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\noindent\parbox{2.85cm}{\includegraphics*[keepaspectratio=true,scale=1.75]{BJMA.jpg}}
\noindent\parbox{4.85in}{\hspace{0.1mm}\\[1.5cm]\noindent
Banach J. Math. Anal. 5 (2011), no. 2, 25--32\\
$\frac{\rule{4.55in}{0.05in}}{{}}$\\
{\footnotesize
\textcolor[rgb]{0.65,0.00,0.95}{\textsc{\textbf{\large{B}}anach
\textbf{\large{J}}ournal of \textbf{\large{M}}athematical
\textbf{\large{A}}nalysis}}\\
ISSN: 1735-8787 (electronic)\\
\textcolor[rgb]{0.00,0.00,0.84}{\textbf{www.emis.de/journals/BJMA/ }}\\
$\frac{{}}{\rule{4.55in}{0.05in}}$}\\[.5in]}


\title{EP elements in Banach algebras}

\author[D. Mosi\' c, D.S. Djordjevi\'c]{Dijana Mosi\' c$^1$ and Dragan S. Djordjevi\'c$^2$$^{*}$}

\address{$^{1}$ Faculty of Sciences and Mathematics,
University of Ni\v s, P. O. Box 224, Vi\v segradska 33,18000 Ni\v
s, Serbia.}

\email{\textcolor[rgb]{0.00,0.00,0.84}{dijana@pmf.ni.ac.rs}}

\address{$^{2}$ Faculty of Sciences and Mathematics,
University of Ni\v s, P. O. Box 224, Vi\v segradska 33,18000 Ni\v
s, Serbia.}

\email{\textcolor[rgb]{0.00,0.00,0.84}{dragan@pmf.ni.ac.rs}}

\dedicatory{{\rm Communicated by D. R. Larson}}


\subjclass[2010]{Primary 46L05; Secondary 47A05.}

\keywords{EP elements, Moore--Penrose inverse, group inverse, Banach algebra.}

\date{Received: 15 November 2010; Accepted: 19 November 2010.
\newline \indent $^{*}$ Corresponding author.}

\begin{abstract}
An element of a Banach algebra is EP, if it commutes with its
Moore-Penrose inverse. We present a number of new characterizations of EP
elements in Banach algebra.
\end{abstract} \maketitle

\section{Introduction and preliminaries}


 Generalized inverses of matrices have important roles in theoretical and numerical methods of  linear
 algebra. The most significant fact is that we can use generalized
 inverses of matrices, in the case when  ordinary inverses do not exist, in order to
 solve some matrix equations. Similar reasoning can be applied to
 linear (bounded or unbounded) operators on Banach and Hilbert
 spaces. Then, it is interesting to consider generalized elements
 of Banach and $C^*$-algebras, or more general, in rings with or without
 involution. For general theory of generalized inverses see
\cite{BIG} and  \cite{DjR}.


 Let $\mathcal A$ be a complex  unital Banach algebra. An element $a\in\mathcal A$ is
 generalized (or inner) invertible, if there exists some $b\in\mathcal A$ such
 that $aba=a$ holds. In this case $b $ is a generalized (or inner)
 inverse of $a$. If $aba=a$, then take $c=bab$ to obtain the
 following: $aca=a$ and $cac=c$. Such $c$ is called a reflexive
 (or normalized) generalized inverse of $a$. Finally, if $aba=a$,
 then $ab$ and $ba$ are idempotents. In the case of the
 $C^*$-algebra, we can require that $ab$ and $ba$ are Hermitian.
We arrive to the definition of the Moore-Penrose inverse in
$C^*$-algebras (see \cite{HM}).

\begin{definition} Let $\mathcal A$ be a unital $C^*$-algebra. An element
$a\in\mathcal A$ is Moore-Penrose invertible, if there exists some
$b\in\mathcal A$ such that
$$aba=a,\ bab=b,\ (ab)^*=ab,\ (ba)^*=ba$$hold. Such $b$ is
uniquely determined if it exists \cite{HM} and known as the {\it
Moore-Penrose inverse} of $a$, writtern $a^\dag$.
\end{definition}

