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\ihead{Similar matrices and equivalent polynomial matrices}
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\begin{document}

\title{Similar matrices and equivalent polynomial matrices}
\author{Darij Grinberg}
\date{\today}
\maketitle

\section*{***}

The purpose of this note is to prove a classical result in linear algebra:
that two $m\times m$-matrices $a$ and $b$ over a commutative ring $K$ are
similar if and only if the polynomial matrices $tI_{m}-a$ and $tI_{m}-b$
(where $I_{m}$ is the identity matrix, and $t$ is a polynomial indeterminate)
are equivalent (i.e., satisfy $\left(  tI_{m}-a\right)  p=q\left(
tI_{m}-b\right)  $ for two invertible polynomial matrices $p$ and $q$).

Even better, we shall prove a generalization of this result, replacing the
matrices $a$ and $b$ by two elements $a$ and $b$ of a (not necessarily
commutative) ring $R$, and replacing the polynomial matrices $tI_{m}-a$ and
$tI_{m}-b$ by the polynomials $t-a$ and $t-b$ in $R\left[  t\right]  $. Here,
$R\left[  t\right]  $ denotes the polynomial ring over $R$ in a single
indeterminate $t$; we will define this object precisely in Definition
\ref{def.Rt}.

Neither this generalization nor our proof is really new. The generalization
was observed by @user20948 in a comment at MathOverflow (
\url{https://mathoverflow.net/questions/66269/#comment866429_96046} ), who
proved it tersely but nicely using a commutative diagram of $R\left[
t\right]  $-modules and their quotients. The proof I give below is merely an
elementary rewording of @user20948's proof -- much less slick, but fully
elementary and self-contained. A similar proof appears in \cite[Chapter VI,
\S 4--\S 5]{Gantma77}.

\section{Notations and definitions regarding polynomials}

\begin{convention}
In the following, rings are always understood to be associative and with
unity, but not necessarily commutative.
\end{convention}

We will use the concept of a polynomial ring $R\left[  t\right]  $ over a ring
$R$ that is not necessarily commutative. This notion is widely known in the
case when $R$ is commutative. The definition in the general case is more or
less the same, except that certain shortcuts requiring are not available
(e.g., the polynomial ring $R\left[  t\right]  $ is not an $R$-algebra in
general, and we have to distinguish between left and right multiplication).
Here is the general definition:

\begin{definition}
\label{def.Rt}Let $R$ be a ring. Then, $R\left[  t\right]  $ shall denote the
ring of polynomials in a single indeterminate $t$ over $R$. The definition of
this ring is well-known when $R$ is commutative; we use the same definition in
the general case: A polynomial $p\in R\left[  t\right]  $ means an infinite
sequence $\left(  p_{0},p_{1},p_{2},\ldots\right)  $ of elements of $R$ such
that all but finitely many $n\geq0$ satisfy $p_{n}=0$. We define sums and
products of such polynomials by the usual formulas:%
\begin{align*}
\left(  p_{0},p_{1},p_{2},\ldots\right)  +\left(  q_{0},q_{1},q_{2}%
,\ldots\right)   &  :=\left(  p_{0}+q_{0},\ p_{1}+q_{1},\ p_{2}+q_{2}%
,\ \ldots\right)  ;\\
\left(  p_{0},p_{1},p_{2},\ldots\right)  \cdot\left(  q_{0},q_{1},q_{2}%
,\ldots\right)   &  :=\left(  r_{0},r_{1},r_{2},\ldots\right)  ,
\end{align*}
where $r_{n}:=\sum_{i=0}^{n}p_{i}q_{n-i}$ for each $n\geq0$.

