η=1
(Aiη Xη Biη + Ciη Xη Diη ) = Fi , i ∈ I[1, N],
(4)
C. Song et al. / Filomat 26:4 (2012), 809–826
811
where Aiη , Ciη ∈ Cmi ×rη , Biη , Diη ∈ Csη ×ni , Fi ∈ Cmi ×ni , i ∈ I[1, N], η ∈ I[1, p] are the given known matrices, and
Xη ∈ Crη ×sη , η ∈ I[1, p] are the matrices to be determined.At the same time, Wu et al. [40] proposed a finite
iterative method for the so-called Sylvester-conjugate matrix equation (4). In [19, 41], the following linear
equations
r
s
∑
∑
Ai XBi +
C j XT D j = E,
(5)
i=1
j=1
where Ai , Bi , C j , D j , i = 1, · · · , r; j = 1, · · · , s and E are some known constant matrices of appropriate
dimensions and X is a matrix to be determined, was considered. In [35], the special case of equation (5)
AXB + CXT D = E was considered by the iterative algorithm. A more special case of (5), namely, the matrix
equation AX + XT B = C, was investigated by Piao et al. [22]. The Moore-Penrose generalized inverse was
used in [22] to find explicit solutions to this matrix equation.
However, it is not considered how to computer the iterative solution of the above matrix equation over
quaternion field. So in this paper, we extend the conjugate graduate iterative algorithm from complex field
to the quaternion field and give the iterative solution of the following quaternion matrix equations based
on the complex representation of a quaternion matrix. It should be remarked that all the aforementioned
matrix equations can all be rewritten as the following quaternion matrix equations system:
 p

p
p
∑
∑
∑


 = [M1 , M2 , · · · , Mp ],
T
(X
),
T
(X
),
·
·
·
,
T
(X
)
1i
i
2i
i
pi
i


i=1
i=1
i=1
where Tsi ∈ LQmi ×ni ,ps ×qs and Ms ∈ Qps ×qs , i = 1, 2, · · · , p; s = 1, 2, · · · , p. In the present paper, we mainly
consider the following two problems:
Problem I. For given Tsi ∈ LQmi ×ni ,ps ×qs , Ms ∈ Qps ×qs , find Xi , i = 1, 2, · · · , p such that
 p

p
p
∑
∑
∑


T1i (Xi ),
T2i (Xi ), · · · ,
Tpi (Xi ) = [M1 , M2 , · · · , Mp ].
(6)

i=1
i=1
i=1
bi ∈ Qmi ×ni , find Xi ∈ S, i = 1, 2, · · · , p, such
Problem II. Let S denote the solution set of Problem I, for given X
that
 p

p


∑
∑



2
2
bi ∥ = min 
bi ∥ 
∥Xi − X
∥Xi − X
.
(7)




(X1 ,··· ,Xp )∈S 
i=1
i=1
The rest of the paper is organized as follows. In Section 2, we give some preliminaries. In Section 3, we
propose an iterative algorithm to obtain a solution or the least Frobenius norm solution of Problem I and
present some basic properties of the algorithm. Some numerical examples are given in Section 4 to show
the efficiency of the proposed iterative method. Finally, we put some conclusions in Section 5.
2. Preliminary
In this section, some concepts on quaternion matrix are given. A new real inner product is firstly defined
for quaternion matrix space over the real field R. This inner product will play a very important role in this
paper.
2.1. Complex representation of a quaternion matrix
Some well-known equalities for complex and real matrices hold for quaternion matrices , whereas some
others are no more valid. Here is a short list of relations that hold for quaternion matrices A ∈ QN×M and
B ∈ QM×P :
(1) (AB)H = BH AH ;
(2) AB , A B in general;
(3) (AH )−1 = (A−1 )H ;
C. Song et al. / Filomat 26:4 (2012), 809–826
812
(4) (A)−1 , A−1 in general.
See [42] for a more complete list. The most common way to study quaternion matrix is to use its complex
representation of a quaternion matrix introduced in [42]. For any quaternion matrix A = B1 + B2 j ∈ Qm×n ,
the author [42] defined a complex representation
[
]
B1 B2
Aσ =
∈ C2m×2n .
(8)
−B2 B1
The complex matrix Aσ was called complex representation of a quaternion matrix A. The complex
representation of a quaternion matrix has the following results, which can be found in [1].
Proposition 2.1. (1) If A, B ∈ Qm×n , a ∈ R, then (A + B)σ = Aσ + Bσ , (aA)σ = aAσ ;
(2) If A ∈ Qm×n , B ∈ Qn×s , then (AB)σ = Aσ Bσ , (AH )σ = (Aσ )H ;
(3) If A ∈ Qm×m , then A is nonsingular if and only if Aσ is nonsingular, and (Aσ )−1 = (A−1 )σ ;
(4) If A ∈ Qm×m , then A is unitary if[ and only ]if Aσ is unitary;
0 It
(5) Aσ = Q−1
is a unitary matrix, It is a t × t identity matrix.
m Aσ Qn , in which Qt =
−It 0
From Proposition 2.1, we know that σ : Qm×n → σ(Qm×n ) is an isomorphism of vector space, and
σ : Qm×n → σ(Qm×n ) is an isomorphism of algebra.
2.2. Quaternion matrix norm
Definition 2.2. A function ν : A ∈ Qm×n → R is a quaternion matrix norm on Qm×n if it satisfies the following
conditions:
(1) definiteness, A , 0 =⇒ ν(A) > 0;
(2) homogeneity, ν(aA) = |a|ν(A);
(3) the triangle inequality, ν(A + B) ≤ ν(A) + ν(B),
where A, B ∈ Qm×n are arbitrary quaternion matrices, and a is an arbitrary quaternion.
For A = A1 + A2 i + A3 j + A4 k, At ∈ Rm×n , t = 1, · · · , 4, we define
√
1
ν1 (A) = ∥A1 ∥2F + ∥A2 ∥2F + ∥A3 ∥2F + ∥A4 ∥2F = √ ∥Aσ ∥F .
2
Obviously, ν1 satisfies (1) and (3).
Let α = α1 + α2 i + α3 j + α4 k ∈ Q, αl ∈ R, (l = 1, 2, 3, 4). Then by carefully computation, we obtain that
αA =
(α1 A1 − α2 A2 − α3 A3 − α4 A4 ) + (α1 A2 + α2 A1 + α3 A4 − α4 A3 )i + (α1 A3 − α2 A4 + α3 A1 + α4 A2 )j
+(α1 A4 + α2 A3 − α3 A2 + α4 A1 )k,
and ν21 (αA) = (α21 + α22 + α23 + α24 )(∥A1 ∥2F + ∥A2 ∥2F + ∥A3 ∥2F + ∥A4 ∥2F ) = |α|2 ν21 (A).
Similarly, ν1 (αA) = |α|ν1 (A), that is, ν1 also satisfy (2), so ν1 is a quaternion matrix norm, denoted by
∥.∥(F) , which has the following properties:
1
(1) ∥A∥(F) = √ ∥Aσ ∥F ;
2
(2) ∥AB∥(F) ≤ ∥A∥F ∥B∥F ;
(3) Unitary invariant norm;
√∑
σ2i (A),where σi (A)s are singular value of A;
(4) ∥A∥(F) =
√
(5) ∥A∥(F) = trace(AH A);
√∑
(6) ∥A∥(F) =
|ai j |2 .
Therefore, it is a natural generality of Frobenius norm of complex matrix.
C. Song et al. / Filomat 26:4 (2012), 809–826
813
2.3. Real inner product of Hilbert quaternion matrix space
Definition 2.3. A real inner product space is a vector space V over the real field R together with an inner
product defined by a real-valued function ⟨·⟩ : V × V → R, satisfying the following three axioms for all
vectors x, y ∈ V and all scalars a ∈ R:
(1) Symmetry: ⟨x, y⟩ = ⟨y, x⟩;
(2) Linearity in the first argument: ⟨ax, y⟩ = a⟨x, y⟩, ⟨x + y, z⟩ = ⟨x, z⟩ + ⟨y, z⟩;
(3) Positive-definiteness: ⟨x, x⟩ > 0 for all x , 0.
Two vectors u, v ∈ V are said to be orthogonal if ⟨u, v⟩ = 0.
The following theorem defines a real inner product on quaternion space Qm×n over the real field R.
Theorem 2.4. In the quaternion Hilbert space Qm×n over the real filed R, an inner product can be defined as
⟨A, B⟩ = Re[tr(AH B)],
(9)
for A, B ∈ Qm×n . This inner product space is denoted as (Qm×n , R, ⟨·, ·⟩).
Proof. (1) For A, B ∈ Qm×n , according to the properties of the trace, we have
⟨A, B⟩ = Re[tr(AH B)] =
1
1
Re[tr((Aσ )H Bσ )] = Re[tr((Bσ )H Aσ )] = Re[tr(BH A)] = ⟨B, A⟩.
2
2
(2) For a ∈ R, we have
⟨aA, B⟩ = Re[tr((aA)H B)] = Re[tr(aAH B)] = aRe[tr(AH B)] = a⟨A, B⟩,
⟨A + B, C⟩ = Re[tr((A + B)H C)] = Re[tr(AH C)] + Re[tr(BH C)] = ⟨A, C⟩ + ⟨B, C⟩.
(3) It is well known that tr(AH A) > 0 for all A , 0. Thus, ⟨A, A⟩ = Re(tr(AH A)) > 0 for all A , 0.
According to Definition 2.3, all the above arguments reveal that the space Qm×n over real field R with
the inner product defined in (9) is an inner product space. This completes the proof.
2.4. Relationship between quaternion matrix equations and its complex representation matrix equations
The complex representation matrix equations of quaternion matrix equations (6) can be stated as the
following matrix equations
 p