More generally, an extension \cite{R1,R2} of the Moore-Penrose
inverse to more general unital Banach algebra flows from the
following definition of {\it Hermitian elements}:


\begin{definition} An element $a\in \mathcal A$
is said to be Hemitian if $\|\exp(ita)\|=1$ for all $t\in R$.
\end{definition}


\begin{definition} Let $\mathcal A$ be a complex unital Banach algebra and
$a\in\mathcal A$. If there exists $b\in\mathcal A$ such that
$$aba=a, \quad  bab=b, \quad ab\ {\rm and}\ ba\ {\rm are\ Hermitian},$$ then
the element $b$ is the Moore-Penrose inverse of $a$, and it will
be denoted by $a^\dagger$.
\end{definition}

For an element $a$ in a Banach algebra, if $a^\dagger$ exists, it
is uniquely determined (see \cite{R1}).

When the Moore-Penrose inverse $a^\dagger$ exists we shall say
that $a\in\mathcal A$ is MP. The Moore-Penrose inverse has many
nice approximation properties, but does not in general commute
with the original element. An element with a commuting generalized
inverse is very special:

\begin{definition} Let $\mathcal A$ be a unital Banach algebra and
$a\in\mathcal A$. An element $b\in\mathcal A$ is the group inverse
of $a$, if the following conditions are satisfied:
$$aba=a, \quad  bab=b, \quad ab=ba.$$
\end{definition}

The group inverse of $a\in\mathcal A$ is unique if it exists,
written $a^\#$. When a Moore-Penrose inverse for $a\in\mathcal A$
is also a group inverse then $a\in\mathcal A$ is said to be EP.


Let $X$ be a Banach space and $\mathcal L(X)$ the Banach algebra
of all linear  bounded operators on $X$. In addition, if
$T\in\mathcal L(X)$, then $N(T)$ and $R(T)$  stand for the null
space and the range of $T$, respectively. The ascent of $T$ is
defined as ${\rm asc}(T)=\inf\{n\geq 0:N(T^n)=N(T^{n+1})\}$, and
the descent of $T$ is defined as ${\rm dsc}(T)=\inf\{n\geq
0:R(T^n)=R(T^{n+1})\}$. In both cases the infimum of the empty set
is equal to $\infty$. If ${\rm asc}(T)<\infty$ and ${\rm
dsc}(T)<\infty$, then ${\rm asc}(T)={\rm dsc}(T)$.

If $T\in\mathcal L(X)$ a closed range operator, then it is not
difficult to prove that the following statements are equivalent
(see \cite{R} where they were proved for linear transformations):

(i) $T^\#$ exists,

(ii) $X = N (T) \oplus R(T )$,

(iii) $R(T^2)= R(T)$ and $N(T^2) = N(T )$

(iv) $T|_{R(T)}:R(T)\rightarrow R(T)$ is a Banach space
isomorphism,

In addition, if $T^\#$ exists, then, according to \cite[Theorem
1]{R},

(v) $N (T^\#) = N (T )$ and $R(T^\#) = R(T )$,

(vi) $R(TT^\#) = R(T)$ and $N(TT^\#)= N(T)$.

Note that, according to (iii), given $T\in\mathcal L(X)$,
necessary and sufficient for $T^\#$ to exists is the fact that
${\rm asc}(T )={\rm dsc}(T ) \leq  1$. Obviously,
$R((T^\#)^n)=R(T^\#) =R(T)=R(T^n)$ and
$N((T^\#)^n)=N(T^\#)=N(T)=N(T^n)$ for every non-negative integer
$n$.

Finally, if $a\in\mathcal A$ is an EP element then the group
inverse $a^\#$ clearly exists, and coincides with the
Moore-Penrose inverse $a^\dagger$. On the other hand if $a^\#$
exists then necessary and sufficient for $a$ to be EP is that
$aa^\#(=a^\#a)$ is Hermitian, so that also $a$ is MP, with
$a^\dagger=a^\#$.


