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\ihead{The trace Cayley-Hamilton theorem}
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\begin{document}

\title{The trace Cayley-Hamilton theorem}
\author{Darij Grinberg}
\date{
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\maketitle
\tableofcontents

\section{Introduction}

Let $\mathbb{K}$ be a commutative ring. The famous Cayley-Hamilton theorem
says that if $\chi_{A}=\det\left(  tI_{n}-A\right)  \in\mathbb{K}\left[
t\right]  $ is the characteristic polynomial of an $n\times n$-matrix
$A\in\mathbb{K}^{n\times n}$, then $\chi_{A}\left(  A\right)  =0$. Speaking
more explicitly, it means that if we write this polynomial $\chi_{A}$ in the
form $\chi_{A}=\sum_{i=0}^{n}c_{n-i}t^{i}$ (with $c_{n-i}\in\mathbb{K}$), then
$\sum_{i=0}^{n}c_{n-i}A^{i}=0$. Various proofs of this theorem are well-known
(we will present one in this note, but it could not be any farther from being
new). A less standard fact, which I call the \textit{trace Cayley-Hamilton
theorem}, states that
\begin{equation}
kc_{k}+\sum_{i=1}^{k}\operatorname*{Tr}\left(  A^{i}\right)  c_{k-i}%
=0\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\mathbb{N} \label{eq.trach.intro}%
\end{equation}
(where $\sum_{i=0}^{n}c_{n-i}t^{i}$ is $\chi_{A}$ as before, and where we set
$c_{n-i}=0$ for every $i<0$). In the case of $k\geq n$, this can easily be
obtained from the Cayley-Hamilton theorem $\sum_{i=0}^{n}c_{n-i}A^{i}=0$ by
multiplying by $A^{k-n}$ and taking traces\footnote{The details are left to
the interested reader. The $kc_{k}$ term on the left hand side appears off,
but it actually is harmless: In the $k=n$ case, it can be rewritten as
$\operatorname*{Tr}\left(  A^{0}\right)  c_{n}$ and incorporated into the sum,
whereas in the $k>n$ case, it simply vanishes.}; no such simple proof exists
in the general case, however. The result itself is not new (the $k\leq n$
case, for example, is \cite[Chapter III, Exercise 14]{LomQui16}), and is
well-known e.g. to algebraic combinatorialists; however, it is hard to find an
expository treatment.

When the ground ring $\mathbb{K}$ is a field, it is possible to prove the
trace Cayley-Hamilton theorem by expressing both $\operatorname*{Tr}\left(
A^{i}\right)  $ and the $c_{j}$ through the eigenvalues of $A$ (indeed,
$\operatorname*{Tr}\left(  A^{i}\right)  $ is the sum of the $i$-th powers of
these eigenvalues, whereas $c_{j}$ is $\left(  -1\right)  ^{j}$ times their
$j$-th elementary symmetric function); the identity (\ref{eq.trach.intro})
then boils down to the Newton identities for said eigenvalues. However, of
course, the use of eigenvalues in this proof requires $\mathbb{K}$ to be a
field. There are ways to adapt this proof to the case when $\mathbb{K}$ is a
commutative ring. One is to apply the \textquotedblleft method of universal
identities\textquotedblright\ (see, e.g., \cite[Chapter III, Exercise
14]{LomQui16}; the method is also explained in \cite{Conrad09}) to reduce the
general case to the case when $\mathbb{K}$ is a field\footnote{This relies on
the observation that (\ref{eq.trach.intro}) (for a given $k$) is a polynomial
identity in the entries of $A$.}. Another is to build up the theory of
eigenvalues for square matrices over an arbitrary commutative ring
$\mathbb{K}$; this is not as simple as for fields, but doable (see
\cite{Laksov13}).

In this note, I shall give a proof of both the Cayley-Hamilton and the trace
Cayley-Hamilton theorems via a trick whose use in proving the former is
well-known (see, e.g., \cite[Chapter Five, Section IV, Lemma 1.9]{Heffer14}).
The trick is to observe that the adjugate matrix $\operatorname*{adj}\left(
tI_{n}-A\right)  $ can be written as $D_{0}t^{0}+D_{1}t^{1}+\cdots
+D_{n-1}t^{n-1}$ for some $n$ matrices $D_{0},D_{1},\ldots,D_{n-1}%
\in\mathbb{K}^{n\times n}$; then, a telescoping sum establishes the
Cayley-Hamilton theorem. The same trick can be used for the trace
Cayley-Hamilton theorem, although it requires more work; in particular, an
intermediate step is necessary, establishing that the derivative of the
characteristic polynomial $\chi_{A}=\det\left(  tI_{n}-A\right)  $ is
$\operatorname*{Tr}\left(  \operatorname*{adj}\left(  tI_{n}-A\right)
\right)  $. I hope that this writeup will have two uses: making the trace
Cayley-Hamilton theorem more accessible, and demonstrating that the trick just
mentioned can serve more than one purpose. Next, I shall show an application
of the trace Cayley-Hamilton theorem, answering a question from
\cite{m.se1798703}. Finally, I shall discuss several other properties of the
adjugate matrix as well as further applications of polynomial matrices in
proving determinant identities.

