Advances in Pure Mathematics
Vol.2 No.6(2012), Article ID:24364,4 pages DOI:10.4236/apm.2012.26054
A Note on Nilpotent Operators
Department of Mathematics, Duquesne University, Pittsburgh, USA
Email: gaura@duq.edu
Received July 16, 2012; revised September 11, 2012; accepted September 19, 2012
Keywords: Numerical Range; Numerical Radius; Nilpotent Operator Weighted Shift; Eigenvalues
ABSTRACT
We find that a bounded linear operator T on a complex Hilbert space H satisfies the norm relation,
for any vector a in H such that
. 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.
1. Introduction
The motivation for this note is provided by the results obtained in [1-4]. 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.
Lemma 1.1., where T* is the adjoint operator of T and
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
such that
implies that
and
. In Theorem 2.1, we present a different proof of their result in [2] and show that
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
if
. Example 3.3 justifies why
fails to increase until and unless
.
2. Main Results
Theorem 2.1. The following statements are true for a bounded linear operator T on a Hilbert space H with.
1) such that
,
.
2) If for some integer n, then
and
.
3) The set 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
. Then the inner product relation
implies that
That is,
Hence,
Since
it follows that
Dividing the above inequality by, we have
Let be the following block-diagonal matix of order n and
If γn denotes the determinant of such that
then the value of γn is positive because all principal minors of
are nonnegative. Suppose that
(2.1)
We consider the following cases:
Case 1. Iffor the least
then
and
converges to zero.
Case 2. Let for all
. Then
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),
as
. Thus
. Obviously, q = 1 only if
.
2) By the assumption, for some positive integer n. Now fom Equation (2.1), we obtain:
and so that
. The equality,
now follows from (a) and thus
. Also,
which gives
since
.
3) To prove this case, we assume that if the vector is orthogonal to the spanning set
then
. Let
, for
. Then
Hence, for
and the spanning set
is a non-trivial invariant subspace on T.
In [2, p. 1052], an example of an operator T on and an element x in H with
, is given where
. Theorem 2.1 above establishes that
is the best constant in this case.
Remark 2.2. An operator A on H is hyponormal if
. Let
then
if A is a hyponormal operator. Hence,
,
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 and
then
. 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
with
. Such a polynomial is positive if
for all
. Here, we present a simplified proof of Theorem 1 in [1].
Theorem 2.3. For an operator N on H with and
, we have
.
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:
and
for
The matrix for S gives a dialation for T. Let A be the matrix for S and
If is a unitary operator on
with diagonal
then
. By Lemma 1, we have:
This helps to define the characteristic function of a contraction.
For the operator N on H, let then
is a positive operator and
depends on N. Let the range of
be denoted by
. Then the tensor product,
, is a Hilbert space. We define the map
so that F is an isometry.
For λ, let where
I is the identity operator, and
is an operator on
.
Therefore and
.
Now, we claim that, for we hope that
By Lemma 1.1
That is,.
Since, we have:
and
where is the spectral radius of
. By the definition of the spectral radius, we have the characteristic polynomial f such that
by [5, p. 179, Example 9], the roots of
are given by
,
and
and
.
Karaev in [3] has proved, using Theorem 1 in [1] and the Sz.-Nagy-Foias model in [6] that the numerical range 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
,
, we have
as a closed or an open disc centered at zero with radius equal to
. In fact, we have the following theorem.
Theorem 2.4. For a nilpotent operator N on H with,
and
, the numerical range
is a disc centered at zero with radius
.
Proof. For any we must claim that
, for
and
is a vector in
.
From [1, p. 374, Proposition 2], we have. Also, for some
,
. Now by [1] [P.375, Lemma 2], we obtain:
and
Let. Then:
and the theorem follows from above since is arbitrarily chosen.
3. An Application
An operator A is a unilateral weghted shift if there is an orthonormal basis and a sequence of scalers
such that
for all
. It is easy to see that
where S is the unilateral shift and D is the diagonal operator with
, for all n.
Thus, and
for all n. So
is the basis of eigenvectors for
. Also, note that A is bounded if
is bounded.
If A is a unilateral shift then and
for
. Consequently, for a hyponormal operator A,
and
for
. A wighted shift is hyponormal if and only if its weight sequence is increasing.
Example 3.1. Let be an operator on
such that
and
for
and
. Here, we show that
is not an eigenvalue of
if
. We prove our claim by contradiction Let
be an eigenvalue of
. Then, there exists
with
and
, n = 2, 3, ···. It is not hard to see that:
For, we have
and thus
, which shows that
, contrary to our assumption. Thus,
is not an eigenvalue of
if
.
Remark 3.2. Following [2], if then
Therefore, the numerical radius, is equal to 1.
The example below shows that there exists an operator such that
for
.
Example 3.3. Let be a unilateral shift. If
is the orthogonal projection of
onto the spanning set of vectors
then
and
has the usual matrix representation. Let
Then the characteristic polynomial of is given by a Chebyshev polynomial
of the first kind. Let
where
. Then:
(easily proven by trigonometric identities) and for
is a linear combination of powers of xk. Also, det
. If
then the 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
, for
.
Suppose that
then. Hence,
if
.
REFERENCES
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- M. T. Karev, “The Numerical Range of a Nilpotent Operator on a Hilbert Space,” Proceedings of the American Mathematical Society, Vol. 132, 2004, pp. 2321-2326.
- J. P. Williams and T. Crimmins, “On the Numerical Radius of a Linear Operator,” American Mathematical Monthly, Vol. 74, No. 7, 1967, pp. 832-833. doi:10.2307/2315808
- J. T. Scheick, “Linear Algebra with Applications,” International Series in Pure and Applied Mathematics, McGraw-Hill, New York, 1997.
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