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EXTREME RANKS OF (SKEW-)HERMITIAN SOLUTIONS TO A

QUATERNION MATRIX EQUATION∗

QING WEN WANG† AND JING JIANG

Abstract. The extreme ranks, i.e., the maximal and minimal ranks, are established for the general Hermitian solution as well as the general skew-Hermitian solution to the classical matrix equationAXA∗+BY B=Cover the quaternion algebra. Also given in this paper are the formulas

of extreme ranks of real matrices Xi, Yi, i = 1,· · ·,4,in a pair (skew-)Hermitian solutionX =

X1+X2i+X3j+X4k, Y =Y1+Y2i+Y3j+Y4k.Moreover, the necessary and sufficient conditions for the existence of a real (skew-)symmetric solution, a complex (skew-)Hermitian solution, and a pure imaginary (skew-)Hermitian solution to the matrix equation mentioned above are presented in this paper. Also established are expressions of such solutions to the equation when corresponding solvability conditions are satisfied. The findings of this paper widely extend the known results in the literature.

Key words. Quaternion matrix equation, Minimal rank, Maximal rank, Hermitian solution, Skew-Hermitian solution.

AMS subject classifications. 15A03, 15A09, 15A24, 15A33.

1. Introduction. Throughout this paper, we denote the real number field by

R, the complex number field byC,the set of allm×nmatrices over the quaternion

algebra

H={a0+a1i+a2j+a3k|i2=j2=k2=ijk=−1,a0, a1, a2, a3 R}

byHm×n,the identity matrix with the appropriate size byI.For a quaternion matrix

A,we denote the column right space, the row left space ofAbyR(A),N(A), respec-tively, the dimension ofR(A) by dimR(A).By [9], dimR(A) = dimN(A),which is called the rank ofA,and denoted byr(A).The Moore-Penrose inverse of a matrix

A∈Hm×n,denoted by A, is defined to be the unique matrix X Hn×msatisfying

Received by the editors February 14, 2010. Accepted for publication July 31, 2010. Handling

Editor: Michael Neumann.

Department of Mathematics, Shanghai University, Shanghai 200444, P.R. China

(wqw858@yahoo.com.cn). Supported by the grants from the Ph.D. Programs Foundation of Ministry of Education of China (20093108110001), the Scientific Research Innovation Foundation of Shanghai Municipal Education Commission (09YZ13), Natural Science Foundation of China (60672160), Singapore MoE Tier 1 Research Grant RG60/07, and Shanghai Leading Academic Discipline Project (J50101).

Department of Mathematics, Shanghai University, Shanghai 200444, P.R. China, and

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the following four matrix equations

(1)AXA=A, (2)XAX=X, (3) (AX)∗=AX, (4) (XA)=XA.

Moreover, LA andRA stand for the two projectors LA =I−A†A, RA =I−AA†

induced byA.

We know that matrix equation is one of the topics of very active research in matrix theory and applications, and a large number of papers have presented several methods for solving several matrix equations (e.g. [4]-[6], [10], [23]-[26], [29]-[31], [37], [39], [42]). As a very classical linear matrix equation,

(1.1) AXA∗+BY B∗=C

has been investigated by many authors from different aspects. For example, using gen-eralized singular value decomposition, Chang and Wang [2] derived the expressions for the general symmetric and minimum-2-norm symmetric solutions to (1.1) within the real setting. Xu, Wei, and Zheng [38] obtained the general form of all least-squares Hermitian (skew-Hermitian) solutions to (1.1). Liao and Bai [11] used generalized singular value decomposition to investigate the symmetric positive semidefinite solu-tion to (1.1). Zhang [40] gave necessary and sufficient condisolu-tions for the existence of a Hermitian nonnegative-definite solution to (1.1). Wang and Zhang [32] gave a neces-sary and sufficient condition for the existence and an expression for the re-nonnegative definite solution to (1.1) overHby using the decomposition of pairwise matrices.

Research on extreme ranks, i.e., maximal and minimal ranks, of solutions to linear matrix equations have been actively ongoing for more than 30 years (see [15]-[17], [19], [20], [27], [28], [33]-[35]). It is worthy to say that minimal and maximal ranks of a general solution to a matrix equation are very useful in linear programming computations (see [15]-[17]). In 2009, Liu, Tian and Takane [13] presented formulas for the maximal and minimal ranks of a Hermitian solution and a skew-Hermitian solution to the special case of the matrix equation (1.1) in whichB = 0.The following matrix equation:

(1.2) AXA∗=C

overC, has been well examined by many authors (see [1], [3], [7], [8], [32], [36], [41]).

Note that, to our knowledge, there has been little information on extreme ranks of the (skew-)Hermitian solution to the matrix equation (1.1). This paper aims to consider the formulas of extremal ranks of the general (skew-)Hermitian solution to (1.1).

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general Hermitian solution to (1.1). We also give the corresponding results on the skew-Hermitian solution to (1.1). In Section 3, we first present the maximal and minimal ranks of the real matrices X1, X2, X3, X4 and Y1, Y2, Y3, Y4 in a Hermitian

solutionX =X1+X2i+X3j+X4k, Y =Y1+Y2i+Y3j+Y4k to (1.1), then give

some necessary and sufficient conditions for (1.1) to have a real symmetric solution, a complex Hermitian solution, and a pure imaginary Hermitian solution. The corre-sponding results on the skew-Hermitian solution to (1.1) are also considered. Some special cases of (1.1) are also considered in Section 4.