p
p
∑
∑
∑


 = [(M1 )σ , (M2 )σ , · · · , (Mp )σ ],
B
(X
)
,
B
(X
)
,
·
·
·
,
B
(X
)
(10)
1i
i
σ
2i
i
σ
pi
i
σ


i=1
i=1
2mi ×2ni ,2ps ×2qs
i=1
2ps ×2qs
where Bsi ∈ LC
and (Ms )σ ∈ C
, i = 1, 2, · · · , p; s = 1, 2, · · · , p.
Meanwhile, we define the following matrix equations
 p

p
p
∑
∑
∑


B1i (Yi ),
B2i (Yi ), · · · ,
Bpi (Yi ) = [(M1 )σ , (M2 )σ , · · · , (Mp )σ ],

i=1
i=1
(11)
i=1
where Bsi ∈ LC2mi ×2ni ,2ps ×2qs and (Ms )σ ∈ C2ps ×2qs , s = 1, 2, · · · , p.
So we have the following relationship between the solutions Yi to the matrix equations (11) and the solutions
Xi to the matrix equations (6). It can be stated as the following Theorem 2.5.
Theorem 2.5. The quaternion matrix equations (6) has a solution Xi ∈ Qmi ×ni if and only if the complex representation
matrix equations (11) has a solution Yi ∈ C2mi ×2ni , in which case, if Yi is a solution to (11), then the following matrices
are a solution to the quaternion matrix equations (6):
[
]
]
1[
Ip
−1
In −jIn (Yi + Qn Yi Qp )
Xi =
, i = 1, 2, · · · , p.
(12)
jIp
4
C. Song et al. / Filomat 26:4 (2012), 809–826
814
Proof. We only prove that if matrix
[
Yi =
Yi11
Yi21
Yi12
Yi22
]
, Yi ∈ C2mi ×2ni , i = 1, 2, · · · , p.
(13)
is a solution to the complex representation matrix equations (11), then Xi in (12) is a solution to the quaternion
matrix equations (6).
−1
By (5) of Proposition 2.1, we have Aσ = Qm Aσ Q−1
n , i.e. if Yi is a solution to (11), then Qn Yi Qp is also a
solution to (11). Thus the following matrix
Ŷi =
1
(Yi + Qn Yi Q−1
p ),
2
(14)
is also a solution to (11).
It is easy to get by direct calculation

(1)
 Ŷi
Ŷi = 
(2)
−Ŷi
(2)
Ŷi
(1)
Ŷi


 ,

(15)
where
22
21
1
ˆ(1) 1
Yi = (Yi11 + Yi ), Yˆ2 = (Yi12 − Yi ).
2
2
From (15) we construct a quaternion matrix
]
[
]
1[
Ip
(1)
(2)
I
−jI
Xi = Ŷi + Ŷi j =
.
Ŷi
n
n
jIp
2
(16)
(17)
Clearly (Xi )σ = Ŷi . So Xi is a solution to (6).
3. Main result
In this section, we first propose an iterative algorithm to solve Problem I, then present some basic
properties of the algorithm. We also consider finding the least Frobenius norm solution of Problem I. In the
sequel, the least norm solution always means the least Frobenius norm solution.
Algorithm 1.
Step 1. Input Tsi ∈ LQmi ×ni ,ps ×qs , Ms ∈ Qps ×qs , i, s = 1, 2, · · · , p., and arbitrary [X1 (0), · · · , Xp (0)] ∈ S, i =
1, 2, · · · , p; s = 1, 2, · · · , p;
Step 2. Compute
R(1)
=
dia1((M1 −
p
∑
i=1
Pi (1)
=
p
∑
T1i (Xi (1)), M2 −
p
∑
T2i (Xi (1)), . . . Mp −
i=1
p
∑
Tpi (Xi (1))) = dia1(R1 (1), R2 (1), . . . Rp (1)),
i=1
T∗si (Rs (1)), i = 1, 2, · · · , p; k := 0;
s=1
Step 3. If R(k) = 0 then stop and (X1 (k), X2 (k), · · · , Xp (k)) is the constraint solution group;
else if R(k) , 0 but Pm (k) = 0 for all m = 1, 2, · · · , p, then stop and the coupled quaternion matrix
equations (6) are not consistent;
p
∑
else k := k + 1; ∥Ri (k)∥2(F) , 0 but Pk = 0, then stop; else, k := k + 1;
i=1
Step 4. Compute
C. Song et al. / Filomat 26:4 (2012), 809–826
Xi (k) = Xi (k − 1) +
815
∥R(k − 1)∥(F) 2
Pi (k − 1);
p
∑
2
∥Pi (k − 1)∥(F)
t=1
R(k) =
=
=
Zi (k) =


