# Drazin invertibility of linear operators on quaternionic Banach spaces

El Hassan Benabdi and Mohamed Barraa  Department of Mathematics, Faculty of
Sciences Semlalia, University Cadi Ayyad, Marrakesh, Morocco
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

Let $A$ be a right linear operator on a two-sided quaternionic Banach space
$X$. The paper studies the Drazin inverse for right linear operators on a
quaternionic Banach space. It is shown that if $A$ is Drazin invertible then
the Drazin inverse of $A$ is given by $f(A)$ where $f$ is $0$ in an axially
symmetric neighborhood of $0$ and $f(q)=q^{-1}$ in an axially symmetric
neighborhood of the nonzero spherical spectrum of $A$. Some results analogous
to the ones concerning the Drazin inverse of operators on complex Banach
spaces are proved in the quaternionic context.

###### Key words and phrases:

Drazin inverse; quaternionic Banach space; slice function; S-spectrum

###### 2020 Mathematics Subject Classification:

46S10; 47A60; 47C15; 30G35

##  1\. Introduction and preliminaires

We denote by $\mathbb{H}$ the algebra of quaternions, introduced by Hamilton
in 1843. An element $q$ of $\mathbb{H}$ is of the form

$q=a+b\mathrm{i}+c\mathrm{j}+d\mathrm{k};a,b,c,d\in\mathbb{R}$

where $\mathrm{i},\mathrm{j}$ and $\mathrm{k}$ are imaginary units. By
definition, they satisfy

$\mathrm{i}^{2}=\mathrm{j}^{2}=\mathrm{k}^{2}=\mathrm{i}\mathrm{j}\mathrm{k}=-1.$

Given $q=a+b\mathrm{i}+c\mathrm{j}+d\mathrm{k}$, then

  * •

the conjugate quaternion of $q$ is
$\bar{q}=a-b\mathrm{i}-c\mathrm{j}-d\mathrm{k}$;

  * •

the norm of $q$ is $|q|=\sqrt{q\bar{q}}=\sqrt{a^{2}+b^{2}+c^{2}+d^{2}}$;

  * •

the real and the imaginary parts of $q$ are respectively
$\text{Re}(q):=\frac{1}{2}(q+\bar{q})=a$ and
$\text{Im}(q):=\frac{1}{2}(q-\bar{q})=b\mathrm{i}+c\mathrm{j}+d\mathrm{k}$.

The unit sphere of imaginary quaternions is given by

$\mathbb{S}=\\{q\in\mathbb{H}:q^{2}=-1\\}.$

Let $p$ and $q$ be two quaternions. $p$ and $q$ are said to be conjugated, if
there is $s\in\mathbb{H}\setminus\\{0\\}$ such that $p=sqs^{-1}$. The set of
all quaternions conjugated with $q$, is equal to the 2-sphere

$[q]=\\{\text{Re}(q)+|\text{Im}(q)|j:j\in\mathbb{S}\\}=\text{Re}(q)+|\text{Im}(q)|\mathbb{S}.$

For every $j\in\mathbb{S}$, denote by $\mathbb{C}_{j}$ the real subalgebra of
$\mathbb{H}$ generated by $j$; that is,

$\mathbb{C}_{j}:=\\{u+vj\in\mathbb{H}:u,v\in\mathbb{R}\\}.$

We say that $U\subseteq\mathbb{H}$ is axially symmetric if $[q]\subset U$ for
every $q\in U$.

For a thorough treatment of the algebra of quaternions $\mathbb{H}$, the
reader is referred, for instance, to [4].

######  Definition 1.1 ([1, Definition 2.1.2], Slice hyperholomorphic
functions).

Let $U\subseteq\mathbb{H}$ be an axially symmetric open set and let
$\mathcal{U}=\\{(u,v)\in\mathbb{R}^{2}:u+v\mathbb{S}\subset U\\}$.
A function $f:U\rightarrow\mathbb{H}$ is called a left slice function if there
exist two functions $f_{0},f_{1}:\mathcal{U}\rightarrow\mathbb{H}$ such that:

$f(q)=f_{0}(u,v)+jf_{1}(u,v)\;\;\;\text{ for every }q=u+vj\in U$

and if $f_{0},f_{1}$ satisfy the compatibility conditions

$f_{0}(u,-v)=f_{0}(u,v),\;\;\;\;\;\;f_{1}(u,-v)=-f_{1}(u,v).$ (1.1)

If in addition $f_{0}$ and $f_{1}$ satisfy the Cauchy-Riemann equations

$\begin{split}\frac{\partial}{\partial u}f_{0}(u,v)-\frac{\partial}{\partial
v}f_{1}(u,v)=0,\\\ \frac{\partial}{\partial
v}f_{0}(u,v)+\frac{\partial}{\partial u}f_{1}(u,v)=0,\end{split}$ (1.2)

then $f$ is called left slice hyperholomorphic. We denote the set of all left
slice functions on $U$ by $\mathcal{SF}_{L}(U)$ and the set of all left slice
hyperholomorphic functions on $U$ by $\mathcal{SH}_{L}(U)$.
A function $f:U\rightarrow\mathbb{H}$ is called a right slice function if
there exist two functions $f_{0},f_{1}:\mathcal{U}\rightarrow\mathbb{H}$ such
that:

$f(q)=f_{0}(u,v)+f_{1}(u,v)j\;\;\;\text{ for every }q=u+vj\in U$

and if $f_{0},f_{1}$ satisfy the compatibility conditions $(\ref{3eq6})$. If
in addition $f_{0}$ and $f_{1}$ satisfy the Cauchy-Riemann equations
$(\ref{3eq7})$, then $f$ is called right slice hyperholomorphic. We denote the
set of all right slice functions on $U$ by $\mathcal{SF}_{R}(U)$ and the set
of all right slice hyperholomorphic functions on $U$ by $\mathcal{SH}_{R}(U)$.
If $f$ is a left $($or right$)$ slice function such that $f_{0}$ and $f_{1}$
are real-valued, then $f$ is called intrinsic. The set of all intrinsic slice
functions on $U$ will be denoted by $\mathcal{FN}(U)$ and the set of all
intrinsic slice hyperholomorphic functions on $U$ will be denoted by
$\mathcal{N}(U)$.

######  Lemma 1.2 ([1, Lemma 2.1.6], Splitting lemma).

Let $U\subseteq\mathbb{H}$ be an axially symmetric open set and let
$i,j\in\mathbb{S}$ with $ij=-ji$. If $f\in\mathcal{SH}_{L}(U)$, then the
restriction $f_{j}=f|_{U\cap\mathbb{C}_{j}}$ satisfies

$\frac{1}{2}\left(\frac{\partial}{\partial
u}f_{j}(z)+j\frac{\partial}{\partial v}f_{j}(z)\right)=0,$

for all $z=u+vj\in U\cap\mathbb{C}_{j}$. Hence

$f_{j}(z)=F_{0}(z)+F_{1}(z)i$

with holomorphic functions
$F_{0},F_{1}:U\cap\mathbb{C}_{j}\rightarrow\mathbb{C}_{j}.$
If $f\in\mathcal{SH}_{R}(U)$, then the restriction
$f_{j}=f|_{U\cap\mathbb{C}_{j}}$ satisfies

$\frac{1}{2}\left(\frac{\partial}{\partial u}f_{j}(z)+\frac{\partial}{\partial
v}f_{j}(z)j\right)=0,$

for all $z=u+vj\in U\cap\mathbb{C}_{j}$. Hence

$f_{j}(z)=F_{0}(z)+iF_{1}(z)$

with holomorphic functions
$F_{0},F_{1}:U\cap\mathbb{C}_{j}\rightarrow\mathbb{C}_{j}.$

######  Theorem 1.3 ([1, Theorem 2.1.21], Cauchy’s integral theorem).

Let $U\subseteq\mathbb{H}$ be open, $j\in\mathbb{S}$ and $D_{j}\subset
U\cap\mathbb{C}_{j}$ be a bounded open subset of the complex plane
$\mathbb{C}_{j}$ with $\overline{D_{j}}\subset U\cap\mathbb{C}_{j}$ such that
$\partial D_{j}$ is a finite union of piecewise continuously differentiable
Jordan curves. Then for all $f\in\mathcal{SH}_{L}(U)$ and all
$g\in\mathcal{SH}_{R}(U)$

$\int_{\partial D_{j}}g(s)ds_{j}f(s)=0,$

where $ds_{j}=ds(-j)$.