Hence, if $H$ is a Hilbert space and $T\in \mathcal L(H)$ is EP,
then $N(T^n)=N(T)$ and $R(T^n)=R(T)$ for all positive integers
$n$. Moreover, the decomposition $H=R(T)\oplus N(T)$ is
orthogonal. It follows that the orthogonal projection on every
$R(T^n)$ is unique. This is the derivation of the term EP, which
stands for "equal projections". The
 interest for EP matrices can be found in \cite{BT,CM,
CT,HK}. EP operators on Hilbert spaces are considered in
\cite{Dj1,Dj,DjK,DKP,K1,L}. Results on EP elements in
$C^*$-algebras and rings with involution can be found in
\cite{DKS,K2,MDjK,PP}. Finally, and the most interesting results
related to our work, EP elements in Banach algebras are
investigated in \cite{B,BR,M}.


In the following Banach space theorem the range and null space of
the Hilbert adjoint are replaced by those of the Moore-Penrose
inverse:


\begin{theorem}{\rm \cite{B}}\label{teBoassoEP} Let $X$ be a Banach space, and
consider $T\in\mathcal L(X)$ such that $T^\dagger$ exists. Then,
the following statements are equivalent:
\begin{itemize}
\item[\rm(i)] $T$ is an EP operator;

\item[\rm(ii)] $N(T)=N(T^\dagger)$;

\item[\rm(iii)] $R(T)=R(T^\dagger)$.
\end{itemize}
\end{theorem}

Finally, we formulate the essential corollary for the proof of our
main theorem.

\begin{corollary}\label{corBanachEP}
Let $X$ be a Banach space, and let $T\in B(X)$ such that $T^\dag$
and $T^\#$ exist. Then the following statements are equivalent:

\begin{itemize}

\item[\rm (i)] $T$ is EP;

\item[\rm(ii)] $R(T^\dag)\subset R(T)$;

\item[\rm(iii)] $R(T)\subset R(T^\dag)$;

\item[\rm(iv)] $N(T)\subset N(T^\dag)$;

\item[\rm(v)] $N(T^\dag)\subset N(T)$.
\end{itemize}
\end{corollary}

\begin{proof} According to \cite[Theorem 6]{B}, the
Banach space $X$ can be decomposed using the idempotents $T T^\#$,
$T^\dagger T$, and $TT^\dagger$, that is respectively:
$$X = N (T )\oplus R(T ) = N (T )\oplus R(T^\dagger) = N (T^\dagger)\oplus
R(T ).$$

(i) $\Rightarrow$ (ii): Obviously, from Theorem \ref{teBoassoEP}.

(ii) $\Rightarrow$ (i): Using the hypothesis $R(T^\dag)\subset
R(T)$ and the decomposition of $X$ derived from the idempotents
$TT^\#$ and $T^\dagger T$, we have $R(T^\dagger) = R(T)$. By
Theorem \ref{teBoassoEP}, we deduce that $T$ is  EP.

The rest of the proof follows in a similar way.
\end{proof}


\section{EP elements in Banach algebras}

There are many characterizations of EP elements in $C^*$-algebras,
or rings, as well as EP operators on Hilbert spaces. The following
result relies on characterization of the Moore-Penrose inverse in
Banach algebras. We use $N$ to denote the set of positive
 integers.

\begin{theorem} \label{te1EPb} Let $X$ be a Banach space, and
consider $T\in\mathcal L(X)$ such that $T^\dagger$ and $T^\#$
exist. Then, $T$ is EP if and only if for some $m,n\in N$ one of
the following equivalent conditions holds:

\begin{itemize}

\item[\rm(i)] $(T^\#)^{n+m-1}=(T^\dagger)^m (T^\#)^{n-1}$;

\item[\rm(ii)] $(T^\#)^{n+m-1}=(T^\#)^{n-1}(T^\dagger)^m$;

\item[\rm(iii)] $TT^\dagger (T^\#)^{n+m-1}=(T^\dagger)^n
(T^\#)^mT$;