p
p
p
∑
∑
∑



dia1 M1 −
T1i (Xi (k)), M2 −
T2i (Xi (k)), . . . , Mp −
Tpi (Xi (k))
i=1
i=1
i=1

 p
p
p
2
∑
∑
∑


∥R(k − 1)∥(F)

T1i (Pi (k − 1)),
T2i (Pi (k − 1)), . . . ,
Tpi (Pi (k − 1))
dia1 
R(k − 1) − p
∑
i=1
i=1
i=1
∥Pi (k − 1)∥2(F)
t=1
(
)
dia1 R1 (k), R2 (k), . . . Rp (k) ,
p
∑
T∗si (Rs (k)),
i=1
Pi (k) =
Zi (k) +
∥R(k)∥2(F)
∥R(k − 1)∥2(F)
p
∑
Pi (k − 1) =
T∗si (Rs (k))
+
s=1
∥R(k)∥2(F)
∥R(k − 1)∥2(F)
Pi (k − 1), i = 1, 2, · · · , p;
Step 5. Go to Step 3.
Some basic properties of Algorithm 1 are listed in the following lemmas.
Lemma 3.1. For the sequences {Ri (k)}, {Pi (k)} and {Zi (k)}, i = 1, 2, · · · , p generated by Algorithm 1, they follow that
p
∑
⟨Ri (m + 1), Ri (n)⟩ =
i=1
p
∑
p
∑
⟨Pi (m), Zi (n)⟩, m, n = 1, · · · , k; m , n.
∥R(m)∥2(F)
⟨Ri (m), Ri (n)⟩ −
p
∑
i=1
t=1
∥Pt (m)∥2(F)
i=1
Proof. By Algorithm 1, we have
p
∑
⟨Ri (m + 1), Ri (n)⟩
=
i=1
p ⟨
∑
Ri (m) −
i=1
∥R(m)∥2(F)
p
∑
t=1
=
∥Pt (m)∥2(F)
p
∑
⟨Ri (m), Ri (n)⟩ −
i=1
=
i=1
p
∑
=
⟨Ri (m), Ri (n)⟩ −
i=1
p
∑
p
∑
=
i=1
t=1
This completes the proof.
p ∑
p
∑
∥Pt (m)∥2(F) i=1 t=1
⟨Tit (Pt (m)), Ri (n)⟩
p
∑
⟨Pt (m), T∗it (Ri (n))⟩
⟨Pt (m), Zt (n)⟩
∥Pt (m)∥2(F) t=1
∥R(m)∥2(F)
p
∑
p ∑
p
∑
∥Pt (m)∥2(F) i=1 t=1
∥R(m)∥2(F)
t=1
p
∑
⟨Ri (m), Ri (n)⟩ −
t=1
∥R(m)∥2(F)
t=1
p
∑
⟩
Tit (Pt (m)), Ri (n)
∥R(m)∥2(F)
t=1
p
∑
⟨Ri (m), Ri (n)⟩ −
p
∑
∥Pt (m)∥2(F)
p
∑
i=1
⟨Pi (m), Zi (n)⟩.
(18)
C. Song et al. / Filomat 26:4 (2012), 809–826
816
Lemma 3.2. Assume that Problem I is consistent. If there exists a positive integer m such that ∥Ri (m)∥2(F) , 0 for all
m = 1, 2, · · · , k, then the sequences {Xi (m)}, {Ri (m)} and {Pi (m)} generated by Algorithm 1 satisfy
p
∑
⟨Ri (m), Ri (n)⟩ = 0
p
∑
⟨Pi (m), Pi (n)⟩ = 0, (m, n = 1, 2, . . . , k, m , n).
and
i=1
(19)
i=1
Proof. Note that ⟨A, B⟩ = ⟨B, A⟩ holds for arbitrary quaternion matrices A and B. We only need to prove the
conclusion holds for all 0 ≤ n < m ≤ k. For m = 1, by Lemma 3.1, we have
p
∑
⟨Ri (2), Ri (1)⟩
p
∑
⟨Ri (1), Ri (1)⟩ −
=
i=1
r
∑
i=1
t=1
p
∑
=
∥Ri (1)∥2(F) −
i=1
p
∑
∥R(1)∥2(F)
∥Pt (1)∥2(F)
p
∑
∥R(1)∥2(F)
p
∑
⟨Pi (1), Zi (1)⟩
i=1
⟨Pi (1), Pi (1)⟩ = 0,
∥Pt (1)∥2(F) i=1
t=1
and
p
∑
p ⟨∑
p
∑
⟨Pi (2), Pi (1)⟩ =
i=1
p ⟨
∑
i=1
T∗si (Rs (2))
+
s=1
∥R(2)∥2(F)
∥R(1)∥2(F)
⟩
Pi (1), Pi (1)
⟩ ∑
⟩
p ⟨
∥R(2)∥2(F)
=
Rs (2),
Tsi (Pi (1)) +
Pi (1), Pi (1)
∥R(1)∥2(F)
s=1
i=1
i=1
⟩ ∥R(2)∥2 ∑
p ⟨
p
p
p
∑
∑
∥R(1)∥2(F) ∑
(F)
=
Rs (1) − p
∥Pi (1)∥2(F)
Tsi (Pi (1)),
Tsi (Pi (1)) +
2
∑
∥R(1)∥
2
(F)
s=1
i=1
i=1
∥Pt (1)∥(F) i=1
p
∑
t=1
⟩
p ⟨
p
∑
∑
=
Rs (1),
Tsi (Pi (1)) −
s=1
p
∑
i=1
⟩
p ⟨
p
∑
∑
=
Rs (1),
Tsi (Pi (1)) −
s=1
p ⟨∑
p
∑
∥R(1)∥2(F)
t=1
∥Pt (1)∥2(F) s=1
∥R(1)∥2(F)
p
∑
i=1
t=1
∥Pt (1)∥2(F)
Tsi (Pi (1)),
p
∑
i=1
i=1
p ⟨∑
p
∑
p
∑
s=1
Tsi (Pi (1)),
i=1
i=1
⟩ ∥R(2)∥2 ∑
p
(F)
∥Pi (1)∥2(F)
Tsi (Pi (1)) +
∥R(1)∥2(F) i=1
⟩ ∥R(2)∥2 ∑
p
(F)
Tsi (Pi (1)) +
∥Pi (1)∥2(F)
∥R(1)∥2(F) i=1
p
∑
∥Pt (1)∥2(F) ⟩
p ⟨
∑
t=1
=
Rs (1), (Ri (1) − Ri (2))
∥R(1)∥2(F)
s=1
−
∥R(1)∥2(F)
p
∑
t=1
p
∑
=
=
∥Pt (1)∥2(F)
p ⟨
∑
p
∑
t=1
t=1
(Ri (1) − Ri (2)),
⟨Ri (1), Ri (1) − Ri (2)⟩ −
i=1
−
p
∑
p
∑
p
∑
∥Pt (1)∥2(F)
∥Pt (1)∥2(F)
∥R(1)∥2(F)
s=1
∥Pt (1)∥2(F) ∑
p
t=1
∥R(1)∥2(F)
p
∑
t=1
t=1
∥Pt (1)∥2(F)
∥R(1)∥2(F)
∥Pt (1)∥2(F)
∥R(1)∥2(F)
⟩ ∥R(2)∥2 ∑
p
(F)
∥Pi (1)∥2(F)
(Ri (1) − Ri (2) +
∥R(1)∥2(F) i=1
⟨Ri (1) − Ri (2), Ri (1) − Ri (2)⟩ +
∥Pt (1)∥2(F) (
p
) ∥R(2)∥2(F) ∑
2
2
∥Pi (1)∥2(F) = 0.
∥R(1)∥
+
∥R(2)∥
+
(F)
(F)
∥R(1)∥2(F)
∥R(1)∥2(F) i=1
t=1
p
∥R(2)∥2(F) ∑
∥R(1)∥2(F)
i=1
∥Pi (1)∥2(F)
C. Song et al. / Filomat 26:4 (2012), 809–826
p
∑
Assume that
p
∑
⟨Ri (m), Ri (n)⟩ = 0 and
i=1
⟨Ri (m + 1), Ri (n)⟩ = 0 and
i=1
p
∑
p
∑
817
⟨Pi (m), Pi (n)⟩ = 0 for all 0 ≤ n < m, 0 < m ≤ k, we shall show that
i=1
⟨Pi (m + 1), Pi (n)⟩ = 0 hold for all 0 ≤ n < m + 1, 0 < m + 1 ≤ k.
i=1
By the hypothesis and Lemma 3.1, for the case where 0 ≤ n < m, we have
p
∑
⟨Ri (m + 1), Ri (n)⟩ =
i=1
p
∑
⟨Ri (m), Ri (n)⟩ −
i=1
=−
p
∑
t=1
=−
p ⟨
∑
Pi (m), Pi (n) −
∥R(m)∥2(F)
∥Pt (m)∥2(F) i=1
p
∑
∥R(m)∥2(F)
p
∑
t=1
∥Pt (m)∥2(F)
∥R(m)∥2(F)
p
∑
t=1
⟩
∥R(n)∥2(F)
∥R(n − 1)∥2(F)
Pi (n − 1)
∥R(n)∥2(F)
p
∑
∥R(n − 1)∥2(F)
i=1
∥R(m)∥2(F)
⟨Pi (m), Pi (n)⟩ +
p
∑
i=1
t=1
⟨Pi (m), Zi (n)⟩
∥Pt (m)∥2(F)
∥Pt (m)∥2(F)
⟨Pi (m), Pi (n − 1)⟩ = 0,
and
p
∑
p
∑
⟨Pi (m + 1), Pi (n)⟩ =
i=1
i=1
p
∑
=
i=1
p
∑
=
⟨Zi (m + 1) +
∥R(m + 1)∥2(F)
∥R(m)∥2(F)