We identify each $a\in R$ with the polynomial $\left(  a,0,0,0,\ldots\right)
\in R\left[  t\right]  $. This makes $R$ into a subring of $R\left[  t\right]
$. Hence, a polynomial
\[
p=\left(  p_{0},p_{1},p_{2},\ldots\right)  \in R\left[  t\right]
\]
can be multiplied by an element $a\in R$ both from the left and from the
right: namely,%
\begin{align}
ap  &  =\left(  ap_{0},ap_{1},ap_{2},\ldots\right)
\ \ \ \ \ \ \ \ \ \ \text{and}\nonumber\\
pa  &  =\left(  p_{0}a,p_{1}a,p_{2}a,\ldots\right)  . \label{darij1.eq.1}%
\end{align}


Now, define the indeterminate $t$ as the sequence $\left(  0,1,0,0,0,\ldots
\right)  \in R\left[  t\right]  $ (only the second entry is $1$). Then, each
polynomial $p=\left(  p_{0},p_{1},p_{2},\ldots\right)  \in R\left[  t\right]
$ satisfies%
\[
p=\sum_{n\geq0}p_{n}t^{n}=\sum_{n\geq0}t^{n}p_{n}.
\]
This is, of course, the standard way of writing polynomials.

Note that the indeterminate $t$ commutes with every polynomial $p\in R\left[
t\right]  $, since multiplying $p$ by $t$ (in either order) is tantamount to
shifting all coefficients of $p$ by one position to the right: If $p=\left(
p_{0},p_{1},p_{2},\ldots\right)  \in R\left[  t\right]  $, then%
\begin{equation}
pt=tp=\left(  0,p_{0},p_{1},p_{2},\ldots\right)  . \label{darij1.eq.2}%
\end{equation}

\end{definition}

\begin{remark}
The main difference between the general case (i.e., the case when $R$ is
arbitrary ring) and the classical commutative case (i.e., the case when $R$ is
a commutative ring) is that in the general case, it is not clear how to
evaluate a polynomial $p=\left(  p_{0},p_{1},p_{2},\ldots\right)  \in R\left[
t\right]  $ at a given element $a\in R$. Indeed, we can define the
\textquotedblleft left evaluation\textquotedblright\ $\sum_{n\geq0}p_{n}a^{n}$
and the \textquotedblleft right evaluation\textquotedblright\ $\sum_{n\geq
0}a^{n}p_{n}$, but these are in general not the same (unless $a$ belongs to
the center of $R$), and neither of them is as well-behaved as in the
commutative case. (More on that below.)

Another difference is that, as mentioned above, $R\left[  t\right]  $ is not
an $R$-algebra if $R$ is not commutative (since the notion of an $R$-algebra
does not exist in this case).
\end{remark}

\section{The claim}

Recall that an element $p$ of a ring $R$ is called \emph{invertible} if and
only if it has an inverse (i.e., if there is an element $q\in R$ such that
$pq=qp=1$).

We need one more definition before we can state the main result:

\begin{definition}
Let $R$ be a ring. Let $a$ and $b$ be two elements of $R$.

\begin{enumerate}
\item[\textbf{(a)}] We say that $a$ and $b$ are \emph{conjugate} in $R$ if
there exists an invertible element $p\in R$ such that $ap=pb$.

\item[\textbf{(b)}] We say that $a$ and $b$ are \emph{equivalent} in $R$ if
there exist invertible elements $p,q\in R$ such that $ap=qb$.
\end{enumerate}
\end{definition}

I'm not sure how standard the word \textquotedblleft
equivalent\textquotedblright\ is; I think other authors use \textquotedblleft
unit-equivalent\textquotedblright\ or \textquotedblleft
associate\textquotedblright\ for the same notion.

Our goal is to prove the following:

\begin{theorem}
\label{thm.1}Let $R$ be a ring. Let $a,b\in R$ be two elements. Then, $a$ and
$b$ are conjugate in $R$ if and only if the polynomials $t-a$ and $t-b$ are
equivalent in $R\left[  t\right]  $.
\end{theorem}

