uÀ“‰ŒÆÆ(g,‰Æ‡)

14Ï 2012 c 7 

Journal of East China Normal University (Natural Science)

No. 4 Jul. 2012

Article ID: 1000-5641(2012)04-0012-06

Young type inequalities for matrices HU Xing-kai (Faculty of Science, Kunming University of Science and Technology, Kunming

650500, Yunnan, China)

Abstract: First, some Young type inequalities for scalars were given. Then on the base of them, corresponding Young type inequalities for matrices were established. Key words: unitarily invariant norms; trices;

Young type inequality;

positive semidefinite ma-

singular values

CLC number: O151.21

Document code: A

DOI: 10.3969/j.issn.1000-5641.2012.04.002

Ý  Young .ت ,p (&²nóŒÆ nÆ, H &²

Á‡: Äk‰Ñ eZIþ Young .ت. ,

650500)

3dÄ:þ, ïá ƒAÝ Young .Ø

ª. '…c: jØC‰ê;

0

Young .ت;

Œ½Ý ;

ÛɊ

Introduction Let Mm,n be the space of m × n complex matrices and Mn = Mn,n . Let k·k denote

any unitarily invariant norm on Mn . So, kU AV k = kAk for all A ∈ Mn and for all unitary matrices U, V ∈ Mn . For A = (aij ) ∈ Mn , s the Hilbert-Schmidt norm, the trace norm, and the n n P P s2j (A), kAk1 = sj (A), and kAksp = s1 (A), spectral norm of A are defined by kAk2 = j=1

j=1

respectively, where s1 (A) > s2 (A) > · · · > sn−1 (A) > sn (A) are the singular values of A, that 1

is, the eigenvalues of the positive semidefinite matrix |A| = (AA∗ ) 2 , arranged in decreasing order and repeated according to multiplicity. The classical Young inequality for nonnegative real numbers says that if a, b > 0 and 0 6 v 6 1, then av b1−v 6 va + (1 − v) b ÂvFÏ: 2011-10 Šö{0: ,p, I, a¬, Ï, ïЕêŠ‚5“ê. E-mail: [email protected].

(0.1)

14Ï

,pµÝ Young.ت

with equality if and only if a = b. If p, q > 0 such that be written as ab 6

a p

p

+

b q

q

1 p

+

1 q

13

= 1, then the inequality (0.1) can

.

Several matrix versions of the Young inequality(0.1) have been recently established[1-9] . It seems that matrix versions of the Young inequality have aroused considerable interest. The main purpose of this paper is to give some Young type inequalities for matrices.

1

Young type inequalities for scalars We begin this section with the Young type inequalities for scalars. Lemma 1.1 Suppose that a, b > 0. If 0 6 v 6 21 , then

If

1 2



6 v 6 1, then

v

(va) b1−v

2

2

2

+ v 2 (a − b) 6 v 2 a2 + (1 − v) b2 .

(1.1)

{av [(1 − v) b]1−v }2 + (1 − v)2 (a − b)2 6 v 2 a2 + (1 − v)2 b2 . Proof

If 0 6 v 6 21 . Then, by inequality (0.1), we have 2

and so

If

1 2

(1.2)

2

v 2 a2 + (1 − v) b2 − v 2 (a − b) = b [2v (va) + (1 − 2v) b]  2 2v v > b (va) b1−2v = (va) b1−v ,  2 2 v 2 v 2 a2 + (1 − v) b2 > (va) b1−v + v 2 (a − b) .

6 v 6 1, then 2

2

2

2

v 2 a2 + (1 − v) b2 − (1 − v) (a − b) = a[(2v − 1) a + 2 (1 − v) b] 2−2v

> aa2v−1 [(1 − v) b]

1−v 2

= {av [(1 − v) b]

} ,

and so 2

1−v 2

v 2 a2 + (1 − v) b2 > {av [(1 − v) b]

2

2

} + (1 − v) (a − b) .