⟨Zi (m + 1), Pi (n)⟩ +
p
∥R(m + 1)∥2(F) ∑
∥R(m)∥2(F)
⟨Pi (m), Pi (n)⟩
i=1
⟨Zi (m + 1), Pi (n)⟩
i=1
p
∑
p
∑
i=1
i=1
⟨Ri (m + 1), Ri (n)⟩ −
=
Pi (m), Pi (n)⟩
⟨Ri (m + 1), Ri (n + 1)⟩
= 0.
∥R(m)∥2(F)
p
∑
t=1
∥Pt (m)∥2(F)
For the case n = m, we have
p
∑
⟨Ri (m + 1), Ri (m)⟩ =
i=1
p
∑
⟨Ri (m), Ri (m)⟩ −
i=1
∥R(m)∥2(F)
p
∑
t=1
=
p
∑
∥Ri (m)∥2(F)
i=1
−
∥R(m)∥2(F)
p
∑
t=1
=
∥R(m)∥2(F)
−
∥Pt (m)∥2(F)
∥R(m)∥2(F)
+
p
∑
i=1
t=1
∥Pt (m)∥2(F)
⟨Pi (m), Zi (m)⟩
∥Pt (m)∥2(F) i=1
p ⟨
∑
Pi (m), Pi (m) −
∥R(m)∥2(F)
p
∑
∥R(m)∥2(F)
∥R(m − 1)∥2(F)
∥R(m)∥2(F)
p
∑
∥R(m − 1)∥2(F)
i=1
⟩
Pi (m − 1)
⟨Pi (m), Pi (m − 1)⟩ = 0,
C. Song et al. / Filomat 26:4 (2012), 809–826
818
and
p
∑
⟨Pi (m + 1), Pi (m)⟩ =
p
∑
⟨Zi (m + 1) +
∥R(m + 1)∥2(F)
i=1
i=1
∥R(m)∥2(F)
Pi (m), Pi (m)⟩
p
p
∑
∥R(m + 1)∥2(F) ∑
=
⟨Zi (m + 1), Pi (m)⟩ +
⟨Pi (m), Pi (m)⟩
∥R(m)∥2(F) i=1
i=1
p
∑
=
⟨Ri (m + 1), Ri (m)⟩ −
i=1
p
∑
i=1
⟨Ri (m + 1), Ri (m + 1)⟩
+
∥R(m)∥2(F)
p
∑
t=1
p
∥R(m + 1)∥2(F) ∑
∥R(m)∥2(F)
∥Pi (m)∥2(F) = 0.
i=1
∥Pt (m)∥2(F)
Then by the principle of induction, we complete the proof.
Lemma 3.3. Assume that X∗ = [X1∗ , X2∗ , · · · , Xp∗ ] is a solution of Problem I. Let X(k) = [X1 (k), X2 (k), · · · , Xp (k)],
then the sequences {Xi (k)}, {R(k)} and {P(k)} generated by Algorithm 1 satisfy the following equality:
p
∑
⟨Pi (n), Xi∗ − Xi (m)⟩ = ∥R(n)∥2(F) , n ≥ m,
(20)
i=1
p
∑
⟨Pi (n), Xi∗ − Xi (m)⟩ = 0, n < m.
i=1
Proof. By induction we first demonstrate the conclusion
p
∑
⟨Pi (m), Xi∗ − Xi (m)⟩ = ∥R(m)∥2(F) ; m = 1, 2, · · · .
i=1
For m = 1, by Algorithm 1 we have
p
∑
⟨Pi (1), Xi∗
− Xi (1)⟩ =
p ∑
p
∑
⟨
T∗si (Rs (1)), Xi∗ − Xi (1)⟩
=
p ∑
p
∑
⟨
T∗si (Ri (1)), Xi∗ − Xi (1)⟩
i=1
i=1
i=1
=
=
=
=
p
∑
i=1
p
∑
s=1
p
∑
s=1
p
∑
s=1
s=1
Rs (1),
Tsi (Xi∗ − Xi (1))⟩
s=1
⟨Rs (1), Ms −
p
∑
i=1
⟨Rs (1), Rs (1)⟩
∥Rs (1)∥2(F)
s=1
=
p
∑
∥R(1)∥2(F) .
Tsi (X1 )⟩
(21)
C. Song et al. / Filomat 26:4 (2012), 809–826
819
Assume that the conclusion holds for 1 ≤ m ≤ u. For m = u + 1, we have
p
∑
⟨Pi (u +
p ∑
p
∑
∥R(u + 1)∥2(F)
∗
Pi (u), Xi∗ − Xi (u + 1)⟩
− Xi (u + 1)⟩ =
⟨
Tsi (Rs (u + 1)) +
2
∥R(u)∥
(F)
i=1 s=1
1), Xi∗
i=1
=
=
=
p ∑
p
p
∑
∥R(u + 1)∥2(F) ∑
⟨
T∗si (Rs (u + 1)), Xi∗ − Xi (u + 1)⟩ +
⟨Pi (u), Xi∗ − Xi (u + 1)⟩
2
∥R(u)∥
(F)
s=1
i=1 s=1
p
∑
⟨Rs (u + 1),
p
∑
s=1
i=1
p
∑
p
∑
⟨Rs (u + 1),
s=1
Tsi (Xi∗ − Xi (u + 1))⟩ +
Tsi (Xi∗ − Xi (u + 1))⟩ +
i=1
p
∥R(u + 1)∥2(F) ∑
∥R(u)∥2(F)
⟨Pi (u), Xi∗ − Xi (u + 1)⟩
i=1
2 ∑
∥R(u + 1)∥(F) p
⟨Pi (u), Xi∗
∥R(1)∥2(F) i=1
− Xi (u)⟩ −
p
∥R(u + 1)∥2(F) ∑
p
∑
t=1
= ∥R(u + 1)∥2(F) +
∥R(u + 1)∥2(F)
∥R(u)∥2(F)
⟨Pi (u), Pi (u)⟩
∥Pt (u)∥2(F) i=1
∥R(u)∥2(F) − ∥R(u + 1)∥2(F)
= ∥R(u + 1)∥2(F) .
By the principle of introduction, the conclusion holds for m = 1, 2. · · · . Now we assume that
p
∑
⟨Pi (m + r), Xi∗ − Xi (m)⟩ = ∥R(m + r)∥2(F) ,
i=1
For r = 0, 1, 2, · · · , k using the previous results, gives us
p
∑
⟨Pi (m + r +
1), Xi∗
i=1
p ∑
p
∑
∥R(m + r + 1)∥2(F)
− Xi (m)⟩ =
⟨
T∗si (Rs (m + r + 1)) +
Pi (m + r), Xi∗ − Xi (m + r + 1)⟩
2
∥R(m
+
r)∥
(F)
i=1 s=1
p ∑
p
p
∑
∥R(m + r + 1)∥2(F) ∑
∗
∗
Tsi (Rs (m + r + 1)), Xi − Xi (m)⟩ −
=
⟨
⟨Pi (m + r), Xi∗ − Xi (m)⟩
2
∥R(m
+
r)∥
(F)
i=1 s=1
i=1
=
p
∑
⟨Rs (m + r + 1)),
s=1
=
=
p
∑
s=1
p
∑
p
∑
Tsi (Xi∗
− Xi (m))⟩ +
s=1
⟨Rs (m + r + 1), Ms −
p
∑
∥R(m + r + 1)∥2(F)
∥R(m + r)∥2(F)
∥R(m + r)∥2(F)
Tsi (Xi (m))⟩ + ∥R(m + r + 1)∥2(F)
i=1
⟨Rs (m + r + 1), Rs (m)⟩ + ∥R(m + r + 1)∥2(F)
s=1
= ∥R(m + r + 1)∥2(F) .
By the principle of induction, the conclusion (20) holds. It follows from Algorithm 1 that
p
∑
⟨Pi (m), Xi∗ − Xi (m + 1)⟩ =
p
∑
⟨Pi (m), Xi∗ − Xi (m) −
i=1
i=1
p
∑
=
⟨Pi (m), Xi∗ − Xi (m)⟩ −
∥R(m)∥2(F)
∥R(m)∥2(F)
p
∑
t=1
i=1
p
∑
t=1
= ∥R(m)∥2(F) − ∥R(m)∥2(F) = 0.
p
∑
∥Pt (m)∥2(F) t=1
∥Pt (m)∥2(F)
⟨Pi (m), Pi (m)⟩
Pi (m)⟩
C. Song et al. / Filomat 26:4 (2012), 809–826
820
p
∑
Hence suppose that
⟨Pi (m), Xi∗ − Xi (m + r)⟩ = 0 for r = 1, 2, · · · . By using (19), we have
i=1
p
∑
⟨Pi (m), Xi∗ − Xi (m + r + 1)⟩ =
i=1
p
∑
⟨Pi (m), Xi∗ − Xi (m + r) −
i=1
∥R(m + r)∥2(F)
p
∑
t=1
=
p
∑
⟨Pi (m), Xi∗ − Xi (m + r)⟩ −
i=1
∥R(m + r)∥2(F)
p
∑
t=1
∥Pt (m + r)∥2(F)
∥Pt (m +
Pi (m + r)⟩
r)∥2(F)
p
∑
⟨Pi (m), Pi (m + r)⟩ = 0.
t=1
Hence the conclusion (21) holds by the principle of induction. The proof is completed.
Remark 3.4. Lemma 3.3 implies that if Problem I is consistent, then ∥R(k)∥2(F) , 0 implies that
