- Email: [email protected]
The minimal rank of A-BX with respect to Hermitian matrix
Applied Mathematics and Computation 233 (2014) 55–61
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
The minimal rank of A BX with respect to Hermitian matrix q Hongxing Wang Department of Mathematics, Huainan Normal University, Anhui, PR China
a r t i c l e
i n f o
a b s t r a c t In this paper, we discuss the minimal rank of A BX when X is Hermitian by applying singular value decomposition and some rank equalities of matrices, and obtain a representation of the minimal rank. Based on the representation, we obtain necessary and sufficient conditions for the matrix equation BX ¼ A to have Hermitian solutions. Ó 2014 Elsevier Inc. All rights reserved.
Keywords: Singular value decomposition Hermitian matrix Rank Inertia
1. Introduction Throughout this paper, we adopt the following notation. The symbol Cmn is the set of all m n complex matrices, Hmm (Umm ) is the set of all m m Hermitian matrices (unitary matrices). The conjugate transpose of A is denoted by AH . The inertia of a Hermitian matrix A is defined to be the triplet lnðAÞ ¼ fnþ ðAÞ; n ðAÞ; n0 ðAÞg, where nþ ðAÞ, n ðAÞ and n0 ðAÞ are the numbers of the positive, negative and zero eigenvalues of A counted with multiplicities, respectively. The two numbers nþ ðAÞ and n ðAÞ are called the positive and negative index of inertia, respectively. The symbol Ip denotes the p-by-p identity matrix. The row block matrix consisting of A and B is denoted by ½ A B and its rank is denoted by rankð½ A B Þ. The symbols nþ fA; Bg and n fA; Bg denote the positive and negative index of the inertia of the block Hermitian matrix
" # BH i A AH ; B 0
respectively. The Moore–Penrose inverse of A 2 Cmn is defined as the unique X 2 Cnm satisfying
ð1Þ AXA ¼ A; ð2Þ XAX ¼ X; ð3Þ ðAX ÞH ¼ AX; ð4Þ ðXAÞH ¼ XA; and is usually denoted by X ¼ Ay (see [1]). Furthermore, the symbols EA and F A stand for the two orthogonal projections:
EA ¼ Im AAy and F A ¼ In Ay A: Consider the following well-known linear matrix equation
BX ¼ A; mn
ð1:1Þ mn
where A 2 C and B 2 C are given matrices. It is obviously that (1.1) has a Hermitian solution if and only if the minimal rank of A BX with respect to X ¼ X H equals 0. The problem for algebraic properties of Hermitian solutions of (1.1) has been widely studied. Khatri and Mitra[7] deduced a necessary and sufficient condition for the existence of a Hermitian solution to (1.1). They show that a Hermitian solution X exists if and only if
BBy A ¼ A and BAH ¼ ABH :
q
This work was supported by the NSFC under grant 11171226 and the Foundation of Anhui Educational Committee under grant KJ2012B175. E-mail address: [email protected]
http://dx.doi.org/10.1016/j.amc.2014.01.116 0096-3003/Ó 2014 Elsevier Inc. All rights reserved.
ð1:2Þ
56
H. Wang / Applied Mathematics and Computation 233 (2014) 55–61
Horn et al. [6] studied (1.1) in a given ⁄congruence class. Li et al.[8] studied the submatrices in a solution X of (1.1) and given some necessary and sufficient conditions for the existence of solutions to be local positive (negative) semidefinite by elementary block matrix operations. Guo and Huang[5] studied the problem for determining the extremal ranks of one matrix expression with respect to Hermitian matrix by applying the quotient singular value decomposition, especially when A is a skew-Hermitian matrix,
min rankðA X Þ ¼ max fnþ ðiAÞ; n ðiAÞg: X¼X H
ð1:3Þ
It is well known that rank of matrix is an important tool in matrix theory and its applications, and many problems are closely related to the ranks of some matrix expressions under some restrictions. Recently, the extremal ranks of some matrix expressions have found many applications in control theory, statistics and economics[2,3,9], etc., and hence the problems for determining the extremal ranks of matrix expressions have been widely studied [4,10,11,13–17],etc. In this paper, we will study the minimal rank of A BX with respect to X ¼ X H by applying singular value decomposition (SVD), Sylvester’s law of inertia and some rank equalities of matrices. 2. Preliminaries In order to find the minimal rank of A BX with respect to Hermitian matrix, we need the following results on minimal ranks of two special matrix expressions. Denote