2. Ranks of the general Hermitian solution to (1.1). In this section, we consider the maximal and minimal ranks of the general (skew-)Hermitian solution to (1.1) overH.

We begin with the following lemma that is due to Tian [18], and can be generalized toH.

Lemma 2.1. Suppose that the matrix equation

(2.1) AXB+CY D=E

is consistent overH,whereX∈Hp×q, Y ∈Hs×tunknown. Then the general solution of (2.1)can be expressed by

X =X0+S1LGU RHT1+LAV1+V2RB,

Y =Y0−S2LGU RHT2+LCW1+W2RD,

whereX0 andY0 are a special pair solution of (2.1),

S1= (Ip,0), S2= (0, Is), T1= (Iq,0)∗, T2= (0, It)∗, G= (A, C), H =

µ

B D

;

U, V1, V2, W1, andW2 are arbitrary quaternion matrices with suitable sizes.

Lemma 2.2. Consider the matrix equation(1.1) where A ∈Hm×n, B ∈ Hm×p, C∈Hm×mare given, andX Hn×n, Y Hp×p unknown.

1. If C =C∗, and (1.1)has a Hermitian solution, then the general Hermitian

solution to(1.1) can be expressed as

X =X0+S1LGZLGS1∗+LAV +V∗LA,

Y =Y0−S2LGZLGS2∗+LBW +W∗LB,

(2.2)

whereX0 andY0 are a special pair Hermitian solution of (1.1),

(2.3) S1= (In,0), S2= (0, Ip), G= (A, B) ;

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2. IfC=−C ,and(1.1)has a Hermitian solution, then the general skew-Hermitian solution can be expressed as

X =X0+S1LGZLGS1∗+LAV −V∗LA,

Y =Y0−S2LGZLGS2∗+LBW −W∗LB,

whereX0 andY0 are a special pair skew-Hermitian solution of(1.1), and S1, S2,andGare the same as(2.3);Z is an arbitrary skew-Hermitian quaternion matrix with consistent size, and V andW are arbitrary quaternion matrices with suitable sizes.

Proof. We here only prove 1. The proof of 2 can be similarly finished.

Suppose that X1 = X1∗, Y1 = Y1∗ are an arbitrary pair Hermitian solution of

(1.1). It follows from Lemma 2.1 andR(A,B)∗ =L(A,B), RA∗ =LA, RB∗ =LB that

X1= ˜X0+S1LGU LGS∗1+LAV1+V2LA,

Y1= ˜Y0−S2LGU LGS2∗+LBW1+W2LB,

where ˜X0 and ˜Y0are a special pair solution of (1.1), andU, V1, V2, W1, andW2 are

arbitrary matrices with suitable sizes. Since³X1+X1∗ 2 ,

Y1+Y1∗ 2

´

is also a pair Hermitian solution of (1.1), we get that

X1=

1

2(X1+X ∗

1) =

1

2( ˜X0+ ˜X ∗

0) +

1

2S1LG(U+U ∗)L

GS1∗+

1

2LA(V1+V ∗

2)

+1 2(V1+V

2)∗LA,

Y1=

1

2(Y1+Y ∗

1) =

1

2( ˜Y0+ ˜Y ∗

0)−

1

2S2LG(U+U ∗)L

GS2∗+

1

2LB(W1+W ∗

2)

+1

2(W1+W ∗

2)∗LB.

Putting

X0=1

2( ˜X0+ ˜X ∗

0), Y0= 1

2( ˜Y0+ ˜Y ∗

0), V =

1

2(V1+V ∗

2),

W =1

2(W1+W ∗

2), Z=

1 2(U+U

),

and noting that X0 and Y0 are a special pair Hermitian solution of (1.1), Z is an

arbitrary Hermitian matrix, we get that any pair Hermitian solution (X1, Y1) of (1.1)

has the form of (2.2).

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Tian and Liu in [21] and [12] gave the following extremal ranks of matrix expres-sionsA−BXB∗andA−BX−X∗B∗over a field. The results can be generalized to

H.

Lemma 2.3. Let AHm×m, BHm×n.

1. If A=A∗,then max

X=X∗Hn×nr(A−BXB

) =r[A, B], min

X=X∗Hn×nr(A−BXB

) = 2r[A, B]r

·

A B B∗ 0

¸

;

max

X∈Hn×mr(A+BX+X

B) = min

½

m, r

·

A B B∗ 0

¸¾

,

min

X∈Hn×mr(A+BX+X

B) =r

·

A B B∗ 0

¸

−2r(B).

2. If A=−A∗,then max

X=−X∗Hn×nr(A−BXB

) =r[A, B], min

X=−X∗Hn×nr(A−BXB

) = 2r[A, B]r

·

A B

−B∗ 0

¸

;

max

X∈Hn×mr(A+BX−X

B) = min

½

m, r

·

A B

−B∗ 0

¸¾

,

min

X∈Hn×mr(A+BX−X

B) =r

·

A B

−B∗ 0

¸

−2r(B).

The following lemma is due to Marsaglia and Styan [14], which can be generalized toH.

Lemma 2.4. Let A Hm×n, B Hm×k, C Hl×n, D Hj×k andE Hl×i.