Before we prove this theorem, let us see how it can be used to prove the
result promised at the beginning of this note:

\begin{corollary}
\label{cor.2}Let $K$ be a ring. Let $a,b\in K^{m\times m}$ be two $m\times
m$-matrices over $K$. Then, the matrices $a$ and $b$ are similar if and only
if the matrices $tI_{m}-a$ and $tI_{m}-b$ in $\left(  K\left[  t\right]
\right)  ^{m\times m}$ are equivalent in $\left(  K\left[  t\right]  \right)
^{m\times m}$. (Here, $I_{m}$ denotes the $m\times m$-identity matrix.)
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.2}.]The polynomial ring $K^{m\times m}\left[
t\right]  $ is known to be isomorphic to the matrix ring $\left(  K\left[
t\right]  \right)  ^{m\times m}$. Indeed, there is a ring isomorphism%
\[
\rho:K^{m\times m}\left[  t\right]  \rightarrow\left(  K\left[  t\right]
\right)  ^{m\times m}%
\]
that sends each polynomial $\sum_{n\geq0}A_{n}t^{n}\in K^{m\times m}\left[
t\right]  $ (with $A_{n}\in K^{m\times m}$ for all $n\geq0$) to the matrix
$\sum_{n\geq0}t^{n}A_{n}\in\left(  K\left[  t\right]  \right)  ^{m\times m}$
(where each $A_{n}$ is now regarded as a matrix over $K\left[  t\right]  $).
This isomorphism $\rho$ sends the polynomial $t-a\in K^{m\times m}\left[
t\right]  $ to the matrix $tI_{m}-a\in\left(  K\left[  t\right]  \right)
^{m\times m}$; that is, we have $\rho\left(  t-a\right)  =tI_{m}-a$.
Similarly, $\rho\left(  t-b\right)  =tI_{m}-b$.

Conjugate elements of $K^{m\times m}$ are better known as similar matrices.
Thus, we have the following chain of logical equivalences:%
\begin{align*}
&  \ \left(  \text{the matrices }a\text{ and }b\text{ are similar}\right) \\
&  \Longleftrightarrow\ \left(  \text{the matrices }a\text{ and }b\text{ are
conjugate in }K^{m\times m}\right) \\
&  \Longleftrightarrow\ \left(  \text{the polynomials }t-a\text{ and
}t-b\text{ are equivalent in }K^{m\times m}\left[  t\right]  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem
\ref{thm.1}, applied to }R=K^{m\times m}\right) \\
&  \Longleftrightarrow\ \left(  \text{the matrices }\rho\left(  t-a\right)
\text{ and }\rho\left(  t-b\right)  \text{ are equivalent in }\left(  K\left[
t\right]  \right)  ^{m\times m}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\rho\text{ is a ring isomorphism, and thus}\\
\text{two elements }c\text{ and }d\text{ of }K^{m\times m}\left[  t\right] \\
\text{are equivalent in }K^{m\times m}\left[  t\right]  \text{ if and only
if}\\
\text{their images }\rho\left(  c\right)  \text{ and }\rho\left(  d\right)
\text{ are equivalent in }\left(  K\left[  t\right]  \right)  ^{m\times m}%
\end{array}
\right) \\
&  \Longleftrightarrow\ \left(  \text{the matrices }tI_{m}-a\text{ and }%
tI_{m}-b\text{ are equivalent in }\left(  K\left[  t\right]  \right)
^{m\times m}\right)
\end{align*}
(since $\rho\left(  t-a\right)  =tI_{m}-a$ and $\rho\left(  t-b\right)
=tI_{m}-b$). This proves Corollary \ref{cor.2}.
\end{proof}

\section{Right evaluations}

The trick to the proof of Theorem \ref{thm.1} is the following definition:

\begin{definition}
Let $R$ be a ring, and let $a\in R$ be arbitrary. Then, we let $r_{a}:R\left[
t\right]  \rightarrow R$ be the map that sends each polynomial $\left(
p_{0},p_{1},p_{2},\ldots\right)  \in R\left[  t\right]  $ (with $p_{0}%
,p_{1},p_{2},\ldots\in R$) to $\sum_{n\geq0}a^{n}p_{n}$.
\end{definition}

This map $r_{a}$ can be called the \emph{right evaluation map at }$a$, since
it \textquotedblleft evaluates\textquotedblright\ polynomials at $t=a$. But
this shouldn't be taken too literally; in particular, it is not always true
that any two polynomials $p,q\in R\left[  t\right]  $ satisfy $r_{a}\left(
pq\right)  =r_{a}\left(  p\right)  \cdot r_{a}\left(  q\right)  $. (However,
this equality still holds if $a$ lies in the center of $R$.) \medskip