Else if there exists a positive number m such that ∥R(m)∥2(F) , 0 but
p
∑
p
∑
Pi (k) , 0.
i=1
Pi (m) = 0, then Problem I must
i=1
be inconsistent. Hence the solvability of Problem I can be determined by Algorithm 1 in the absence of
roundoff errors.
Theorem 3.5. Suppose that Problem I is consistent, then for any arbitrary initial quaternion matrix pair [X1 (1),
X2 (1), · · · , Xp (1)], a quaternion matrix solution pair of Problem I can be obtained within finite number of iterations
in the absence of roundoff errors.
Proof. Firstly, in the space Qp1 ×q1 × Qp2 ×q2 × · · · × Qpp ×qp , we define an inner product as
⟨(A1 , A2 , · · · , Ap ), (B1 , B2 , · · · , Bp )⟩ = Re[
p
∑
tr(BH
i Ai )]
(22)
i=1
for (A1 , A2 , · · · , Ap ), (B1 , B2 , · · · , Bp ) ∈ Qp1 ×q1 × Qp2 ×q2 × · · · × Qpp ×qp .
Denote d =
p
∑
mi ni .
From above
i=1
it is known that the space Qp1 ×q1 × Qp2 ×q2 × · · · × Qpp ×qp with the inner product defined in (22) is a
p
∑
4d−dimentional. According to Lemma 3.2, ⟨Ri (m), Ri (n)⟩ = 0, for m, n = 0, 1, · · · , 4d − 1, and m , n.
i=1
Thus (R1 (k), · · · , Rp (k)), k = 0, 1, 2, · · · , 4d − 1 is a group orthogonal basis of the previously defined inner
p
∑
product space. In addition, it follows from Lemma 3.2 that ⟨Ri (4d), Ri (m)⟩ = 0, for k = 0, 1, · · · , 4d − 1.
i=1
Consequently, it is derived from the property of an inner product space that (R1 (4d), · · · , Rp (4d)) = 0. So
(X1 (4d), · · · , Xp (4d)) are a group of exact solution to the Problem I.
Next we consider finding the least norm solution of Problem I. The following lemmas are needed for
our derivation.
Lemma 3.6. ([31]) Suppose that the consistent system of linear equation Ax = b has a solution x∗ ∈ R(AH ), then x∗
is the unique least norm solution of linear equation.
Lemma 3.7. ([27]) For A ∈ LCm×n,p×q , there exists a unique matrix M ∈ Cpq×mn , such that vec(A(X)) = Mvec(X)
for all X ∈ Cm×n .
According to Lemma 3.7 and the definition of self-adjoint operator, one can easily obtain the following
corollary.
Corollary 3.8. Let A and M be the same as those in Lemma 4, and A∗ be the self-adjoint operator of A, then
vec(A∗ (Y)) = MH vec(Y) for all Y ∈ Cp×q .
C. Song et al. / Filomat 26:4 (2012), 809–826
Theorem 3.9. Assume that Problem I is consistent. If we choose the initial quaternion matrix Xi (1) =
821
p
∑
T∗si (Hi ),
s=1
for i = 1, 2, · · · , p, where Hi is an arbitrary matrix, or more especially, let Xi (1) = 0, i = 1, 2, · · · , p, then the solution
[X1∗ , X2∗ , · · · , Xp∗ ] obtained by Algorithm 1 is the least norm solution.
Proof. We only need to prove that the solution [(X1∗ )σ , (X2∗ )σ , · · · , (Xp∗ )σ ] = [Y1∗ , Y2∗ , · · · , Yp∗ ] is the least norm
solution to its complex representation matrix equations (11). By (12) and (6) has a solution if and only if its
complex representation matrix equations (11) has a solution. So the Problem I is equivalent to the following
′
Problem I .
′
Problem I . For given Bsi ∈ LC2mi ×2ni ,2ps ×2qs and Ms ∈ C2ps ×2qs , i = 1, 2, · · · , p; s = 1, 2, · · · , p, find
Yi , i = 1, 2, · · · , p such that
p
∑
[
i=1
B1i (Xi ),
p
∑
B2i (Yi ), · · · ,
i=1
p
∑
Bpi (Yi )] = [(M1 )σ , (M2 )σ , · · · , (Mp )σ ].
(23)
i=1
′
If Problem I has a solution [Y1 , Y2 , · · · , Yp ], then [Y1 , Y2 , · · · , Yp ] must be a solution of the system (11).
Conversely, if the system (11) has a solution [Y1 , Y2 , · · · , Yp ], then it is easy to verify that [X1 , X2 , · · · , Xp ] is
′
a solution of Problem I. Therefore, the solvability of Problem I is equivalent to that of the system (11).
p
∑
If we choose the initial matrix Yi (1) = (Xi (1))σ =
B∗si ((Hi )σ ), i = 1, 2, · · · , p; s = 1, 2, · · · , p, where Hi
i=1
is an arbitrary quaternion matrix in Qmi ×ni , by Algorithm 1 and Theorem 3.5, we can obtain the solution
′
[Y1 , Y2 , · · · , Yp ] of Problem I within finite iteration steps, which can be represented as Yi (1)∗ = (Xi∗ )σ =
p
∑
′
B∗si (Yi ), i, s = 1, 2, · · · , p. Since the solution set of Problem I is a subset of that of the system (11),
i=1
next we shall prove that [Y1 , Y2 , · · · , Yp ] is the least norm solution of the system (11), which implies that
′
[Y1 , Y2 , · · · , Yp ] is the least norm solution of Problem I .