\section{Notations and theorems}

\subsection{Notations}

Before we state the theorems that we will be occupying ourselves with, let us
agree on the notations.

\begin{definition}
\label{def.notations}Throughout this note, the word \textquotedblleft
ring\textquotedblright\ will mean \textquotedblleft associative ring with
unity\textquotedblright. We will always let $\mathbb{K}$ denote a commutative
ring with unity. The word \textquotedblleft matrix\textquotedblright\ shall
always mean \textquotedblleft matrix over $\mathbb{K}$\textquotedblright,
unless explicitly stated otherwise.

As usual, we let $\mathbb{K}\left[  t\right]  $ denote the polynomial ring in
the indeterminate $t$ over $\mathbb{K}$.

If $f\in\mathbb{K}\left[  t\right]  $ is a polynomial and $n$ is an integer,
then $\left[  t^{n}\right]  f$ will denote the coefficient of $t^{n}$ in $f$.
(If $n$ is negative or greater than the degree of $f$, then this coefficient
is understood to be $0$.)

Let $\mathbb{N}$ denote the set $\left\{  0,1,2,\ldots\right\}  $.

If $n\in\mathbb{N}$ and $m\in\mathbb{N}$, and if we are given an element
$a_{i,j}\in\mathbb{K}$ for every $\left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  \times\left\{  1,2,\ldots,m\right\}  $, then we use the
notation $\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}$ for the
$n\times m$-matrix whose $\left(  i,j\right)  $-th entry is $a_{i,j}$ for all
$\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  \times\left\{
1,2,\ldots,m\right\}  $.

For every $n\in\mathbb{N}$, we denote the $n\times n$ identity matrix by
$I_{n}$.

For every $n\in\mathbb{N}$ and $m\in\mathbb{N}$, we denote the $n\times m$
zero matrix by $0_{n\times m}$.

If $A$ is any $n\times n$-matrix, then we let $\det A$ denote the determinant
of $A$, and we let $\operatorname*{Tr}A$ denote the trace of $A$. (Recall that
the trace of $A$ is defined to be the sum of the diagonal entries of $A$.)

We consider $\mathbb{K}$ as a subring of $\mathbb{K}\left[  t\right]  $. Thus,
for every $n\in\mathbb{N}$, every $n\times n$-matrix in $\mathbb{K}^{n\times
n}$ can be considered as a matrix in $\left(  \mathbb{K}\left[  t\right]
\right)  ^{n\times n}$.
\end{definition}

\subsection{The main claims}

We shall now state the results that we will prove further below. We begin with
a basic fact:

\begin{proposition}
\label{prop.ta+b}Let $n\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times n}$ and
$B\in\mathbb{K}^{n\times n}$ be two $n\times n$-matrices. Consider the matrix
$tA+B\in\left(  \mathbb{K}\left[  t\right]  \right)  ^{n\times n}$.

\textbf{(a)} Then, $\det\left(  tA+B\right)  \in\mathbb{K}\left[  t\right]  $
is a polynomial of degree $\leq n$ in $t$.

\textbf{(b)} We have $\left[  t^{0}\right]  \left(  \det\left(  tA+B\right)
\right)  =\det B$.

\textbf{(c)} We have $\left[  t^{n}\right]  \left(  \det\left(  tA+B\right)
\right)  =\det A$.
\end{proposition}

\begin{definition}
\label{def.chiA}Let $n\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times n}$ be an
$n\times n$-matrix. Then, we consider $A$ as a matrix in $\left(
\mathbb{K}\left[  t\right]  \right)  ^{n\times n}$ as well (as explained
above); thus, a matrix $tI_{n}-A\in\left(  \mathbb{K}\left[  t\right]
\right)  ^{n\times n}$ is defined. We let $\chi_{A}$ denote the polynomial
$\det\left(  tI_{n}-A\right)  \in\mathbb{K}\left[  t\right]  $; we call
$\chi_{A}$ the \textit{characteristic polynomial} of $A$.
\end{definition}

We notice that the notion of the characteristic polynomial is not standardized
across the literature. Our definition of $\chi_{A}$ is identical with the
definition in \cite[\S V.3]{Knapp2016} (except that we use $t$ instead of $X$
as the indeterminate), but the definition in \cite[Chapter Five, Section II,
Definition 3.9]{Heffer14} is different (it defines $\chi_{A}$ to be
$\det\left(  A-tI_{n}\right)  $ instead). The two definitions differ merely in
a sign (namely, one version of the characteristic polynomial is $\left(
-1\right)  ^{n}$ times the other), whence any statement about one of them can
easily be translated into a statement about the other; nevertheless this
discrepancy creates some occasions for confusion. I shall, of course, use
Definition \ref{def.chiA} throughout this note.

\begin{corollary}
\label{cor.chiA}Let $n\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times n}$.

\textbf{(a)} Then, $\chi_{A}\in\mathbb{K}\left[  t\right]  $ is a polynomial
of degree $\leq n$ in $t$.

\textbf{(b)} We have $\left[  t^{0}\right]  \chi_{A}=\left(  -1\right)
^{n}\det A$.