######  Definition 1.4 ([1, Definition 2.1.23]).

We define the left slice hyperholomorphic Cauchy kernel as:

$S_{L}^{-1}(s,q):=-(q^{2}-2\text{Re}(s)q+|s|^{2})^{-1}(q-\bar{s});\;\;\;\;\;\;\;\;q\notin[s],$

and the right slice hyperholomorphic Cauchy kernel as:

$S_{R}^{-1}(s,q):=-(q-\bar{s})(q^{2}-2\text{Re}(s)q+|s|^{2})^{-1};\;\;\;\;\;\;\;\;q\notin[s].$

######  Lemma 1.5 ([1, Lemma 2.1.27]).

Let $q,s\in\mathbb{H}$ with $s\notin[q]$.
The left slice hyperholomorphic Cauchy kernel $S_{L}^{-1}(s,q)$ is left slice
hyperholomorphic in $q$ and right slice hyperholomorphic in $s$.
The right slice hyperholomorphic Cauchy kernel $S_{R}^{-1}(s,q)$ is left slice
hyperholomorphic in $s$ and right slice hyperholomorphic in $q$.

######  Definition 1.6 ([1, Definition 2.1.30], Slice Cauchy domain).

An axially symmetric open set $U\subset\mathbb{H}$ is called a slice Cauchy
domain if $U\cap\mathbb{C}_{j}$ is a Cauchy domain in $\mathbb{C}_{j}$ for
every $j\in\mathbb{S}$. More precisely, $U$ is a slice Cauchy domain if for
every $j\in\mathbb{S}$ the boundary $\partial(U\cap\mathbb{C}_{j})$ of
$U\cap\mathbb{C}_{j}$ is the union a finite number of nonintersecting
piecewise continuously differentiable Jordan curves in $\mathbb{C}_{j}$.

######  Theorem 1.7 ([1, Theorem 2.1.32], Cauchy’s formulas).

Let $U\subset\mathbb{H}$ be a bounded slice Cauchy domain, let
$j\in\mathbb{S}$, and set $ds_{j}=ds(-j)$. If $f$ is a left slice
hyperholomorphic function on a set that contains $\overline{U}$, then

$f(q)=\frac{1}{2\pi}\int_{\partial(U\cap\mathbb{C}_{j})}S_{L}^{-1}(s,q)ds_{j}f(s)\text{,
for every }q\in U.$

If $f$ is a right slice hyperholomorphic function on a set that contains
$\overline{U}$, then

$f(q)=\frac{1}{2\pi}\int_{\partial(U\cap\mathbb{C}_{j})}f(s)ds_{j}S_{R}^{-1}(s,q)\text{,
for every }q\in U.$

These integrals depend neither on $U$ nor on the imaginary unit
$j\in\mathbb{S}$.

######  Definition 1.8 ([1, Definition 2.3.1]).

Let $(X,+)$ be an abelian group.

  * •

$X$ is a right quaternionic vector space denoted by $X_{R}$ if it is endowed
with a right quaternionic multiplication $(X,\mathbb{H})\rightarrow X$,
$(u,q)\mapsto uq$ such that for all $u,v\in X$ and all $p,q\in\mathbb{H}$,

$u(p+q)=up+uq,(u+v)q=uq+vq,(up)q=u(pq)\text{ and }u1=u.$

  * •

$X$ is a left quaternionic vector space denoted by $X_{L}$ if it is endowed
with a left quaternionic multiplication $(\mathbb{H},X)\rightarrow X$,
$(q,u)\mapsto qu$ such that for all $u,v\in X$ and all $p,q\in\mathbb{H}$,

$(p+q)u=pu+qu,q(u+v)=qu+qv,q(pu)=(qp)u\text{ and }1u=u.$

  * •

$X$ is a two-sided quaternionic vector space if it is endowed with a left and
a right quaternionic multiplication such that $X$ is both a left and a right
quaternionic vector space and such that $ru=ur$ for all $r\in\mathbb{R}$, and
$(pu)q=p(uq)$ for all $p,q\in\mathbb{H}$ and all $u\in X$.

######  Definition 1.9.

Let $X_{R}$ be a right quaternionic vector space. A function
$\|\cdot\|:X_{R}\rightarrow[0;+\infty)$ is called a norm on $X_{R}$, if it
satisfies

  * (i)

$\|u\|=0$ if and only if $u=0$;

  * (ii)

$\|uq\|=\|u\||q|$ for all $u\in X_{R}$ and all $q\in\mathbb{H}$;

  * (iii)

$\|u+v\|\leq\|u\|+\|v\|$ for all $u,v\in X_{R}$.

If $X_{R}$ is complete with respect to the metric induced by $\|\cdot\|$, we
call $X_{R}$ a right quaternionic Banach space.
Let $X_{L}$ be a left quaternionic vector space. A function
$\|\cdot\|:X_{L}\rightarrow[0;+\infty)$ is called a norm on $X_{L}$, if it
satisfies $(i)$, $(iii)$ and

  * (ii’)

$\|qu\|=|q|\|u\|$ for all $u\in X_{L}$ and $q\in\mathbb{H}$.

If $X_{L}$ is complete with respect to the metric induced by $\|\cdot\|$, we
call $X_{L}$ a left quaternionic Banach space.
Finally, a two-sided quaternionic vector space $X$ is called a two-sided
quaternionic Banach space if it is endowed with a norm $\|\cdot\|$ such that
it is both a left and a right quaternionic Banach space.

######  Remark 1.10.

_If $X$ is a two-sided quaternionic Banach space, then
$\|qu\|=\|uq\|=|q|\|u\|$ for all $u\in X$ and all $q\in\mathbb{H}$._

######  Definition 1.11 ([1, Definition 2.3.9], Slice hyperholomorphic vector-
valued functions).

Let $U\subseteq\mathbb{H}$ be an axially symmetric open set and let
$\mathcal{U}=\\{(u,v)\in\mathbb{R}^{2}:u+v\mathbb{S}\subset U\\}$. A function
$f:U\rightarrow X_{L}$ with values in a left quaternionic Banach space $X_{L}$
is called a left slice function if it is of the form:

$f(q)=f_{0}(u,v)+jf_{1}(u,v)\text{ for every }q=u+vj\in U$

with two functions $f_{0},f_{1}:\mathcal{U}\rightarrow X_{L}$ that satisfy the
compatibility conditions $(\ref{3eq6})$.
If in addition $f_{0}$ and $f_{1}$ satisfy the Cauchy-Riemann equations
$(\ref{3eq7})$, then $f$ is called left slice hyperholomorphic.
A function $f:U\rightarrow X_{R}$ with values in a right quaternionic Banach
space $X_{R}$ is called a right slice function if it is of the form:

$f(q)=f_{0}(u,v)+f_{1}(u,v)j\text{ for every }q=u+vj\in U$

with two functions $f_{0},f_{1}:\mathcal{U}\rightarrow X_{R}$ that satisfy the
compatibility conditions $(\ref{3eq6})$.
If in addition $f_{0}$ and $f_{1}$ satisfy the Cauchy-Riemann equations
$(\ref{3eq7})$, then $f$ is called right slice hyperholomorphic.

######  Theorem 1.12 ([1, Theorem 2.3.19], Vector-valued Cauchy formulas).

Let $U\subset\mathbb{H}$ be a bounded slice Cauchy domain, let
$j\in\mathbb{S}$, and set $ds_{j}=ds(-j)$. If $f$ is a left slice
hyperholomorphic function with values in a left quaternionic Banach space
$X_{L}$ that is defined on a set that contains $\overline{U}$, then

$f(q)=\frac{1}{2\pi}\int_{\partial(U\cap\mathbb{C}_{j})}S_{L}^{-1}(s,q)ds_{j}f(s)\text{,
for every }q\in U.$

If $f$ is a right slice hyperholomorphic function with values in a right
quaternionic Banach space $X_{R}$ that is defined on a set that contains
$\overline{U}$, then

$f(q)=\frac{1}{2\pi}\int_{\partial(U\cap\mathbb{C}_{j})}f(s)ds_{j}S_{R}^{-1}(s,q)\text{,
for every }q\in U.$

These integrals depend neither on $U$ nor on the imaginary unit
$j\in\mathbb{S}$.