The following property of $r_{a}$ is straightforward:

\begin{proposition}
\label{prop.3}Let $R$ be a ring, and let $a\in R$ be arbitrary. Then, the map
$r_{a}:R\left[  t\right]  \rightarrow R$ is a right $R$-linear map, i.e., a
homomorphism of right $R$-modules.
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.3}.]If $p=\left(  p_{0},p_{1},p_{2}%
,\ldots\right)  $ and $q=\left(  q_{0},q_{1},q_{2},\ldots\right)  $ are two
polynomials in $R\left[  t\right]  $ (with $p_{0},p_{1},p_{2},\ldots\in R$ and
$q_{0},q_{1},q_{2},\ldots\in R$), then their sum is $p+q=\left(  p_{0}%
+q_{0},\ p_{1}+q_{1},\ p_{2}+q_{2},\ \ldots\right)  $, and thus the definition
of $r_{a}$ yields%
\begin{align*}
r_{a}\left(  p+q\right)   &  =\sum_{n\geq0}\underbrace{a^{n}\left(
p_{n}+q_{n}\right)  }_{=a^{n}p_{n}+a^{n}q_{n}}=\sum_{n\geq0}\left(  a^{n}%
p_{n}+a^{n}q_{n}\right) \\
&  =\underbrace{\sum_{n\geq0}a^{n}p_{n}}_{\substack{=r_{a}\left(  p\right)
\\\text{(by the definition of }r_{a}\text{)}}}+\underbrace{\sum_{n\geq0}%
a^{n}q_{n}}_{\substack{=r_{a}\left(  q\right)  \\\text{(by the definition of
}r_{a}\text{)}}}\\
&  =r_{a}\left(  p\right)  +r_{a}\left(  q\right)  .
\end{align*}
Thus, the map $r_{a}$ respects addition.

If $p=\left(  p_{0},p_{1},p_{2},\ldots\right)  $ is a polynomial in $R\left[
t\right]  $ (with $p_{0},p_{1},p_{2},\ldots\in R$), and if $c\in R$, then
\begin{align*}
pc  &  =\left(  p_{0},p_{1},p_{2},\ldots\right)  c\ \ \ \ \ \ \ \ \ \ \left(
\text{since }p=\left(  p_{0},p_{1},p_{2},\ldots\right)  \right) \\
&  =\left(  p_{0}c,p_{1}c,p_{2}c,\ldots\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by \eqref{darij1.eq.1}, applied to }c\text{ instead of }a\right)  ,
\end{align*}
and thus the definition of $r_{a}$ yields%
\[
r_{a}\left(  pc\right)  =\sum_{n\geq0}a^{n}p_{n}c=\underbrace{\sum_{n\geq
0}a^{n}p_{n}}_{\substack{=r_{a}\left(  p\right)  \\\text{(by the definition of
}r_{a}\text{)}}}\cdot c=r_{a}\left(  p\right)  \cdot c.
\]
Thus, the map $r_{a}$ is right $R$-linear (since we already know that $r_{a}$
respects addition). This proves Proposition \ref{prop.3}.
\end{proof}

We will also need the following properties of $r_{a}$:

\begin{proposition}
\label{prop.4}Let $R$ be a ring, and let $a\in R$ be arbitrary. Let $s\in
R\left[  t\right]  $. Then:

\begin{enumerate}
\item[\textbf{(a)}] We have $r_{a}\left(  sc\right)  =r_{a}\left(  s\right)
c$ for any $c\in R$.

\item[\textbf{(b)}] We have $r_{a}\left(  st\right)  =ar_{a}\left(  s\right)
$.

\item[\textbf{(c)}] We have $r_{a}\left(  \left(  t-a\right)  s\right)  =0$.