 vec((H1 )σ ) 
∑p
∑p


∑p

..
 for all Ei ∈ C s=1 4ps qs × i=1 4mi ni ,
Let Es be the matrices such that vec( i=1 B∗si ((Hs )σ )) = Ei 

.


vec((Hp )σ )
s = 1, 2, · · · , p; (Hi )σ ∈ C2mi ×2ni , i = 1, · · · , p, are arbitrary matrices. Then the system (11) is equivalent to the
following system of linear equations:
∑p




∗
 vec(∑i=1 B1i ((Hi )σ )) 
 E1   vec((H1 )σ ) 


p
∗


 vec( i=1 B2i ((Hi )σ )) 
 E2  




 
.

.
..

 =  ..  

.


 .  

.
∑p




vec((H
)
)
p σ
vec( i=1 B∗pi ((Hi )σ )) 
Ep


vec((H1 )σ ) 
((
((
)H ) 
) )

..
H
H
H
H
H H


EH
E
·
·
·
E
E
E
·
·
·
E
=
∈
R
.


p
p
2
2


1
1
.


vec((Hp )σ )
Obviously, if we consider Yi (1) =
p
∑
s=1
B∗si ((Hi )σ ), for i = 1, 2, · · · , p, where Hi is an arbitrary matrix, or more
especially, let Yi (1) = 0, i = 1, 2, · · · , p, then all Yi (k), i = 1, 2, · · · , p, generated by Algorithm 1 satisfy


 vec(Y1 (k)) 
(
)


..