######  Definition 1.13.

Let $X$ be a two-sided quaternionic Banach space. A right $($resp. left$)$
linear operator on $X$ is a map $T:X\rightarrow X$ such that:

$T(up+v)=(Tu)p+Tv(\text{ resp. }T(pu+v)=p(Tu)+Tv)\text{ for all }u,v\in
X\text{ and all }p\in\mathbb{H}.$

A right or left linear operator $T$ on $X$ is called bounded if

$\|T\|:=\sup\\{\|Tu\|:u\in X,\|u\|=1\\}<\infty.$

The set of all right $($resp. left$)$ linear bounded operators on $X$ is
denoted by $\mathcal{B}_{R}(X)$ $($resp. $\mathcal{B}_{L}(X))$.
$\mathcal{B}_{R}(X)$ $($resp. $\mathcal{B}_{L}(X))$ is viewed as a two-sided
quaternionic vector space equipped with the metric
$\mathcal{B}_{R}(X)\times\mathcal{B}_{R}(X)\ni(A,B)\mapsto\|A-B\|$ $($resp.
$\mathcal{B}_{L}(X)\times\mathcal{B}_{L}(X)\ni(A,B)\mapsto\|A-B\|)$.

In a two-sided quaternionic Banach space $X$, we can define a left and a right
quaternionic multiplication on $\mathcal{B}_{R}(X)$ $($resp.
$\mathcal{B}_{L}(X))$ by

$(Tq)u=T(qu)\text{ and }(qT)(u)=q(T(u))\text{ for all }q\in\mathbb{H},u\in
X\text{ and all }T\in\mathcal{B}_{R}(X)$ $(\text{resp. }(Tq)u=T(u)q\text{ and
}(qT)(u)=T(uq)\text{ for all }q\in\mathbb{H},u\in X\text{ and all
}T\in\mathcal{B}_{L}(X)).$

The spectral theory over quaternionic Hilbert spaces has been developed in [4]
and [6].

In the remainder of this paper, $X$ will be a two-sided quaternionic Banach
space. We will consider just right linear operators on $X$. The theory we
develop here also applies in the case of left linear operators with obvious
modifications.

Let $T\in\mathcal{B}_{R}(X)$. We want to define the notion of spectrum for
right linear operators on $X$ such that this notion generalize the known
results on the spectrum in the complex case (for instance, the compactness of
the spectrum, the spectrum of self-adjoint operators is real). Note that if
$q\in\mathbb{H}$, $T-Iq=T-qI$, where $I$ is the identity operator on $X$. Take
$X=\mathbb{H}\oplus\mathbb{H}$ equipped with the standard scalar product:

$\langle\begin{bmatrix}p\\\ q\end{bmatrix};\begin{bmatrix}p^{\prime}\\\
q^{\prime}\end{bmatrix}\rangle=\bar{p}p^{\prime}+\bar{q}q^{\prime}\text{ for
all }p,q,p^{\prime},q^{\prime}\in\mathbb{H}.$

Then $T:=\begin{bmatrix}0&\mathrm{i}\\\ -\mathrm{i}&0\end{bmatrix}$ is self-
adjoint. Let $u=\begin{bmatrix}1\\\ -\mathrm{k}\end{bmatrix}$, then
$(T-\mathrm{j}I)u=0$, hence $\mathrm{j}$ is an eigenvalue of $T$, thus the
spectrum of $T$ is not real, and so the operators $T-Iq$ and $T-qI$ should not
be used to define the spectrum of $T$. F. Colombo et al. [2] extended the
definitions of the spectrum and resolvent in quaternionic Banach spaces as
follows.

######  Definition 1.14.

Let $T\in\mathcal{B}_{R}(X)$. For $q\in\mathbb{H}$, we set

$Q_{q}(T):=T^{2}-2\text{Re}(q)T+|q|^{2}I.$

Where $I$ is the identity operator on $X$. We define the S-resolvent set
$\rho_{S}(T)$ of $T$ as:

$\rho_{S}(T):=\\{q\in\mathbb{H}:Q_{q}(T)\text{ is invertible in
}\mathcal{B}_{R}(X)\\},$

and we define the S-spectrum $\sigma_{S}(T)$ of $T$ as:

$\sigma_{S}(T):=\mathbb{H}\setminus\rho_{S}(T).$

######  Proposition 1.15 ([1, Proposition 3.1.8]).

Let $T\in\mathcal{B}_{R}(X)$. The sets $\sigma_{S}(T)$ and $\rho_{S}(T)$ are
axially symmetric.

######  Theorem 1.16 ([1, Theorem 3.1.13], Compactness of the S-spectrum).

Let $T\in\mathcal{B}_{R}(X)$. The S-spectrum $\sigma_{S}(T)$ of $T$ is a
nonempty compact set contained in the closed ball
$\\{q\in\mathbb{H}:|q|\leq\|T\|\\}$.

Let $T\in\mathcal{B}_{R}(X)$. Then the S-spectral radius of $T$ is defined to
be the nonnegative real number

$r_{S}(T):=\sup\\{|q|:q\in\sigma_{S}(T)\\}.$

######  Theorem 1.17 ([1, Theorem 4.2.3]).

For $T\in\mathcal{B}_{R}(X)$, we have

$r_{S}(T)=\lim_{n\rightarrow+\infty}\|T^{n}\|^{\frac{1}{n}}.$

######  Theorem 1.18.

Let $T\in\mathcal{B}_{R}(X)$ and $q\in\mathbb{H}$ with $r_{S}(T)<|q|$. Then

$(T^{2}-2\text{Re}(q)T+|q|^{2}I)^{-1}=\sum_{n=0}^{+\infty}T^{n}\sum_{k=0}^{n}\bar{q}^{-k-1}q^{-n+k-1},$

where this series converges in the operator norm.

###### Proof.

Let

$a_{n}:=\sum_{k=0}^{n}\bar{q}^{-k-1}q^{-n+k-1},$

then

$|a_{n}|\leq(n+1)|q|^{-n-2}.$

Hence

$\|T^{n}\sum_{k=0}^{n}\bar{q}^{-k-1}q^{-n+k-1}\|\leq\|T^{n}\|(n+1)|q|^{-n-2}.$

We have

$\lim_{n\rightarrow\infty}\left(\|T^{n}\|(n+1)|q|^{-n-2}\right)^{\frac{1}{n}}=\frac{r_{S}(T)}{|q|}.$

Thus the series
$\sum_{n=0}^{+\infty}T^{n}\sum_{k=0}^{n}\bar{q}^{-k-1}q^{-n+k-1}$ converges in
the operator norm. The rest follows from the proof of [1, Theorem 3.1.5]. ∎

######  Definition 1.19 ([1, Definition 3.2.5], S-functional calculus).

Let $T\in\mathcal{B}_{R}(X)$. Let $U\supset\sigma_{S}(T)$ be a bounded slice
Cauchy domain, let $j\in\mathbb{S}$, and set $ds_{j}=ds(-j)$. For every
$f\in\mathcal{SH}_{L}(\bar{U})$, we define

$f(T):=\frac{1}{2\pi}\int_{\partial(U\cap\mathbb{C}_{j})}S_{L}^{-1}(s,T)ds_{j}f(s).$

For every $f\in\mathcal{SH}_{R}(\bar{U})$, we define

$f(T):=\frac{1}{2\pi}\int_{\partial(U\cap\mathbb{C}_{j})}f(s)ds_{j}S_{R}^{-1}(s,T).$

Let $K\subseteq\mathbb{H}$. In the following, we mean by $\mathcal{SH}_{R}(K)$
$($resp. $\mathcal{SH}_{L}(K)$, $\mathcal{N}(K))$, the set of all right
$($resp. left, intrinsic$)$ slice hyperholomorphic functions on an open
axially symmetric set $U$ that contains $K$.