\item[\textbf{(d)}] There exists a polynomial $\overline{s}\in R\left[
t\right]  $ such that $s=r_{a}\left(  s\right)  +\left(  t-a\right)
\overline{s}$.
\end{enumerate}
\end{proposition}

\begin{proof}
[Proof of Proposition \ref{prop.4}.]Write the polynomial $s\in R\left[
t\right]  $ in the form $s=\left(  s_{0},s_{1},s_{2},\ldots\right)  $. Thus,
the definition of $r_{a}$ yields $r_{a}\left(  s\right)  =\sum_{n\geq0}%
a^{n}s_{n}$. \medskip

\textbf{(a)} This follows immediately from Proposition \ref{prop.3}. \medskip

\textbf{(b)} From $s=\left(  s_{0},s_{1},s_{2},\ldots\right)  $, we obtain
$st=\left(  0,s_{0},s_{1},s_{2},\ldots\right)  $ (by \eqref{darij1.eq.2},
applied to $p=s$). Setting $s_{-1}:=0$, we can rewrite this as%
\[
st=\left(  s_{-1},s_{0},s_{1},s_{2},\ldots\right)  .
\]
Hence, the definition of $r_{a}$ yields
\begin{align*}
r_{a}\left(  st\right)   &  =\sum_{n\geq0}a^{n}s_{n-1}=a^{0}%
\underbrace{s_{0-1}}_{=s_{-1}=0}+\sum_{n\geq1}\underbrace{a^{n}}_{=aa^{n-1}%
}s_{n-1}\\
&  =\underbrace{a^{0}0}_{=0}+\sum_{n\geq1}aa^{n-1}s_{n-1}=\sum_{n\geq
1}aa^{n-1}s_{n-1}\\
&  =\sum_{n\geq0}aa^{n}s_{n}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have substituted }n\\
\text{for }n-1\text{ in the sum}%
\end{array}
\right)  \\
&  =a\underbrace{\sum_{n\geq0}a^{n}s_{n}}_{\substack{=r_{a}\left(  s\right)
\\\text{(by the definition of }r_{a}\text{)}}}=ar_{a}\left(  s\right)  .
\end{align*}
This proves Proposition \ref{prop.4} \textbf{(b)}. \medskip

\textbf{(c)} From $s=\left(  s_{0},s_{1},s_{2},\ldots\right)  $, we obtain
$as=\left(  as_{0},as_{1},as_{2},\ldots\right)  $. Thus, the definition of
$r_{a}$ yields%
\[
r_{a}\left(  as\right)  =\sum_{n\geq0}\underbrace{a^{n}a}_{\substack{=a^{n+1}%
\\=aa^{n}}}s_{n}=\sum_{n\geq0}aa^{n}s_{n}=a\underbrace{\sum_{n\geq0}a^{n}%
s_{n}}_{\substack{=r_{a}\left(  s\right)  \\\text{(by the definition of }%
r_{a}\text{)}}}=ar_{a}\left(  s\right)  .
\]


Now, \eqref{darij1.eq.2} (applied to $p=s$) yields $st=ts$. Hence, $\left(
t-a\right)  s=\underbrace{ts}_{=st}-as=st-as$. Therefore,%
\begin{align*}
r_{a}\left(  \left(  t-a\right)  s\right)   &  =r_{a}\left(  st-as\right)  \\
&  =\underbrace{r_{a}\left(  st\right)  }_{\substack{=ar_{a}\left(  s\right)
\\\text{(by Proposition \ref{prop.4} \textbf{(b)})}}}-\underbrace{r_{a}\left(
as\right)  }_{=ar_{a}\left(  s\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Proposition \ref{prop.3}}\right)  \\
&  =ar_{a}\left(  s\right)  -ar_{a}\left(  s\right)  =0.
\end{align*}
This proves Proposition \ref{prop.4} \textbf{(c)}. \medskip