\textbf{(c)} We have $\left[  t^{n}\right]  \chi_{A}=1$.
\end{corollary}

Of course, combining parts \textbf{(a)} and \textbf{(c)} of Corollary
\ref{cor.chiA} shows that, for every $n\in\mathbb{N}$ and $A\in\mathbb{K}%
^{n\times n}$, the characteristic polynomial $\chi_{A}$ is a monic polynomial
of degree $n$.

Let me now state the main two theorems in this note:

\begin{theorem}
[Cayley-Hamilton theorem]\label{thm.CH}Let $n\in\mathbb{N}$. Let
$A\in\mathbb{K}^{n\times n}$. Then, $\chi_{A}\left(  A\right)  =0_{n\times n}%
$. (Here, $\chi_{A}\left(  A\right)  $ denotes the result of substituting $A$
for $t$ in the polynomial $\chi_{A}$. It does \textbf{not} denote the result
of substituting $A$ for $t$ in the expression $\det\left(  tI_{n}-A\right)  $;
in particular, $\chi_{A}\left(  A\right)  $ is an $n\times n$-matrix, not a determinant!)
\end{theorem}

\begin{theorem}
[trace Cayley-Hamilton theorem]\label{thm.TCH}Let $n\in\mathbb{N}$. Let
$A\in\mathbb{K}^{n\times n}$. For every $j\in\mathbb{Z}$, define an element
$c_{j}\in\mathbb{K}$ by $c_{j}=\left[  t^{n-j}\right]  \chi_{A}$. Then,%
\[
kc_{k}+\sum_{i=1}^{k}\operatorname*{Tr}\left(  A^{i}\right)  c_{k-i}%
=0\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\mathbb{N}.
\]

\end{theorem}

Theorem \ref{thm.CH} is (as has already been said) well-known and a
cornerstone of linear algebra. It appears (with proofs) in \cite{Bernha11},
\cite[Theorem 7.23]{Brown93}, \cite[Theorem 2.16]{Camero08}, \cite[Theorem
23.1]{Climen13}, \cite[Theorem 4.6.12]{Ford22}, \cite[\S 28.10]{Garrett09},
\cite[Chapter Five, Section IV, Lemma 1.9]{Heffer14}, \cite[Theorem
5.9]{Knapp2016}, \cite[\S 5.15]{Loehr14}, \cite[\S 4, Theorem 1]{Mate16},
\cite[Theorem I.8]{McDona84}, \cite[\S 6.1, Theorem 1.1]{Moore68},
\cite[Seconde m\'{e}thode (\S 3)]{Sage08}, \cite{Shurma15}, \cite[Theorem
3.1]{Stoll17}, \cite{Straub83}, \cite[Theorem 7.10]{Willia89}, \cite[\S 3]%
{Zeilbe} and in many other sources\footnote{All the sources we are citing
(with the possible exception of \cite[\S 28.10]{Garrett09}) prove Theorem
\ref{thm.CH} in full generality, although some of them do not \textbf{state}
Theorem \ref{thm.CH} in full generality (indeed, they often state it under the
additional requirement that $\mathbb{K}$ be a field). There are other sources
which only prove Theorem \ref{thm.CH} in the case when $\mathbb{K}$ is a
field. The note \cite{Sage08} gives four proofs of Theorem \ref{thm.CH} for
the case when $\mathbb{K}=\mathbb{C}$; the first of these proofs works for
every field $\mathbb{K}$, whereas the second works for any commutative ring
$\mathbb{K}$, and the third and the fourth actually require $\mathbb{K}%
=\mathbb{C}$.
\par
Note that some authors decline to call Theorem \ref{thm.CH} the
Cayley-Hamilton theorem; they instead use this name for some related result.
For instance, Hefferon, in \cite{Heffer14}, uses the name \textquotedblleft
Cayley-Hamilton theorem\textquotedblright\ for a corollary.}. The proof we
will give below will essentially repeat the proof in \cite[Chapter Five,
Section IV, Lemma 1.9]{Heffer14}.

Theorem \ref{thm.TCH} is a less known result. It appears in \cite[Chapter III,
Exercise 14]{LomQui16} (with a sketch of a proof), in \cite[($C-H$)]{Zeilbe93}
(with a beautiful short proof using exterior algebra) and in \cite[Exercise
5]{Zeilbe} (without proof); its particular case when $\mathbb{K}$ is a field
also tends to appear in representation-theoretical literature (mostly left as
an exercise to the reader). We will prove it similarly to Theorem
\ref{thm.CH}; this proof, to my knowledge, is new.

\section{The proofs}

\subsection{Proposition \ref{prop.ta+b} and Corollary \ref{cor.chiA}}

Let us now begin proving the results stated above. As a warmup, we will prove
the (rather trivial) Proposition \ref{prop.ta+b}.