######  Theorem 1.20 ([1, Theorem 4.1.3], Product rule).

Let $T\in\mathcal{B}_{R}(X)$, $f\in\mathcal{N}(\sigma_{S}(T))$ and
$g\in\mathcal{SH}_{L}(\sigma_{S}(T))$ or
$g\in\mathcal{SH}_{R}(\sigma_{S}(T))$. Then

$(fg)(T)=f(T)g(T).$

######  Theorem 1.21 ([1, Theorem 4.2.1], The spectral mapping theorem).

Let $T\in\mathcal{B}_{R}(X)$ and $f\in\mathcal{N}(\sigma_{S}(T))$. Then

$\sigma_{S}(f(T))=f(\sigma_{S}(T)):=\\{f(q):q\in\sigma_{S}(T)\\}.$

######  Theorem 1.22 ([1, Theorem 4.2.4], Composition rule).

Let $T\in\mathcal{B}_{R}(X)$ and $f\in\mathcal{N}(\sigma_{S}(T))$. If
$g\in\mathcal{SH}_{L}(\sigma_{S}(f(T)))$, then $g\circ
f\in\mathcal{SH}_{L}(\sigma_{S}(T))$, and if
$g\in\mathcal{SH}_{R}(f(\sigma_{S}(T)))$, then $g\circ
f\in\mathcal{SH}_{R}(\sigma_{S}(T))$. In both cases,

$g(f(T))=(g\circ f)(T).$

A bounded right projection $P\in\mathcal{B}_{R}(X)$ (or simply projection when
no confusion can arise) is such that $P^{2}=P$. If $P$ is a projection, then
so is $I-P$, and their null spaces and ranges are related as follows:

$\mathcal{R}(P)=\mathcal{N}(I-P)\text{ and }\mathcal{N}(P)=\mathcal{R}(I-P).$

The range and the kernel form a pair of algebraic complements,

$\mathcal{R}(P)+\mathcal{N}(P)=X\text{ and
}\mathcal{R}(P)\cap\mathcal{N}(P)=\\{0\\}.$

##  2\. Generalized inverse

In this section, we study the generalized invertibility of right linear
operators on quaternionic Banach spaces.

######  Definition 2.1.

An operator $B\in\mathcal{B}_{R}(X)$ is called a generalized inverse of
$A\in\mathcal{B}_{R}(X)$ if $ABA=A$ and $BAB=B$.

######  Remark 2.2.

_Let $A,B\in\mathcal{B}_{R}(X)$._

  * 1)

_If_ $B$ is a generalized inverse of $A$, then $AB$ and $BA$ are projections.
Indeed

$(AB)^{2}=(ABA)B=AB\text{, }(BA)^{2}=B(ABA)=BA.$

_Clearly,_ $\mathcal{R}(AB)\subseteq\mathcal{R}(A)$ and
$\mathcal{N}(A)\subseteq\mathcal{N}(BA)$. Since $ABA=A$, we have
$\mathcal{R}(A)=\mathcal{R}(ABA)\subseteq\mathcal{R}(AB)$ and
$\mathcal{N}(BA)\subseteq\mathcal{N}(ABA)=\mathcal{N}(A)$. It follows that
$\mathcal{R}(AB)=\mathcal{R}(A)$ and $\mathcal{N}(BA)=\mathcal{N}(A)$.
Similarly, $\mathcal{R}(BA)=\mathcal{R}(B)$ and
$\mathcal{N}(AB)=\mathcal{N}(B)$. Since
$X=\mathcal{R}(AB)\oplus\mathcal{N}(AB)$,
$X=\mathcal{R}(A)\oplus\mathcal{N}(B)$.

  * 2)

_If_ $ABA=A$, then $T:=BAB$ is a generalized inverse of $A$.

  * 3)

_If_ $A$ is left $($resp. right$)$ invertible, then it is generalized
invertible.

######  Theorem 2.3.

An operator $A\in\mathcal{B}_{R}(X)$ has a generalized inverse if and only if
both the range $\mathcal{R}(A)$ and the null space $\mathcal{N}(A)$ are closed
complemented subspaces of $X$.

###### Proof.

Let $Y$ and $Z$ be two closed complemented subspaces of $X$. If
$X=\mathcal{R}(A)\oplus Y=\mathcal{N}(A)\oplus Z$, then the operator defined
by $B(Az+y)=z$ with $z\in Z$ is right linear on $X$. It is easy to check that
$B$ is a generalized inverse of $A$.
Conversely, if $A$ has a generalized inverse $B\in\mathcal{B}_{R}(X)$, then by
Remark 2.2, $\mathcal{R}(AB)=\mathcal{R}(A)$ and
$\mathcal{N}(BA)=\mathcal{N}(A)$. Since $AB$ and $BA$ are bounded right
projections, both the range $\mathcal{R}(AB)$ and the null space
$\mathcal{N}(BA)$ are closed complemented subspaces of $X$, then so is
$\mathcal{R}(A)$ and $\mathcal{N}(A)$. ∎

The generalized inverse is not unique in general, the following theorem
describes all generalized inverses of $A\in\mathcal{B}_{R}(X)$.

######  Theorem 2.4.

Suppose $B\in\mathcal{B}_{R}(X)$ is a generalized inverse of
$A\in\mathcal{B}_{R}(X)$. Then the set of all generalized inverses of $A$
consists of all operators of the form:

$T=PBQ,$

where $Q$ is a projection onto $\mathcal{R}(A)$ and $P$ is a projection whose
kernel coincides with $\mathcal{N}(A)$.

###### Proof.

Let $B$ and $T$ be two generalized inverses of $A$, then $ABA=ATA=A$ and
$TAT=T$, hence $T(ABA)T=(TA)B(AT)=T$. By Remark 2.2, $TA$ and $AT$ are
projections and such that $\mathcal{N}(TA)=\mathcal{N}(A)$ and
$\mathcal{R}(A)=\mathcal{R}(AT)$.
Conversely, if $B$, $P$ and $Q$ are as in Theorem 2.4, then $QA=A=AP$, hence
$A(PBQ)A=A$ and $(PBQ)A(PBQ)=PBQ$. ∎

######  Lemma 2.5.

Let $A\in\mathcal{B}_{R}(X)$. Then $A$ has a generalized inverse $B$ such that
$AB=BA$ if and only if $X$ can be written as
$X=\mathcal{R}(A)\oplus\mathcal{N}(A)$. In such a case, $B$ is unique.

###### Proof.

By Remark 2.2, $X=\mathcal{R}(A)\oplus\mathcal{N}(B)$ and since $AB=BA$,
$\mathcal{N}(B)=\mathcal{N}(AB)=\mathcal{N}(BA)=\mathcal{N}(A)$, then
$X=\mathcal{R}(A)\oplus\mathcal{N}(A)$.
Conversely, since $X=\mathcal{R}(A)\oplus\mathcal{N}(A)$, the restriction
$T:=A/\mathcal{R}(A)$ is invertible, then $B=T^{-1}\oplus 0$ is a generalized
inverse of $A$ which commutes with $A$. If $B$ and $C$ are two generalized
inverses of $A$ that commute with $A$, then
$B=AB^{2}=CA^{2}B^{2}=CAB=C^{2}A^{2}B=C^{2}A=C$. ∎

######  Theorem 2.6.

Suppose $A\in\mathcal{B}_{R}(X)$ with generalized inverse $B$ such that
$AB=BA$. Then

$\sigma_{S}(B)\setminus\\{0\\}=\\{q^{-1}:q\in\sigma_{S}(A)\setminus\\{0\\}\\}.$

###### Proof.

By Lemma 2.5, $X=\mathcal{R}(A)\oplus\mathcal{N}(A)$, then $A=T\oplus 0$ on
$\mathcal{R}(A)\oplus\mathcal{N}(A)$ and $B=T^{-1}\oplus 0$. We have
$Q_{q}(B)=Q_{q}(T^{-1})\oplus Q_{q}(0)$, for all $q\in\mathbb{H}$. Then we
have $\sigma_{S}(B)=\sigma_{S}(T^{-1})\cup\sigma_{S}(0)$, since
$Q_{q}(0)=|q|^{2}I$ is always invertible where $I$ is the identity operator on
$\mathcal{N}(A)$,

$\sigma_{S}(B)\setminus\\{0\\}=\sigma_{S}(T^{-1})\setminus\\{0\\}.$

The function $f:\mathbb{H}\setminus\\{0\\}\ni q\mapsto q^{-1}$ is intrinsic
slice hyperholomorphic (because $q^{-1}=\frac{\bar{q}}{|q|^{2}}$), then by
Theorem 1.21,
$\sigma_{S}(T^{-1})=\sigma_{S}(f(T))=\\{q^{-1}:q\in\sigma_{S}(T)\\}$. Thus

$\sigma_{S}(B)\setminus\\{0\\}=\\{q^{-1}:q\in\sigma_{S}(A)\setminus\\{0\\}\\}.$

∎

##  3\. Drazin inverse

In this section, we study the Drazin invertibility of right linear operators
acting on a quaternionic Banach space.

######  Definition 3.1.

Let $A\in\mathcal{B}_{R}(X)$. An element $B\in\mathcal{B}_{R}(X)$ is a Drazin
inverse of $A$, written $B=A^{d}$, if

$AB=BA\text{, }AB^{2}=B\text{, }A^{k+1}B=A^{k},$ (3.1)

for some nonnegative integer $k$. The least nonnegative integer $k$ for which
these equations hold is the Drazin index $i(A)$ of $A$.

