An eigenvalue majorization inequality for positive semidefinite block

Linear and Multilinear Algebra
2012, 1–4, iFirst
An eigenvalue majorization inequality for positive semidefinite
block matricesyz
Minghua Lin and Henry Wolkowicz*
Department of Combinatorics and Optimization, University of Waterloo,
Waterloo, ON N2L 3G1, Canada
Downloaded by [University of Waterloo] at 07:59 16 January 2012
Communicated by R. Loewy
(Received 26 July 2011; final version received 17 December 2011)
Let H ¼ KM KN be a Hermitian matrix. It is known that the vector of
diagonal elements of H, diag(H), is majorized by the vector of the
eigenvalues of H, (H), and that this majorization can be extended to
the eigenvalues of diagonal blocks of H. Reverse majorization results for
the eigenvalues are our goal. Under the additional assumptions that H is
positive semidefinite and the block K is Hermitian, the main result of this
article provides a reverse majorization inequality for the eigenvalues.
This results in the following majorization inequalities when combined with
known majorization inequalites on the left:
diagðHÞ ðM NÞ ðHÞ ððM þ NÞ 0Þ:
Keywords: majorization; eigenvalue inequalities
AMS Subject Classifications: 15A42; 06A06
1. Introduction
An early result concerning eigenvalue majorization is the fundamental result due to I.
Schur (see, e.g. [1,5,6]), which states that the diagonal entries of a Hermitian matrix
A are majorized by its eigenvalues. i.e., diag(A) a (A).
This
result can be extended
to block Hermitian matrices. For example, if H ¼ KM KN is Hermitian, then
diagðHÞ ðM NÞ ðHÞ:
To see the second inequality, let M ¼ U*D1U, N ¼ V*D2V, where D1, D2 are diagonal
matrices,
the spectral decomposition of M, N, respectively. Then ðHÞ ¼
D1 UKVbe
VK
ðD1 D2 Þ ¼ ðM NÞ.
U D
2
Reverse majorization results are our goal. Here and throughout, K* denotes the
conjugate transpose
of
K; M N denotes the direct sum of M and N, i.e., the block
0
diagonal matrix M
0 N and 0 is a zero block matrix of compatible size.
*Corresponding author. Email: [email protected]
yIn memory of Ky Fan.
zResearch Report CORR 2011-04.
ISSN 0308–1087 print/ISSN 1563–5139 online
ß 2012 Taylor & Francis
http://dx.doi.org/10.1080/03081087.2011.651723
http://www.tandfonline.com
2
M. Lin and H. Wolkowicz
Majorization inequalities are useful and important; see, e.g. [6]. The main result
of this article is the following reverse majorization inequality for a Hermitian positive
semidefinite 2 2 block matrix. (We delay the proof until Section 2.)
THEOREM 1.1 Let H ¼ KM KN be a Hermitian positive semidefinite matrix. If, in
addition, the block K is Hermitian, then the following majorization inequality holds:
Downloaded by [University of Waterloo] at 07:59 16 January 2012
ðHÞ ððM þ NÞ 0Þ:
ð1:1Þ
1.1. Preliminary results
Let Mmn(C) be the space of all complex matrices of size m n with
Mn(C) ¼ Mnn(C). For A 2 Mn(C), the vector of eigenvalues of A is denoted by
(A) ¼ (1(A), 2(A), . . . , n(A)). If A is Hermitian, we arrange the eigenvalues of A in
nonincreasing order: 1(A) 2(A) n(A).
For two sequences of real numbers arranged in nonincreasing order,
x ¼ ðx1 , x2 , . . . , xn Þ,
y ¼ ð y1 , y2 , . . . , yn Þ,
we say that x is majorized by y, denoted by x a y (or y x), if
k
X
j¼1
xj k
X
yj
ðk ¼ 1, . . . , n 1Þ,
and
n
X
j¼1
j¼1
xj ¼
n
X
yj :
j¼1
We make use of the following lemmas in our proof of Theorem 1.1.