This completes the proof. Hirzallah and Kittaneh[7] obtained a refinement of the scalar Young’s inequality as follows: av b1−v


2

2

2

+ r02 (a − b) 6 (va + (1 − v) b) ,

(1.3)

where r0 = min {v, 1 − v}. In Kittaneh and Manasrah’s paper[9] , the following related refinement of the scalar Young’s inequality was obtained av b1−v


2

+ r0 (a − b)2 6 va2 + (1 − v) b2 .

(1.4)

When comparing Lemma 1.1 with the inequalities (1.3) and (1.4), it is easy to observe that both the left-hand and the right-hand sides of Lemma 1.1 are greater than or equal to the corresponding sides in (1.3) and (1.4), respectively. It should be noticed that neither Lemma 1.1 nor (1.3) and (1.4) is uniformly better than the other.

uÀ“‰ŒÆÆ(g,‰Æ‡)

14

2

2012 c

Young type inequalities for matrices A matrix Young inequality in [3] says that if A, B ∈ Mn are positive semidefinite, then
sj Av B 1−v 6 sj (vA + (1 − v) B)

for j = 1, · · · , n. The above singular value inequality of Ando entailed the norm inequality

v 1−v

A B

6 kvA + (1 − v) Bk .

Kosaki[4], Bhatia and Parthasarathy[6] proved that if A, B, X ∈ Mn such that A and B are positive semidefinite and if 0 6 v 6 1, then

v

A XB 1−v 2 6 kvAX + (1 − v) XBk2 . 2 2

(2.1)

Based on the refined Young inequality (1.3), it has been shown in [7] that if A, B, X ∈ Mn such that A and B are positive semidefinite and if 0 6 v 6 1, then

v

A XB 1−v 2 + r02 kAX − XBk2 6 kvAX + (1 − v) XBk2 . 2 2 2

(2.2)

Obviously, (2.2) is an improvement of (2.1) for the Hilbert-Schmidt norm. In Kittaneh and Manasrah’s paper[9] , the following Young inequalities for matrices were obtained p p

v

A XB 1−v + A1−v XB v + 2r0 ( kAXk2 − kXBk2 ) 6 kAX + XBk2 , 2

v

A XB 1−v + A1−v XB v 2 + 2r0 kAX − XBk2 6 kAX + XBk2 . 2 2 2

In this section, we give the trace norm, the Hilbert-Schmidt norm, and determinant versions of Young type inequalities based on the Young type inequalities (1.1) and (1.2). To do this, we need the following lemma. Lemma 2.1[10]

Let A, B ∈ Mn . Then n X j=1

Theorem 2.2

If

1 2

sj (AB) 6

n X

sj (A) sj (B).

j=1

Let A, B ∈ Mn be positive semidefinite. If 0 6 v 6 21 , then

q

2 2 2 2 v v Av B 1−v 1 6 v 2 kAk2 + (1 − v) kBk2 − v 2 (kAk2 − kBk2 ) .

(2.3)

6 v 6 1, then 1−v

(1 − v) Proof

q

v 1−v

A B

6 v 2 kAk2 + (1 − v)2 kBk2 − (1 − v)2 (kAk − kBk )2 . 2 2 2 2 1

If 0 6 v 6 21 , by the inequality (1.1), we have

[(vsj (A))v sj (B)1−v ]2 + v 2 (sj (A) − sj (B))2 6 v 2 s2j (A) + (1 − v)2 s2j (B)

(2.4)

14Ï

,pµÝ Young.ت

15

for j = 1, · · · , n. Thus, by the Cauchy-Schwarz inequality, we have 2

2

tr(v 2 A2 + (1 − v) B 2 ) = v 2 trA2 + (1 − v) trB 2 =

n X

2

(v 2 s2j (A) + (1 − v) s2j (B))

j=1

>

n X

v

1−v 2

[(vsj (A)) sj (B)

]

j=1

+ v2

X n

s2j (A) +

j=1

>

n X j=1

n X

s2j (B) − 2

n X j=1

j=1

 sj (A) sj (B)

 v
2 v sj (Av ) sj B 1−v

 X  12 X  21  n n 2 2 2 2 + v kAk2 + kBk2 − 2 sj (A) sj (B) 2

j=1

= v 2v

n X 

sj (Av ) sj B 1−v

j=1


2

j=1

2

+ v 2 (kAk2 − kBk2 ) .

