Advances in Linear Algebra & Matrix Theory
Vol.3 No.4(2013), Article ID:40527,5 pages DOI:10.4236/alamt.2013.34007
Singular Value Inequalities for Compact Normal Operators
Department of Basic Sciences, Petra University, Amman, Jordan
Email: waudeh@uop.edu.jo
Copyright © 2013 Wasim Audeh. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Received September 25, 2013; revised October 28, 2013; accepted November 7, 2013
Keywords: Compact Operator; Inequality; Normal Operator; Self-Adjoint Operator; Singular Value
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
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