\item [{\rm (iv)}] $TT^\dagger (T^\#)^{n+m-1}=(T^\#)^m
T(T^\dagger)^n$;

\item [{\rm (v)}] $T(T^\#)^n(T^\dagger)^m=T^\dagger
T(T^\#)^{n+m-1}$;

\item [{\rm (vi)}] $T(T^\#)^n(T^\dagger)^m=(T^\#)^{n+m-1}T^\dagger
T$;

\item [{\rm (vii)}] $T^\dagger T(T^\#)^n=(T^\#)^{n}T^\dagger T$;

\item [{\rm (viii)}] $(T^\dagger)^2(T^\#)^n=T^\dagger
(T^\#)^nT^\dagger$;

\item [{\rm (ix)}] $T^\dagger
(T^\#)^nT^\dagger=(T^\#)^n(T^\dagger)^2$;

\item [{\rm (x)}] $T^\dagger(T^\#)^{n}=T^\#T^\dagger
(T^\#)^{n-1}$;

\item [{\rm (xi)}] $T^\dagger(T^\#)^{n}=(T^\#)^{n}T^\dagger$;

\item [{\rm (xii)}] $(T^\#)^{n}T^\dagger=(T^\#)^{n-1}T^\dagger
T^\#$;

\item [{\rm (xiii)}] $T(T^\dagger)^{n+1}=(T^\#)^n$;

\item [{\rm (xiv)}] $(T^\dagger)^{n+1}=(T^\#)^nT^\dagger$;

\item [{\rm (xv)}] $(T^\dagger)^{n+1}=T^\dagger (T^\#)^n$;

\item [{\rm (xvi)}] $(T^\dagger)^n=(T^\#)^n$;

\item [{\rm (xvii)}] $T(T^\#)^nT^\dagger=(T^\#)^n$;

\item [{\rm (xviii)}] $T^\dagger T(T^\#)^n=(T^\dagger)^n$;

\item [{\rm (xix)}] $(T^\#)^nT^\dagger T=(T^\dagger)^n$;

\item [{\rm (xx)}] $T^\dagger (T^\#)^nT+TT^\#(T^\dagger)^n
=2(T^\dagger)^n$;

\item [{\rm (xxi)}] $T^nTT^\dagger+T^\dagger TT^n=2T^n$;

\item[{\rm (xxii)}]  $T^n=T^nTT^\dagger$;

\item[{\rm (xxiii)}] $T^n=T^\dagger TT^n$;

\item[{\rm (xxiv)}] $T^nT^\dagger=T^\dagger T^n$.
\end{itemize}
\end{theorem}

\begin{proof} If $a$ is EP, then it is easy to check that all  statements  hold.

On the other hand, to prove that $T$ is EP, we show that one of
the conditions of Theorem \ref{teBoassoEP} or Corollary
\ref{corBanachEP} is satisfied.


(i) From the assumption $(T^\#)^{n+m-1}=(T^\dagger)^m
(T^\#)^{n-1}$, we obtain
$$R(T)=R((T^\#)^{n+m-1})=R((T^\dagger)^m
(T^\#)^{n-1})\subseteq R(T^\dagger).$$ Hence, by Corollary
\ref{corBanachEP}, $T$ is EP.

(ii) Suppose that $~(T^\#)^{n+m-1} = (T^\#)^{n-1}(T^\dagger)^m$. If
$x \in N(T^\dagger)$, then $~(T^\#)^{n+m-1}x \newline = 0$, i.e. $x\in
N((T^\#)^{n+m-1})=N(T)$. Thus, $N(T^\dagger)\subseteq N(T)$ and,
by Corollary \ref{corBanachEP}, $T$ is EP.

(iii) Assume that $TT^\dagger (T^\#)^{n+m-1}=(T^\dagger)^n
(T^\#)^mT$. Since
$$R(TT^\dagger (T^\#)^{n+m-1})=TT^\dagger R((T^\#)^{n+m-1})=TT^\dagger R(T)
=R(TT^\dagger T)=R(T),$$ and $R((T^\dagger)^n (T^\#)^mT)\subseteq
R(T^\dagger)$, we conclude $R(T)\subseteq R(T^\dagger)$. So, $T$
is EP.