######  Definition 3.2.

An element $A$ of $\mathcal{B}_{R}(X)$ is called quasinilpotent if
$\sigma_{S}(A)=\\{0\\}$. The set of all quasinilpotent elements in
$\mathcal{B}_{R}(X)$ will be denoted by $QN(\mathcal{B}_{R}(X))$.

######  Proposition 3.3.

An element $A$ of $\mathcal{B}_{R}(X)$ is quasinilpotent if and only if, for
every $T$ commuting with $A$, we have $I-TA$ is invertible.

###### Proof.

Let $A\in\mathcal{B}_{R}(X)$, assume that for every $T\in\mathcal{B}_{R}(X)$
commuting with $A$, we have $I-TA$ is invertible. Let
$T=\frac{-1}{|q|^{2}}A+\frac{2\text{Re}(q)}{|q|^{2}}I$ with
$q\in\mathbb{H}\setminus\\{0\\}$, clearly $T$ commutes with $A$ and
$I-TA=\frac{1}{|q|^{2}}[A^{2}-2\text{Re}(q)A+|q|^{2}I]$ is invertible, hence
$\sigma_{S}(A)=\\{0\\}$.
Conversely, if $\sigma_{S}(A)=\\{0\\}$. Let $T\in\mathcal{B}_{R}(X)$ commutes
with $A$, then by Theorem 1.17, $r_{S}(TA)\leq r_{S}(T)r_{S}(A)=0$ and hence
$\sigma_{S}(TA)=\\{0\\}$. Then by Theorem 1.21, $\sigma_{S}(I-TA)=\\{1\\}$ and
hence $I-TA$ is invertible. ∎

An operator $A\in\mathcal{B}_{R}(X)$ is said to be nilpotent if there exists
$k\in\mathbb{N}$ such that $A^{k}=0$. The least nonnegative integer $k$ for
which $A^{k}=0$ is called the nilpotency index of $A$ and the set of all
nilpotent elements in $\mathcal{B}_{R}(X)$ is denoted by
$N(\mathcal{B}_{R}(X))$.

######  Lemma 3.4.

In $\mathcal{B}_{R}(X)$, $(\ref{3eq0})$ is equivalent to

$AB=BA,AB^{2}=B,A-A^{2}B\in N(\mathcal{B}_{R}(X)).$ (3.2)

The Drazin index $i(A)$ is equal to the nilpotency index of $A-A^{2}B$.

###### Proof.

If $AB=BA$ and $AB^{2}=B$, then $I-AB$ is a projection. Hence the equivalence
and the last statement are given by this equalities
$(A-A^{2}B)^{k}=A^{k}(I-AB)^{k}=A^{k}(I-AB)=A^{k}-A^{k+1}B$. ∎

Koliha [5, Definition 2.3] generalized the notion of Drazin invertibility in a
complex Banach algebra. According to this definition one can generalize the
notion of Drazin invertibility in $\mathcal{B}_{R}(X)$.

######  Definition 3.5.

Let $A\in\mathcal{B}_{R}(X)$. An element $B\in\mathcal{B}_{R}(X)$ is a
generalized Drazin inverse of $A$, written $B=A^{D}$, if

$AB=BA,AB^{2}=B,A-A^{2}B\in QN(\mathcal{B}_{R}(X)).$ (3.3)

######  Lemma 3.6.

In $\mathcal{B}_{R}(X)$, an element $A$ has a Drazin $($resp. generalized
Drazin$)$ inverse if and only if there is a projection $P$ commuting with $A$
such that

$AP\in N(\mathcal{B}_{R}(X))(\text{resp. }AP\in QN(\mathcal{B}_{R}(X)))\text{
and }A+P\text{ is invertible}.$ (3.4)

A Drazin $($resp. generalized Drazin$)$ inverse of $A$ is given by

$A^{d}=(A+P)^{-1}(I-P)(\text{resp. }A^{D}=(A+P)^{-1}(I-P)).$ (3.5)

###### Proof.

Suppose that there is a projection $P$ commuting with $A$ and satisfying
(3.4). Set $B=(A+P)^{-1}(I-P)$, then $AB=BA$, $AB^{2}=B$ and $A-A^{2}B=AP\in
N(\mathcal{B}_{R}(X))\;(\text{resp. }A-A^{2}B=AP\in QN(\mathcal{B}_{R}(X)))$.
Conversely, suppose that $B$ satisfies (3.2) (resp. (3.3)) and set $P=I-AB$.
Since $AB^{2}=B$, $P$ is a projection commuting with $A$ and $AP=A-A^{2}B$,
then $AP\in N(\mathcal{B}_{R}(X))\text{ (resp. }AP\in
QN(\mathcal{B}_{R}(X))\text{)}$. Furthermore $(A+P)(B+P)=(B+P)(A+P)=I+AP$ and
$I+AP$ is invertible because $\sigma_{S}(I+AP)=\\{1\\}$, then $A+P$ is
invertible in $\mathcal{B}_{R}(X)$. ∎

######  Definition 3.7.

Let $A\in\mathcal{B}_{R}(X)$. For $s\in\rho_{S}(T)$, we define the left
S-resolvent operator as

$S^{-1}_{L}(s,A)=-Q_{s}(T)^{-1}(T-\bar{s}I).$

######  Theorem 3.8 ([1, Theorem 4.1.5]).

Let $A\in\mathcal{B}_{R}(X)$ and assume that
$\sigma_{S}(A)=\sigma_{1}\cup\sigma_{2}$ with

$dist(\sigma_{1},\sigma_{2})>0.$

We choose an open axially symmetric set $O$ with $\sigma_{1}\subset O$ and
$\overline{O}\cap\sigma_{2}=\emptyset$, and define a function
$\chi_{\sigma_{1}}$ on $\mathbb{H}$ by $\chi_{\sigma_{1}}(s)=1$ for $s\in O$
and $\chi_{\sigma_{1}}(s)=0$ for $s\notin O$. Then
$\chi_{\sigma_{1}}\in\mathcal{N}(\sigma_{S}(A))$, and for an arbitrary
imaginary unit $j$ in $\mathbb{S}$ and an arbitrary bounded slice Cauchy
domain $U\subset\mathbb{H}$ such that $\bar{U}\subset O$, we have

$P_{\sigma_{1}}:=\chi_{\sigma_{1}}(A)=\frac{1}{2\pi}\int_{\partial(U\cap\mathbb{C}_{j})}S^{-1}_{L}(s,A)ds_{j}$

is a continuous projection that commutes with $A$. Hence $P_{\sigma_{1}}(X)$
is a right linear subspace of $X$ that is invariant under $A$.

######  Remark 3.9.

_Let $q\in\mathbb{H}$. If $\sigma_{1}=\\{q\\}$, we say that the projection
$P_{\sigma_{1}}$ is the Riesz’s projection of $A$ corresponding to $q$._

Denote by $acc\;U$ (resp. $iso\;U$) the set of all accumulation (resp.
isolated) points of a set $U\subseteq\mathbb{H}$.

######  Theorem 3.10.

Let $A\in\mathcal{B}_{R}(X)$. Then $0\notin acc\;\sigma_{S}(A)$ if and only if
there is a projection $P\in\mathcal{B}_{R}(X)$ commuting with $A$ such that

$AP\in QN(\mathcal{B}_{R}(X))\text{ and }A+P\text{ is invertible in
}\mathcal{B}_{R}(X).$ (3.6)

Moreover, $0\in iso\;\sigma_{S}(A)$ if and only if $P\neq 0$, in which a case
$P$ is the Riesz’s projection of $A$ corresponding to $q=0$.

