A bounded linear operator T on a Hilbert space is an isometric N -Jordan operator if it can be written as A + Q, where A is an isometry and Q is a nilpotent of order N such that AQ = QA. In this paper, we will show that the only isometric N -Jordan weighted shift operators are isometries. This answers a question recently raised. | Turk J Math (2016) 40: 1114 – 1117 ¨ ITAK ˙ c TUB ⃝ Turkish Journal of Mathematics doi: Research Article Isometric N -Jordan weighted shift operators 1 Saeed YARMAHMOODI1 , Karim HEDAYATIAN2,∗ Department of Mathematics, Marvdasht University, Islamic Azad University, Marvdasht, Iran 2 Department of Mathematics College of Sciences, Shiraz University, Shiraz, Iran Received: • Accepted/Published Online: • Final Version: Abstract: A bounded linear operator T on a Hilbert space is an isometric N -Jordan operator if it can be written as A + Q , where A is an isometry and Q is a nilpotent of order N such that AQ = QA . In this paper, we will show that the only isometric N -Jordan weighted shift operators are isometries. This answers a question recently raised. Key words: Isometric N -Jordan operator, nilpotent, weighted shift operator 1. Introduction and preliminaries Let H be a Hilbert space and B(H) stand for the space of all bounded linear operators on H . An operator T in B(H) is called an isometric N -Jordan operator if T = A + Q, where A is an isometry and Q is a nilpotent operator of order N , that is, QN = 0 but QN −1 ̸= 0 , and AQ = QA. Note that the notions of isometric 1 -Jordan and isometry coincide. It follows from Proposition of [11] that the operator T is injective. The dynamic and spectral properties of T have been studied in [11]. We note that T ∗ T is invertible. Indeed, by Corollary of [11] the operator T is bounded below, and so for every h ∈ H , ∥T ∗ T h∥∥h∥ ≥ |⟨T ∗ T h, h⟩| = ∥T h∥2 ≥ c∥h∥2 for some c > 0 , which implies that T ∗ T is also bounded below and so is injective and has closed range. However, H = (ker(T ∗ T ))⊥ = ran(T ∗ T ) = ran(T ∗ T ) implies that T ∗ T is invertible. It is easy to see that if A is a unitary operator then (T ∗ T )−1 = 3I − 3T T ∗ + T 2 T ∗2 . For a positive integer m an operator S ∈ B(H) is an m-isometry .