F ðX ; Y Þ ¼ H X YJ : In the following subsection, we will study the minimal rank of F ðX ; Y Þ when X ¼ X H and Y are arbitrary matrices. 2.1. The minimal rank of F ðX ; Y Þ when X ¼ X H and Y are arbitrary matrices Suppose that the matrix H 2 Cm1 m1 and J 2 Cl1 m1 are given. Write
HH ¼
H þ HH H HH ; HS ¼ ; 2 2
then
H ¼ HH þ HS ; where HH is an m1 m1 Hermitian matrix, and HS is an m1 m1 skew-Hermitian matrix. Notice that for any X 2 Hm1 m1
min rankðH X Þ ¼ min rankðHS X Þ;
X ¼X H
X ¼X H
therefore, it follows from applying (1.3) that
min rankðH X Þ ¼ max nþ i H HH ; n i H HH :
X ¼X H
ð2:1Þ
The following formula for inertias of block matrix can be derived from Theorem 2.3 [12]
n fH; J g ¼ rankðJ Þ þ n F J i H HH F J :
ð2:2Þ
We have the following result. Lemma 2.1. Suppose that H 2 Cm1 m1 and J 2 Cl1 m1 are given, and X 2 Hm1 m1 and Y 2 Cm1 l1 are arbitrary matrices. Then the minimal rank of the matrix expression H X YJ is
min rankðH X YJ Þ ¼ max fnþ fH; J g; n fH; J gg rankðJ Þ:
X ¼X H ;Y
Proof. Let the SVD of J be
" J ¼U
b 0 R
# VH ;
0
0
H1
H2
H3
H4
and partition
V H HV ¼
;
ð2:3Þ
57
H. Wang / Applied Mathematics and Computation 233 (2014) 55–61
where H4 2 Cðm1 rÞðm1 rÞ and rankðJ Þ ¼ r. Then
V H F J HF J V ¼
0
0
0 H4
:
ð2:4Þ
It follows from applying (2.2) that
n i H4 HH4 ¼ n F J i H HH F J ¼ n fH; J g rankðJ Þ:
ð2:5Þ
Furthermore, partition
V H XV ¼
X1
X2
X H2
X4
; V H YU ¼
Y1
Y2
Y3
Y4
;
where X 4 2 Cðm1 rÞðm1 rÞ and Y 4 2 Cðm1 rÞðl1 rÞ . Since X is Hermitian, X 4 is Hermitian. From the SVD of J and (2.4), we obtain
"
H X YJ ¼ V V H HV V H XV V H YU
b 0 R 0
#!
0
"
VH ¼ V
b H1 X 1 Y 1 R H b H3 X Y 3 R 2
H2 X 2 H4 X 4
#
VH :
It follows that
" min rankðH X YJ Þ ¼
X ¼X H ;Y
min
X 4 ¼X H ;X 1 ¼X H ;X 2 ;Y 1 ;Y 3 4 1
rank
b H1 X 1 Y 1 R H b H3 X Y 3 R 2
H2 X 2
#!
H4 X 4
P min rankðH4 X 4 Þ X 4 ¼X H 4
¼ max nþ i H4 HH4 ; n i H4 HH4 ¼ max fnþ fH; J g; n fH; J gg rankðJ Þ: 1 b . Therefore, the formulas in (2.3) b 1 and Y 3 ¼ H3 X H R The lower bound is obtained if we set X 2 ¼ H2 , Y 1 ¼ ðH1 X 1 Þ R 2 hold, and we complete the proof of the theorem. h Denote
f ðX; Y Þ ¼
HX
Y
J
K
:
In the following subsection, we will study the minimal rank of f ðX; YÞ when X ¼ X H and Y are arbitrary matrices. 2.2. The minimal rank of f ðX; YÞ when X ¼ X H and Y are arbitrary matrices Suppose that the matrices H 2 Cm2 m2 , J 2 Ck2 m2 , K 2 Ck2 l2 are given with rankðK Þ ¼ r1 and rankðEK J Þ ¼ r 2 . Let the SVD of K be
K ¼ U1
R1 0 0
0
V H1 ;
ð2:6Þ
and partition
U H1 J ¼
J1 J2
;
ð2:7Þ
where J 1 2 Cr1 m2 and J 2 2 Cðk2 r1 Þm2 . It follows that
rankðJ 2 Þ ¼ rankðEK J Þ ¼ rankð½ K
J Þ rankðK Þ:
ð2:8Þ
Furthermore, let the SVD of J 2 be
J2 ¼ U2
R2 0 0
0
V H2 ;
ð2:9Þ
and partition
V H2 HV 2 ¼
H1
H2
H3
H4
; J 1 V 2 ¼ ½ J 11
J 12 ;
ð2:10Þ
where H4 2 Cðm2 r2 Þðm2 r2 Þ and J 12 2 Cr1 ðm2 r2 Þ . It follows that
V H2 F J2 HF J2 V 2 ¼
0 0 ; J 1 F J2 V 2 ¼ ½ 0 J 12 0 H4
ð2:11Þ
58
H. Wang / Applied Mathematics and Computation 233 (2014) 55–61
and
rankðJ 12 Þ ¼ rankðJ Þ rankðJ 2 Þ ¼ rankðJ Þ þ rankðK Þ rankð½ K
J Þ:
ð2:12Þ
Therefore, applying (2.2) and Sylvester’s law of inertia to n fH; J g gives
02 31 02 31 02 J H1 J H2 i H HH H H iF J2 H HH F J2 B6 7C i HH J B 6 7 C @ 4 5 A @ 4 ¼ n @4 n J1 0 0 5A ¼ n J 1 F J2 J 0 J2 0 0 02 31 H iV 2 F J2 H HH F J2 V 2 V H2 F J2 J H1 5A þ rankðJ 2 Þ ¼ n @4 J 1 F J2 V 2 0 02 31 i H4 HH4 J H12 @ 4 5A þ rankðJ Þ rankðJ 12 Þ; ¼ n J 12 0
F J2 J H1 0
31 5A þ rankðJ 2 Þ
that is,
n fH4 ; J 12 g rankðJ 12 Þ ¼ n fH; J g rankðJ Þ:
ð2:13Þ
We have the following result. Lemma 2.2. Suppose that H 2 Cm2 m2 , J 2 Ck2 m2 , K 2 Ck2 l2 be given, and X 2 Hm2 m2 and Y 2 Cm2 l2 are arbitrary matrices. Then
HX min rank J X¼X H ;Y
Y K
¼ rankð½ K
J Þ rankðJ Þ þ max fnþ fH; J g; n fH; J gg:
ð2:14Þ
Proof. Let the SVDs of K and J 2 , and the partitions of U H1 J, V H2 F J2 HF J2 V 2 and J 1 V 2 are given in (2.6)–(2.11), respectively. Partition
YV 1 ¼ ½ Y 1
Y 2 ;
where Y 1 2 Cm2 r1 and Y 2 2 Cm2 ðl2 r1 Þ . Denote
2
Im2 6 P¼4 0 0
3 " 7 I m2 0 5 0
Y 1 R1 1
0
Ir1 0
Ik2 r1
0 U H1
# ;
Q¼
I m2 0
2 I m2 0 6 1 R 4 1 J1 V1 0
3
0
0
Ir1
7 0 5:
0
Il2 r1
It is obvious that P and Q are both nonsingular matrices. Therefore, 02 31 " #!