###### Proof.

Clearly, $0\notin\sigma_{S}(A)$ if and only if (3.6) holds with $P=0$.
Assume that $0\in iso\;\sigma_{S}(A)$. Let $P$ be the spectral projection of
$A$ corresponding to $q=0$, then $P\neq 0$, commutes with $A$ and
$AP=id(A)\chi_{\\{0\\}}(A)=(id\chi_{\\{0\\}})(A)$ where
$id:\mathbb{H}\rightarrow\mathbb{H},q\mapsto q$. Hence
$\sigma_{S}(AP)=id\chi_{\\{0\\}}(\sigma_{S}(A))=\\{0\\}$, thus $AP\in
QN(\mathcal{B}_{R}(X))$. Similarly
$A+P=id(A)+\chi_{\\{0\\}}(A)=(id+\chi_{\\{0\\}})(A)$, then
$0\notin\sigma_{S}(A+P)=(id+\chi_{\\{0\\}})\sigma_{S}(A)$, hence $A+P$ is
invertible.
Conversely, assume that there is a nonzero projection $P$ commuting with $A$
such that (3.6) holds. For any $q\in\mathbb{H}$, we have

$\displaystyle A^{2}-2\text{Re}(q)A+|q|^{2}I$
$\displaystyle=P((AP)^{2}-2\text{Re}(q)AP+|q|^{2}I)+(I-P)((A+P)^{2}-2\text{Re}(q)(A+P)+|q|^{2}I).$

There is $r>0$ such that if $|q|<r$ then
$(A+P)^{2}-2\text{Re}(q)(A+P)+|q|^{2}I$ is invertible. Since $AP\in
QN(\mathcal{B}_{R}(X))$, $(AP)^{2}-2\text{Re}(q)AP+|q|^{2}I$ is invertible of
all $q\neq 0$. Hence for all $0<|q|<r$, it is easy to check that

$\displaystyle(A^{2}-2\text{Re}(q)A+|q|^{2}I)^{-1}=$ $\displaystyle
P((AP)^{2}-2\text{Re}(q)AP+|q|^{2}I)^{-1}$
$\displaystyle+(I-P)((A+P)^{2}-2\text{Re}(q)(A+P)+|q|^{2}I)^{-1}.$

That is,

$Q_{q}(A)^{-1}=PQ_{q}(AP)^{-1}+(I-P)Q_{q}(A+P)^{-1}.$

Since $S^{-1}_{L}(q,A)=-Q_{q}(A)^{-1}(A-\bar{q}I)$, it is easy to see that

$S^{-1}_{L}(q,A)=PS^{-1}_{L}(q,AP)+(I-P)S^{-1}_{L}(q,A+P).$ (3.7)

Since $P\neq 0$, $0\in iso\;\sigma_{S}(A)$. Indeed, if $A$ is invertible, then
$A^{-1}AP=P$, so that $r_{S}(A^{-1}AP)\leq r_{S}(A^{-1})r_{S}(AP)=0$ and
$r_{S}(P)=1$. To show that $P$ is the Riesz’s projection of $A$ corresponding
to $q=0$. Let $j$ and $U$ as in Theorem 3.21, then

$\chi_{\\{0\\}}(A)=\frac{1}{2\pi}\int_{\partial(U\cap\mathbb{C}_{j})}S^{-1}_{L}(s,A)ds_{j}.$

If we take $U$ such that $U\subset\\{q\in\mathbb{H}:|q|<\frac{r}{2}\\}$, then
by (3.7)

$\displaystyle\chi_{\\{0\\}}(A)$
$\displaystyle=\frac{1}{2\pi}\int_{\partial(U\cap\mathbb{C}_{j})}S^{-1}_{L}(s,A)ds_{j}$
$\displaystyle=\frac{1}{2\pi}\int_{\partial(U\cap\mathbb{C}_{j})}PS^{-1}_{L}(s,AP)+(I-P)S^{-1}_{L}(s,A+P)ds_{j}$
$\displaystyle=\frac{1}{2\pi}\int_{\partial(U\cap\mathbb{C}_{j})}PS^{-1}_{L}(s,AP)ds_{j}+\frac{1}{2\pi}\int_{\partial(U\cap\mathbb{C}_{j})}(I-P)S^{-1}_{L}(s,A+P)ds_{j}$
$\displaystyle=P\frac{1}{2\pi}\int_{\partial(U\cap\mathbb{C}_{j})}S^{-1}_{L}(s,AP)ds_{j}+(I-P)\frac{1}{2\pi}\int_{\partial(U\cap\mathbb{C}_{j})}S^{-1}_{L}(s,A+P)ds_{j}.$

Since $S^{-1}_{L}(\cdot,A+P)$ is right slice hyperholomorphic function on $U$
(see, [1, Lemma 3.1.11]),

$\int_{\partial(U\cap\mathbb{C}_{j})}S^{-1}_{L}(s,A+P)ds_{j}=0.$

On the other hand,

$\frac{1}{2\pi}\int_{\partial(U\cap\mathbb{C}_{j})}S^{-1}_{L}(s,AP)ds_{j}=I,$

because $\sigma_{S}(AP)=\\{0\\}\subset U$. Hence $\chi_{\\{0\\}}(A)=P$. This
completes the proof. ∎

######  Corollary 3.11.

The Drazin $($resp. generalized Drazin$)$ inverse of an operator
$A\in\mathcal{B}_{R}(X)$ is uniquely determined.

###### Proof.

If $A$ is invertible, then it has a unique Drazin inverse which coincides with
its inverse $A^{-1}$. Assume that $A$ is not invertible and let $B$ and $C$ be
two Drazin (resp. generalized Drazin) inverses of $A$, then by the proof of
Lemma 3.6, $I-BA$ and $I-AC$ are two projections commuting with $A$ and
satisfying (3.6). Then by Theorem 3.10, $I-BA=I-AC$, and so $BA=AC$, thus
$B=B^{2}A=BAC=AC^{2}=C$. Hence the Drazin (resp. generalized Drazin) inverse
of $A$ is unique. ∎

######  Theorem 3.12.

Let $A\in\mathcal{B}_{R}(X)$. If $0\in iso\;\sigma_{S}(A)$, then

$A^{D}=f(A),$

where $f\in\mathcal{N}(\sigma_{S}(A))$ is such that $f$ is $0$ in an axially
symmetric neighborhood of $0$ and $f(q)=q^{-1}$ in an axially symmetric
neighborhood of $\sigma_{S}(A)\setminus\\{0\\}$, and

$\sigma_{S}(A^{D})\setminus\\{0\\}=\\{q^{-1}:q\in\sigma_{S}(A)\setminus\\{0\\}\\}.$

###### Proof.

Let $O_{1}$ be an axially symmetric open neighborhood of $0$ and $O_{2}$ be an
axially symmetric open neighborhood of $\sigma_{S}(A)\setminus\\{0\\}$ with
$\overline{O_{1}}\cap\overline{O_{2}}=\emptyset$. Define $f$ by $f(q)=0$ if
$q\in O_{1}$ and $f(q)=q^{-1}$ if $q\in O_{2}$, clearly
$f\in\mathcal{N}(\sigma_{S}(A))$. By Theorem 1.20 and Theorem 1.21, it is easy
to see that (3.3) holds for $A$ and $f(A)$.
By Theorem 1.21, it follows that
$\sigma_{S}(A^{D})\setminus\\{0\\}=\sigma_{S}(f(A))\setminus\\{0\\}=\\{f(q):q\in\sigma_{S}(A)\setminus\\{0\\}\\}=\\{q^{-1}:q\in\sigma_{S}(A)\setminus\\{0\\}\\}.$
∎

######  Theorem 3.13.

Let $A\in\mathcal{B}_{R}(X)$. The following conditions are equivalent:

  * (i)

$A$ is generalized Drazin invertible;

  * (ii)

$0\notin acc\;\sigma_{S}(A)$;

  * (iii)

$A=A_{1}\oplus A_{2}$, where $A_{1}$ is invertible on some closed subspace
$X_{1}$ of $X$ and $A_{2}$ is quasinilpotent on some complemented subspace
$X_{1}$ of $X$.

###### Proof.

$(i)\Leftrightarrow(ii)$ Already proved in Lemma 3.6 and Theorem 3.10.
$(i)\Rightarrow(iii)$ By Lemma 3.6 there exists a projection $P=I-AA^{D}$ such
that $AP$ is quasinilpotent and $AP=PA$, then $\mathcal{R}(P)$ and
$\mathcal{N}(P)$ are invariant under $A$, that is
$A\mathcal{R}(P)\subset\mathcal{R}(P)$ and
$A\mathcal{N}(P)\subset\mathcal{N}(P)$. Let $u\in\mathcal{N}(P)$, then
$u=AA^{D}u$, thus the restriction of $A$ to the kernel of $P$ is injective and
surjective, and so invertible. If we write $A=A_{1}\oplus A_{2}$ on
$X=\mathcal{N}(P)\oplus\mathcal{R}(P)$, then $A_{2}\in\mathcal{B}_{R}(X_{1})$
is quasinilpotent and $A_{1}\in\mathcal{B}_{R}(X_{2})$ is invertible.
$(iii)\Rightarrow(i)$ It is easy to check that $A^{D}=A_{1}^{-1}\oplus 0$. ∎

######  Corollary 3.14.

Let $A\in\mathcal{B}_{R}(X)$. The following conditions are equivalent:

  * (i)

$A$ is Drazin invertible;

  * (iii)

$A=A_{1}\oplus A_{2}$, where $A_{1}$ is invertible on some closed subspace
$X_{1}$ of $X$, $A_{2}$ is nilpotent on some complemented subspace $X_{1}$ of
$X$ and the nilpotency index of $A_{2}$ is the Drazin index of $A$.