Then they satisfy the following rank equalities:

(a) r(CLA) =r

·

A C

¸

−r(A),

(b) r£ B ALC

¤

=r

·

B A

0 C

¸

−r(C),

(c) r

·

C RBA

¸

=r

·

C 0

A B

¸

−r(B).

Now we give one of the main theorems in this paper.

Theorem 2.5. Suppose that the matrix equation (1.1), where A∈ Hm×n, B ∈

Hm×p, C Hm×m, C=C, XHn×n, andY Hp×p, has a Hermitian solution.

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are given by:

max

AXA∗+BY B∗=C

X=X∗

r(X) = min{n, r[B, C] + 2n−r(A)−r(G)},

(2.4)

min

AXA∗+BY B=C

X=X∗

r(X) = 2r[B, C]−r

·

C B

B∗ 0

¸

; (2.5)

max

AXA∗+BY B=C

Y=Y∗

r(Y) = min{p, r[A, C] + 2p−r(B)−r(G)},

(2.6)

min

AXA∗+BY B=C

Y=Y∗

r(Y) = 2r[A, C]−r

·

C A A∗ 0

¸

.

(2.7)

(b) The rank of the general Hermitian solutionX to(1.1)is invariant if and only if

2r[B, C]−r

·

C B

B∗ 0

¸

=n

or

r[B, C] +r(A) +r(G) =r

·

C B B∗ 0

¸

+ 2n.

The rank of the general Hermitian solutionY to(1.1)is invariant if and only if

2r[A, C]−r

·

C A A∗ 0

¸

=p

or

r[A, C] +r(B) +r(G) =r

·

C A A∗ 0

¸

+ 2p.

Proof. (a) Applying Lemma 2.2 and Lemma 2.3 toX of (2.2), we get that

max

AXA∗+BY B=C

X=X∗

r(X) = max

Z=Z∗,Vr(X0+S1LGZLGS

1 +LAV +V∗LA)

= min

½

n, max

Z=Z∗r ·

X0+S1LGZLGS∗1 LA

LA 0

¸¾

= min

½

n, max

Z=Z∗r µ·

X0 LA

LA 0

¸

+

·

S1LG

0

¸

Z[ LGS1∗ 0 ]

¶¾

= min

½

n, r

·

X0 LA S1LG

LA 0 0

¸¾

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min

AXA∗+BY B∗=C

X=X∗

r(X) = min

Z=Z∗,Vr(X0+S1LGZLGS

1+LAV +V∗LA)

= min

Z=Z∗r ·

X0+S1LGZLGS1∗ LA

LA 0

¸

−2r(LA)

= min

Z=Z∗r µ·

X0 LA

LA 0

¸

+

·

S1LG

0

¸

Z[ LGS1∗ 0 ]

−2r(LA)

= 2r

·

X0 LA S1LG

LA 0 0

¸

−r

 

X0 LA S1LG

LA 0 0

LGS1∗ 0 0

 

−2r(LA).

(2.9)

By Lemma 2.4, block Gaussian elimination, andAX0A∗+BY0B∗=C,we have that r(LA) =n−r(A),

r

·

X0 S1LG LA

LA 0 0

¸

=r

   

X0 In S1 0 In 0 0 A∗

0 A 0 0

0 0 G 0

  

−r(A)−r(A∗)−r(G)

=r[B, C] + 2n−r(A)−r(G),

r

 

X0 LA S1LG

LA 0 0

LGS1∗ 0 0

 =r

      

X0 In S1 0 0 In 0 0 A∗ 0

S∗

1 0 0 0 G∗

0 A 0 0 0

0 0 G 0 0

      

−r(A)−r(A∗)2r(G)

=r

·

C B B∗ 0

¸

+ 2n−2r(G).

Substituting above two equalities into (2.8) and (2.9) yields (2.4) and (2.5), respec-tively.

Similarly, we can get the corresponding results onY.

(b) The ranks ofX, Y,expressed as (2.2), in the general pair Hermitian solution to (1.1) are invariant if and only if

(2.10) maxr(X)−minr(X) = 0, maxr(Y)−minr(Y) = 0.

Hence result (b) follows from (2.4)-(2.7), and (2.10).

Similarly, we can get the following.

Theorem 2.6. Let AHm×n, BHm×p, CHm×m,C =C, X Hn×n,

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(a) The maximal and minimal ranks of the general skew-Hermitian solution to (1.1) are given by

max

AXA∗+BY B∗=C

X=−X∗

r(X) = min{n, r[B, C] + 2n−r(A)−r(G)},

min

AXA∗+BY B=C

X=−X∗

r(X) = 2r[B, C]−r

·

C B

−B∗ 0

¸

.

max

AXA∗+BY B=C

Y=−Y∗

r(Y) = min{p, r[A, C] + 2p−r(B)−r(G)};

min

AXA∗+BY B=C

Y=−Y∗

r(Y) = 2r[A, C]−r

·

C A

−A∗ 0

¸

.

(b) The rank of the general skew-Hermitian solution X to (1.1) is invariant if and only if

2r[B, C]−r

·

C B

−B∗ 0

¸

=n,

orr[B, C] +r(A) +r(G) =r

·

C B

−B∗ 0

¸

+ 2n.