LEMMA 1.2
If A, B 2 Mn(C) are Hermitian, then
2ðAÞ ðA þ BÞ þ ðA BÞ:
ð1:2Þ
Proof The lemma is equivalent to Ky Fan’s eigenvalue inequality, i.e., (A þ B) a
(A) þ (B), [2]. A proof can be found in [4, Theorem 4.3.27] and [7, Theorem
7.15].
g
LEMMA 1.3
Let A 2 Mmn(C) with m n, then we have
ðAA Þ ¼ ðA A 0Þ:
ð1:3Þ
2. Proof of main result: corollaries
Before we prove Theorem 1.1, we show by an example that the requirement K being
Hermitian is necessary.
Example 2.1 Let M ¼ 10 04 , N ¼ 21 11 and K ¼ 12 02 . Then
pffiffiffi
pffiffiffi
ððM þ NÞ 0Þ ¼ ð4 þ 2, 4 2, 0, 0Þ,
pffiffiffi
pffiffiffi
M K
¼ ð4 þ 5, 4 5, 0, 0Þ:
K N
M K Therefore K N ˛ ðM þ NÞ 0Þ.
3
Linear and Multilinear Algebra
K
Proof of Theorem 1.1 Since H :¼ M
is positive semidefinite, we may write
K N
H ¼ P*P, where P ¼ [X Y ], for some X, Y 2 M2nn(C). Therefore, we have M ¼ X*X,
N¼ Y*Y and K ¼ X*Y ¼ Y*X. Note that by Lemma 1.3, we have
K
¼ ðPP Þ. The conclusion (1.1) is then equivalent to showing
M
K N
fX Y ¼ Y Xg ¼) ððX X þ Y YÞ 0Þ ðXX þ YY Þ :
ð2:1Þ
First, note that
ðX þ iYÞ ðX þ iYÞ ¼ X X þ Y Y þ iðX Y Y XÞ
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¼ X X þ Y Y
ðX iYÞ ðX iYÞ ¼ X X þ Y Y iðX Y Y XÞ
¼ X X þ Y Y
ðX þ iYÞðX þ iYÞ ¼ XX þ YY iðXY YX Þ
ðX iYÞðX iYÞ ¼ XX þ YY þ iðXY YX Þ:
Therefore we see that
1
ððX þ iYÞ ðX þ iYÞ 0Þ þ ððX iYÞ ðX iYÞ 0Þ
2
1
¼ ðððX þ iYÞðX þ iYÞ Þ þ ððX iYÞðX iYÞ ÞÞ
2
ðXX þ YY Þ,
ððX X þ Y YÞ 0Þ ¼
where the second equality is by Lemma 1.3 and the majorization follows from
applying Lemma 1.2 with A ¼ (XX* þ YY*), B ¼ i(XY* YX*).
g
As we can see from the above proof, a special case of Theorem 1.1 can be stated
as follows.
COROLLARY 2.2
Let X, Y 2 Mn(C) with X*Y Hermitian. Then we have
ðXX þ YY Þ ðX X þ Y YÞ:
COROLLARY 2.3
ð2:2Þ
n
Let k 1 be an integer. If A, B 2 M (C) are Hermitian, then we have
ðA2 þ ðABÞk ðBAÞk Þ ðA2 þ ðBAÞk ðABÞk Þ:
k
ð2:3Þ
k
Proof Let X ¼ A and Y ¼ (BA) . Then XY ¼ A(BA) is Hermitian. The result now
follows from Corollary 2.2.
g
COROLLARY 2.4 Let k 1 be an integer, p 2 [0, 1); and let A, B 2 Mn(C) be
Hermitian. Then we have
(1) trace[(A2 þ (AB)k(BA)k)p] trace[(A2 þ (BA)k(AB)k)p], for p 1;
(2) trace[(A2 þ (AB)k(BA)k)p] trace[(A2 þ (BA)k(AB)k)p], for 0 p 1.
Proof Since f(x) ¼ xp is a convex function for p 1 and concave for 0 p 1,
Corollary 2.4 follows from Corollary 2.3 and a general property of majorization
[5, p. 56].
g
Remark 2.5
The case where k ¼ 1 in Corollary 2.4 was proved in [3].
4
M. Lin and H. Wolkowicz
Acknowledgements
The authors thank the anonymous referees for valuable suggestions that improved the
presentation of this article. This research was supported by The Natural Sciences and
Engineering Research Council of Canada.
Downloaded by [University of Waterloo] at 07:59 16 January 2012
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