###### Proof.

Assume that $A$ is Drazin invertible, then by Theorem 3.13 (iii),
$A=A_{1}\oplus A_{2}$ and $A^{d}=A_{1}^{-1}\oplus 0$. Hence, by (3.1),
$A^{k+1}A^{d}=A^{k}$, then $A_{1}^{k}\oplus 0=A_{1}^{k}\oplus A_{2}^{k}$, thus
$A_{2}^{k}=0$, so that the nilpotency index of $A_{2}$ is less than the Drazin
index of $A$.
Conversely, let $B=A_{1}^{-1}\oplus 0$, where $A_{1}$ is invertible and
$A_{2}$ is nilpotent, then (3.1) holds for $A$, $B$ and the nilpotency index
of $A_{2}$. Hence $A$ is Drazin invertible and the Drazin index of $A$ is less
than the nilpotency index of $A_{2}$. ∎

######  Theorem 3.15.

Suppose that $A\in\mathcal{B}_{R}(X)$ has the generalized Drazin inverse
$A^{D}$. Then

  * (i)

$(A^{k})^{D}=(A^{D})^{k}$ for all $k\in\mathbb{N}$;

  * (ii)

$(A^{D})^{D}=A^{2}A^{D}$;

  * (iii)

$((A^{D})^{D})^{D}=A^{D}$;

  * (iv)

$A^{D}(A^{D})^{D}=AA^{D}$.

###### Proof.

Let $f\in\mathcal{N}(\sigma_{S}(A))$ such that $f$ is $0$ in an axially
symmetric neighborhood of $0$ and $f(q)=q^{-1}$ in an axially symmetric
neighborhood of $\sigma_{S}(A)\setminus\\{0\\}$. By Theorem 3.12,
$A^{D}=f(A)$. Let $k\in\mathbb{N}$ and $g_{k}\in\mathcal{N}(\mathbb{H})$ such
that $g_{k}(q)=q^{k}$ if $q\in\mathbb{H}$. Clearly $f\circ g_{k}=g_{k}\circ f$
for all $k\in\mathbb{N}$, $f\circ f=g_{2}\circ f$, $f\circ f\circ f=f$ and
$f(f\circ f)=g_{1}f$. The above assertions are easily verified by using the
previous equalities, Theorem 1.20 and Theorem 1.22. ∎

######  Theorem 3.16.

Let $A,B\in\mathcal{B}_{R}(X)$ be commuting elements such that $A^{d}$ and
$B^{d}$ exist. Then $(AB)^{d}$ exists and

$(AB)^{d}=A^{d}B^{d}.$

###### Proof.

By [3, Theorem 1], $A,B,A^{d}$ and $B^{d}$ commute mutually, then the result
follows by Definition 3.1. ∎

Given a right linear operator $T$ on $X$. $T$ is said to have finite ascent if
there is an integer $k$ such that $\mathcal{N}(T^{k})=\mathcal{N}(T^{k+1})$,
the smallest such positive integer $k$ is called the ascent of $T$ and denoted
by $a(T)$. If there is no such integer we set $a(T):=\infty$. Analogously, $T$
is said to have finite descent if there is an integer $k$ such that
$T^{k+1}(X)=T^{k}(X)$, the smallest such positive integer $k$ is called the
descent of $T$ and denoted by $d(T)$. If there is no such integer we set
$d(T):=\infty$.

As in the complex case, we have the following result.

######  Theorem 3.17.

Suppose that $T$ is a right linear operator on $X$ and let $k\in\mathbb{N}$.
Then $a(T)=d(T)\leq k$ if and only if, we have the decomposition

$X=\mathcal{R}(T^{k})\oplus\mathcal{N}(T^{k}).$

###### Proof.

Let $k=a(T)=d(T)$, then $\mathcal{R}(T^{k})\cap\mathcal{N}(T^{k})=\\{0\\}$.
Indeed, let $u\in\mathcal{R}(T^{k})\cap\mathcal{N}(T^{k})$, then there is a
vector $v\in X$ such that $u=T^{k}v$, hence $T^{k}u=T^{2k}v=0$. Since
$\mathcal{N}(T^{2k})=\mathcal{N}(T^{k})$, $u=T^{k}v=0$. On the other hand, we
have $\mathcal{R}(T^{k})+\mathcal{N}(T^{k})=X$. Indeed, let $u\in X$, since
$\mathcal{R}(T^{k})=\mathcal{R}(T^{2k})$, there is a vector $v\in X$ such that
$T^{k}u=T^{2k}v$, hence $u=T^{k}v+u-T^{k}v$. Thus
$X=\mathcal{R}(T^{k})\oplus\mathcal{N}(T^{k}).$
Conversely, let $u\in\mathcal{N}(T^{k+1})$. Since
$\mathcal{R}(T^{k})\cap\mathcal{N}(T^{k})=\\{0\\}$, $T^{k}u=0$. Hence
$\mathcal{N}(T^{k+1})=\mathcal{N}(T^{k})$. On the other hand, let
$u\in\mathcal{R}(T^{k})$, then there is a vector $v\in X$ such that
$u=T^{k}v$. Since $v\in\mathcal{R}(T^{k})+\mathcal{N}(T^{k})$,
$u=T^{k}v\in\mathcal{R}(T^{k+1})$. Hence
$\mathcal{R}(T^{k})=\mathcal{R}(T^{k+1})$. Let now $p=a(T)$ and $q=d(T)$, we
can suppose that $p>0$ and $q>0$, assume that $p\leq q$, let
$u\in\mathcal{R}(T^{p})$, then there is a vector $v\in X$ such that
$u=T^{p}v$. Since $v\in\mathcal{R}(T^{q})\oplus\mathcal{N}(T^{p})$,
$u=T^{p}v\in\mathcal{R}(T^{p+q})\subseteq\mathcal{R}(T^{p+1})$. Thus $p=q\leq
k$. Assume that $q\leq p$, let $u\in\mathcal{N}(T^{q+1})$, then $T(T^{q}u)=0$,
that is $T^{q}u\in\mathcal{R}(T^{q})\cap\mathcal{N}(T)$. Since
$X=\mathcal{R}(T^{q})\oplus\mathcal{N}(T^{p})$, $T^{q}u=0$, and so
$u\in\mathcal{N}(T^{q})$. Thus $p=q\leq k$. Hence $a(T)=d(T)\leq k$. ∎

######  Theorem 3.18.

An operator $A$ in $\mathcal{B}_{R}(X)$ has a Drazin inverse if and only if it
has finite ascent and descent. In such a case, the Drazin index of $A$ is
equal to the common value of $a(A)$ and $d(A)$.

###### Proof.

By Corollary 3.14, $A$ is Drazin invertible if and only if $A=A_{1}\oplus
A_{2}$ with $A_{1}$ is invertible and $A_{2}$ is nilpotent. Let $k$ be the
nilpotency index of $A_{2}$, then $k$ is the least integer such that
$X=\mathcal{R}(T^{k})\oplus\mathcal{N}(T^{k})$, hence $a(A)=d(A)=k$. By
Corollary 3.14, again, $k=i(A)$, thus $a(A)=d(A)=i(A)$. ∎

######  Definition 3.19.

A two-sided quaternionic Banach algebra is a two-sided quaternionic Banach
space $\mathcal{A}$ that is endowed with a product
$\mathcal{A}\times\mathcal{A}\rightarrow\mathcal{A}$ such that:

  * (i)

The product is associative and distributive over the sum in $\mathcal{A}$;

  * (ii)

one has $(qx)y=q(xy)$ and $x(yq)=(xy)q$ for all $x,y\in\mathcal{A}$ and all
$q\in\mathbb{H}$;

  * (iii)

one has $\|xy\|\leq\|x\|\|y\|$ for all $x,y\in\mathcal{A}$.

If in addition there exists $e\in\mathcal{A}$ such that $ex=xe=x$ for all
$x\in\mathcal{A}$, then $\mathcal{A}$ is called a two-sided quaternionic
Banach algebra with unit.