The rank of the general skew-Hermitian solution Y to (1.1) is invariant if and only if

2r[A, C]−r

·

C A

−A∗ 0

¸

=p, orr[A, C]+r(B)+r(G) =r

·

C A

−A∗ 0

¸

+2p.

3. Extreme ranks of the real matrices in a Hermitian solution to (1.1) overH. In this section, we consider the maximal and minimal ranks of real matrices Xi, Yi in a pair Hermitian solution X = X1+X2i+X3j+X4k ∈ Hn×n and Y = Y1+Y2i+Y3j+Y4k∈Hp×p to (1.1) where

A=A1+A2i+A3j+A4k, B=B1+B2i+B3j+B4k, C=C1+C2i+C3j+C4k,

Ai, Bi, Ci, i= 1,· · · ,4,are real matrices with suitable sizes.

For an arbitrary quaternion matrixM =M1+M2i+M3j+M4k,we now define

a mapφ(·), fromHm×n to R4m×4n, by

(3.1) φ(M) =

   

M1 −M2 −M3 −M4 M2 M1 −M4 M3 M3 M4 M1 −M2 M4 −M3 M2 M1

   .

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(a) M =N ⇐⇒ φ(M) =φ(N).

(b) φ(M+N) =φ(M) +φ(N), φ(M N) =φ(M)φ(N), φ(kM) =kφ(M), k∈R.

(c) φ(M∗) =φT(M), φ(M) =φ(M). (d) φ(M) =T−1

m φ(M)Tn=R−m1φ(M)Rn =S−m1φ(M)Sn, where fort=m, n,

Rt=

·

0 −I2t

I2t 0

¸

, St=

   

0 0 0 −It

0 0 It 0

0 −It 0 0

It 0 0 0

   ,

Tt=

   

0 −It 0 0

It 0 0 0

0 0 0 It

0 0 −It 0

   .

(e) r[φ(M)] = 4r(M).

(f) M∗=M φT(M) =φ(M), andM=M φT(M) =φ(M).

In the following theorems and corollaries,X0, Y0, S1, S2,andGare defined as in

Lemma 2.2.

Theorem 3.1. The matrix equation (1.1) has a Hermitian solution over H if and only if the matrix equation

(3.2) φ(A) (Xij)4×4φT(A) +φ(B) (Yij)4×4φT(B) =φ(C), i, j= 1,2,3,4,

has a symmetric solution overR. In this case, the general Hermitian solution of(1.1) overHcan be written as:

X =X1+X2i+X3j+X4k

=1

4(X11+X22+X33+X44) + 1 4(X

T

12−X12+X34T −X34)i

+1 4(X

T

13−X13+X24−X24T)j+

1 4(X

T

14−X14+X23T −X23)k,

(3.3)

Y =Y1+Y2i+Y3j+Y4k

=1

4(Y11+Y22+Y33+Y44) + 1 4(Y

T

12−Y12+Y34T−Y34)i

+1 4(Y

T

13−Y13+Y24−Y24T)j+

1 4(Y

T

14−Y14+Y23T −Y23)k,

(3.4)

where Xtt =XttT, Ytt =YttT, t= 1,2,3,4; X1Tj =Xj1,Y1Tj =Yj1, j = 2,3,4; X2Tj =

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.Written in explicit forms,Xi, Yi, i= 1,2,3,4, in(3.3)and(3.4)are

X1=

1

4P1φ(X0)P

T

1 +

1

4P2φ(X0)P

T

2 +

1

4P3φ(X0)P

T

3 +

1

4P4φ(X0)P

T

4

+1

4[P1, P2, P3, P4]E1ZE

T

1 [P1, P2, P3, P4]T

+ [P1, P2, P3, P4]Lφ(A)

    V1 V2 V3 V4    +     V1 V2 V3 V4     T

LTφ(A)[P1, P2, P3, P4]T,

(3.5)

X2=

1

4P2φ(X0)P

T

1 −

1

4P1φ(X0)P

T

2 +

1

4P4φ(X0)P

T

3 −

1

4P3φ(X0)P

T

4

−1

4[P1,−P2, P3,−P4]E1ZE

T

1 [P2, P1, P4, P3]T

−[−P2, P1,−P4, P3]Lφ(A)

    V1 V2 V3 V4    +     V1 V2 V3 V4     T

LTφ(A)[−P2, P1,−P4, P3]T,

(3.6)

X3=

1

4P3φ(X0)P

T

1 −

1

4P1φ(X0)P

T

3 +

1

4P2φ(X0)P

T

4 −

1

4P4φ(X0)P

T

2

−1

4[P1,−P2,−P3, P4]E1ZE

T

1 [P3, P4, P1, P2]T

−[−P3, P1,−P2, P4]Lφ(A)

    V1 V2 V3 V4    +     V1 V2 V3 V4     T

LTφ(A)[−P3, P1,−P2, P4]T,

(3.7)

X4=1

4P4φ(X0)P

T

1 −

1

4P1φ(X0)P

T

4 +

1

4P3φ(X0)P

T

2 −

1

4P2φ(X0)P

T

3

−1

4[−P1,−P2, P3, P4]E1ZE

T

1 [P4, P3, P2, P1]T

−[−P4, P1,−P3, P2]Lφ(A)

