Some inequalities for unitarily invariant norms of matrices

This article aims to discuss inequalities involving unitarily invariant norms. We obtain a refinement of the inequality shown by Zhan. Meanwhile, we give an improvement of the inequality presented by Bhatia and Kittaneh for the Hilbert-Schmidt norm.

Mathematical Subject Classification

MSC (2010) 15A60; 47A30; 47B15

Let M_m,n be the space of m × n complex matrices and M_n = M_n,n. Let denote any unitarily invariant norm on M_n. So, for all A∈M_n and for all unitary matrices U,V∈M_n. For A = (a_ij)∈M_n, the Hilbert-Schmidt norm of A is defined by

where tr is the usual trace functional and s₁(A) ≥ s₂(A) ≥ ... ≥ s_n-1(A) ≥ s_n(A) are the singular values of A, that is, the eigenvalues of the positive semidefinite matrix , arranged in decreasing order and repeated according to multiplicity. The Hilbert-Schmidt norm is in the class of Schatten norms. For 1 ≤ p < ∝, the Schatten p-norm is defined as

For k = 1,...,n, the Ky Fan k-norm is defined as

It is known that these norms are unitarily invariant, and it is evident that each unitarily invariant norm is a symmetric guage function of singular values [1, p. 54-55].

Bhatia and Davis proved in [2] that if A,B,X∈M_n such that A and B are positive semidefinite and if 0 ≤ r ≤ 1, then

Let A,B,X∈M_n such that A and B are positive semidefinite. In [3], Zhan proved that

for any unitarily invariant norm and real numbers r,t satisfying 1 ≤ 2r ≤ 3,-2 < t ≤ 2. The case r = 1,t = 0 of this result is the well-known arithmetic-geometric mean inequality

Meanwhile, for r∈[0,1], Zhan pointed out that he can get another proof of the following well-known Heinz inequality

by the same method used in the proof of (1.2).

Let A,B,X∈M_n such that A and B are positive semidefinite and suppose that

Then ψ is a convex function on [-1,1] and attains its minimum at v = 0 [4, p. 265].

In [5], for positive semidefinite n × n matrices, the inequality

was shown to hold for every unitarily invariant norm. Meanwhile, Bhatia and Kittaneh [5] asked the following.

Question

Let A,B∈M_n be positive semidefinite. Is it true that

, ?

The case n = 2 is known to be true [5]. (See also, [1, p. 133], [6, p. 2189-2190], [7, p. 198].)

Obviously, if A,B∈M_n are positive semidefinite and AB = BA, then we have , .

In this section, we first utilize the convexity of the function

to obtain an inequality for unitarily invariant norms that leads to a refinement of the inequality (1.2). To do this, we need the following lemmas on convex functions.

Lemma 2.1

Let A,B,X∈M_n such that A and B are positive semidefinite. Then, for each unitarily invariant norm, the function

is convex on [0,2] and attains its minimum at r = 1.

Proof

Replace v+1 by r in (1.3).□

Lemma 2.2

Let ψ be a real valued convex function on an interval [a,b] which contains (x₁,x₂). Then for x₁ ≤ x ≤ x₂, we have

Proof

Since ψ is a convex function on [a,b], for a ≤ x₁ ≤ x ≤ x₂ ≤ b, we have

This is equivalent to the inequality (2.1).□

Theorem 2.1

Let A,B,X∈M_n such that A and B are positive semidefinite. If 1 ≤ 2r ≤3 and -2 <t ≤ 2, then

where r₀ = min{r,2-r}.

Proof

If , then by Lemma 2.1 and Lemma 2.2, we have

That is

It follows from (1.2) and (2.3) that

If , then by Lemma 2.1 and Lemma 2.2, we have

That is

It follows from (1.2) and (2.4) that

It is equivalent to the following inequality

This completes the proof.□

Now, we give a simple comparison between the upper bound in (1.2) and the upper bound in (2.2).

Therefore, Theorem 2.1 is a refinement of the inequality (1.2).

Let A,B,X∈M_n such that A and B are positive semidefinite. Then, for each unitarily invariant norm, the function

is a continuous convex function on [0,1] and attains its minimum at . See [4, p. 265]. Then, by the same method above, we have the following result.

Theorem 2.2.[8]

Let A,B,X∈M_n such that A and B are positive semidefinite. If 0 ≤ v ≤ 1, then

where r₀ = min{v,1-v}. This is a refinement of the second inequality in (1.1).

Next, we will obtain an improvement of the inequality (1.4) for the Hilbert-Schmidt norm. To do this, we need the following lemma.

Lemma 2.3.[9]

Let A,B,X∈M_n such that A and B are positive semidefinite. If 0 ≤ v ≤ 1, then

Theorem 2.3

Let A,B,X∈M_n such that A and B are positive semidefinite. If 0 ≤ v ≤ 1, then

Proof

Let

So,

By Lemma 2.3, we have

That is,

Hence,

This completes the proof.□

Let A,B,X∈M_n such that A and B are positive semidefinite, for Hilbert-Schmidt norm, the following equality holds:

Taking in Theorem 2.3, and then we have the following result.

Theorem 2.4.[10]

Let A,B,X∈M_n such that A and B are positive semidefinite. Then

Bhatia and Kittaneh proved in [5] that if A,B∈M_n are positive semidefinite, then

Now, we give an improvement of the inequality (1.4) for the Hilbert-Schmidt norm.

Theorem 2.5

Let A,B∈M_n be positive semidefinite. Then

Proof

Let

Then, by Theorem 2.4, we have

It follows form (2.5) and (2.6) that

That is,

This completes the proof.□

The authors declare that they have no competing interests.

SW and LZ designed and performed all the steps of proof in this research and also wrote the paper. YJ participated in the design of the study and suggest many good ideas that made this paper possible and helped to draft the first manuscript. All authors read and approved the final manuscript.

The authors wish to express their heartfelt thanks to the referees and Professor Vijay Gupta for their detailed and helpful suggestions for revising the manuscript. At the same time, we are grateful for the suggestions of Yang Peng. This research was supported by Natural Science Foundation Project of Chongqing Science and Technology Commission (No. CSTC, 2010BB0314), Natural Science Foundation of Chongqing Municipal Education Commission (No. KJ101108), and Scientific Research Project of Chongqing Three Gorges University (No. 10ZD-16).

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