H X Y 1 R1 1 J1 0 Y 2 HX Y HX Y H X Y 1 R1 B6 7C 1 J1 Y 2 rank ¼ rank ð K Þ þ rank ¼ rank P Q ¼ rank@4 : 5 A 0 R1 0 J K J K J2 0 J2 0 0 ð2:15Þ Furthermore, partition
" Hb b Y 1 R1 1 ¼ Y ; V2 Y ¼
b 2C where Y
m2 r 1
" rank
# b1 X1 Y 21 Y ; V H2 Y 2 ¼ ; V H2 XV 2 ¼ b2 X H2 Y 22 Y
b2 2 C ,Y
ðm2 r 2 Þr 1
bJ HXY 1
Y2
J2
0
Y 22 2 C
ðm2 r 2 Þðl2 r 2 Þ
#!
X4
;
ðm2 r 2 Þðm2 r 2 Þ
and X 4 2 C
" ¼ rankðJ2 Þ þ rank
X2
b 1J H2 X 2 Y 12 b 2J H4 X 4 Y 12
. Obviously, X 4 is an Hermitian matrix and
Y 21 Y 22
#!
b 2J ; P rankðJ 2 Þ þ rank H4 X 4 Y 12
Applying Lemma 2.1, (2.12) and (2.13), we have
b 2J min rank H4 X 4 Y 12 ¼ max fnþ fH4 ; J 12 g; n fH 4 ; J 12 gg rankðJ 12 Þ b X 4 ¼X H 4 ;Y 2 ¼ rankðJ Þ þ rankðK Þ rank½ K J þ max fnþ fH; J g; n fH; J gg:
ð2:16Þ
59
H. Wang / Applied Mathematics and Computation 233 (2014) 55–61
From Eqs. (2.8), (2.15) and (2.16), we obtain
HX min rank J X¼X H ;Y
Y K
"
¼
min
X 2 ;b Y 1 ;b Y 2 ;Y 21 ;Y 22 ;X 4 ¼X H 4
rank
b 1J H2 X 2 Y 12 b 2J H4 X 4 Y 12
Y 21
#!
Y 22
þ rankðK Þ þ rankðJ 2 Þ
b 2J P min rank H4 X 4 Y 12 þ rankðK Þ þ rankðJ 2 Þ b X 4 ¼X H 4 ;Y 2 ¼ rankð½ K
J Þ rankðJ Þ þ max fnþ fH; J g; n fH; J gg:
b 1 J ; Y 21 ¼ 0 and Y 22 ¼ 0. Therefore, the formula in (2.14) hold, and we The lower bound is obtained if we set X 2 ¼ H2 Y 12 complete the proof of the theorem. h
3. The minimal rank of A BX when X is Hermitian In this section, we give a closed form representation of the minimal rank of A BX when X is Hermitian. Theorem 3.1. Suppose that A 2 Cmn and B 2 Cmn , and X ¼ X H is an arbitrary matrix. Then the minimal rank of the matrix expression A BX is
min rankðA BX Þ ¼ rankð½ A B Þ rank ABH X¼X
B
H
n o : þ max nþ i ABH BAH ; n i ABH BAH
ð3:1Þ
Proof. Let the SVD of B be
B¼U
R 0 0
0
V H;
ð3:2Þ
and partition
U H AV ¼
A1
A2
A3
A4
;
ð3:3Þ
where A1 2 CrB rB , A2 2 CrB ðnrB Þ , rankð½ A3 A4 Þ ¼ rankðEB AÞ and
A3 2 CðmrB ÞrB ,
and
A4 2 CðmrB ÞðnrB Þ .