    V1 V2 V3 V4    +     V1 V2 V3 V4     T

LTφ(A)[−P4, P1,−P3, P2]T,

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Y1=

1

4Q1φ(Y0)Q

T

1 +

1

4Q2φ(Y0)Q

T

2 +

1

4Q3φ(Y0)Q

T

3 +

1

4Q4φ(Y0)Q

T

4

−1

4[Q1, Q2, Q3,Q4]E2ZE

T

2 [Q1, Q2, Q3, Q4]T

+ [Q1, Q2, Q3, Q4]Lφ(B)

    W1 W2 W3 W4    +     W1 W2 W3 W4     T LT

φ(B)[Q1, Q2, Q3, Q4]T,

(3.9)

Y2= 1

4Q2φ(Y0)Q

T

1 −

1

4Q1φ(Y0)Q

T

2 +

1

4Q4φ(Y0)Q

T

3 −

1

4Q3φ(Y0)Q

T

4

+1

4[Q1,−Q2, Q3,−Q4]E2ZE

T

2 [Q2, Q1, Q4, Q3]T

−[−Q2, Q1,−Q4, Q3]Lφ(B)

    W1 W2 W3 W4    +     W1 W2 W3 W4     T

LTφ(B)[−Q2, Q1,−Q4, Q3]T,

(3.10)

Y3= 1

4Q3φ(Y0)Q

T

1 −

1

4Q1φ(Y0)Q

T

3 +

1

4Q2φ(Y0)Q

T

4 −

1

4Q4φ(Y0)Q

T

2

+1

4[Q1,−Q2,−Q3, Q4]E2ZE

T

2 [Q3, Q4, Q1, Q2]T

−[−Q3, Q1,−Q2, Q4]Lφ(B)

    W1 W2 W3 W4    +     W1 W2 W3 W4     T

LTφ(B)[−Q3, Q1,−Q2, Q4]T,

(3.11)

Y4=

1

4Q4φ(Y0)Q

T

1 −

1

4Q1φ(Y0)Q

T

4 +

1

4Q3φ(Y0)Q

T

2 −

1

4Q2φ(Y0)Q

T

3

+1

4[−Q1,−Q2, Q3, Q4]E2ZE

T

2 [Q4, Q3, Q2, Q1]T

−[−Q4, Q1,−Q3, Q2]Lφ(B)

    W1 W2 W3 W4    +     W1 W2 W3 W4     T

LTφ(B)[−Q4, Q1,−Q3, Q2]T,

(12)

where

P1= [In,0,0,0], P2= [0, In,0,0], P3= [0,0, In,0], P4= [0,0,0, In],

Q1= [Ip,0,0,0], Q2= [0, Ip,0,0], Q3= [0,0, Ip,0], Q4= [0,0,0, Ip],

E1=φ(S1)Lφ(G), E2=φ(S2)Lφ(G);

Z is arbitrary real symmetric matrix, and V1, V2, V3, V4, W1, W2, W3, and W4 are arbitrary real matrices with compatible sizes.

Proof. Suppose that (1.1) has a Hermitian solution

X=X1+X2i+X3j+X4k, Y =Y1+Y2i+Y3j+Y4k

overH. Applying properties (a) and (b) ofφ(·) to (1.1) yields

φ(A)φ(X)φT(A) +φ(B)φ(Y)φT(B) =φ(C), φT(X) =φ(X), φT(Y) =φ(Y),

which implies that φ(X), φ(Y) are real symmetric solutions to (3.2). Conversely, suppose that (3.2) has a real symmetric solution

X =

   

X11 X12 X13 X14 X21 X22 X23 X24 X31 X32 X33 X24 X41 X42 X43 X44

   , Y =

   

Y11 Y12 Y13 Y14 Y21 Y22 Y23 Y24 Y31 Y32 Y33 Y24 Y41 Y42 Y43 Y44

   ,

i.e.

φ(A)XφT(A) +φ(B)Y φT(B) =φ(C), XT =X, YT =Y.

By property (d) ofφ(·), we have that

T−1

m φ(A)TnXTn−1φT(A)Tm+Tm−1φ(B)TpY Tp−1φT(B)Tm=Tm−1φ(C)Tm,

R−m1φ(A)RnXRn−1φT(A)Rm+R−m1φ(B)RpY Rp−1φT(B)Rm=R−m1φ(C)Rm,

Sm−1φ(A)SnXSn−1φT(A)Sm+Sm−1φ(B)SpY Sp−1φT(B)Sm=Sm−1φ(C)Sm.

Hence

φ(A)TnXTn−1φT(A) +φ(B)TpY Tp−1φT(B) =φ(C),

φ(A)RnXRn−1φT(A) +φ(B)RpY R−p1φT(B) =φ(C),

φ(A)SnXSn−1φT(A) +φ(B)SpY Sp−1φT(B) =φ(C),

implying TnXTn−1, TpY Tp−1, RnXR−n1, RpY R−p1, SnXSn−1, and SpY Sp−1 are also

symmetric solutions of (3.2) overR. Thus,

1

4(X+TnXT −1

n +RnXR−n1+SnXSn−1),

1

4(Y +TpY T −1

(13)

are symmetric solutions of (3.2), where

X+TpXTp−1+RpXR−p1+SpXSp−1= (gXij)4×4, i, j= 1,2,3,4

and

g

X11=X11+X22+X33+X44, Xg12=X12−X12T +X34−X34T,

g

X13=X13−X13T +X24T −X24, Xg14=X14−X14T +X23−X23T,

g

X21=X12T −X12+X34T −X34, Xg22=X11+X22+X33+X44,

g

X23=X14−X14T +X23−X23T, Xg24=X13T −X13+X24−X24T,

g

X31=X13T −X13+X24−X24T, Xg32=X14T −X14+X23T −X23,

g

X33=X11+X22+X33+X44, Xg34=X12−X12T +X34−X34T,

g

X41=X14T −X14+X23T −X23, Xg42=X13−X13T +X24T −X24,

g

X43=X12T −X12+X34T −X34, Xg44=X11+X22+X33+X44.