We first recall how the determinant of a matrix is defined: For any
$n\in\mathbb{N}$, let $S_{n}$ denote the $n$-th symmetric group (i.e., the
group of all permutations of $\left\{  1,2,\ldots,n\right\}  $). If
$n\in\mathbb{N}$ and $\sigma\in S_{n}$, then $\left(  -1\right)  ^{\sigma}$
denotes the sign of the permutation $\sigma$. If $n\in\mathbb{N}$, and if
$A=\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$ is an $n\times
n$-matrix, then%
\begin{equation}
\det A=\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{n}a_{i,\sigma\left(  i\right)  }. \label{eq.det}%
\end{equation}


We prepare for the proof of Proposition \ref{prop.ta+b} by stating a simple lemma:

\begin{lemma}
\label{lem.ta+b.prod}Let $n\in\mathbb{N}$. Let $x_{1},x_{2},\ldots,x_{n}$ be
$n$ elements of $\mathbb{K}$. Let $y_{1},y_{2},\ldots,y_{n}$ be $n$ elements
of $\mathbb{K}$. Define a polynomial $f\in\mathbb{K}\left[  t\right]  $ by
$f=\prod_{i=1}^{n}\left(  tx_{i}+y_{i}\right)  $.

\textbf{(a)} Then, $f$ is a polynomial of degree $\leq n$.

\textbf{(b)} We have $\left[  t^{n}\right]  f=\prod_{i=1}^{n}x_{i}$.

\textbf{(c)} We have $\left[  t^{0}\right]  f=\prod_{i=1}^{n}y_{i}$.
\end{lemma}

\begin{vershort}
\begin{proof}
[Proof of Lemma \ref{lem.ta+b.prod}.]Obvious by multiplying out the product
$\prod_{i=1}^{n}\left(  tx_{i}+y_{i}\right)  $ (or, if one desires a formal
proof, by a straightforward induction over $n$).
\end{proof}
\end{vershort}

\begin{verlong}
Before we come to the proof of this lemma, let us recall the following
well-known facts about polynomials:

\begin{lemma}
\label{lem.poly.1}Let $g\in\mathbb{K}\left[  t\right]  $ and $h\in
\mathbb{K}\left[  t\right]  $ be two polynomials.

\textbf{(a)} We have%
\begin{equation}
\left[  t^{k}\right]  \left(  gh\right)  =\sum_{i=0}^{k}\left(  \left[
t^{i}\right]  g\right)  \left(  \left[  t^{k-i}\right]  h\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\mathbb{N}. \label{eq.tkgh}%
\end{equation}
(This is just the definition of the product of two polynomials.)

\textbf{(b)} We have $\left[  t^{0}\right]  \left(  gh\right)  =\left(
\left[  t^{0}\right]  g\right)  \left(  \left[  t^{0}\right]  h\right)  $.

\textbf{(c)} Let $p$ and $q$ be two nonnegative integers. Assume that $g$ is a
polynomial of degree $\leq p$. Assume that $h$ is a polynomial of degree $\leq
q$. Then, $gh$ is a polynomial of degree $\leq p+q$ and satisfies $\left[
t^{p+q}\right]  \left(  gh\right)  =\left(  \left[  t^{p}\right]  g\right)
\cdot\left(  \left[  t^{q}\right]  h\right)  $.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.poly.1}.]\textbf{(a)} Lemma \ref{lem.poly.1} follows
immediately from the definition of the product $gh$ of the two polynomials $g$
and $h$.

\textbf{(b)} The equality (\ref{eq.tkgh}) (applied to $k=0$) yields%
\[
\left[  t^{0}\right]  \left(  gh\right)  =\sum_{i=0}^{0}\left(  \left[
t^{i}\right]  g\right)  \left(  \left[  t^{0-i}\right]  h\right)  =\left(
\left[  t^{0}\right]  g\right)  \left(  \underbrace{\left[  t^{0-0}\right]
}_{=\left[  t^{0}\right]  }h\right)  =\left(  \left[  t^{0}\right]  g\right)
\left(  \left[  t^{0}\right]  h\right)  .
\]
This proves Lemma \ref{lem.poly.1} \textbf{(b)}.

\textbf{(c)} The polynomial $g$ has degree $\leq p$. In other words,%
\begin{equation}
\left[  t^{i}\right]  g=0\ \ \ \ \ \ \ \ \ \ \text{for every integer }i>p.
\label{pf.lem.poly.1.c.1}%
\end{equation}


The polynomial $h$ has degree $\leq q$. In other words,%
\begin{equation}
\left[  t^{i}\right]  h=0\ \ \ \ \ \ \ \ \ \ \text{for every integer }i>q.
\label{pf.lem.poly.1.c.2}%
\end{equation}


