vol18_pp21-29. 131KB Jun 04 2011 12:06:02 AM

Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 18, pp. 21-29, January 2009

ELA

http://math.technion.ac.il/iic/ela

THE ANTI-SYMMETRIC ORTHO-SYMMETRIC SOLUTIONS OF
THE MATRIX EQUATION AT XA = D∗
QING-FENG XIAO† , XI-YAN HU† , AND LEI ZHANG†
Abstract. In this paper, the following problems are discussed.
n such that AT XA = D,
Problem I. Given matrices A ∈ Rn×m and D ∈ Rm×m , find X ∈ ASRP
where
n
ASRP
= {X ∈ ASRn×n |P X ∈ SRn×n for given P ∈ ORn×n satisfying P T = P }.

˜ ∈ Rn×n , find X
ˆ ∈ SE such that

Problem II. Given a matrix X
˜ − Xk
ˆ =
kX

˜ − Xk,
inf kX

X∈SE

where k · k is the Frobenius norm, and SE is the solution set of Problem I.
Expressions for the general solution of Problem I are derived. Necessary and sufficient conditions
for the solvability of Problem I are provided. For Problem II, an expression for the solution is given
as well.
Key words. Anti-symmetric ortho-symmetric matrix, Matrix equation, Matrix nearness problem, Optimal approximation, Least-square solutions.
AMS subject classifications. 65F15, 65F20.

1. Introduction. Let Rn×m denote the set of all n × m real matrices, and let
OR
, SRn×n , ASRn×n denote the set of all n × n orthogonal matrices, the set of

all n × n real symmetric matrices, the set of all n × n real skew-symmetric matrices,
respectively. The symbol Ik will stand for the identity matrix of order k, A+ for
the Moore-Penrose generalized inverse of a matrix A, and rank(A) for the rank of
matrix A. For matrices A, B ∈ Rn×m , the expression A ∗ B will be the Hadamard
product of A and B; also k · k will denote the Frobenius norm. Defining the inner
product (A, B) = tr(B T A) for matrices A, B ∈ Rn×m , Rn×m becomes a Hilbert
space. The norm of a matrix generated by this inner product is the Frobenius norm.
If A = (aij ) ∈ Rn×n , let LA = (lij ) ∈ Rn×n be defined as follows: lij = aij whenever
i > j and lij = 0 otherwise (i, j = 1, 2, . . . , n). Let ei be the i-th column of the
identity matrix In (i = 1, 2, . . . , n) and set Sn = (en , en−1 , . . . , e1 ). It is easy to see
that
n×n

SnT = Sn , SnT Sn = In .
An inverse problem [2-6] arising in the structural modification of the dynamic
behaviour of a structure calls for the solution of the matrix equation
(1.1)

AT XA = D,

where A ∈ Rn×m , D ∈ Rm×m , and the unknown X is required to be real and
symmetric, and positive semidefinite or possibly definite. No assumption is made
∗ Received

by the editors October 6, 2008. Accepted for publication on December 30, 2008.
Handling Editor: Harm Bart.
† College of Mathematics and Econometrics, Hunan University, Changsha 410082, P.R. of China
(qfxiao@hnu.cn). Research supported by National Natural Science Foundation of China (under
Grant 10571047) and the Doctorate Foundation of the Ministry of Education of China (under Grant
20060532014).
21

Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 18, pp. 21-29, January 2009

ELA

http://math.technion.ac.il/iic/ela

22

Q.F. Xiao, X.Y. Hu, and L. Zhang

about the relative sizes of m and n, and it is assumed throughout that A 6= 0 and
D 6= 0.
Equation (1.1) is a special case of the matrix equation
(1.2)

AXB = C.

Consistency conditions for equation (1.2) were given by Penrose [7] (see also [1]).
When the equation is consistent, a solution can be obtained using generalized inverses. Khatri and Mitra [8] gave necessary and sufficient conditions for the existence
of symmetric and positive semidefinite solutions as well as explicit formulae using generalized inverses. In [9,10] solvability conditions for symmetric and positive definite
solutions and general solutions of Equation (1.2) were obtained through the use of
the generalized singular value decomposition [11-13].
For important results on the inverse problem AT XA = D associated with several
kinds of different sets S, for instance, symmetric matrices, symmetric nonnegative
definite matrices, bisymmetric (same as persymmetric) matrices, bisymmetric nonnegative definite matrices and so on, we refer the reader to [14-17].
For the case the unknown A is anti-symmetric ortho-symmetric, [18] has discussed

