请问在这个例子中，为什么有一些行不会统计行号？

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\begin{document}
\linenumbers
\title{Jacobson's Lemma for the generalized n-strongly Drazin inverse}

\author{Huanyin Chen}
\author{Marjan Sheibani}


\begin{abstract}
Let $n\in {\Bbb N}$. An element $a\in R$ has generalized n-strongly Drazin inverse if there exists $x\in R$ such that $xax=x, x\in comm^2(a), a^n-ax\in R^{qnil}.$
For any $a,b\in R$, we prove that $1-ab$ has generalized n-strongly Drazin inverse if and only if $1-ba$ has generalized n-strongly Drazin inverse. Extensions in Banach algebra are also obtained.
\end{abstract}

\maketitle

\section{Introduction}

Let $R$ be an associative ring with an identity. The commutant of $a\in R$ is defined by $comm(a)=\{x\in
R~|~xa=ax\}$. The double commutant of $a\in R$ is defined
by $comm^2(a)=\{x\in R~|~xy=yx~\mbox{for all}~y\in comm(a)\}$. Set $R^{qnil}=\{a\in R~|~1+ax\in U(R)~\mbox{for
every}~x\in comm(a)\}$. An element $a\in R$ is quasinilpotent if
$a\in \grave{R^{qnil}}$. For a Banach algebra $\mathcal{A}$ it is well known
 that $$a\in \mathcal{A}^{qnil}\Leftrightarrow
\lim\limits_{n\to\infty}\parallel a^n\parallel^{\frac{1}{n}}=0.$$
An element $a$ in $R$ is said to have generalized Drazin inverse if there exists $x\in R$ such that $$x=xax, x\in comm^2(a), a-a^2x\in R^{qnil}.$$  The preceding $x$ is unique, if such element exists. As usual,
it will be denoted by $a^{d}$, and called the generalized Drazin inverse of $a$. Generalized Drazin inverse is extensively studied in matrix theory and Banach algebra (see~\cite{B, J, LZ} and~\cite{ZC}).

\end{document}

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