Now, let $k$ be an integer such that $k\geq p+q$. From $q\geq0$, we obtain
$p+\underbrace{q}_{\geq0}\geq p\geq0$. Thus, $k\geq p+q\geq p$, so that $p\leq
k$. Also, $p\geq0$, so that $0\leq p$. Every $i\in\left\{  0,1,\ldots
,p-1\right\}  $ satisfies%
\begin{equation}
\left[  t^{k-i}\right]  h=0 \label{pf.lem.poly.1.c.2u}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.poly.1.c.2u}):} Let $i\in\left\{
0,1,\ldots,p-1\right\}  $. Then, $i\leq p-1<p$ and thus $k-\underbrace{i}%
_{<p}>k-p\geq q$ (since $k\geq p+q$). Hence, (\ref{pf.lem.poly.1.c.2})
(applied to $k-i$ instead of $i$) yields $\left[  t^{k-i}\right]  h=0$. This
proves (\ref{pf.lem.poly.1.c.2u}).}. Finally, $k\geq p\geq0$, so that
$k\in\mathbb{N}$. Hence, (\ref{eq.tkgh}) yields%
\begin{align}
\left[  t^{k}\right]  \left(  gh\right)   &  =\sum_{i=0}^{k}\left(  \left[
t^{i}\right]  g\right)  \left(  \left[  t^{k-i}\right]  h\right) \nonumber\\
&  =\sum_{i=0}^{p}\left(  \left[  t^{i}\right]  g\right)  \left(  \left[
t^{k-i}\right]  h\right)  +\sum_{i=p+1}^{k}\underbrace{\left(  \left[
t^{i}\right]  g\right)  }_{\substack{=0\\\text{(by (\ref{pf.lem.poly.1.c.1}%
)}\\\text{(since }i\geq p+1>p\text{))}}}\left(  \left[  t^{k-i}\right]
h\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }0\leq p\leq k\right) \nonumber\\
&  =\sum_{i=0}^{p}\left(  \left[  t^{i}\right]  g\right)  \left(  \left[
t^{k-i}\right]  h\right)  +\underbrace{\sum_{i=p+1}^{k}0\left(  \left[
t^{k-i}\right]  h\right)  }_{=0}=\sum_{i=0}^{p}\left(  \left[  t^{i}\right]
g\right)  \left(  \left[  t^{k-i}\right]  h\right) \nonumber\\
&  =\sum_{i=0}^{p-1}\left(  \left[  t^{i}\right]  g\right)
\underbrace{\left(  \left[  t^{k-i}\right]  h\right)  }%
_{\substack{=0\\\text{(by (\ref{pf.lem.poly.1.c.2u}))}}}+\left(  \left[
t^{p}\right]  g\right)  \left(  \left[  t^{k-p}\right]  h\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}i=p\text{ from the sum}\right) \nonumber\\
&  =\underbrace{\sum_{i=0}^{p-1}\left(  \left[  t^{i}\right]  g\right)
0}_{=0}+\left(  \left[  t^{p}\right]  g\right)  \left(  \left[  t^{k-p}%
\right]  h\right)  =\left(  \left[  t^{p}\right]  g\right)  \left(  \left[
t^{k-p}\right]  h\right)  . \label{pf.lem.poly.1.c.5}%
\end{align}


Now, forget that we fixed $k$. We thus have shown that every integer $k$
satisfying $k\geq p+q$ satisfies (\ref{pf.lem.poly.1.c.5}).

Now, let $k$ be an integer satisfying $k>p+q$. From $k>p+q$, we obtain $k-p>q$
and thus $\left[  t^{k-p}\right]  h=0$ (by (\ref{pf.lem.poly.1.c.2}) (applied
to $i=k-p$)). But from $k>p+q$, we also obtain $k\geq p+q$. Hence,
(\ref{pf.lem.poly.1.c.5}) yields%
\[
\left[  t^{k}\right]  \left(  gh\right)  =\left(  \left[  t^{p}\right]
g\right)  \underbrace{\left(  \left[  t^{k-p}\right]  h\right)  }_{=0}=0.
\]


Now, forget that we fixed $k$. We thus have shown that $\left[  t^{k}\right]
\left(  gh\right)  =0$ for every integer $k>p+q$. In other words, $gh$ is a
polynomial of degree $\leq p+q$.

It remains to prove that $\left[  t^{p+q}\right]  \left(  gh\right)  =\left(
\left[  t^{p}\right]  g\right)  \cdot\left(  \left[  t^{q}\right]  h\right)
$. But (\ref{pf.lem.poly.1.c.5}) (applied to $k=p+q$) yields
\[
\left[  t^{p+q}\right]  \left(  gh\right)  =\left(  \left[  t^{p}\right]
g\right)  \left(  \underbrace{\left[  t^{p+q-p}\right]  }_{=\left[
t^{q}\right]  }h\right)  =\left(  \left[  t^{p}\right]  g\right)  \left(
\left[  t^{q}\right]  h\right)  =\left(  \left[  t^{p}\right]  g\right)
\cdot\left(  \left[  t^{q}\right]  h\right)  .
\]
This completes the proof of Lemma \ref{lem.poly.1} \textbf{(c)}.
\end{proof}

\begin{proof}
[Proof of Lemma \ref{lem.ta+b.prod}.]We shall prove Lemma \ref{lem.ta+b.prod}
by induction over $n$:

\textit{Induction base:} Lemma \ref{lem.ta+b.prod} holds in the case when
$n=0$\ \ \ \ \footnote{\textit{Proof.} Let $n$, $\left(  x_{1},x_{2}%
,\ldots,x_{n}\right)  $, $\left(  y_{1},y_{2},\ldots,y_{n}\right)  $ and $f$
be as in Lemma \ref{lem.ta+b.prod}. Assume that $n=0$. We must prove that
Lemma \ref{lem.ta+b.prod} holds in this case.
\par
We have
\begin{align*}
f  &  =\prod_{i=1}^{n}\left(  tx_{i}+y_{i}\right)  =\prod_{i=1}^{0}\left(
tx_{i}+y_{i}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }n=0\right) \\
&  =\left(  \text{empty product}\right)  =1.
\end{align*}
Hence, $f$ is a polynomial of degree $\leq0$ (since $1$ is clearly a
polynomial of degree $\leq0$). In other words, $f$ is a polynomial of degree
$\leq n$ (since $n=0$). Thus, Lemma \ref{lem.ta+b.prod} \textbf{(a)} holds.
\par
Since $n=0$ and $f=1$, we have $\left[  t^{n}\right]  f=\left[  t^{0}\right]
1=1$. Comparing this with%
\begin{align*}
\prod_{i=1}^{n}x_{i}  &  =\prod_{i=1}^{0}x_{i}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }n=0\right) \\
&  =\left(  \text{empty product}\right)  =1,
\end{align*}
we obtain $\left[  t^{n}\right]  f=\prod_{i=1}^{n}x_{i}$. Thus, Lemma
\ref{lem.ta+b.prod} \textbf{(b)} holds.
\par
Since $f=1$, we have $\left[  t^{0}\right]  f=\left[  t^{0}\right]  1=1$.
Comparing this with%
\begin{align*}
\prod_{i=1}^{n}y_{i}  &  =\prod_{i=1}^{0}y_{i}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }n=0\right) \\
&  =\left(  \text{empty product}\right)  =1,
\end{align*}
we obtain $\left[  t^{0}\right]  f=\prod_{i=1}^{n}y_{i}$. Thus, Lemma
\ref{lem.ta+b.prod} \textbf{(c)} holds.
\par
Now, we have shown that all three parts \textbf{(a)}, \textbf{(b)} and
\textbf{(c)} of Lemma \ref{lem.ta+b.prod} hold. In other words, Lemma
\ref{lem.ta+b.prod} holds. Qed.}. This completes the induction base.

\textit{Induction step:} Let $N$ be a positive integer. Assume that Lemma
\ref{lem.ta+b.prod} holds in the case when $n=N-1$. We now must prove that
Lemma \ref{lem.ta+b.prod} holds in the case when $n=N$. In other words, we
must prove the following claim:

\begin{statement}
\textit{Claim 1:} Let $x_{1},x_{2},\ldots,x_{N}$ be $N$ elements of
$\mathbb{K}$. Let $y_{1},y_{2},\ldots,y_{N}$ be $N$ elements of $\mathbb{K}$.
Define a polynomial $f\in\mathbb{K}\left[  t\right]  $ by $f=\prod_{i=1}%
^{N}\left(  tx_{i}+y_{i}\right)  $.

\textbf{(a)} Then, $f$ is a polynomial of degree $\leq N$.

\textbf{(b)} We have $\left[  t^{N}\right]  f=\prod_{i=1}^{N}x_{i}$.

\textbf{(c)} We have $\left[  t^{0}\right]  f=\prod_{i=1}^{N}y_{i}$.
\end{statement}

But we have assumed that Lemma \ref{lem.ta+b.prod} holds in the case when
$n=N-1$. In other words, the following fact holds:

\begin{statement}
\textit{Fact 2:} Let $x_{1},x_{2},\ldots,x_{N-1}$ be $N-1$ elements of
$\mathbb{K}$. Let $y_{1},y_{2},\ldots,y_{N-1}$ be $N-1$ elements of
$\mathbb{K}$. Define a polynomial $f\in\mathbb{K}\left[  t\right]  $ by
$f=\prod_{i=1}^{N-1}\left(  tx_{i}+y_{i}\right)  $.

\textbf{(a)} Then, $f$ is a polynomial of degree $\leq N-1$.

\textbf{(b)} We have $\left[  t^{N-1}\right]  f=\prod_{i=1}^{N-1}x_{i}$.

\textbf{(c)} We have $\left[  t^{0}\right]  f=\prod_{i=1}^{N-1}y_{i}$.
\end{statement}

Let us now prove Claim 1:

[\textit{Proof of Claim 1:} Define a polynomial $g\in\mathbb{K}\left[
t\right]  $ by $g=\prod_{i=1}^{N-1}\left(  tx_{i}+y_{i}\right)  $. Then,%
\begin{align*}
f  &  =\prod_{i=1}^{N}\left(  tx_{i}+y_{i}\right)  =\underbrace{\left(
\prod_{i=1}^{N-1}\left(  tx_{i}+y_{i}\right)  \right)  }_{=g}\cdot\left(
tx_{N}+y_{N}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the factor}\\
\text{for }i=N\text{ from the product}%
\end{array}
\right) \\
&  =g\cdot\left(  tx_{N}+y_{N}\right)  .
\end{align*}


But recall that $g=\prod_{i=1}^{N-1}\left(  tx_{i}+y_{i}\right)  $. Hence, we
can apply Fact 2 to $g$ instead of $f$.

From Fact 2 \textbf{(a)} (applied to $g$ instead of $f$), we conclude that $g$
is a polynomial of degree $\leq N-1$.