One can prove that $\mathcal{B}_{R}(X)$ and $\mathcal{B}_{L}(X)$ are two-sided
quaternionic Banach algebras with unit.

######  Definition 3.20.

Let $\mathcal{A}$ be a two-sided quaternionic Banach algebra and
$a\in\mathcal{A}$. An element $b\in\mathcal{A}$ is a Drazin inverse of $a$,
written $b=a^{d}$, if

$ab=ba\text{, }ab^{2}=b\text{, }a^{k+1}b=a^{k},$ (3.8)

for some nonnegative integer $k$. The least nonnegative integer $k$ for which
these equations hold is the Drazin index $i(a)$ of $a$.

Let $\mathcal{A}$ be a two-sided quaternionic Banach algebra and
$a\in\mathcal{A}$. For any $a\in\mathcal{A}$ we define the left multiplication
of $a$ by $L_{a}(b)=ab$, for all $b\in\mathcal{A}$. Then
$L_{a}\in\mathcal{B}_{R}(\mathcal{A})$, we have $\|L_{a}\|=\|a\|$.

######  Theorem 3.21.

Let $\mathcal{A}$ be a two-sided quaternionic Banach algebra and
$a\in\mathcal{A}$ with unit. Then $a$ is Drazin invertible if and only if
$L_{a}$ is Drazin invertible. In such a case, $L_{a}^{d}=L_{a^{d}}$ and
$i(L_{a})=i(a)$.

###### Proof.

Let $a\in\mathcal{A}$ such that $a$ is Drazin invertible. For every
$b\in\mathcal{A}$, we have $L_{a}L_{b}=L_{ab}$, hence it is easy to check that
$L_{a^{d}}=L_{a}^{d}$ and then $i(L_{a})\leq i(a)$.
Conversely, assume that $L_{a}$ is Drazin invertible and let $b=L_{a}^{d}(e)$.
Since $L_{a}^{k+1}L_{a}^{d}=L_{a}^{k}$, $a^{k+1}b=a^{k}$. Hence
$L_{a}^{k+1}L_{b}=L_{a}^{k}=L_{a}^{d}L_{a}^{k+1}$, then by [3, Theorem 4] and
its proof, $L_{a}^{d}=L_{a}^{k}L_{b}^{k+1}=L_{a^{k}b^{k+1}}$. Let
$c=a^{k}b^{k+1}$, then $L_{a}L_{c}=L_{c}L_{a}\text{,
}L_{a}L_{c}^{2}=L_{c}\text{, }L_{a}^{k+1}L_{c}=L_{a}^{k}$, hence $ac=ca\text{,
}ac^{2}=c\text{, }a^{k+1}c=a^{k}$. Thus $a$ is Drazin invertible and then
$i(a)\leq i(L_{a})$. ∎

## References

  * [1] F. Colombo, J. Gantner, D.P. Kimsey, Spectral theory on the S-spectrum for quaternionic operators, Operator Theory: Advances and Applications, 270, Springer Basel, 2019.
  * [2] F. Colombo, I. Sabadini, D.C. Struppa, Noncommutative Functional Calculus, Birkhäuser Basel, 2011.
  * [3] M.P. Drazin, Pseudo-inverses in associative rings and semigroups, Amer. Math. Monthly 65 (1958) 506–514.
  * [4] R. Ghiloni, V. Moretti, A. Perotti, Continuous slice functional calculus in quaternionic Hilbert spaces, Rev. Math. Phys. 25 (2013) 1350006, 83 pp.
  * [5] J.J. Koliha, A generalized Drazin inverse, Glasgow Math. J. 38 (1996) 367–381.
  * [6] B. Muraleetharan, K. Thirulogasanthar, Weyl and Browder S-spectra in a right quaternionic Hilbert space, J. Geom. Phys. 135 (2019) 7–20.


”constant” method in which the learning rate, batch size, and momentum were
all constant with a method in which the hyperparameters were updated to reduce
the degree of smoothing $\delta^{\text{\rm{NSHB}}}$. Figures 19 and 19 plot
the accuracy in testing and the loss function value in training with NSHB
versus the number of epochs. We tuned the noise reduction methods so that the
decay rate was
$\gamma_{m}:=\frac{(M-m)^{0.9}}{\left\\{M-(m-1)\right\\}^{0.9}}\
(M=200,m\in[M])$.

Figure 18: Accuracy score for testing and loss function value for training
versus epochs in training of ResNet18 on CIFAR100 dataset with implicit
graduated optimization with NSHB (Algorithm 5).

Figure 19: Accuracy score for testing and loss function value for training
versus epochs in training of WideResNet-28-10 on CIFAR100 dataset with
implicit graduated optimization with NSHB (Algorithm 5).

###  F.4 Experiments on optimal noise scheduling with SHB and NSHB

Sato and Iiduka identified the conditions that the decay rate
$(\gamma_{m})_{m\in[M]}$ of noise $(\delta_{m})_{m\in[M]}$ must satisfy by
considering the conditions necessary for the success of a graduated
optimization algorithm for the new $\sigma$-nice function [53]. They found
that decay rate $\gamma_{m}$ should satisfy

$\displaystyle\frac{\sqrt{(m-M-\sqrt{2})^{2}-1}-1}{-(m-M-\sqrt{2})}\leq\gamma_{m}<1\
\left(m\in[M],p\in(0,1]\right).$ (18)

From (2)-(4) in Section 4.1, the noise level, i.e., the degree of smoothing,
is proportional to the learning rate, which immediately leads to the optimal
decay rate of the decaying learning rate. Figure 21 plots the decay rate of
the existing learning rate scheduler and the regions satisfying inequality
(18) when $M=200$. Figure 21 is an enlarged version of Figure 21.

Figure 20: Decay rate of existing learning rate schedules. Blue-filled area
represents regions satisfying inequality (18) when $M=200$.

Figure 21: An enlarged version of Figure 21.

As shown in Figures 21 and 21, inequality (18) is satisfied by polynomial
decay with $p\in(0,1]$ and not by any other learning rate scheduler. The rate
of polynomial decay can be written as
$\frac{(M-m)^{p}}{\left\\{M-(m-1)\right\\}^{p}}$, which is used in Algorithms
3 and 5. Figures 21 and 21 also show that polynomial decay with $p\in(0,1]$
may give the smallest loss function value through implicit graduated
optimization. Sato and Iiduka confirmed this experimentally for SGD; i.e.,
they showed that polynomial decay with $p\in(0,1]$ can achieve the smallest
training loss function value for SGD, as per theory [53]. We thus clarified
whether polynomial decay with $p\in(0,1]$ can achieve the smallest loss
function value for SGD with momentum as well.

Figure 22: Accuracy score for testing and loss function value for training
versus epochs in training of ResNet18 on CIFAR100 dataset with SHB and
learning rate scheduler.

Figure 23: Accuracy score for testing and loss function value for training
versus epochs in training of ResNet18 on CIFAR100 dataset with NSHB and
learning rate scheduler.

Figures 23 and 23 plot test accuracy and training loss function values versus
batch size for SHB and NSHB with learning rate scheduler in training of
ResNet18 on the CIFAR100 dataset. The initial value of the learning rate was
$0.1$, and the power of the exponential decay was $0.9$. The batch size and
momentum were constants, $256$ and $0.9$, respectively.

Figures 25 and 25 plot test accuracy versus training loss function values for
SHB and NSHB with learning rate scheduler in training of WideResNet-28-10 on
the CIFAR100 dataset.

Figure 24: Accuracy score for testing and loss function value for training
versus epochs in training of WideResNet-28-10 on CIFAR100 dataset with SHB and
learning rate scheduler.

Figure 25: Accuracy score for testing and loss function value for training
versus epochs in training of WideResNet-28-10 on CIFAR100 dataset with NSHB
and learning rate scheduler.

Figures 23-25 show that polynomial decay with $p\in(0,1]$ can achieve the
smallest loss function values among the learning rate schedulers. Note that
when $p=0.1$, the loss function value fell more slowly than when
$p\in\\{0.5,0.9,1\\}$. This can be explained by the convergence rate of
Algorithms 3 and 5 being $\mathcal{O}\left(1/\epsilon^{\frac{1}{p}}\right)\
\left(p\in(0,1]\right)$. That is, the closer $p$ is to $1$, the fewer
iterations are needed, so when $p=0.1$, we need many more iterations than when
$p\in\\{0.5,0.9,1\\}$.