Y +TpY Tp−1+RpY Rp−1+SpY Sp−1 has a form similar to (gXij)4×4. We omit it here

for simplicity.

Let

ˆ

X =1

4(X11+X22+X33+X44) + 1 4(X

T

12−X12+X34T −X34)i

+1 4(X

T

13−X13+X24−X24T)j+

1 4(X

T

14−X14+X23T −X23)k,

ˆ

Y =1

4(Y11+Y22+Y33+Y44) + 1 4(Y

T

12−Y12+Y34T−Y34)i

+1 4(Y

T

13−Y13+Y24−Y24T)j+

1 4(Y

T

14−Y14+Y23T −Y23)k.

Then by (3.1),

φ( ˆX) =1

4(X+TnXT −1

n +RnXR−n1+SnXSn−1),

φ( ˆY) =1

4(Y +TpY T −1

p +RpY R−p1+SpY S−p1),

we have that ˆX,Yˆ are a pair Hermitian solution of (1.1) by the property (a). Observe thatXij andYij, i, j= 1,2,3,4 in (3.2) can be written as

Xij =PiXPjT, Yij =QiY QTj.

From Lemma 2.2, the general solutions to (3.2) can be written as

X=φ(X0) +φ(S1)Lφ(G)ZLφ(G)φT(S1) + 4Lφ(A)[V1, V2, V3, V4]

+ 4 [V1, V2, V3, V4]TLφ(A),

Y =φ(Y0)−φ(S2)Lφ(G)ZLφ(G)φT(S2) + 4Lφ(B)[W1, W2, W3, W4]

(14)

whereZ is an arbitrary real symmetric matrix andV1, V2, V3, V4, W1, W2, W3, and W4 are arbitrary with suitable sizes. Hence, fori, j= 1,2,3,4,

Xij =Piφ(X0)PjT+Piφ(S1)Lφ(G)ZLφ(G)φT(S1)PjT + 4PiLφ(A)Vj+ 4ViTLφ(A)PjT,

Yij =Qiφ(Y0)QTj −Qiφ(S2)Lφ(G)ZLφ(G)φT(S2)QTj + 4QiLφ(B)Wj+ 4WiTLφ(B)QTj.

Substituting them into (3.3) and (3.4) yields real matricesXi andYi, i= 1,2,3,4 in

(3.5)-(3.12).

Now we consider the maximal and minimal ranks of real matrices Xi, Yi in

Her-mitian solutions X = X1 +X2i+X3j+ X4k and Y = Y1 +Y2i+Y3j +Y4k to

(1.1).

Theorem 3.2. Suppose the matrix equation (1.1)has a Hermitian solution over

H, and fori, j= 1,2,3,4,

Ji=

½

Xi∈Rn×n

¯ ¯ ¯

¯ A(X1+X2i+X3j+X4k)A

∗ +B(Y1+Y2i+Y3j+Y4k)B∗=C

¾

,

Tj=

½

Yj∈Rp×p

¯ ¯ ¯

¯ A(X1+X2i+X3j+X4k)A

∗ +B(Y1+Y2i+Y3j+Y4k)B∗=C

¾

.

Then we have the following:

(a) The maximal and minimal ranks of Xi in the general Hermitian solution

X =X1+X2i+X3j+X4kto(1.1)are given by

max

Xi∈Ji

r(Xi) = min

(

n, r

"

0 0 Ai

e

Ai φ(B) φ(C)

#

+ 2n−4r(A)−4r(G)

)

;

min

Xi∈Ji

r(Xi) = 2r

"

0 0 Ai

e

Ai φ(B) φ(C)

#

−r

  

0 0 Ai

0 0 φT(B)

e

Ai φ(B) φ(C)

  

−2r

   

−A2 −A3 −A4 A1 −A4 A3 A4 A1 −A2

−A3 A2 A1

   ,

where

e

A1=

   

A2 A3 −A4

−A1 −A4 −A3 A4 −A1 A2 A3 −A2 −A1

   , Ae2=

   

−A1 A3 −A4

−A2 −A4 −A3

−A3 −A1 A2 A4 −A2 −A1

(15)

e

A3=

   

−A1 A2 −A4

−A2 −A1 −A3

−A3 A4 A2 A4 A3 −A1

   , Ae4=

   

−A1 A2 A3

−A2 −A1 −A4

−A3 A4 −A1 A4 A3 −A2

   ,

A1=

   

−A2 A3 A4

−A1 −A4 −A3

−A4 −A1 −A2 A3 A2 −A1

   

T

, A2=

   

−A1 A3 A4 A2 −A4 −A3

−A3 −A1 −A2 A4 A2 −A1

    T ,

A3=

   

−A1 −A2 A4 A2 −A1 −A3

−A3 −A4 −A2

−A4 A3 −A1

   

T

, A4=

   

−A1 −A2 A3 A2 −A1 A4

−A3 −A4 −A1

−A4 A3 A2

    T .