(iv) Let $TT^\dagger (T^\#)^{n+m-1} = (T^\#)^m T(T^\dagger)^n$ and $x \in N(T^\dagger)$. Then $TT^\dagger (T^\#)^{n+m-1} x\newline = 0$, and
so $(T^\#)^{n+m-1}x\in N(TT^\dagger)\cap
R((T^\#)^{n+m-1})=N(T^\dagger)\cap R(T)=\{0\}$. Hence, $x\in
N((T^\#)^{n+m-1})=N(T)$. Therefore, $N(T^\dagger)\subseteq N(T)$
and, $T$ is EP.

(v)  If the statement $T(T^\#)^n(T^\dagger)^m=T^\dagger
T(T^\#)^{n+m-1}$ holds and $x\in N(T^\dagger)$, then $T^\dagger
T(T^\#)^{n+m-1}x=0$, i.e. $T(T^\#)^{n+m-1}x\in N(T^\dagger)\cap
R(T)=\{0\}$. Thus, $N(T^\dagger)\subseteq N(T)$, and $T$ is EP.

(vi) The assumptions
$T(T^\#)^n(T^\dagger)^m=(T^\#)^{n+m-1}T^\dagger T$ and $x\in
N(T^\dagger)$ imply $(T^\#)^{n+m-1}T^\dagger Tx=0$, that is
$T^\dagger Tx\in N(T)\cap R(T^\dagger)=\{0\}$. So, $x\in
N(T^\dagger T)=N(T)$ and then $N(T^\dagger)\subseteq N(T)$.
Therefore, $T$ is  EP.

(vii) Using the hypothesis $T^\dagger T(T^\#)^{n}=(T^\#)^n
T^\dagger T$ and the equalities
\begin{eqnarray*}R(T^\dagger T(T^\#)^{n})&=&T^\dagger
TR((T^\#)^{n}) =T^\dagger TR(T)\\&=&T^\dagger R(T^2)=T^\dagger
R(T) =R(T^\dagger T)=R(T^\dagger),\end{eqnarray*}
$$R((T^\#)^nT^\dagger T)=(T^\#)^nR(T^\dagger)=(T^\#)^n(R(T^\dagger)\oplus N(T))=R((T^\#)^n)=R(T),$$
 we deduce $R(T^\dagger) = R(T)$. By Theorem \ref{teBoassoEP},
T is  EP.

(viii) Let  we suppose that $(T^\dagger)^2(T^\#)^n=T^\dagger
(T^\#)^nT^\dagger$, and let $x\in N(T^\#)=N(T)$. Now, $T^\dagger
(T^\#)^nT^\dagger x=0$ gives $(T^\#)^nT^\dagger x\in
N(T^\dagger)\cap R(T)=\{0\}$. Hence, $x\in
N((T^\#)^nT^\dagger)\subseteq N(T^\dagger
T^{n+1}(T^\#)^nT^\dagger)=N(T^\dagger TT^\dagger)= N(T^\dagger)$.
Thus, $N(T)\subseteq N(T^\dagger)$ and, by Corollary
\ref{corBanachEP}, $T$ is  EP.

(ix) Assume that $T^\dagger
(T^\#)^nT^\dagger=(T^\#)^n(T^\dagger)^2$. From
\begin{eqnarray*}R(T^\dagger (T^\#)^{n}T^\dagger)&=&T^\dagger (T^\#)^{n}
R(T^\dagger) =T^\dagger (T^\#)^n(R(T^\dagger)\oplus
N(T))\\&=&T^\dagger R((T^\#)^{n})=T^\dagger R(T) =R(T^\dagger
T)=R(T^\dagger),\end{eqnarray*} and
$R((T^\#)^n(T^\dagger)^2)\subseteq R((T^\#)^n)=R(T)$, we deduce
that $R(T^\dagger)\subseteq R(T)$. By Corollary \ref{corBanachEP},
we have that $T$ is  EP.