 ∈ R (EH , EH , · · · , EH
p ) . Then it follows from Lemma 3.6 that [Y1 , · · · , Yp ] is the least norm


.
2
1


vec(Yp (k))
solution of the system (11). It’s showed that [X1 , · · · , Xp ] is the least norm solution of Problem I.
C. Song et al. / Filomat 26:4 (2012), 809–826
822
Now we assume that the Problem I is consistent, and its solution set S is not non-empty, then we have

 p
p
p
∑
∑
∑


Tpi (Xi ) = [F1 , F2 , · · · , Fp ]
T2i (Xi ), · · · ,
T1i (Xi ),

i=1
i=1
i=1
 p

p
p
p
p
∑
∑
∑
∑
∑



e
e
e
e
ei )].


⇐⇒ 
T1i (Xi − Xi ), · · · ,
Tpi (Xi − Xi ) = [F1 −
T1i (Xi ), F2 −
T2i (Xi ), · · · , Fp −
Tpi (X
i=1
i=1
i=1
Define
bi = Xi − X
ei ,
X
Fbj = F j −
p
∑
i=1
i=1
ei ) i = 1, 2, · · · , p; j = 1, 2, · · · , p.
T ji (X
i=1
ei ∈ S, i = 1, 2, · · · , p.
for all X
So the Problem II is equivalent to finding the least norm solution of the following quaternion matrix
equations.
 p

p
p
∑
∑
∑


bi ),
bi ), · · · ,
bi ) = [Fb1 , Fb2 , · · · , Fbp ].
T1i (X
T2i (X
Tpi (X


i=1
i=1
i=1
Therefore, we can apply the Algorithm 1 to find the least norm solution of the quaternion matrix
bi + X
ei , i = 1, 2, · · · , p.
equation. And the least norm solution of the Problem II is Xi = X
4. Applications
Denote qi = −iqi = qa + iqb − jqc − kqd , q j = − jqj = qa − iqb + jqc − kqd , qk = −kqk = qa − iqb − jqc + kqd . Their
conjugate vectors are qi = qa − iqb + jqc + kqd , q j = qa + iqb − jqc + kqd , qk = qa + iqb + jqc − kqd and qa = 12 (q + q),
1
qb = 2i1 (q − qi ), qc = 21j (q − q j ) qd = 2k
(q − qk ). Then q = 12 (qi + q j + qk − q).
In the second-order statistics of quaternion random signals [28], it always occurs to the model qa = Aqr ,
for all qa = [qT , (qi )T , (q j )T , (qk )T ]T , qr ∈ R4n×1 , i.e.

 
 q   I
 qi   I

 
 j  = 
 q   I
 k  
I
q
iI
iI
−iI
−iI
jI
− jI
jI
− jI
kI
−kI
−kI
kI


  qa 
  qb 
 

  qc 
 

qd
for all I ∈ Rn×n is the identity matrix, q = [q1 , q2 , · · · , qn ]T ∈ Qn×1 ; qi , q j , qk ∈ Qn×1 , qa , qb , qc qd ∈ Rn×1 . In linear
mean-squared error (MSE)estimation), it needs to compute complete second-order estimation. Its model in
complex field [23] is b
y = hT x, and its model in quaternion field [29] is
b
y = E[y|x, xi , x j , xk ] + iE[yi |x, xi , x j , xk ] + jE[y j |x, xi , x j , xk ] + kE[yk |x, xi , x j , xk ].
It is to find the the least norm solution of the following quaternion matrix equation [29]:
y = 1H x + hH xi + uH x j + vH xk .
Consider the following general linear vector-sensor signal model [3]
Si (t) =
p
∑
j=1
hij 1 j (t − (i − 1)τp ) + bi (t), i = 1, 2, · · · , n,
C. Song et al. / Filomat 26:4 (2012), 809–826
823
where hij ∈ R, τp , are known scalars, 1 j (t) ∈ QN , bi (t), Si (t) are the determined quaternion vector. We need to
solve the coupled quaternion vector equations
Yi =
p
∑
α j X j + Zi + Ci
j=1
where Yi , i = 1, 2, · · · , n, X j , j = 1, 2, · · · , p, Zi are the determined quaternion vectors, Zi , i = 1, 2, · · · , n, are
the known quaternion vector and α j , j = 1, 2, · · · , p, are the known scalars.
5. Numerical examples
In this section, we shall give a numerical example to illustrate the efficiency of Algorithm 1.
Example 5.1. Consider the matrix equation
{
A11 X1 B11 + A12 X2 B12 = C1
A21 X1 B21 + A22 X2 B22 = C2
where




= 


(25)




 1 + j 2 + k i + j 3 + k 

 5 j
2k
4i
1 


A11
 , B11 = 
,
j−k
1
4k 

 1



2
6j
2
1+ j




4k
2+i
6 + j 
k 
 1 + j
 2 + j 1 + k 3 j
 2 + k


2 + j 6 + i 1 + j + k 
3i
2
6i 

 4 j + k
A12 = 
 , B12 = 
,
2k
3 1 + k 
 3 j + 2k 2 − 4k 1 − j 6 + 2k 
 4j




2 + 4j 3 − 4k
5j
2k
1
2j
3i
0




i
k 1 
4 
 j
 4 + j 3 + k 2 + i
 2i


 4j
2j
2k 4 
1+i
2j
k 

A21 = 
 , B21 = 
,
3 + i 3 + k 1 + 2i 
 1 + i 5 + k 3 1 
 2



k
1+ j k 0
4+k 2+ j 2+k
0




2j
0
1 + k 
12k 11 + i 2k 
 0
 −3j
 0 11 + j 21i


k 
k
2 j 

 21 + k 23 j
A22 = 
 , B22 = 
,
0
2 + j 1 + i 
2j
4k
1 + i 
 4
 1 + j




0
−j
−k
1
2i
11
3i
2
−1
2+k
2+i
2+ j+k

 −58 + i + 88 j + 33k
 −68 + 41i + 53 j + 26k

C1 = 
 −75 + 19i + 98 j + 83k

−121 + 73i − 25 j − 2k

−72 + 21i − 3j + 28k

 −86 + 497i + 582j + 437k

C2 = 
 127 + 121i + 91 j + 90k

36 + 15i + 24 j − 18k
2j
2i
1−k
2k
k
1+i
3j
k
2+ j
2
5j
k
−22 + 24i − 29 j + 71k
−52 + 68i + 21 j − 18k
−42 + 38i + 41 j + 90k
79 + 74i − 17j − 49k
5 − 27i + 71 j − 48k
4 − 4i + 30 j + 3k
30 + 42i + 23 j + 25k
−42 − 11i + 9 j + 26k
−41 + 6i − 43 j − 38k
−402 − 233i − 259j + 812k
−133 + 55i + 199j + 295k
−11 + 18i + 77 j + 54k
52 + 67i + 5j − 5k
−74 + 63i − 22 j − 108k
−87 + 51i + 7 j − 7k
−142 + 15i − 81 j − 110k
−24 − 21i − 9 j − 22k
−352 − 56i + 84 j − 131k
−29 + 113i + 36 j + 73k
−21 + 15i + 15 j + 27k