Then
we
have
rankðA3 Þ ¼ rank EB ABH ,
#! " # " A1 R 0 iðA3 RÞH i A1 R ðA1 RÞH ð A1 R Þ H ð A3 R Þ H H H H U ¼U UH i AB BA ¼ iU A3 R 0 0 0 iA3 R 0 2
3 H R 0 4 i R1 A1 R1 A1 AH3 5 R 0 UH : ¼U 0 iI 0 iI iA3 0 It follows from applying Sylvester’s law of inertia that
rankðA3 Þ ¼ rank ABH rankð½ A3
B
rankðBÞ;
ð3:4Þ
A4 Þ ¼ rankð½ A B Þ rankðBÞ
ð3:5Þ
and
02
31 H H 1 1 A i R A R A 1 1 3 5A ¼ n i ABH BAH : n @4 A3 0
ð3:6Þ
Furthermore, partition
V H XV ¼
X1
X2
X H2
X4
;
where X 2 2 CrB ðnrB Þ and X 1 2 HrB rB . Then
A BX ¼ U
R 0 0
0
"
R1 A1 X 1 R1 A2 X 2 A3
A4
# V H:
ð3:7Þ
60
H. Wang / Applied Mathematics and Computation 233 (2014) 55–61
Applying (2.14) and (3.4)–(3.6) to (3.7) gives
" min rankðA BX Þ ¼ min rank
R1 A1 X 1 R1 A2 X 2
#!
A4 n n o n oo ¼ rankð½ A3 A4 Þ rankðA3 Þ þ max nþ R1 A1 ; A3 ; n R1 A1 ; A3 n o
¼ rankð½ A B Þ rank ABH B þ max nþ i ABH BAH ; n i ABH BAH :
X¼X H
A3
X 1 ¼X H 1 ;X 2
Therefore, the formula in (3.1) hold, and we complete the proof of the theorem. h Some direct consequences of the previous theorem are given below. Corollary 3.2. Let A; B 2 Cmn be given and suppose that RðAÞ # RðBÞ. Then
n o min rankðA BX Þ ¼ max nþ i ABH BAH ; n i ABH BAH : X¼X H
In particular, if the condition RðAÞ # RðBÞ hold, then A ¼ BX has a Hermitian solution if and only if ABH ¼ BAH . Corollary 3.3. Let A; B 2 Cmn be given and suppose that RðAÞ \ RðBÞ ¼ 0. Then
min rankðA BX Þ ¼ rankðAÞ: X¼X H
Corollary 3.4. Let A 2 Cmn and B 2 Cmn be given. Then, the following conditions are equivalent: (1) There exists a Hermitian matrix X such that BX ¼ A. n o
¼ 0. B Þ rank ABH B þ max nþ i ABH BAH ; n i ABH BAH
(2) rankð½ A
(3) BBy A ¼ A and ABH ¼ BAH , [7]. Proof. It is well known that the linear matrix equation BX ¼ A has a Hermitian solution if and only if minX¼X H rankðA BXÞ ¼ 0. Thus setting the both sides of (3.1) to zero, we obtain the equivalence of Conditions (1) and (2). From (3.7), we know that the matrix equation BX ¼ A has a Hermitian solution if and only if
rankð½ A3
H ¼ 0: A4 Þ ¼ 0 and rank R1 A1 R1 A1
rankð½ A3
A4 Þ ¼ rank A BBy A
Since
and
H ¼ rank BBy ABH BAH BBy ; rank R1 A1 R1 A1 we obtain that the matrix equation BX ¼ A has a Hermitian solution if and only if
rank A BBy A ¼ 0 and rank BBy ABH BAH BBy ¼ 0; H i.e., BBy A ¼ A and BBy A BH ¼ B BBy A . Hence Conditions (1) and (3) are equivalent. Example 3.1. Take
2
1 2 3
6 A ¼ 43
i
2 5 Then rankð½ A
3
2
1 0 0
3
7 6 7 0 5 and B ¼ 4 1 0 0 5: i
B Þ ¼ 3, rank ABH
2
0 2 0 B
¼ 3,
3 0 2 2 6 7 0 2 þ 2i 5; AB BA ¼ 4 2 2 2 þ 2i 0 H
H
h
H. Wang / Applied Mathematics and Computation 233 (2014) 55–61
61
¼ 1 and n i ABH BAH ¼ 2. Thus by (3.1), nþ i ABH BAH
min rankðA BX Þ ¼ 2: X¼X H
Acknowledgements The authors are grateful to the referee for his/her valuable comments and suggestions which helped to greatly improve this paper. References [1] A. Ben-Israel, T.N.E. Greville, Generalized Inverses: Theory and Applications, second ed., Springer, Berlin, 2003. [2] D. Chu, H.C. Chan, D.W. Ho, Regularization of singular systems by derivative and proportional output feedback, SIAM J. Matrix Anal. Appl. 19 (1998) 21– 38. [3] D. Chu, L. De Lathauwer, B. De Moor, On the computation of restricted singluar value decomposition via cosine-sine decomposition, SIAM J. Matrix Anal. Appl. 22 (2000) 580–601. [4] D. Chu, Y.S. Hung, H.J. Woerdeman, Inertia and rank characterizations of some matrix expressions, SIAM J. Matrix Anal. Appl. 31 (2009) 1187–1226. [5] W. Guo, T. Huang, Extremal ranks of matrix expression of A BXC with respect to Hermitian matrix, Appl. Math. Comput. 217 (2010) 2381–2389. [6] R.A. Horn, V.V. Sergeichuk, N. Shaked-Monderer, Solution of linear matrix equations in a Hcongruence class, Electron. J. Linear Algebra 13 (2005) 153– 156. [7] C.G. Khatri, S.K. Mitra, Hermitian and nonnegative definite solutions of linear matrix equations, SIAM J. Appl. Math. 31 (1976) 579–585. [8] Y. Li, F. Zhang, W. Guo, J. Zhao, Solutions with special structure to the linear matrix equation AX ¼ B, Comput. Math. Appl. 61 (2011) 374–383. [9] Y. Liu, On equality of ordinary least squares estimator, best linear unbiased estimator and best linear unbiased predictor in the general linear model, J. Stat. Plann. Infer. 