Erratum to ‘Some results about EP modular operators’

Abstract

We give some examples of EP and non-EP operators to show that the main results of Mohammadzadeh Karizaki et al. [Some results about EP modular operators. Linear Multilinear Algebra. DOI:10.1080/03081087.2020.1844613] are not correct even in the case of Hilbert spaces.

Keywords:

• Bounded adjointable operator
• Hilbert C*-module
• closed range

2000 Mathematics Subject Classifications:

• Primary: 47A05
• Secondary: 46L08

This article refers to:

Some results about EP modular operators

K. Sharifi

Faculty of Mathematical Sciences, Shahrood University of Technology, Shahrood, Iran

CONTACT K. Sharifi [email protected] Faculty of Mathematical Sciences, Shahrood University of Technology, P.O. Box 3619995161-316, Shahrood, Iran

1. Main results

A bounded adjointable linear operator T with closed range on a Hilbert C*-module (or complex Hilbert space) H is called an EP operator if T and $T^{\ast }$ have the same range. If a bounded adjointable operator T does not have closed range, then neither $Ker(T)$ nor $Ran(T)$ need to be orthogonally complemented. For the basic theory of Hilbert C*-modules we refer to the book [4] and papers [1, 2]. Let H be a Hilbert module over an arbitrary C*-algebra of coefficients $\mathcal{A}$. An operator $T\in \mathcal{L}(H)$ is called EP if $Ran(T)$ and $Ran(T^{\ast })$ have the same closure [6, Definition 2.1].

Mohammadzadeh Karizaki et al. [5]investigate commuting EP operators and prove that $\overline{Ran(T+S)}=\overline{Ran(T)+Ran(S)}$, when T and S are EP modular operators on H. However, the main results of this paper is not correct even in the case of Hilbert spaces. Indeed, they have utilized some equalities and identities that are generally not valid for operators or matrices.

Example 1.1

Let bounded operators T and S on ${\ell }_2$ be defined by $\begin{array}{rl}T(x_1,x_2,x_3,x_4,x_5,\dots )& =(x_1,x_2,x_2+x_3,x_4,x_5,\dots )\\ S(x_1,x_2,x_3,x_4,x_5,\dots )& =(x_1+x_2,0,0,x_4,x_5,\dots ).\end{array}$Then $\begin{array}{rl}{T}^{\ast }(x_1,x_2,x_3,x_4,x_5,\dots )& =(x_1,x_2+x_3,x_3,x_4,x_5,\dots )\\ S^{\ast }(x_1,x_2,x_3,x_4,x_5,\dots )& =(x_1,x_1,0,x_4,x_5,\dots )\\ (T+S)(x_1,x_2,x_3,x_4,x_5,\dots )& =(2x_1+x_2,x_2,x_2+x_3,x_4,x_5,\dots )\\ (T+S{)}^{\ast }(x_1,x_2,x_3,x_4,x_5,\dots )& =(2x_1,x_1+x_2+x_3,x_3,x_4,x_5,\dots ).\end{array}$One can easily see that T and T + S are EP operators and S is not an EP operator, and so the part ‘’ in [5, Theorem 2.7] is not correct.

Example 1.2

Let bounded operators T and S on ${\ell }_2$ be defined by $\begin{array}{rl}T(x_1,x_2,x_3,x_4,x_5,\dots )& =(x_1-x_3,0,x_3,x_4,x_5,\dots )\\ S(x_1,x_2,x_3,x_4,x_5,\dots )& =(x_3-x_1,0,x_3,x_4,x_5,\dots ).\end{array}$Then $\begin{array}{rl}{T}^{\ast }(x_1,x_2,x_3,x_4,x_5,\dots )& =(x_1,0,x_3-x_1,x_4,x_5,\dots )\\ S^{\ast }(x_1,x_2,x_3,x_4,x_5,\dots )& =(-x_1,0,x_3+x_1,x_4,x_5,\dots )\\ (T+S)(x_1,x_2,x_3,x_4,x_5,\dots )& =(0,0,x_3,x_4,x_5,\dots )\\ (T{T}^{\ast }+S{S}^{\ast })(x_1,x_2,x_3,x_4,x_5,\dots )& =(4x_1,0,2x_3,x_4,x_5,\dots ).\end{array}$One can easily see that T and S are EP operators and S is not an EP operator and $\begin{array}{rl}\overline{Ran(T+S)}& \ne \overline{Ran(T{T}^{\ast })+Ran(S{S}^{\ast })},\\ \overline{Ran(T+S)}& \ne \overline{Ran(T)+Ran(S)},\end{array}$that is, the parts (iii), (iv), (v) and (vi) in [5, Theorem 2.10] are not correct.

Remark 1.3

Some mistakes of this paper arise from the following gaps:

• It is known that $\underset{n=1}{\overset{\mathrm{\infty }}{⨁}}H=\{x=(x(n))\in \prod _{n}H:\sum _n⟨x(n),x(n)⟩\mathrm{converges}\mathrm{in}\mathrm{the}\mathrm{norm}\mathrm{of}\mathcal{A}\}$is a Hilbert $\mathcal{A}$-module. In the proof of Theorem 2.10, the matrix operators $B^{\prime }$ and $C^{\prime }$ take their values in $\underset{n=1}{\overset{\mathrm{\infty }}{⨁}}H$, and so one needs to check that the matrix operators are well defined and $\overline{Ran(B^{\prime })}=\overline{Ran(C^{\prime })}$.

• Let T and S be bounded adjointable operators on H and let $A=\left[\begin{array}{cc}T& 0\\ S& 0\end{array}\right]\in \mathcal{L}(H\oplus H).$The authors have applied the equality $\overline{Ran(T)}\oplus \overline{Ran(S)}=\overline{Ran(A)}$ several times to prove Theorem 2.7, Theorem 2.9 and Theorem 2.10. One should be aware that the equality is not valid even for matrices.

One can consider various types of Hilbert modules and C*-algebras to reinvestigate the range equalities of the paper [5]. In this regards, the papers [1–3, 7] might be useful.