(b) The maximal and minimal ranks of Yj in a Hermitian solutionY1+Y2i+ Y3j+Y4k to(1.1)are given by

max

Yj∈Tj

r(Yj) = min

(

p, r

"

0 0 Bj

e

Bj φ(A) φ(C)

#

+ 2p−4r(B)−4r(G)

)

;

min

Yj∈Tj

r(Yj) = 2r

"

0 0 Bj

e

Bj φ(A) φ(C)

#

−r

  

0 0 Bj

0 0 φT(A)

e

Bj φ(A) φ(C)

   −2r    

−B2 −B3 −B4 B1 −B4 B3 B4 B1 −B2

−B3 B2 B1

   , where e

B1=

   

B2 B3 −B4

−B1 −B4 −B3 B4 −B1 B2 B3 −B2 −B1

   , Be2=

   

−B1 B3 −B4

−B2 −B4 −B3

−B3 −B1 B2 B4 −B2 −B1

   ,

e

B3=

   

−B1 B2 −B4

−B2 −B1 −B3

−B3 B4 B2 B4 B3 −B1

   , Be4=

   

−B1 B2 B3

−B2 −B1 −B4

−B3 B4 −B1 B4 B3 −B2

(16)

B1=

  

−B2 B3 B4

−B1 −B4 −B3

−B4 −B1 −B2 B3 B2 −B1

 

 , B2=

  

−B1 B3 B4 B2 −B4 −B3

−B3 −B1 −B2

−B4 B2 −B1

   ,

B3=

   

−B1 −B2 B4 B2 −B1 −B3

−B3 −B4 −B2

−B4 B3 −B1

   

T

, B4=

   

−B1 −B2 B3 B2 −B1 B4

−B3 −B4 −B1

−B4 B3 B2

   

T

.

Proof. We only derive the maximal and minimal ranks of the matrix X1. The

others can be established similarly. Applying Lemma 2.3 to (3.5), we get the following

max

X1∈J1

r(X1) = max

Z=ZT,Vr(M+

1 4P Zˆ Pˆ

T +P V +VTPT)

= min

½

n, max

Z=ZTr(U)

¾

= min

½

n, r

·

M P Pˆ PT 0 0

¸¾

,

min

X1∈J1

r(X1) = min

Z=ZT,Vr(M+

1 4P Zˆ Pˆ

T +P V +VTPT)

= min

Z=ZTr(U)−2r(P)

= 2r

·

M P Pˆ PT 0 0

¸

−r

  

M P Pˆ PT 0 0

ˆ

PT 0 0

 

−2r(P),

where

U =

·

M +14P Zˆ PˆT P

PT 0

¸

=

·

M P

PT 0

¸

+1 4

· ˆ

P

0

¸

Z

h

ˆ

PT 0 i,

M = 1

4P1φ(X0)P

T

1 +

1

4P2φ(X0)P

T

2 +

1

4P3φ(X0)P

T

3 +

1

4P4φ(X0)P

T

4 ,

(17)

By Lemma 2.4, block Gaussian elimination, andAX0A∗+BY0B∗=C,we have that

r

·

M P Pˆ PT 0 0

¸

=r

   

M D1 D2 0

DT

1 0 0 Ψ(A∗)

0 Ψ(A) 0 0 0 0 Ψ(G) 0

  

−4r(φ(A))

−4r(φ(A∗))4r(φ(G)) =r

"

0 0 A1

e

A1 φ(B) φ(C)

#

+ 2n−4r(A)−4r(G),

where

D1= [P1, P2, P3, P4], D2= [P1, P2, P3, P4]φ(S1),

Ψ(V) =

   

φ(V) 0 0 0 0 φ(V) 0 0 0 0 φ(V) 0 0 0 0 φ(V)

  

, V =A, A∗, G.

Similarly, we can simplify the following

r

  

M P Pˆ PT 0 0

ˆ

PT 0 0

  =r

  

0 0 A1

0 0 φT(B)

e

A1 φ(B) φ(C)

 

+ 2n−4r(G)−4r(G∗),

r(P) =r

   

−A2 −A3 −A4 A1 −A4 A3 A4 A1 −A2

−A3 A2 A1

  

+n−4r(A).

Thus we have the results for extreme ranks of the matrixX1in (a). Similarly, applying

Lemma 2.3 and Lemma 2.4 to (3.6)–(3.12) yields the other results in (a) and (b).

Corollary 3.3. Suppose that the matrix equation(1.1)has a Hermitian solution

overH,then we have the following:

(a) (1.1)has a real symmetric solution X if and only if, fori= 2,3,4,

2r

"

0 0 Ai

e

Ai φ(B) φ(C)

#

=r

  

0 0 Ai

0 0 φT(B)

e

Ai φ(B) φ(C)

  

+ 2r

   

−A2 −A3 −A4 A1 −A4 A3 A4 A1 −A2

−A3 A2 A1

(18)

In that case, the real symmetric solutionX can be expressed as X =X1 in

(3.5).