139 (2009) 1522–1529. [10] Y. Liu, Y. Tian, A simultaneous decomposition of a matrix triplet with applications, Numer. Linear Algebra Appl. 18 (2011) 69–85. [11] Y. Tian, Y. Liu, Extremal ranks of some symmetric matrix expressions with applications, SIAM J. Matrix Anal. Appl. 28 (2006) 890905. [12] Y. Tian, Equalities and inequalities for inertias of Hermitian matrices with applications, Linear Algebra Appl. 433 (2010) 263–296. [13] H. Wang, On least squares solutions subject to a rank restriction, Linear Multilinear Algebra, http://dx.doi.org/10.1080/03081087.2013.860598. [14] Q.W. Wang, Z.H. He, Solvability conditions and general solution for mixed Sylvester equations, Automatica 49 (2013) 2713–2719. [15] M. Wei, Q. Wang, On rank-constrained Hermitian nonnegative-definite least squares solutions to the matrix equation AXAH ¼ B, Int. J. Comput. Math. 84 (2007) 945–952. [16] M. Wei, D. Shen, Minimum rank solutions to the matrix approximation problems in the spectral norm, SIAM J. Matrix Anal. Appl. 33 (2012) 940–957. [17] H.J. Woerdeman, Minimal rank completions for block matrices, Linear Algebra Appl. 121 (1989) 105122.
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
The minimal rank of A BX with respect to Hermitian matrix q Hongxing Wang Department of Mathematics, Huainan Normal University, Anhui, PR China
a r t i c l e
i n f o
a b s t r a c t In this paper, we discuss the minimal rank of A BX when X is Hermitian by applying singular value decomposition and some rank equalities of matrices, and obtain a representation of the minimal rank. Based on the representation, we obtain necessary and sufficient conditions for the matrix equation BX ¼ A to have Hermitian solutions. Ó 2014 Elsevier Inc. All rights reserved.
Keywords: Singular value decomposition Hermitian matrix Rank Inertia
1. Introduction Throughout this paper, we adopt the following notation. The symbol Cmn is the set of all m n complex matrices, Hmm (Umm ) is the set of all m m Hermitian matrices (unitary matrices). The conjugate transpose of A is denoted by AH . The inertia of a Hermitian matrix A is defined to be the triplet lnðAÞ ¼ fnþ ðAÞ; n ðAÞ; n0 ðAÞg, where nþ ðAÞ, n ðAÞ and n0 ðAÞ are the numbers of the positive, negative and zero eigenvalues of A counted with multiplicities, respectively. The two numbers nþ ðAÞ and n ðAÞ are called the positive and negative index of inertia, respectively. The symbol Ip denotes the p-by-p identity matrix. The row block matrix consisting of A and B is denoted by ½ A B and its rank is denoted by rankð½ A B Þ. The symbols nþ fA; Bg and n fA; Bg denote the positive and negative index of the inertia of the block Hermitian matrix
" # BH i A AH ; B 0
respectively. The Moore–Penrose inverse of A 2 Cmn is defined as the unique X 2 Cnm satisfying
ð1Þ AXA ¼ A; ð2Þ XAX ¼ X; ð3Þ ðAX ÞH ¼ AX; ð4Þ ðXAÞH ¼ XA; and is usually denoted by X ¼ Ay (see [1]). Furthermore, the symbols EA and F A stand for the two orthogonal projections:
EA ¼ Im AAy and F A ¼ In Ay A: Consider the following well-known linear matrix equation
BX ¼ A; mn
ð1:1Þ mn
where A 2 C and B 2 C are given matrices. It is obviously that (1.1) has a Hermitian solution if and only if the minimal rank of A BX with respect to X ¼ X H equals 0. The problem for algebraic properties of Hermitian solutions of (1.1) has been widely studied. Khatri and Mitra[7] deduced a necessary and sufficient condition for the existence of a Hermitian solution to (1.1). They show that a Hermitian solution X exists if and only if
BBy A ¼ A and BAH ¼ ABH :
q
This work was supported by the NSFC under grant 11171226 and the Foundation of Anhui Educational Committee under grant KJ2012B175. E-mail address: [email protected]
http://dx.doi.org/10.1016/j.amc.2014.01.116 0096-3003/Ó 2014 Elsevier Inc. All rights reserved.