(x) Let the equality $T^\dagger(T^\#)^{n}=T^\#T^\dagger
(T^\#)^{n-1}$ is satisfied. Because
$$R(T^\dagger(T^\#)^{n})=T^\dagger R((T^\#)^{n})=T^\dagger R(T)=R(T^\dagger
T)=R(T^\dagger),$$ and $R(T^\#T^\dagger (T^\#)^{n-1})\subseteq
R(T^\#)=R(T)$, it follows $R(T^\dagger)\subseteq R(T)$. Hence, $T$
is EP.

(xi) Suppose that $T^\dagger(T^\#)^n=(T^\#)^nT^\dagger$. In
particular,
$$R(T^\dagger(T^\#)^n)=T^\dagger R((T^\#)^n)=T^\dagger R(T^\#)
=T^\dagger R(T)=R(T^\dagger T)=R(T^\dagger).$$ On the other hand,
$$R((T^\#)^nT^\dagger)=(T^\#)^n(R(T^\dagger)\oplus N(T))=R((T^\#)^n)=R(T).$$
So, $R(T^\dagger)=R(T)$ and, by Theorem \ref{teBoassoEP}, T is an
EP operator.

(xii) Assume that $(T^\#)^{n}T^\dagger=(T^\#)^{n-1}T^\dagger
T^\#$. If $x\in N(T^\#)=N(T)$, then $(T^\#)^nT^\dagger x=0$
implies $T^\dagger x\in N(T)\cap R(T^\dagger)=\{0\}$. Now $x\in
N(T^\dagger)$ and $N(T)\subseteq N(T^\dagger)$. Therefore, $T$ is
EP.

(xiii) From $T(T^\dagger)^{n+1}=(T^\#)^n$ and $x\in N(T^\dagger)$,
it follows $(T^\#)^nx=0$, that is $x\in N((T^\#)^n)=N(T)$. Thus,
$N(T^\dagger)\subseteq N(T)$ and $T$ is EP.

(xiv) By the assumption $(T^\dagger)^{n+1}=(T^\#)^nT^\dagger$ and
the equality
$$R((T^\#)^nT^\dagger)=(T^\#)^nR(T^\dagger)=(T^\#)^n(R(T^\dagger)\oplus N(T))=R((T^\#)^n)=R(T),$$
we deduce $R(T)=R((T^\dagger)^{n+1})\subseteq R(T^\dagger)$. As
before, T is an EP operator.

(xv) If $(T^\dagger)^{n+1}=T^\dagger (T^\#)^n$ and $x\in
N(T^\dagger)$, then $T^\dagger(T^\#)^nx=0$. So $(T^\#)^nx\in
N(T^\dagger)\cap R(T)=\{0\}$. Hence, $x\in N(T)$ and
$N(T^\dagger)\subseteq N(T)$, i.e. $T$ is EP.

(xvi) The condition $(T^\dagger)^n=(T^\#)^n$ gives
$$R(T)=R((T^\#)^n)=R((T^\dagger)^n)\subseteq R(T^\dagger).$$
Thus, T is an EP operator.

(xvii) Let $T(T^\#)^nT^\dagger=(T^\#)^n$ and $x\in N(T^\dagger)$.
It follows that  $(T^\#)^n x=0$, i.e. $x\in N((T^\#)^n)=N(T)$. So,
$N(T^\dagger)\subseteq N(T)$ and $T$ is  EP.

(xviii) Suppose that $T^\dagger T(T^\#)^n=(T^\dagger)^n$. If $x\in
N(T^\dagger)$, then  $T^\dagger T(T^\#)^nx=0$ implying
$(T^\#)^nx\in N(T^\dagger T)\cap R((T^\#)^n)=N(T)\cap R(T)=\{0\}$.
Hence, $x\in N((T^\#)^n)=N(T)$ and $N(T^\dagger)\subseteq N(T)$.
Therefore, $ T$ is  EP.