The polynomial $tx_{N}+y_{N}\in\mathbb{K}\left[  t\right]  $ clearly has
degree $\leq1$ and satisfies \newline$\left[  t^{1}\right]  \left(
tx_{N}+y_{N}\right)  =x_{N}$ and $\left[  t^{0}\right]  \left(  tx_{N}%
+y_{N}\right)  =y_{N}$. Now, Lemma \ref{lem.poly.1} \textbf{(c)} (applied to
$h=tx_{N}+y_{N}$, $p=N-1$ and $q=1$) yields that $g\cdot\left(  tx_{N}%
+y_{N}\right)  $ is a polynomial of degree $\leq\left(  N-1\right)  +1$ and
satisfies $\left[  t^{\left(  N-1\right)  +1}\right]  \left(  g\cdot\left(
tx_{N}+y_{N}\right)  \right)  =\left(  \left[  t^{N-1}\right]  g\right)
\cdot\left(  \left[  t^{1}\right]  \left(  tx_{N}+y_{N}\right)  \right)  $.
Since $g\cdot\left(  tx_{N}+y_{N}\right)  =f$, $\left(  N-1\right)  +1=N$ and
$\left[  t^{1}\right]  \left(  tx_{N}+y_{N}\right)  =x_{N}$, this rewrites as
follows: The polynomial $f$ is a polynomial of degree $\leq N$ and satisfies
$\left[  t^{N}\right]  f=\left(  \left[  t^{N-1}\right]  g\right)  \cdot
x_{N}$. In particular, $f$ is a polynomial of degree $\leq N$. This proves
Claim 1 \textbf{(a)}.

Fact 2 \textbf{(b)} (applied to $g$ instead of $f$) shows that $\left[
t^{N-1}\right]  g=\prod_{i=1}^{N-1}x_{i}$. But
\begin{align*}
\prod_{i=1}^{N}x_{i}  &  =\underbrace{\left(  \prod_{i=1}^{N-1}x_{i}\right)
}_{=\left[  t^{N-1}\right]  g}\cdot x_{N}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the factor}\\
\text{for }i=N\text{ from the product}%
\end{array}
\right) \\
&  =\left(  \left[  t^{N-1}\right]  g\right)  \cdot x_{N}.
\end{align*}
Comparing this with $\left[  t^{N}\right]  f=\left(  \left[  t^{N-1}\right]
g\right)  \cdot x_{N}$, we obtain $\left[  t^{N}\right]  f=\prod_{i=1}%
^{N}x_{i}$. Thus, Claim 1 \textbf{(b)} holds.

Fact 2 \textbf{(c)} (applied to $g$ instead of $f$) shows that $\left[
t^{0}\right]  g=\prod_{i=1}^{N-1}y_{i}$. But Lemma \ref{lem.poly.1}
\textbf{(b)} (applied to $h=tx_{N}+y_{N}$) yields $\left[  t^{0}\right]
\left(  g\cdot\left(  tx_{N}+y_{N}\right)  \right)  =\left(  \left[
t^{0}\right]  g\right)  \left(  \left[  t^{0}\right]  \left(  tx_{N}%
+y_{N}\right)  \right)  $. Since $g\cdot\left(  tx_{N}+y_{N}\right)  =f$ and
$\left[  t^{0}\right]  \left(  tx_{N}+y_{N}\right)  =y_{N}$, this rewrites as
$\left[  t^{0}\right]  f=\left(  \left[  t^{0}\right]  g\right)  y_{N}$.
Comparing this with%
\begin{align*}
\prod_{i=1}^{N}y_{i}  &  =\underbrace{\left(  \prod_{i=1}^{N-1}y_{i}\right)
}_{=\left[  t^{0}\right]  g}\cdot y_{N}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the factor}\\
\text{for }i=N\text{ from the product}%
\end{array}
\right) \\
&  =\left(  \left[  t^{0}\right]  g\right)  y_{N},
\end{align*}
we obtain $\left[  t^{0}\right]  f=\prod_{i=1}^{N}y_{i}$. Thus, Claim 1
\textbf{(c)} is proven.

We now have proven all three parts \textbf{(a)}, \textbf{(b)} and \textbf{(c)}
of Claim 1. Thus, Claim 1 holds.]

We have now proven Claim 1. As we have explained, this completes the induction
step. Thus, Lemma \ref{lem.ta+b.prod} is proven by induction.
\end{proof}
\end{verlong}

\begin{proof}
[Proof of Proposition \ref{prop.ta+b}.]Write the $n\times n$-matrix $A$ in the
form $A=\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$. Thus,
$a_{i,j}\in\mathbb{K}$ for every $\left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}$ (since $A\in\mathbb{K}^{n\times n}$).

Write the $n\times n$-matrix $B$ in the form $B=\left(  b_{i,j}\right)
_{1\leq i\leq n,\ 1\leq j\leq n}$. Thus, $b_{i,j}\in\mathbb{K}$ for every
$\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$ (since
$B\in\mathbb{K}^{n\times n}$).