(b) All the solutions of(1.1) forX are real symmetric if and only if

r

"

0 0 Ai

e

Ai φ(B) φ(C)

#

+ 2n= 4r(A) + 4r(G), i= 2,3,4.

In that case, the real symmetric solutionX can be expressed as X =X1 in

(3.5).

(c) (1.1)has a complex Hermitian solutionX if and only if, fori= 3,4,

2r

"

0 0 Ai

e

Ai φ(B) φ(C)

#

=r

  

0 0 Ai

0 0 φT(B)

e

Ai φ(B) φ(C)

  

+ 2r

   

−A2 −A3 −A4 A1 −A4 A3 A4 A1 −A2

−A3 A2 A1

   .

In that case, the complex Hermitian solution X can be expressed as X =

X1+X2i, whereX1, X2 are expressed as (3.5)and(3.6).

(d) All the solutions of(1.1) forX are complex Hermitian if and only if

r

"

0 0 Ai

e

Ai φ(B) φ(C)

#

+ 2n= 4r(A) + 4r(G), i= 3,4.

In that case, the complex Hermitian solution X can be expressed as X =

X1+X2i, whereX1, X2 are expressed as (3.5)and(3.6). (e) (1.1)has a pure imaginary Hermitian solution X if and only if

2r

"

0 0 A1

e

A1 φ(B) φ(C)

#

=r

  

0 0 A1

0 0 φT(B)

e

A1 φ(B) φ(C)

  

+ 2r

   

−A2 −A3 −A4 A1 −A4 A3 A4 A1 −A2

−A3 A2 A1

   .

(19)

(f ) All the solutions of (1.1) forX are pure imaginary Hermitian if and only if

r

"

0 0 A1

e

A1 φ(B) φ(C)

#

+ 2n= 4r(A) + 4r(G).

In that case, the pure imaginary Hermitian solution X can be expressed as X =X2i+X3j+X4k, whereX2, X3, andX4 are expressed as (3.6),(3.7), and(3.8).

Using the same method, we can get the corresponding results onY.

Remark 3.4. Similarly, we can get the corresponding results on the

skew-Hermitian solution of (1.1).

4. Ranks of Hermitian solution to some special cases of (1.1). In this section, we consider some special cases of (1.1) over C. When B vanishes, (1.1)

becomes (1.2) where A ∈ Cm×n, C Cm×m. We get the corresponding results on

(1.2) as follows.

Corollary 4.1. Let A=A1+A2iCm×n, C =C=C

1+C2i ∈Cm×m be given.Then

(a) The matrix equation (1.2) has a Hermitian solution if and only if the real matrix equation

(4.1)

·

A1 −A2 A2 A1

¸ ·

X11 X12 X21 X22

¸ ·

AT1 AT2

−AT

2 AT1

¸

=

·

C1 −C2 C2 C1

¸

has a symmetric solution overR. In this case, the general Hermitian solution of (1.2)overCcan be written as

(4.2) X =X1+X2i=

1

2(X11+X22) + 1 2(X

T

12−X12)i,

where Xtt =XttT,t = 1,2; andX12T =X21 are the general solutions of (4.1) overR.Written in an explicit form, X1,X2 in(4.2) are

X1=

1

2P1φ(X0)P

T

1 +

1

2P2φ(X0)P

T

2 + [P1, P2]Lφ(A)

·

V1 V2

¸

+

·

V1 V2

¸T

LT

φ(A)[P1, P2]T, X2=

1

2P2φ(X0)P

T

1 −

1

2P1φ(X0)P

T

2 −[−P2, P1]Lφ(A)

·

V1 V2

¸

+

·

V1 V2

¸T

(20)

whereX0=A C(A ) , P1= [In,0], P2= [0, In],andV1andV2are arbitrary real matrices with compatible sizes.

(b) Put

J1=

©

X1∈Rn×n | A(X1+X2i)A∗=C

ª

,

J2=©X2∈Rn×n | A(X1+X2i)A∗=Cª. Then we have the following:

(i) The maximal and minimal ranks of X1 in the Hermitian solution X = X1+X2i to(1.2)are given by

max

X1∈J1

r(X1) = min

  n, r

 

C1 −C2 A1 C2 C1 A2 AT

1 AT2 0

+ 2n−4r(A)

  ,

min

X1∈J1

r(X1) =r

 

C1 −C2 A1 C2 C1 A2 AT

1 AT2 0

 −2r

·

A1 A2

¸

.

(ii) The maximal and minimal ranks of X2 in the Hermitian solution X = X1+X2i to(1.2)are given by

max

X2∈J2

r(X2) = min

  n, r

 

C1 −C2 A1 C2 C1 A2 AT

2 −AT1 0

+ 2n−4r(A)

  ,

min

X2∈J2

r(X2) =r

 

C1 −C2 A1 C2 C1 A2 AT

2 −AT1 0

 −2r

·

A1 A2

¸

.

Remark 4.2. Corollary 4.1 is Theorem 2.2 of [13]. Similarly, Theorem 3.2 of

[13] can be regarded as a special case of (1.1) with skew-Hermitian solutions.

Acknowledgment. The authors would like to thank Professor Michael Neumann and a referee very much for their valuable suggestions and comments, which resulted in a great improvement of the original manuscript.

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