 ,


−12 + 2i − 5 j − 6k
−31 − 3i − 27 j + 191k
−19 + 21i + 71 j + 42k
−9i + 10 j
It can be verified that the generalized coupled matrix equation (25) are consistent and have a unique
solution pair (X1∗ , X2∗ ) as follows:

2j
 −1
 2 1 + 2k

∗
X1 = 
2i
 2 j

k
2j
2k
2i
3k
2
1+ j
2k
2i
4i



 1 + k


 ∗  1 + j
 , X2 = 

 1 + i


k
1+i j
2j
k
1+ j 0
1
2
0
1+i
2j
k




 .






 .


C. Song et al. / Filomat 26:4 (2012), 809–826
824
If we let the initial matrix (X1 , X2 ) = 10−6 I4 , applying Algorithm 1, we obtain the solutions, that are

2.0000j
 −0.9999
 2.0000 1.0000 + 2.0000k

X1 (799) = 
2.0000i
 2.0000j

1.0000k
2.0000j

 0.9999 + 1.0000k
 1.0000 + 0.9999j

X2 (799) = 
 1.0000 + 1.0000i

0.9999k
1.9999k
2.0000i
3.0000k
1.9999
1.0000 + 0.9999 j
2.0000k
1.9999i
4.0000i
1.0000 + 0.9999i 1.0000j
2.0000j
1.0000k
1.0000 + 0.9999j 0.0000
1.0000
2.0000




 ,


0.0000
0.9999 + 1.0000i
2.0000j
1.0000k




 ,


with corresponding residual
∥R(799)∥(F) = ∥dia1(C1 −A11 X1 (799)B11 −A12 X2 (799)B12 , C2 −A21 X1 (799)B21 −A22 X2 (799)B22 )∥(F) = 6.2826×10−11 .
The obtained results are presented in Figure 1, where
ek =
∥(X1 (k), X2 (k) − (X1∗ , X2∗ )∥(F)
∥(X1∗ , X2∗ )∥(F)
4
4
2
2
0
0
−2
−2
−4
rk = ∥R(k)∥(F) .
−4
log10rk
−6
log10ek
−8
−10
−10
−12
−12
0
100
200
300
400
500
600
700
−14
800
Figure 1: The relative error of solution and the residual
Now let
log10rk
−6
log10ek
−8
−14
,

 −i
 1 + j

b
X1 = 
 1

0
j
k
1+k 1
1 − j 2k
1 + 2j 1
0
1
1
2j
0
100
200
300
400
500
600
700
800
900
Figure 2: The relative error of solution and the residual






 b

 , X2 = 




2j
2i
2
1 − 4k
2
2j
1−i
2j
2k
2k
1 − 6k
0
0
2i
0
2k




 .


By Algorithm 1 for the generalized coupled Sylvester matrix equation (25),with the initial matrix
pair(X1 , Y1 ) = 0, we have the least Frobenius norm solution of the generalized coupled Sylvester matrix
equation (25) by the following form:

0.9999j
0.9999k
 −1.0000 + 1.0000i
 0.9999 − 1.0000j
∗
0.9999k
−1.0000 + 1.9999i

X1 = X1 (830) = 
0.9999k
 −1.0000 + 2.0000j −0.9999 + 1.9999i + 1.0000j

1.0000k
−0.9999
0.9999
0.9999 + 1.0000 j
−1.0000 + 1.9999k
−1.0000 + 2.0000i
3.9999i − 1.9999j




 ,


C. Song et al. / Filomat 26:4 (2012), 809–826



∗

X2 = X2 (830) = 


0.9999 − 2 j + k
−0.9999 + 0.9999j
−0.9999 + 0.9999i
0.9999 + 1.0000i + 0.9999k
1 − 0.9999i
−1 + 2j + 3.9999k
0.9999 − 0.9999j
0.9999 − 2.0000j
0.9999 − 1.9999k
−0.9999k
−0.9999 + 6k
1.9999
825

0.0000

0.9999 − 0.9999i 

1.9999j


−k
with corresponding residual
∥R(830)∥(F) = ∥dia1(C1 −A11 X1 (830)B11 −A12 X2 (830)B12 , C2 −A21 X1 (830)B21 −A22 X2 (830)B22 )∥(F) = 5.9374×10−11 .
The obtained results are presented in Figure 2. Therefore, the solution of the matrix equation nearness
problem is


1.9999 j
1.9999k 0.9999 + 1.0000j 
 −1.0000


1 + 1.9999k
1.9999i
1.9999k

f1 = X1 ∗ + X
c1 =  1.9999
X
 ,
 2.0000j 0.0001 + 1.9999i 2.9999k
2i


1.0000k
0.0001 + 2j
1.9999 3.9999i + 0.0001j


0.0000

 0.9999 + 1.0000k 1.0000 + 0.0001i 0.9999j + 0.0001k

 1.0001 + 0.9999j 2.0000j − 0.0001k
∗
1.0000k
0.9999
+
1.0001i

f2 = X2 + X
c2 = 
X

 1.0001 + 0.9999i 0.9999 + 1.0001j
0.0001
1.9999j


0.0001 + 0.9999k
0.9999
1.9999
1.0000k
From the Figure 1, we can see that the residual and the relative error of the solution are declining
gradually with the increase of the iterative steps. So we can say that the Algorithm 1 can converge to exact
solution of quaternion matrix equations (25). From the Figure 2, we can see that the Algorithm 1 can solve
the optimal solution of the matrix equation (25). But for any quaternion matrix A = A1 + A2 i + A3 j + A4 k,
i2 = j2 = k2 = −1, complex matrix can be stated as the special case of quaternion matrix. So when we solve
the quaternion matrix equations problem by means of the iterative algorithm, generally its convergence rate
is not fast, such as an operable iterative method called LSQR-Q algorithm [36] for finding the minimumnorm solution of the QLS problem by applying real representation of a quaternion matrix, which is more
appropriate to large scale system.
References
[1] A. Ben-Israel, T.N.E. Greville, Generalized Inverse: Theory and Applications, Springer, New York, 2003.
[2] N.L. Bihan, S.J. Sangwine, Color image decomposition using quaternion singular value decomposition, in: IEEE International
Conference on Visual Information Engineering of quaternion (VIE),Guidford, UK, 2003.
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