(2.5)

On the other hand, we also have 2

2

2

2

v 2 trA2 + (1 − v) trB 2 = v 2 kAk2 + (1 − v) kBk2 .

(2.6)

Therefore, it follows from (2.5) and (2.6) that 2

2

2

2

v 2 kAk2 + (1 − v) kBk2 − v 2 (kAk2 − kBk2 ) > v 2v

n X 
2 sj (Av ) sj B 1−v .

(2.7)

j=1

By Lemma 2.1 and (2.7), we have q

v v Av B 1−v 1 6 v 2 kAk22 + (1 − v)2 kBk22 − v 2 (kAk2 − kBk2 )2 .

If

1 2

6 v 6 1, then by the inequality (1.2) and the same method above, we have the inequality

(2.4). This completes the proof. Theorem 2.3 Let A, B, X ∈ Mn such that A and B are positive semidefinite. If 0 6 v6

1 2,

then kvAX + (1 − v) XBk22 > v 2 kAX − XBk22 + v 2v kAv XB 1−v k22 + 2v (1 − v) kA1/2 XB 1/2 k22 .

If

1 2

(2.8)

6 v 6 1, then 2

2−2v

kvAX + (1 − v) XBk22 > (1 − v) kAX − XBk22 + (1 − v)

kAv XB 1−v k22

+ 2v (1 − v) kA1/2 XB 1/2 k22 . Proof

(2.9)

Since every positive semidefinite matrix is unitarily diagonalizable, it follows that

there are unitary matrices U, V ∈ Mn such that A = U DU ∗ and B = V DV ∗ , where

uÀ“‰ŒÆÆ(g,‰Æ‡)

16

D = diag (λ1 , · · · , λn ) ,

2012 c

E = diag (µ1 , · · · , µn ) , and λi , µi > 0, i = 1, · · · , n.

Let Y = U XV = (yij ). Then ∗

vAX + (1 − v) XB = U (vDY + (1 − v) Y E) V ∗ = U ((vλi + (1 − v) µj ) yij ) V ∗ , 1/2 1/2

AX − XB = U ((λi − µj ) yij ) V ∗ , A1/2 XB 1/2 = U (λi µj yij )V ∗
and Av XB 1−v = U λvi µj1−v yij V ∗ . If 0 6 v 6 12 , by inequality (1.1), we have 2

kvAX + (1 − v) XBk2 = =

n X

2

2

(vλi + (1 − v) µj ) |yij |

i,j=1 n X

2

2

(v 2 λ2i + (1 − v) µ2j + 2v (1 − v) λi µj ) |yij |

i,j=1 n X 2

2

2

(λi − µj ) |yij | + v 2v

>v

i,j=1

+

n X

n X

λvi µj1−v

i,j=1


2

2

|yij |

2v (1 − v) λi µj |yij |2

i,j=1

> v kAX − XBk22 + v 2v kAv XB 1−v k22 + 2v (1 − v) kA1/2 XB 1/2 k22 . 2

If

1 2

6 v 6 1, then by the inequality (1.2) and the same method above, we have the inequality

(2.9). This completes the proof. Remark 2.4 The inequality (2.8) is related to the inequality (2.1). It should be noticed that neither (2.1) nor (2.8) is uniformly better than the other. Theorem 2.5 Let A, B ∈ Mn be positive definite. If 0 6 v 6 12 , then 2

det (vA + (1 − v) B) > v 2nv det Av B 1−v n


2

2

+ v 2n det (A − B)

+ (2v (1 − v)) det B 1/2 AB 1/2 . If

1 2

(2.10)