(xix) By the hypothesis $(T^\#)^nT^\dagger T=(T^\dagger)^n$, if
$x\in N(T^\dagger)$ then $(T^\#)^nT^\dagger Tx=0$, giving
$T^\dagger Tx\in N(T)\cap R(T^\dagger)=\{0\}$. So, $x\in
N(T^\dagger T)=N(T)$. Thus, $N(T^\dagger)\subseteq N(T)$ and $ T$
is  EP.

(xx) Assume that $T^\dagger (T^\#)^nT+TT^\#(T^\dagger)^n
=2(T^\dagger)^n$, and let $x\in N(T^\dagger)$. Now, $T^\dagger
T(T^\#)^nx=T^\dagger (T^\#)^nTx=0$. So, $(T^\#)^nx\in N(T)\cap
R(T)=\{0\}$. Hence, $N(T^\dagger)\subseteq N(T)$ and $T$ is  EP.

(xxi) If  $T^nTT^\dagger+T^\dagger TT^n=2T^n$ and $x\in
N(T^\#)=N(T)$, we have $T^nTT^\dagger x=0$. Hence, $TT^\dagger
x\in N(T)\cap R(T)=\{0\}$. Then we conclude $x\in
N(TT^\dagger)=N(T^\dagger)$. So, $N(T)\subseteq N(T^\dagger)$ and
$T$ is  EP.

(xxii) Let $T^n=T^nTT^\dagger$,  and let $x\in N(T^\dagger)$. Now
$T^nx=0$, that is $x\in N(T^n)=N(T)$. Therefore,
$N(T^\dagger)\subseteq N(T)$ and T is  EP.

(xxiii) If we assume that $T^n=T^\dagger TT^n$, then we have
$$R(T)=R(T^n)=R(T^\dagger TT^n)=T^\dagger R(T)=R(T^\dagger T)=R(T^\dagger).$$
Using Theorem \ref{teBoassoEP}, we get that $T$ is EP.

(xxiv) Let $T^nT^\dagger=T^\dagger T^n$. From
$$R(T^nT^\dagger)=T^{n-1}R(TT^\dagger)=T^{n-1}R(T)=R(T^n)=R(T)$$
and $$R(T^\dagger T^n)=T^\dagger R(T^n)=T^\dagger R(T)=R(T^\dagger
T)=R(T^\dagger),$$ we have $R(T^\dagger)=R(T)$. By Theorem
\ref{teBoassoEP}, T is an EP operator.
\end{proof}

Now, we return to a general case, i.e. $\mathcal A$ is a complex
unital Banach algebra, and $a\in\mathcal A$ is both Moore-Penrose
and group invertible.

\begin{corollary}
Theorem \ref{te1EPb} holds if we change $\mathcal L(X)$ by an
arbitrary complex Banach algebra $\mathcal A $, and if we change
$T$ by an $a\in\mathcal A$ such that $a^\dagger$ and $a^\#$ exist.
\end{corollary}

\begin{proof} The left  multiplication by
$a$ is the mapping $L_a:\mathcal A\to\mathcal A$, which is defined
as $L_a(x)=ax$ for all $x\in\mathcal A$. Then it is easy to see
that $L_{a^\dag}=(L_a)^\dag$ and $L_{a^\#}=(L_a)^\#$ in the Banach
algebra $\mathcal L(\mathcal A)$. Now, if any one of statements
(i)-(xxiv) holds for $a$, then the same statement holds for $L_a$
in $\mathcal L(\mathcal A)$. Hence, by Theorem \ref{te1EPb}, $L_a$
is EP in $\mathcal L(\mathcal A)$. From \cite[Remark 12]{B}, it
follows that $a$ is EP in $\mathcal A$.
\end{proof}

Notice that the right multiplication by $a$, which is considered
as the mapping $R_a:\mathcal A\to\mathcal A$, and defined as
$R_a(x)=xa$ for all $x\in\mathcal A$, achieves the same effect as
$L_a$ in proof of the preceding result.

\smallskip

\textbf{Acknowledgement.} We are grateful to the referee for his/her
very valuable comments concerning the paper. The authors are supported by the Ministry of Science,
Republic of Serbia, grant no. 174007.


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