ð1:2Þ
56
H. Wang / Applied Mathematics and Computation 233 (2014) 55–61
Horn et al. [6] studied (1.1) in a given ⁄congruence class. Li et al.[8] studied the submatrices in a solution X of (1.1) and given some necessary and sufficient conditions for the existence of solutions to be local positive (negative) semidefinite by elementary block matrix operations. Guo and Huang[5] studied the problem for determining the extremal ranks of one matrix expression with respect to Hermitian matrix by applying the quotient singular value decomposition, especially when A is a skew-Hermitian matrix,
min rankðA X Þ ¼ max fnþ ðiAÞ; n ðiAÞg: X¼X H
ð1:3Þ
It is well known that rank of matrix is an important tool in matrix theory and its applications, and many problems are closely related to the ranks of some matrix expressions under some restrictions. Recently, the extremal ranks of some matrix expressions have found many applications in control theory, statistics and economics[2,3,9], etc., and hence the problems for determining the extremal ranks of matrix expressions have been widely studied [4,10,11,13–17],etc. In this paper, we will study the minimal rank of A BX with respect to X ¼ X H by applying singular value decomposition (SVD), Sylvester’s law of inertia and some rank equalities of matrices. 2. Preliminaries In order to find the minimal rank of A BX with respect to Hermitian matrix, we need the following results on minimal ranks of two special matrix expressions. Denote
F ðX ; Y Þ ¼ H X YJ : In the following subsection, we will study the minimal rank of F ðX ; Y Þ when X ¼ X H and Y are arbitrary matrices. 2.1. The minimal rank of F ðX ; Y Þ when X ¼ X H and Y are arbitrary matrices Suppose that the matrix H 2 Cm1 m1 and J 2 Cl1 m1 are given. Write
HH ¼
H þ HH H HH ; HS ¼ ; 2 2
then
H ¼ HH þ HS ; where HH is an m1 m1 Hermitian matrix, and HS is an m1 m1 skew-Hermitian matrix. Notice that for any X 2 Hm1 m1
min rankðH X Þ ¼ min rankðHS X Þ;
X ¼X H
X ¼X H
therefore, it follows from applying (1.3) that
min rankðH X Þ ¼ max nþ i H HH ; n i H HH :
X ¼X H
ð2:1Þ
The following formula for inertias of block matrix can be derived from Theorem 2.3 [12]
n fH; J g ¼ rankðJ Þ þ n F J i H HH F J :
ð2:2Þ
We have the following result. Lemma 2.1. Suppose that H 2 Cm1 m1 and J 2 Cl1 m1 are given, and X 2 Hm1 m1 and Y 2 Cm1 l1 are arbitrary matrices. Then the minimal rank of the matrix expression H X YJ is
min rankðH X YJ Þ ¼ max fnþ fH; J g; n fH; J gg rankðJ Þ:
X ¼X H ;Y
Proof. Let the SVD of J be
" J ¼U
b 0 R
# VH ;
0
0
H1
H2
H3
H4
and partition
V H HV ¼
;
ð2:3Þ
57
H. Wang / Applied Mathematics and Computation 233 (2014) 55–61
where H4 2 Cðm1 rÞðm1 rÞ and rankðJ Þ ¼ r. Then
V H F J HF J V ¼
0
0
0 H4
:
ð2:4Þ
It follows from applying (2.2) that
n i H4 HH4 ¼ n F J i H HH F J ¼ n fH; J g rankðJ Þ:
ð2:5Þ
Furthermore, partition
V H XV ¼
X1
X2
X H2
X4
; V H YU ¼
Y1
Y2
Y3
Y4
;
where X 4 2 Cðm1 rÞðm1 rÞ and Y 4 2 Cðm1 rÞðl1 rÞ . Since X is Hermitian, X 4 is Hermitian. From the SVD of J and (2.4), we obtain
"
H X YJ ¼ V V H HV V H XV V H YU
b 0 R 0
#!
0
"
VH ¼ V
b H1 X 1 Y 1 R H b H3 X Y 3 R 2
H2 X 2 H4 X 4
#
VH :
It follows that
" min rankðH X YJ Þ ¼
X ¼X H ;Y
min
X 4 ¼X H ;X 1 ¼X H ;X 2 ;Y 1 ;Y 3 4 1
rank
b H1 X 1 Y 1 R H b H3 X Y 3 R 2
H2 X 2
#!
H4 X 4
P min rankðH4 X 4 Þ X 4 ¼X H 4
