A Note on Nilpotent Operators

We find that a bounded linear operator T on a complex Hilbert space H satisfies the norm relation 2 n T a q  1,2, , n   , for any vector a in H such that   2 1 4 1 Ta  q Ta    . A partial converse to Theorem 1 by Haagerup and Harpe in [1] is suggested. We establish an upper bound for the numerical radius of nilpotent operators.


Introduction
The motivation for this note is provided by the results obtained in [1][2][3][4].Let T be a bounded linear operator on a complex Hilbert space H.The numerical range of T, denoted by W(T), is the subset of the complex plane and The numerical radius of T is defined as, The following lemma is known and is an easy consequence of the definitions involved.

   
, where T * is the adjoint operator of T and z z is the complex conjugate of .
Berger and Stampfli in [2] have proved that if and , for some n, then  .Also, they gave an example of an operator T and an element x H  such that w T implies that . In Theorem 2.1, we present a different proof of their result in [2] and show that 2 q T T is indeed the best constant.
Theorem 2.1 also generalizes the result in [4] and provides a partial converse to Theorem 1 in [1, p. 372].
Our next main result in Theorem 2.3 gives an alternative and shorter proof of Theorem 1 in [1].
Applying Lemma 2 and Proposition 2 of [1], a new result on the numerical range of nilpotent operators on H is obtained in Theorem 2.4.This gives a restricted version of Theorem 1 in [3].
Finally, two examples are discussed.Example 3.1 deals with the operator , where 1 is not the eigenvalue of q if 1 < .Example 3.3 justifies why 2 q   q w T fails to increase until and unless 2 q  .

Main Results
Theorem 2.1.The following statements are true for a bounded linear operator T on a Hilbert space H with 3) The set   , , , n a Ta T a  forms a nontrivial subspace of T so that its orthogonal complement is invariant.
Proof. 1) For each real number  and a postive integer, n, let n .Then the inner product relation e ,e i i , e e d d , Dividing the above inequality by , we have Let be the following block-diagonal matix of order n and We consider the following cases: and by induction Further, the inequality implies that converges to q as n goes to infinity for some q ≥ 0. Therefore from Equation (2.1), 2) By the assumption, for some positive integer n.Now fom Equation (2.1), we obtain: 3) To prove this case, we assume that if the vector Re , , In [2, p. 1052], an example of an operator T on H and an element x in H with , is given where

and the set of vectors forms a reducing subspace of A.
A natural connection between Feijer's inequality and the numerical radius of a nilpotent operator was estaplished by Haagerup and Harpe in [1].They proved, using positive definite kernals, that for a bounded linear operator T on a Hilbert space H such that T and 1 . The external operator is shown  to be a truncated shift with a suitable choice of the vector in H.The inequality is related to a result from Feijer about trigonometric polynomials of the form Proof.We will follow the notations of Theorem 1 in [1].Let S be the operator on and be the basis in .We define the operator S as follows: C , is a Hilbert space.We define the map so that F is an isometry.
For λ, let and w N and Karaev in [3] has proved, using Theorem 1 in [1] and the Sz.-Nagy-Foias model in [6] that the numerical range   W N of an arbitrary nilpotent operator N on a complex Hilbert space H is an open or closed disc centered at zero with radius less than or equal to Using Theorem 2 and the assumption that closed or an open disc centered at zero with radius equal . In fact, we have the following theorem.

Theorem 2.4. For a nilpotent operator N on H with
, W N is a disc centered at zero with radius


Proof.For any  we must claim that . Now by [1] [P.375, Lemma 2], we obtain:   T q for and .Here, we show that 1 is not an eigenvalue of if We prove our claim by contradiction Let be an eigenvalue of T .Then, there exists with 1 , contrary to our assumption.Thus, is not an eigenvalue of if Therefore, the numerical radius, is equal to 1. q The example below shows that there exists an operator such that for   1 -trivial invariant subspace on T.
The matrix for S gives a dialation for T. Let A be the matrix for S and operator and  depends on N. Let the range of be denoted by .Then the tensor product, radius of  S  y the definition of the spectral radius, we have the characteristic polynomial f such that   and the theorem follows from above since  is arbitrarily chosen.

Example 3 . 3 .
Let A be a unilateral shift.If E is the orthogonal projection of 2 roots are given by the Chebyshev polynomial of the first kind.The roots can be found by finding the eigenvalues of matrix B. By [2, p. 179, Example 9], the eigenvalues of B are given by [1]  R .Here, we present a  simplified proof of Theorem 1 in[1].