\textbf{(d)} The element $t$ of $R\left[  t\right]  $ commutes with $a$ (since
$ta=\left(  0,a,0,0,0,\ldots\right)  =at$). Hence, the equality%
\[
x^{n}-y^{n}=\left(  x-y\right)  \sum_{k=0}^{n-1}x^{k}y^{n-1-k}%
\]
(which holds for any two commuting elements $x$ and $y$ and any $n\geq0$) can
be applied to $x=t$ and $y=a$. Thus, for any $n\geq0$, we have%
\begin{equation}
t^{n}-a^{n}=\left(  t-a\right)  \sum_{k=0}^{n-1}t^{k}a^{n-1-k}.
\label{darij1.pf.prop4.d.1}%
\end{equation}


Subtracting the equality $r_{a}\left(  s\right)  =\sum_{n\geq0}a^{n}s_{n}$
from the equality $s=\left(  s_{0},s_{1},s_{2},\ldots\right)  =\sum_{n\geq
0}t^{n}s_{n}$, we obtain%
\begin{align*}
s-r_{a}\left(  s\right)   &  =\sum_{n\geq0}t^{n}s_{n}-\sum_{n\geq0}a^{n}%
s_{n}=\sum_{n\geq0}\underbrace{\left(  t^{n}s_{n}-a^{n}s_{n}\right)
}_{=\left(  t^{n}-a^{n}\right)  s_{n}}\\
&  =\sum_{n\geq0}\underbrace{\left(  t^{n}-a^{n}\right)  }_{\substack{=\left(
t-a\right)  \sum_{k=0}^{n-1}t^{k}a^{n-1-k}\\\text{(by
\eqref{darij1.pf.prop4.d.1})}}}s_{n}=\sum_{n\geq0}\left(  t-a\right)  \left(
\sum_{k=0}^{n-1}t^{k}a^{n-1-k}\right)  s_{n}\\
&  =\left(  t-a\right)  \underbrace{\sum_{n\geq0}\left(  \sum_{k=0}^{n-1}%
t^{k}a^{n-1-k}\right)  s_{n}}_{\substack{\text{This sum is well-defined,}%
\\\text{since all but finitely many}\\\text{of its addends are }0\text{
(because}\\\text{all but finitely many }n\text{ satisfy }s_{n}=0\text{)}}}.
\end{align*}
Thus, there exists a polynomial $\overline{s}\in R\left[  t\right]  $ such
that $s-r_{a}\left(  s\right)  =\left(  t-a\right)  \overline{s}$ (namely,
$\overline{s}=\sum_{n\geq0}\left(  \sum_{k=0}^{n-1}t^{k}a^{n-1-k}\right)
s_{n}$). In other words, there exists a polynomial $\overline{s}\in R\left[
t\right]  $ such that $s=r_{a}\left(  s\right)  +\left(  t-a\right)
\overline{s}$. This proves Proposition \ref{prop.4} \textbf{(d)}.
\end{proof}

\section{Proof of Theorem \ref{thm.1}}

We are now ready to prove Theorem \ref{thm.1}:

\begin{proof}
[Proof of Theorem \ref{thm.1}.]$\Longrightarrow:$ Assume that $a$ and $b$ are
conjugate in $R$. We must show that the polynomials $t-a$ and $t-b$ are
equivalent in $R\left[  t\right]  $.

Since $a$ and $b$ are conjugate in $R$, there exists an invertible element
$r\in R$ such that $ar=rb$ (by the definition of \textquotedblleft
conjugate\textquotedblright). Consider this $r$. Then, $r\in R\subseteq
R\left[  t\right]  $, and furthermore the element $r$ is invertible in
$R\left[  t\right]  $ (since $r$ is invertible in $R$). In the ring $R\left[
t\right]  $, we have%
\[
\left(  t-a\right)  r=\underbrace{tr}_{\substack{=rt\\\text{(by
\eqref{darij1.eq.2}, applied}\\\text{to }p=r\text{)}}}-\underbrace{ar}%
_{=rb}=rt-rb=r\left(  t-b\right)  .
\]
Thus, there exist invertible elements $p,q\in R\left[  t\right]  $ such that
$\left(  t-a\right)  p=q\left(  t-b\right)  $ (namely, $p=r$ and $q=r$). In
other words, the polynomials $t-a$ and $t-b$ are equivalent in $R\left[
t\right]  $ (by the definition of \textquotedblleft
equivalent\textquotedblright). This proves the \textquotedblleft%
$\Longrightarrow$\textquotedblright\ direction of Theorem \ref{thm.1}.
\medskip

$\Longleftarrow:$ Assume that the polynomials $t-a$ and $t-b$ are equivalent
in $R\left[  t\right]  $. We must show that $a$ and $b$ are conjugate in $R$.