¼ max nþ i H4 HH4 ; n i H4 HH4 ¼ max fnþ fH; J g; n fH; J gg rankðJ Þ: 1 b . Therefore, the formulas in (2.3) b 1 and Y 3 ¼ H3 X H R The lower bound is obtained if we set X 2 ¼ H2 , Y 1 ¼ ðH1 X 1 Þ R 2 hold, and we complete the proof of the theorem. h Denote
f ðX; Y Þ ¼
HX
Y
J
K
:
In the following subsection, we will study the minimal rank of f ðX; YÞ when X ¼ X H and Y are arbitrary matrices. 2.2. The minimal rank of f ðX; YÞ when X ¼ X H and Y are arbitrary matrices Suppose that the matrices H 2 Cm2 m2 , J 2 Ck2 m2 , K 2 Ck2 l2 are given with rankðK Þ ¼ r1 and rankðEK J Þ ¼ r 2 . Let the SVD of K be
K ¼ U1
R1 0 0
0
V H1 ;
ð2:6Þ
and partition
U H1 J ¼
J1 J2
;
ð2:7Þ
where J 1 2 Cr1 m2 and J 2 2 Cðk2 r1 Þm2 . It follows that
rankðJ 2 Þ ¼ rankðEK J Þ ¼ rankð½ K
J Þ rankðK Þ:
ð2:8Þ
Furthermore, let the SVD of J 2 be
J2 ¼ U2
R2 0 0
0
V H2 ;
ð2:9Þ
and partition
V H2 HV 2 ¼
H1
H2
H3
H4
; J 1 V 2 ¼ ½ J 11
J 12 ;
ð2:10Þ
where H4 2 Cðm2 r2 Þðm2 r2 Þ and J 12 2 Cr1 ðm2 r2 Þ . It follows that
V H2 F J2 HF J2 V 2 ¼
0 0 ; J 1 F J2 V 2 ¼ ½ 0 J 12 0 H4
ð2:11Þ
58
H. Wang / Applied Mathematics and Computation 233 (2014) 55–61
and
rankðJ 12 Þ ¼ rankðJ Þ rankðJ 2 Þ ¼ rankðJ Þ þ rankðK Þ rankð½ K
J Þ:
ð2:12Þ
Therefore, applying (2.2) and Sylvester’s law of inertia to n fH; J g gives
02 31 02 31 02 J H1 J H2 i H HH H H iF J2 H HH F J2 B6 7C i HH J B 6 7 C @ 4 5 A @ 4 ¼ n @4 n J1 0 0 5A ¼ n J 1 F J2 J 0 J2 0 0 02 31 H iV 2 F J2 H HH F J2 V 2 V H2 F J2 J H1 5A þ rankðJ 2 Þ ¼ n @4 J 1 F J2 V 2 0 02 31 i H4 HH4 J H12 @ 4 5A þ rankðJ Þ rankðJ 12 Þ; ¼ n J 12 0
F J2 J H1 0
31 5A þ rankðJ 2 Þ
that is,
n fH4 ; J 12 g rankðJ 12 Þ ¼ n fH; J g rankðJ Þ:
ð2:13Þ
We have the following result. Lemma 2.2. Suppose that H 2 Cm2 m2 , J 2 Ck2 m2 , K 2 Ck2 l2 be given, and X 2 Hm2 m2 and Y 2 Cm2 l2 are arbitrary matrices. Then
HX min rank J X¼X H ;Y
Y K
¼ rankð½ K
J Þ rankðJ Þ þ max fnþ fH; J g; n fH; J gg:
ð2:14Þ
Proof. Let the SVDs of K and J 2 , and the partitions of U H1 J, V H2 F J2 HF J2 V 2 and J 1 V 2 are given in (2.6)–(2.11), respectively. Partition
YV 1 ¼ ½ Y 1
Y 2 ;
where Y 1 2 Cm2 r1 and Y 2 2 Cm2 ðl2 r1 Þ . Denote
2
Im2 6 P¼4 0 0
3 " 7 I m2 0 5 0
Y 1 R1 1
0
Ir1 0
Ik2 r1
0 U H1
# ;
Q¼
I m2 0
2 I m2 0 6 1 R 4 1 J1 V1 0
3
0
0
Ir1
7 0 5:
0
Il2 r1
It is obvious that P and Q are both nonsingular matrices. Therefore, 02 31 " #!
H X Y 1 R1 1 J1 0 Y 2 HX Y HX Y H X Y 1 R1 B6 7C 1 J1 Y 2 rank ¼ rank ð K Þ þ rank ¼ rank P Q ¼ rank@4 : 5 A 0 R1 0 J K J K J2 0 J2 0 0 ð2:15Þ Furthermore, partition
" Hb b Y 1 R1 1 ¼ Y ; V2 Y ¼
b 2C where Y
m2 r 1
" rank
# b1 X1 Y 21 Y ; V H2 Y 2 ¼ ; V H2 XV 2 ¼ b2 X H2 Y 22 Y
b2 2 C ,Y
ðm2 r 2 Þr 1
bJ HXY 1
Y2
J2
0
Y 22 2 C
ðm2 r 2 Þðl2 r 2 Þ
#!
X4
;
ðm2 r 2 Þðm2 r 2 Þ
and X 4 2 C
" ¼ rankðJ2 Þ þ rank
X2
b 1J H2 X 2 Y 12 b 2J H4 X 4 Y 12
. Obviously, X 4 is an Hermitian matrix and
Y 21 Y 22
#!
b 2J ; P rankðJ 2 Þ þ rank H4 X 4 Y 12
Applying Lemma 2.1, (2.12) and (2.13), we have
b 2J min rank H4 X 4 Y 12 ¼ max fnþ fH4 ; J 12 g; n fH 4 ; J 12 gg rankðJ 12 Þ b X 4 ¼X H 4 ;Y 2 ¼ rankðJ Þ þ rankðK Þ rank½ K J þ max fnþ fH; J g; n fH; J gg:
ð2:16Þ
59
H. Wang / Applied Mathematics and Computation 233 (2014) 55–61
From Eqs. (2.8), (2.15) and (2.16), we obtain
HX min rank J X¼X H ;Y
Y K
"
¼
min
X 2 ;b Y 1 ;b Y 2 ;Y 21 ;Y 22 ;X 4 ¼X H 4
rank
b 1J H2 X 2 Y 12 b 2J H4 X 4 Y 12
Y 21
#!
