Positivity 13 (2009), 273–276
2008 Birkh¨auser Verlag Basel/Switzerland
1385-1292/010273-4, published online September 1, 2008
On positive transitive operators
Abstract. In this short paper we prove that even though the latest transitive
operator has only one negative entry, we cannot get rid of it by simple opera-
tions. The research is motivated by, still unsolved, invariant subspace problem
for positive operators.
Mathematics Subject Classiﬁcation (2000). 47A15.
Keywords. Operator, invariant subspace, transitive operator.
The study of operators without non-trivial closed invariant subspaces has been
going on for about thirty years. Most of the variety [2,4, 6-11] of examples is due
to C. J. Read. These are an operator with every non-zero vector being hypercyclic
, quasinilpotent , and strictly singular such operators . None of known
examples acts on a reﬂexive Banach space. The ﬁrst operator acts on W ⊕
any separable Banach space W . The quasinilpotent one was constructed on
The latest, which turned out to be ﬁnitely strictly singular , was deﬁned on
, where J
are James spaces with strictly increasing numbers
. Nevertheless, investigation of order properties of these examples is not that
developed. See  for a survey of the invariant subspace problem for positive
operators. In 1998 V. G. Troitsky  showed that Read’s operator in isa
diﬀerence of two positive operators one of which represents a compact operator
. Recently  the author constructed a simpler operator on
trivial invariant subspaces. This example is a diﬀerence of two positive operators
one of which has only one non-zero entry in its matrix. Professor Troitsky in
personal communications proposed to conjugate this new transitive operator by
relatively simple operators in attempt to get rid of that single negative entry.
In the present paper we consider the operator T constructed in and
conjugate it by operators of the form S = Id + R, where matrix representation of
R in canonical basis has only ﬁnitely many non-zero entries. Conjugation of T by a
continuous invertible operator preserves transitivity. So does addition of a multiple