We have assumed that the polynomials $t-a$ and $t-b$ are equivalent in
$R\left[  t\right]  $. In other words, there exist invertible elements $p,q\in
R\left[  t\right]  $ such that
\begin{equation}
\left(  t-a\right)  p=q\left(  t-b\right)  \label{darij1.pf.t1.0}%
\end{equation}
(by the definition of \textquotedblleft equivalent\textquotedblright).
Consider these $p$ and $q$. Note that $p$ and $q$ are invertible; thus,
$p^{-1}$ and $q^{-1}$ are invertible as well.

We now claim that%
\begin{equation}
r_{a}\left(  q\right)  \cdot r_{b}\left(  q^{-1}\right)  =1.
\label{darij1.pf.t1.1}%
\end{equation}


[\textit{Proof of \eqref{darij1.pf.t1.1}:} Proposition \ref{prop.4}
\textbf{(a)} (applied to $s=q$ and $c=r_{b}\left(  q^{-1}\right)  $) yields%
\begin{equation}
r_{a}\left(  qr_{b}\left(  q^{-1}\right)  \right)  =r_{a}\left(  q\right)
\cdot r_{b}\left(  q^{-1}\right)  .\label{darij1.pf.t1.1.pf.1}%
\end{equation}
However, Proposition \ref{prop.4} \textbf{(d)} (applied to $b$ and $q^{-1}$
instead of $a$ and $s$) yields that there exists a polynomial $\overline{s}\in
R\left[  t\right]  $ such that $q^{-1}=r_{b}\left(  q^{-1}\right)  +\left(
t-b\right)  \overline{s}$. Consider this $\overline{s}$. Then, solving the
equality $q^{-1}=r_{b}\left(  q^{-1}\right)  +\left(  t-b\right)  \overline
{s}$ for $r_{b}\left(  q^{-1}\right)  $, we find%
\[
r_{b}\left(  q^{-1}\right)  =q^{-1}-\left(  t-b\right)  \overline{s}.
\]
Thus,%
\[
q\underbrace{r_{b}\left(  q^{-1}\right)  }_{=q^{-1}-\left(  t-b\right)
\overline{s}}=q\left(  q^{-1}-\left(  t-b\right)  \overline{s}\right)
=1-\underbrace{q\left(  t-b\right)  }_{\substack{=\left(  t-a\right)
p\\\text{(by \eqref{darij1.pf.t1.0})}}}\overline{s}=1-\left(  t-a\right)
p\overline{s}.
\]
Applying the map $r_{a}$ to this equality, we find%
\begin{align*}
r_{a}\left(  qr_{b}\left(  q^{-1}\right)  \right)   &  =r_{a}\left(  1-\left(
t-a\right)  p\overline{s}\right)  \\
&  =\underbrace{r_{a}\left(  1\right)  }_{\substack{=1\\\text{(this follows
easily}\\\text{from the}\\\text{definition of }r_{a}\text{)}}%
}-\underbrace{r_{a}\left(  \left(  t-a\right)  p\overline{s}\right)
}_{\substack{=0\\\text{(by Proposition \ref{prop.4} \textbf{(c)}%
,}\\\text{applied to }p\overline{s}\text{ instead of }s\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition \ref{prop.3}}\right)  \\
&  =1-0=1.
\end{align*}
Comparing this with \eqref{darij1.pf.t1.1.pf.1}, we obtain $r_{a}\left(
q\right)  \cdot r_{b}\left(  q^{-1}\right)  =1$. Thus, \eqref{darij1.pf.t1.1}
is proven.] \medskip