Y 22
þ rankðK Þ þ rankðJ 2 Þ
b 2J P min rank H4 X 4 Y 12 þ rankðK Þ þ rankðJ 2 Þ b X 4 ¼X H 4 ;Y 2 ¼ rankð½ K
J Þ rankðJ Þ þ max fnþ fH; J g; n fH; J gg:
b 1 J ; Y 21 ¼ 0 and Y 22 ¼ 0. Therefore, the formula in (2.14) hold, and we The lower bound is obtained if we set X 2 ¼ H2 Y 12 complete the proof of the theorem. h
3. The minimal rank of A BX when X is Hermitian In this section, we give a closed form representation of the minimal rank of A BX when X is Hermitian. Theorem 3.1. Suppose that A 2 Cmn and B 2 Cmn , and X ¼ X H is an arbitrary matrix. Then the minimal rank of the matrix expression A BX is
min rankðA BX Þ ¼ rankð½ A B Þ rank ABH X¼X
B
H
n o : þ max nþ i ABH BAH ; n i ABH BAH
ð3:1Þ
Proof. Let the SVD of B be
B¼U
R 0 0
0
V H;
ð3:2Þ
and partition
U H AV ¼
A1
A2
A3
A4
;
ð3:3Þ
where A1 2 CrB rB , A2 2 CrB ðnrB Þ , rankð½ A3 A4 Þ ¼ rankðEB AÞ and
A3 2 CðmrB ÞrB ,
and
A4 2 CðmrB ÞðnrB Þ .
Then
we
have
rankðA3 Þ ¼ rank EB ABH ,
#! " # " A1 R 0 iðA3 RÞH i A1 R ðA1 RÞH ð A1 R Þ H ð A3 R Þ H H H H U ¼U UH i AB BA ¼ iU A3 R 0 0 0 iA3 R 0 2
3 H R 0 4 i R1 A1 R1 A1 AH3 5 R 0 UH : ¼U 0 iI 0 iI iA3 0 It follows from applying Sylvester’s law of inertia that
rankðA3 Þ ¼ rank ABH rankð½ A3
B
rankðBÞ;
ð3:4Þ
A4 Þ ¼ rankð½ A B Þ rankðBÞ
ð3:5Þ
and
02
31 H H 1 1 A i R A R A 1 1 3 5A ¼ n i ABH BAH : n @4 A3 0
ð3:6Þ
Furthermore, partition
V H XV ¼
X1
X2
X H2
X4
;
where X 2 2 CrB ðnrB Þ and X 1 2 HrB rB . Then
A BX ¼ U
R 0 0
0
"
R1 A1 X 1 R1 A2 X 2 A3
A4
# V H:
ð3:7Þ
60
H. Wang / Applied Mathematics and Computation 233 (2014) 55–61
Applying (2.14) and (3.4)–(3.6) to (3.7) gives
" min rankðA BX Þ ¼ min rank
R1 A1 X 1 R1 A2 X 2
#!
A4 n n o n oo ¼ rankð½ A3 A4 Þ rankðA3 Þ þ max nþ R1 A1 ; A3 ; n R1 A1 ; A3 n o
¼ rankð½ A B Þ rank ABH B þ max nþ i ABH BAH ; n i ABH BAH :
X¼X H
A3
X 1 ¼X H 1 ;X 2
Therefore, the formula in (3.1) hold, and we complete the proof of the theorem. h Some direct consequences of the previous theorem are given below. Corollary 3.2. Let A; B 2 Cmn be given and suppose that RðAÞ # RðBÞ. Then
n o min rankðA BX Þ ¼ max nþ i ABH BAH ; n i ABH BAH : X¼X H
In particular, if the condition RðAÞ # RðBÞ hold, then A ¼ BX has a Hermitian solution if and only if ABH ¼ BAH . Corollary 3.3. Let A; B 2 Cmn be given and suppose that RðAÞ \ RðBÞ ¼ 0. Then
min rankðA BX Þ ¼ rankðAÞ: X¼X H
Corollary 3.4. Let A 2 Cmn and B 2 Cmn be given. Then, the following conditions are equivalent: (1) There exists a Hermitian matrix X such that BX ¼ A. n o
¼ 0. B Þ rank ABH B þ max nþ i ABH BAH ; n i ABH BAH
(2) rankð½ A
(3) BBy A ¼ A and ABH ¼ BAH , [7]. Proof. It is well known that the linear matrix equation BX ¼ A has a Hermitian solution if and only if minX¼X H rankðA BXÞ ¼ 0. Thus setting the both sides of (3.1) to zero, we obtain the equivalence of Conditions (1) and (2). From (3.7), we know that the matrix equation BX ¼ A has a Hermitian solution if and only if
rankð½ A3
H ¼ 0: A4 Þ ¼ 0 and rank R1 A1 R1 A1
rankð½ A3
A4 Þ ¼ rank A BBy A
Since
and
H ¼ rank BBy ABH BAH BBy ; rank R1 A1 R1 A1 we obtain that the matrix equation BX ¼ A has a Hermitian solution if and only if
rank A BBy A ¼ 0 and rank BBy ABH BAH BBy ¼ 0; H i.e., BBy A ¼ A and BBy A BH ¼ B BBy A . Hence Conditions (1) and (3) are equivalent. Example 3.1. Take
2
1 2 3
6 A ¼ 43
i
2 5 Then rankð½ A
3
2
1 0 0
3
7 6 7 0 5 and B ¼ 4 1 0 0 5: i
B Þ ¼ 3, rank ABH
2
0 2 0 B
¼ 3,
3 0 2 2 6 7 0 2 þ 2i 5; AB BA ¼ 4 2 2 2 þ 2i 0 H
H
h
H. Wang / Applied Mathematics and Computation 233 (2014) 55–61
61
¼ 1 and n i ABH BAH ¼ 2. Thus by (3.1), nþ i ABH BAH
min rankðA BX Þ ¼ 2: X¼X H
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