the inverse problem AX = B. However, for this case, the inverse problem AT XA = D
has not been dealt with yet. This problem will be considered here.
Definition 1.1. A matrix P ∈ Rn×n is said to be a symmetric orthogonal matrix
T
if P = P, P T P = In .
In this paper, without special statement, we assume that P is a given symmetric
orthogonal matrix.
Definition 1.2. A matrix X ∈ Rn×n is said to be a anti-symmetric orthosymmetric matrix if X T = −X, (P X)T = P X. We denote the set of all n × n
n
anti-symmetric ortho-symmetric matrices by ASRP
.
The problem studied in this paper can now be described as follows.
Problem I. Given matrices A ∈ Rn×m and D ∈ Rm×m , find an anti-symmetric
ortho-symmetric matrix X such that
AT XA = D.
In this paper, we discuss the solvability of this problem and an expression for its
solution is presented.
The optimal approximation problem of a matrix with the above-given matrix
restriction comes up in the processes of test or recovery of a linear system due to
˜ of the unknown

incomplete data or revising given data. A preliminary estimate X
matrix X can be obtained by the experimental observation values and the information
ˆ that satisfies the
of statistical distribution. The optimal estimate of X is a matrix X
˜
given matrix restriction for X and is the best approximation of X, see [19-21].
In this paper, we will also consider the so-called optimal approximation problem
associated with AT XA = D. It reads as follows.
˜ ∈ Rn×n , find X
ˆ ∈ SE such that
Problem II. Given matrix X
˜ − Xk,
˜ − Xk
ˆ = inf kX
kX
X∈SE

where SE is the solution set of Problem I.
We point out that if Problem I is solvable, then Problem II has a unique solution,
and in this case an expression for the solution can be derived.

Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 18, pp. 21-29, January 2009

ELA

http://math.technion.ac.il/iic/ela

The Anti-symmetric Ortho-symmetric Solutions of the Matrix Equation AT XA = D

23

The paper is organized as follows. In Section 2, we obtain the general form of SE
and the sufficient and necessary conditions under which Problem I is solvable mainly
n
by using the structure of ASRP
and orthogonal projection matrices. In Section 3,
the expression for the solution of the matrix nearness problem II will be provided.
2. The expression of the general solution of problem I. In this section we

first discuss some structure properties of symmetric orthogonal matrices. Then, given
n
such a matrix P , we consider structural properties of the subset ASRP
of Rn×n .
Finally, we present necessary and sufficient conditions for the existence of and the
expressions for the anti-symmetric ortho-symmetric (with respect to the given P )
solutions of problem I.
Lemma 2.1. Assume P is a symmetric orthogonal matrix of size n, and let
(2.1)

P1 =

1
1
(In + P ), P2 = (In − P ).
2
2

Then P1 and P2 are orthogonal projection matrices satisfying P1 +P2 = In , P1 P2 = 0.
Proof. By direct computation.

Lemma 2.2. Assume P1 and P2 are defined as (2.1) and rank(P1 ) = r. Then
rank(P2 ) = n − r, and there exist unit column orthogonal matrices U1 ∈ Rn×r and
U2 ∈ Rn×(n−r) such that
P1 = U1 U1T , P2 = U2 U2T , P = U1 U1T − U2 U2T , U1T U2 = 0.
Proof. Since P1 and P2 are orthogonal projection matrices satisfying P1 + P2 =
In , P1 P2 = 0, the column space R(P2 ) of the matrix P2 is the orthogonal complement
of the column space R(P1 ) of the matrix P1 , in other words, Rn = R(P1 ) ⊕ R(P2 ).
Hence, if rank(P1 ) = r, then rank(P2 ) = n − r. On the other hand, rank(P1 ) = r,
rank(P2 ) = n − r, and P1 , P2 are orthogonal projection matrices. Thus there exist
unit column orthogonal matrices U1 ∈ Rn×r and U2 ∈ Rn×(n−r) such that P1 =
U1 U1T , P2 = U2 U2T . Using Rn = R(P1 ) ⊕ R(P2 ), we have U1T U2 = 0. Substituting
P1 = U1 U1T , P2 = U2 U2T into (2.1), we have P = U1 U1T − U2 U2T .
Elaborating on Lemma 2.2 and its proof, we note that U = (U1 , U2 ) is an orthogonal matrix and that the symmetric orthogonal matrix P can be expressed as
¶
µ
Ir
0
(2.2)
UT .
P =U

0 −In−r
n
Lemma 2.3. The matrix X ∈ ASRP
if and only if X can be expressed as
¶
µ
0
F
(2.3)
UT ,
X=U
−F T 0

where F ∈ Rr×(n−r) and U is the same as (2.2).
n
n
Proof. Assume X ∈ ASRP
. By Lemma 2.2 and the definition of ASRP
, we have
P1 XP1 =

P2 XP2 =

I +P I +P
1
1
X
= (X + P X + XP + P XP ) = (P X + XP ),
2
2
4
4
1
1
I −P I −P
X
= (X − P X − XP + P XP ) = − (P X + XP ).
2
2
4
4

Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 18, pp. 21-29, January 2009

ELA

http://math.technion.ac.il/iic/ela

24

Q.F. Xiao, X.Y. Hu, and L. Zhang

Hence,
X = (P1 + P2 )X(P1 + P2 ) = P1 XP1 + P1 XP2 + P2 XP1 + P2 XP2
= P1 XP2 + P2 XP1 = U1 U1T XU2 U2T + U2 U2T XU1 U1T .
Let F = U1T XU2 , G = U2T XU1 , it is easy to verify that F T = −G. Then we
have
¶
µ
¶
µ
0
F
0 F
T
T
T
UT
X = U1 F U2 + U2 GU1 = U
U =U
G 0
−F T 0
Conversely, for any F ∈ Rr×(n−r) , let
µ
0
X=U
−F T

¶

F
0

UT .

It is easy to verify that X T = −X. Using (2.2), we have
P XP = P U

µ

µ

0
−F T

¶
µ
0
0
UT U
−In−r
−F T
¶
µ
0 −F
U T = −X.
=U
FT
0
=U

Ir
0

This implies that X = U

µ

0
−F T

F
0

¶

F
0

F
0
¶

¶

UT P

UT U

µ

Ir
0

0
−In−r

¶

UT

n
.
U T ∈ ASRP

Lemma 2.4. Let A ∈ Rn×n , D ∈ ASRn×n and assume A − AT = D. Then there
is precisely one G ∈ SRn×n such that
A = LD + G,
and G = 12 (A + AT ) − 12 (LD + LTD ) .
Proof. For given A ∈ Rn×n , D ∈ ASRn×n and A − AT = D, it is easy to
verify that there exist unique G = 21 (A + AT ) − 12 (LD + LTD ) ∈ SRn×n , and we have
A = 21 (A − AT ) + 21 (A + AT ) = 21 (LD − LTD ) + 12 (A + AT ) = LD + 12 (A + AT ) − 21 (LD +
LTD ) = LD + G.
Let A ∈ Rn×m and D ∈ Rm×m , U defined in (2.2). Set
(2.4)

UT A =

µ

A1
A2

¶

, A1 ∈ Rr×m , A2 ∈ R(n−r)×m .

The generalized singular value decomposition (see [11,12,13]) of the matrix pair
[AT1 , AT2 ] is
(2.5)

AT1 = M ΣA1 W T , AT2 = M ΣA2 V T ,

Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 18, pp. 21-29, January 2009

ELA

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The Anti-symmetric Ortho-symmetric Solutions of the Matrix Equation AT XA = D

25

where W ∈ C m×m is a nonsingular matrix, W ∈ ORr×r , V ∈ OR(n−r)×(n−r) and


Ik
k


s
S
1


 t−k−s
O
(2.6)
,
ΣA1 = 
1




···············
m−t
O


(2.7)

ΣA2



O2
S2



=



It−k−s
···············
O

k

s

 t−k−s


m−t

,

where
t = rank(AT1 , AT2 ), k = t − rank(AT2 ),
s = rank(AT1 ) + rank(AT2 ) − t
S1 = diag(α1 , · · · , αs ), S2 = diag(β1 , · · · , βs ),
with 1 > α1 ≥ · · · ≥ αs > 0, 0 < β1 ≤ · · · ≤ βs < 1, and αi2 + βi2 = 1, i = 1, · · · , s.
O, O1 and O2 are corresponding zero submatrices.
Then we can immediately obtain the following theorem about the general solution
to Problem I.
Theorem 2.5. Given A ∈ Rn×m and D ∈ Rm×m , U defined in (2.2), and U T A
has the partition form of (2.4), the generalized singular value decomposition of the
−T
matrix pair [AT1 , AT2 ] as (2.5). Partition
the matrix M −1 DM
as


D11 D12 D13 D14
k


D
D
D
D
s
21
22
23
24 
M −1 DM −T = 
 D31 D32 D33 D34  t − k − s ,
m−t
D41 D42 D43 D44
k s t−k−s m−t
,
n
then the problem I has a solution X ∈ ASRP
if and only if
(2.8)

DT = −D, D11 = O, D33 = O, D41 = O, D42 = O, D43 = O, D44 = O.

In that case it has the general solution
µ
(2.9)
X=U

0
−F T

F
0

¶

UT ,

where
(2.10)



X11

F =W
X21
X31

D12 S2−1
S1−1 (LD22 + G)S2−1
X32


D13
S1−1 D23  V T ,
X33

with X11 ∈ Rr×(n−r+k−t) , X21 ∈ Rs×(n−r+k−t) , X31 ∈ R(r−k−s)×(n−r+k−t) , X32 ∈
R(r−k−s)×s , X33 ∈ R(r−k−s)×(t−k−s) and G ∈ SRs×s are arbitrary matrices.

Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 18, pp. 21-29, January 2009

ELA

http://math.technion.ac.il/iic/ela

26

Q.F. Xiao, X.Y. Hu, and L. Zhang

n
Proof. The necessity. Assume Eq.(1.1) has a solution X ∈ ASRP
. By the
n
T
definition of ASRP , it is easy to verify that D = −D, and we have from Lemma 2.3
that X can be expressed as
¶
µ
0
F
(2.11)
UT .
X=U
−F T 0

where F ∈ Rr×(n−r) .
Note that U is an orthogonal matrix, and the definition of Ai (i = 1, 2), Eq.(1.1)
is equivalent to
(2.12)

AT1 F A2 − AT2 F A1 = D.

Substituting (2.5) into (2.12), then we have
(2.13)

ΣA1 (W T F V )ΣTA2 − ΣA2 (W T F V )T ΣTA1 = M −1 DM −T ,

Partition the matrix W T F V as
(2.14)



X11
W T F V =  X21
X31

X12
X22
X32


X13
X23  ,
X33

where X11 ∈ Rr×(n−r+k−t) , X22 ∈ Rs×s , X33 ∈ R(r−k−s)×(t−k−s) .
Taking W T F V and M −1 DM −T into (2.13), we have
 

0
X12 S2
X13
0
D11 D12
T
T


 −S2 X21
S
X
S
−
(S
X
S
)
S
X
0
1 22 2
1 22 2
1 23
 =  D21 D22

T
T
 −X13
−X23
S1
0
0   D31 D32
D41 D42
0
0
0
0
(2.15)

D13
D23
D33
D43


D14
D24 
.
D34 
D44

Therefore (2.15) holds if and only if (2.8) holds and
X12 = D12 S2−1 , X13 = D13 , X23 = S1−1 D23
and
S1 X22 S2 − (S1 X22 S2 )T = D22 .
It follows from Lemma 2.4 that X22 = S1−1 (LD22 + G)S2−1 , where G ∈ SRs×s is
arbitrary matrix. Substituting the above into (2.14), (2.11), thus we have formulation
(2.9) and (2.10).
The sufficiency. Let


D13
X11
D12 S2−1
FG = W  X21 S1−1 (LD22 + G)S2−1 S1−1 D23  V T .
X31
X32
X33
Obviously, FG ∈ Rr×(n−r) . By Lemma 2.3 and
¶
µ
0
FG
UT ,
X0 = U
−FGT
0

Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 18, pp. 21-29, January 2009

ELA

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The Anti-symmetric Ortho-symmetric Solutions of the Matrix Equation AT XA = D

27

n
we have X0 ∈ ASRP
. Hence
T

T

T

T

( AT1

A X0 A = A U U X0 U U A =

µ

AT2

)

µ

0
−FG

FG
0

¶µ

A1
A2

¶

¶

A1
= ( AT2 (−FG ) AT1 FG )
= AT2 (−FG )A1 + AT1 FG A2
A2


0
D12 S2T
D13 0
T
 −S2 D12
LD22 − LTD22 D23 0 
 M T = D.
=M
T
T
 −D13
−D23
0
0 
0
0
0
0
¶
µ
0
FG
n
is the anti-symmetric orthoU T ∈ ASRP
This implies that X0 = U
−FGT
0
symmetric solution of Eq. (1.1). The proof is completed.
3. The expression of the solution of Problem II. To prepare for an explicit
expression for the solution of the matrix nearness problem II, we first verify the
following lemma.
Lemma 3.1. Suppose that E, F ∈ Rs×s , and let Sa = diag(a1 , · · · , as ) > 0,
Sb = diag(b1 , · · · , bs ) > 0. Then there exist a unique Ss ∈ SRs×s and a unique
Sr ∈ ASRs×s such that
kSa SSb − Ek2 + kSa SSb − F k2 = min

(3.1)
and
(3.2)

Ss = Φ ∗ [Sa (E + F )Sb + Sb (E + F )T Sa ],

(3.3)

Sr = Φ ∗ [Sa (E + F )Sb − Sb (E + F )T Sa ],

where
(3.4)

Φ = (ψij ) ∈ SRs×s , ψij =

2(a2i b2j

1
, 1 ≤ i, j ≤ s.
+ a2j b2i )

Proof. We prove only the existence of Sr and (3.3). For any S = (sij ) ∈ ASRs×s ,
E = (eij ), F = (fij ) ∈ Rs×s , since sii = 0, sij = −sji ,
X
kSa SSb − Ek2 + kSa SSb − F k2 =
[(ai bj sij − eij )2 + (ai bj sij − fij )2 ]
1≤i,j≤s

=

X

[(ai bj sij − eij )2 + (−aj bi sij − eji )2 + (ai bj sij − fij )2 + (−aj bi sij − fji )2 ] +

1≤i