6 v 6 1, then 2

2n(1−v)

det (vA + (1 − v) B) > (1 − v)

det Av B 1−v

n


2

+ (1 − v)

2n

+ (2v (1 − v)) det B 1/2 AB 1/2 . Proof

2

det (A − B)

(2.11)

To prove the determinant inequality, note that by the inequality (1.1), we have

v 2v [svj (B −1/2 AB −1/2 )]2 + v 2 (sj (B −1/2 AB −1/2 ) − 1)2 6 v 2 s2j (B −1/2 AB −1/2 ) + (1 − v)2 for j = 1, · · · , n. Therefore, det(vB −1/2 AB −1/2 + (1 − v)I)2 =

n Y

(vsj (B −1/2 AB −1/2 ) + 1 − v)2

j=1

=

n Y

j=1

(v 2 s2j (B −1/2 AB −1/2 ) + (1 − v)2 + 2v(1 − v)sj (B −1/2 AB −1/2 ))

14Ï

>

n Y

,pµÝ Young.ت

17

−1/2 (v 2v s2v AB −1/2 ) + v 2 (sj (B −1/2 AB −1/2 ) − 1)2 + 2v(1− v)sj (B −1/2 AB −1/2 )) j (B

j=1

> v 2nv

n Y

−1/2 s2v AB −1/2 ) + v 2n j (B

j=1

n Y

(sj (B −1/2 AB −1/2 ) − 1)2

j=1

+ (2v(1 − v))n

n Y

sj (B −1/2 AB −1/2 )

j=1

=v

2nv

det(B

−1/2

AB −1/2 )2v +v 2n det(B −1/2 AB −1/2 −I)2 +(2v(1 − v))n det B −1/2 AB −1/2 .

Thus, we have det (vA + (1 − v) B)2 > v 2nv det Av B 1−v If

1 2


2

+ v 2n det (A − B)2 + (2v (1 − v))n det B 1/2 AB 1/2 .

6 v 6 1, then by the inequality (1.2) and the same method above, we have the inequality

(2.11). This completes the proof. Remark 2.6 If A, B ∈ Mn are positive definite, a determinant version of the arithmeticgeometric mean inequality is known[11] : det



A+B 2

2

> det (AB) .

(2.12)

Obviously, the inequality (2.10) or (2.11) is a generalization of the inequality (2.12).

[ References ] [1]

BHATIA R, KITTANEH F. On singular values of a product of operators [J]. SIAM J Matrix Anal Appl, 1990, 11: 271-277.

[2] [3] [4]

KITTANEH F. On some operator inequalities [J]. Linear Algebra Appl, 1994, 208/209: 19-28.

[5] [6]

ZHAN X. Inequalities for unitarily invariant norms [J]. SIAM J Matrix Anal Appl, 1998, 20: 466-470.

[7]

HIRZALLAH O, KITTANEH F. Matrix Young inequalities for the Hilbert-Schmidt norm [J]. Linear Algebra Appl, 2000, 308: 77–84.

[8]

BHATIA R, KITTANEH F. Notes on matrix arithmetic–geometric mean inequalities [J]. Linear Algebra Appl, 2000, 308: 203-211.

[9]

KITTANEH F, MANASRAH Y. Improved Young and Heinz inequalities for matrices [J]. J Math Anal Appl, 2010, 361: 262-269.

[10] [11]

ZHAN X. Matrix theory [M]. Beijing: Higher Education Press, 2008. (in Chinese)

ANDO T. Matrix Young inequality [J]. Oper Theory Adv Appl, 1995, 75: 33-38. KOSAKI H. Arithmetic-geometric mean and related inequalities for operators [J]. J Funct Anal, 1998, 156: 429-451. BHATIA R, PARTHASARATHY K R. Positive definite functions and operator inequalities [J]. Bull London Math Soc, 2000, 32: 214-228.

HORN R A, JOHNSON C R. Matrix analysis [M]. Cambridge: Cambridge University Press, 1985: 467.