Now, we notice a symmetry slightly hidden in our setting: If we multiply both
sides of the equality \eqref{darij1.pf.t1.0} by $q^{-1}$ on the left and by
$p^{-1}$ on the right, then we obtain $q^{-1}\left(  t-a\right)
pp^{-1}=q^{-1}q\left(  t-b\right)  p^{-1}$. This simplifies to $q^{-1}\left(
t-a\right)  =\left(  t-b\right)  p^{-1}$ (since $q^{-1}\left(  t-a\right)
\underbrace{pp^{-1}}_{=1}=q^{-1}\left(  t-a\right)  $ and $\underbrace{q^{-1}%
q}_{=1}\left(  t-b\right)  p^{-1}=\left(  t-b\right)  p^{-1}$). In other
words,
\[
\left(  t-b\right)  p^{-1}=q^{-1}\left(  t-a\right)  .
\]
This equality has the same form as \eqref{darij1.pf.t1.0}, but with the
elements $b$, $a$, $p^{-1}$ and $q^{-1}$ playing the roles of $a$, $b$, $p$
and $q$. Hence, we can prove the equality%
\[
r_{b}\left(  q^{-1}\right)  \cdot r_{a}\left(  \left(  q^{-1}\right)
^{-1}\right)  =1
\]
using the same reasoning that we used to prove \eqref{darij1.pf.t1.1} (but
with $a$, $b$, $p$ and $q$ replaced by $b$, $a$, $p^{-1}$ and $q^{-1}$). Since
$\left(  q^{-1}\right)  ^{-1}=q$, this equality rewrites as
\[
r_{b}\left(  q^{-1}\right)  \cdot r_{a}\left(  q\right)  =1.
\]
Combining this equality with \eqref{darij1.pf.t1.1}, we conclude that the
elements $r_{a}\left(  q\right)  $ and $r_{b}\left(  q^{-1}\right)  $ are
mutually inverse in $R$. Thus, the element $r_{a}\left(  q\right)  \in R$ is invertible.

Finally, applying the map $r_{a}$ to both sides of \eqref{darij1.pf.t1.0}, we
obtain%
\begin{align*}
r_{a}\left(  \left(  t-a\right)  p\right)   &  =r_{a}\left(
\underbrace{q\left(  t-b\right)  }_{=qt-qb}\right)  =r_{a}\left(
qt-qb\right)  \\
&  =\underbrace{r_{a}\left(  qt\right)  }_{\substack{=ar_{a}\left(  q\right)
\\\text{(by Proposition \ref{prop.4} \textbf{(b)},}\\\text{applied to
}s=q\text{)}}}-\underbrace{r_{a}\left(  qb\right)  }_{\substack{=r_{a}\left(
q\right)  b\\\text{(by Proposition \ref{prop.4} \textbf{(a)},}\\\text{applied
to }s=q\text{ and }c=b\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
Proposition \ref{prop.3}}\right)  \\
&  =ar_{a}\left(  q\right)  -r_{a}\left(  q\right)  b.
\end{align*}
Hence,%
\[
ar_{a}\left(  q\right)  -r_{a}\left(  q\right)  b=r_{a}\left(  \left(
t-a\right)  p\right)  =0
\]
(by Proposition \ref{prop.4} \textbf{(c)}, applied to $s=p$). In other words,
$ar_{a}\left(  q\right)  =r_{a}\left(  q\right)  b$. Since $r_{a}\left(
q\right)  \in R$ is invertible, this shows that there exists an invertible
element $z\in R$ such that $az=zb$ (namely, $z=r_{a}\left(  q\right)  $). In
other words, $a$ and $b$ are conjugate in $R$. This proves the
\textquotedblleft$\Longleftarrow$\textquotedblright\ direction of Theorem
\ref{thm.1}. The proof of Theorem \ref{thm.1} is thus complete.
\end{proof}

\begin{thebibliography}{99999999}                                                                                         %
\bibitem[Gantma77]{Gantma77}F. R. Gantmacher, \textit{The Theory of Matrices,
volume 1}, AMS Chelsea Publishing 1977.
\end{thebibliography}


\end{document}