ABSTRACT

We give singular value inequality to compact normal operators, which states that if is compact normal operator on a complex separable Hilbert space, where is the cartesian decomposition of, then Moreover, we give inequality which asserts that if is compact normal operator, then Several inequalities will be proved.

1. Introduction

Let denote the space of all bounded linear operators on a complex separable Hilbert space H, and let denote the two-sided ideal of compact operators in. For, the singular values of, denoted by are the eigenvalues of the positive operator as and repeated according to multiplicity. Note that It follows Weyl’s monotonicity principle (see, e.g., [1, p. 63] or [2, p. 26]) that if are positive and, then The singular values of and are the same, and they consist of those of together with those of . Here, we use the direct sum notation for the blockdiagonal operator defined on.

The Jordan decomposition for self-adjoint operators asserts that every self-adjoint operator can be expressed as the difference of two positive operators. In fact, if is self-adjoint, then where

are the positive operators given by

and, see [1].

Let be any operator, we can write in the form

, where and are self-adjoint operators, this is called the Cartesian decomposition of the operator. If is normal, then.

Audeh and Kittaneh have proved in [3] that if such that, then

(1.1)

Also, Audeh and Kittaneh have proved in [3] that if such that is self-adjoint, , then

(1.2)

In addition to this, Audeh and Kittaneh have proved in [3] that if be self-adjoint operators, then

(1.3)

Zhan has proved in [4] that if are positive, then

(1.4)

Moreover, it has proved in [3] that (1.3) is a generalization of (1.4).

Hirzallah and Kittaneh have proved in [5] that if, then

(1.5)

In this paper, we will give singular value inequalities for normal operators:

Let be normal operator in. Then

(1.6)

We will give singular value inequality to the normal operator, where is normal:

Let be normal operator in. Then

(1.7)

2. Main Results

We will begin by presenting the following theorem for complex numbers Theorem 2.1. Let be complex number. Then

(2.1)

Also,

(2.2)

Proof. The right hand side of the inequalities is well known. To prove the left hand side,

Moreover,

Now, we will present operator version of Theorem 2.1, inequality (2.1).

Theorem 2.2. Let be normal operator in, where be the Cartesian decomposition of. Then

Proof. Let be the Cartesian decomposition of the normal operator, which implies that. Now, , it follows that

In fact for

By using Weyl’s monotonicity principle [1]

and the inequality, we get the right hand side of the theorem. To prove the left hand side of the inequality, we will use the inequality which is well known for commuting self-adjoint operators and it asserts that

(2.3)

This implies that

(2.4)

But it is known that it follows Weyl’s monotonicity principle [1] and the inequality (2.4) that

(2.5)

Inequality (2.5) is equivalent to saying that

Remark 1. (i) Equality holds in the right hand side of Theorem 2.2 if either or.

(ii) Equality holds in the left hand side of theorem 2.2 if.

We will present operator version of Theorem 2.1, inequality (2.2).

Remark 2. Let

where is normal operator. Then is normal operator with is the Cartesian decomposition of.

and.

It follows that, and.

Now, by direct calculations and applying Theorem 2.2 we get

(2.6)

Remark 3. We note that the right hand side of the inequality (2.6) is the same as the inequality (1.6), but the left hand side of the inequalities (1.6) and (2.6) says that the singular value of the addition or subtraction of the Cartesian decomposition for the normal operator divided by is less than or equal to the singular value of the normal operator itself.

As an application of the Theorem 2.2, we will determine upper and lower bounds for singular values of the normal operator, where is normal.

Theorem 2.3. Let be normal operator, where is the Cartesian decomposition of

. Then

Proof. Note that is normal operator, so we can write the Cartesian decomposition of as

where, and

where the cartesian decomposition of is given by. By making comparison of and we see easily that. It follows that . Moreover,

Similarly,. Now, apply Theorem 2.2 to get

(2.7)

This is equivalent to saying that

We will give simple and new proof to the inequality

(1.2).

Theorem 2.4. Let such that is self-adjoint, , then

Proof. Since is self-adjoint operator, we can write

in the form apply the inequality (1.4) we get

which is equivalent to saying that

Audeh and Kittaneh separates Jordan of self-adjoint operator in the inequality (1.3). Here we will give a shorter proof.

Theorem 2.5. Let be self-adjoint operators. Then

Proof. Since and are self-adjoint operators, we can write in the form and similarly we will write in the form. Apply the inequality (1.4) we get

We will present the following two theorems as an application to the inequality (1.5).

Theorem 2.6. Let be self-adjoint operator. Then

(2.8)

Proof. It was proved in Theorem 2.2 that if  is normal operator with Cartesian decomposition , then from this, it follows that

The following theorem is the second application of the inequality (1.5).

Theorem 2.7. Let be self-adjoint operator. Then

(2.9)

Moreover,

(2.10)

Proof. It is well known that, so using the inequality (1.5) we get

Similarly, so using the inequality (1.5) we get

Bhatia and Kittaneh have proved in [6] that if, then

For related Cauchy-Schwarz type inequalities, we refer to [2] and references therein. Here, we will present similar new inequality.

Theorem 2.8. Let be operators. Then

(2.11)

Proof. Suppose and This implies that

, and

On the other hand, we have

, and.

Since and are positive operators, then

is positive operator. Now by applying the inequality (1.1), we get

REFERENCES