The norm of an elementary operator has been studied by many mathematicians. Varied results have been established especially on the lower bound of this norm. Here, we attempt the same problem for finite dimensional operators.

Bounded Linear Operator Elementary Operator
1. Introduction

Let be a complex Hilbert space and be the set of bounded operators on. A basic elementary operator, , is defined as:

for and fixed.

An elementary operator, , is a finite sum of the basic elementary operators, de-

fined as, , for all, where are fixed, for.

When, we have, for all and fixed, for

Given the elementary operator on, the question on whether the equation

, holds remains an area of interest to many mathematicians. This paper attempts to an-

swer this question for finite dimensional operators.

For a complex Hilbert space, with dual, we define a finite rank operator by, for all, where, and is a unit vector, with:

In this paper, we use finite rank operators to determine the norm of. We first review some known results on the norm of the Jordan elementary operator, , for all with fixed. We will then proceed to show that for an operator with and for all unit vectors, then:

Some mathematicians have attempted to determine the norm of. Timoney, used (matrix) numerical ranges and the tracial geometric mean to obtain an approximation of  , while Nyamwala and Agure used the spectral resolution theorem to calculate the norm of induced by normal operators in a finite dimensional Hilbert space  .

The study of the norm of the Jordan elementary operator has also attracted many researchers in operator theory. Mathieu  , in 1990, proved that in the case of a prime C*-algebra, the lower bound of the norm of

can be estimated by In 1994, Cabrera and Rodriguez  , showed that for prime JB*-algebras.

On their part, Stacho and Zalar  , in 1996 worked on the standard operator algebra which is a sub-algebra of, that contains all finite rank operators. They first showed that the operator actually represents a Jordan triple structure of a C*-algebra. They also showed that if is a standard operator algebra acting on a Hilbert space, and, then They later (1998), proved that

for the algebra of symmetric operators acting on a Hilbert space. They attached a family of Hil-

bert spaces to standard operator algebra, using the inner products on them to obtain their results.

In 2001, Barraa and Boumazguor  , used the concept of the maximal numerical range and finite rank operators to show that if with, then:

where,

is the maximal numerical range of relative to , and is the Hilbert adjoint of.

Okelo and Agure  used the finite rank operators to determine the norm of the basic elementary operator. Their work forms the basis of the results in this paper.

2. The Norm of Elementary Operator

In this section, we present some of the known results on elementary operators and proceed to determine norm of the elementary operator.

In the following theorem Okelo and Agure  , determined the norm of the basic elementary operator.

Theorem 2.1  : Let be a complex Hilbert space and the algebra of bounded linear operators on. Let be defined by for all with as fixed elements in. If for all with, we have for all unit vectors, then;

Proof: Since, we have,;

Therefore:

Letting, we obtain:

On the other hand, we have:

with:

So, setting, and, we have:

, with fixed in.

obtaining;

Hence, from (1) and (2), we obtain

.

For any vectors, the rank one operator, , is defined by, for all.

In the following three results Baraa and Boumazgour give three estimations to the lower bound of the norm of

the Jordan elementary operator. See  . Recall that the Jordan elementary operator is the operator

, for all with fixed.

Theorem 2.2. Let be the Jordan elementary operator with fixed, and with. Then

where, is the maximal numerical range of relative to, as defined earlier.

Proof: Let. Then there exists a sequence of unit vectors in such that and. Consider unit vectors, and recall the rank one operator, , defined as, for all unit vectors. For fixed operators, we have;

That is.

Thus we have:

Hence

Letting, we obtain:

and this is true for any, and for any unit vector.

Now, consider the set.

We have:

But.

Therefore:

and this completes the proof.

Corollary 2.3: Let be a complex Hilbert space and be bounded linear operators on. Let . Then we have Proof: Let. Then , or, and therefore, either there is a sequence of unit vectors in such that and or, there is a sequence of unit vectors in such that and.

Recall that in the previous theorem (Inequality (3)), we obtained:

This is equivalent to:

considering the sequence,

Taking limits in either (3) or (4), we obtain

and this is true for any unit vector.

Now, consider the set.

We have:

But.

Therefore:

and this completes the proof.

Proposition 2.4: Let be a complex Hilbert space and be bounded linear operators on. If

Proof: Suppose. Then and, and therefore we can find two sequences and of unit vectors in such that:

, and,.

Since and, then and

For each, we have:

Now, we have:

Therefore:

Letting we obtain:

That is and this implies that.

Clearly, and therefore we obtain.

We recall that an elementary operator, , is defined as, for all where are fixed, for. When, we have, for all and fixed, for

The following result gives the norm of.

Theorem 2.5: Let be a complex Hilbert space and be the algebra of all bounded linear operators on: Let be the elementary operator on defined above. If for an operator with, we have for all unit vectors, then:

Proof: Recall that is defined as, for all and fixed, for

We have:

Therefore, for all with.

So, for all, for all with.

Therefore,.

Letting, we obtain:

Next, we show that.

Since, then we have for all. But.

Now, let be functionals for

Choose unit vectors and define finite rank operators and on, for by for all with, for, and, for with, for.

Observe that the norm of for is,

That is for any unit vector with, for.

Likewise, the norm of is for any unit vector with, for.

Therefore, for all with, we have

Since, we have:

Now, since and are all positive real numbers, we have

and.

Thus and hence we have.

That is,

Now, (5) and (6) implies that:

and this completes the proof.

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