For every $\sigma\in S_{n}$, define a polynomial $f_{\sigma}\in\mathbb{K}%
\left[  t\right]  $ by%
\begin{equation}
f_{\sigma}=\prod_{i=1}^{n}\left(  ta_{i,\sigma\left(  i\right)  }%
+b_{i,\sigma\left(  i\right)  }\right)  . \label{pf.prop.ta+b.fsigma=}%
\end{equation}
The following holds:

\begin{statement}
\textit{Fact 1:} For every $\sigma\in S_{n}$, the polynomial $f_{\sigma}$ is a
polynomial of degree $\leq n$.
\end{statement}

[\textit{Proof of Fact 1:} Let $\sigma\in S_{n}$. Then, Lemma
\ref{lem.ta+b.prod} \textbf{(a)} (applied to $a_{i,\sigma\left(  i\right)  }$,
$b_{i,\sigma\left(  i\right)  }$ and $f_{\sigma}$ instead of $x_{i}$, $y_{i}$
and $f$) shows that $f_{\sigma}$ is a polynomial of degree $\leq n$. This
proves Fact 1.]

\begin{vershort}
From $A=\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$ and
$B=\left(  b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$, we obtain
$tA+B=\left(  ta_{i,j}+b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$.
Hence,%
\begin{align*}
\det\left(  tA+B\right)   &  =\sum_{\sigma\in S_{n}}\left(  -1\right)
^{\sigma}\underbrace{\prod_{i=1}^{n}\left(  ta_{i,\sigma\left(  i\right)
}+b_{i,\sigma\left(  i\right)  }\right)  }_{\substack{=f_{\sigma}\\\text{(by
(\ref{pf.prop.ta+b.fsigma=}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{eq.det}), applied to }\mathbb{K}\left[  t\right]  \text{,
}tA+B\text{ and }ta_{i,j}+b_{i,j}\\
\text{instead of }\mathbb{K}\text{, }A\text{ and }a_{i,j}%
\end{array}
\right) \\
&  =\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}f_{\sigma}.
\end{align*}
Hence, $\det\left(  tA+B\right)  $ is a $\mathbb{K}$-linear combination of the
polynomials $f_{\sigma}$ for $\sigma\in S_{n}$. Since all of these polynomials
are polynomials of degree $\leq n$ (by Fact 1), we thus conclude that
$\det\left(  tA+B\right)  $ is a $\mathbb{K}$-linear combination of
polynomials of degree $\leq n$. Thus, $\det\left(  tA+B\right)  $ is itself a
polynomial of degree $\leq n$. This proves Proposition \ref{prop.ta+b}
\textbf{(a)}.
\end{vershort}

\begin{verlong}
We have%
\begin{align*}
t\underbrace{A}_{=\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}%
}+\underbrace{B}_{=\left(  b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}}
&  =\underbrace{t\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}%
}_{=\left(  ta_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}}+\left(
b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}\\
&  =\left(  ta_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}+\left(
b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}\\
&  =\left(  ta_{i,j}+b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}.
\end{align*}
Hence, (\ref{eq.det}) (applied to $\mathbb{K}\left[  t\right]  $, $tA+B$ and
$ta_{i,j}+b_{i,j}$ instead of $\mathbb{K}$, $A$ and $a_{i,j}$) shows that%
\[
\det\left(  tA+B\right)  =\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma
}\underbrace{\prod_{i=1}^{n}\left(  ta_{i,\sigma\left(  i\right)
}+b_{i,\sigma\left(  i\right)  }\right)  }_{\substack{=f_{\sigma}\\\text{(by
(\ref{pf.prop.ta+b.fsigma=}))}}}=\sum_{\sigma\in S_{n}}\left(  -1\right)
^{\sigma}f_{\sigma}.
\]
Hence, $\det\left(  tA+B\right)  $ is a $\mathbb{K}$-linear combination of the
polynomials $f_{\sigma}$ for $\sigma\in S_{n}$. Since all of these polynomials
are polynomials of degree $\leq n$ (by Fact 1), we thus conclude that
$\det\left(  tA+B\right)  $ is a $\mathbb{K}$-linear combination of
polynomials of degree $\leq n$. Thus, $\det\left(  tA+B\right)  $ is itself a
polynomial of degree $\leq n$ (since any $\mathbb{K}$-linear combination of
polynomials of degree $\leq n$ is itself a polynomial of degree $\leq n$).
This proves Proposition \ref{prop.ta+b} \textbf{(a)}.
\end{verlong}

\textbf{(b)} We have%
\begin{align*}
\left[  t^{0}\right]  \underbrace{\left(  \det\left(  tA+B\right)  \right)
}_{=\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}f_{\sigma}}  &  =\left[
t^{0}\right]  \left(  \sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma
}f_{\sigma}\right)  =\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma
}\underbrace{\left[  t^{0}\right]  f_{\sigma}}_{\substack{=\prod_{i=1}%
^{n}b_{i,\sigma\left(  i\right)  }\\\text{(by Lemma \ref{lem.ta+b.prod}
\textbf{(c)} (applied to}\\a_{i,\sigma\left(  i\right)  }\text{, }%
b_{i,\sigma\left(  i\right)  }\text{ and }f_{\sigma}\\\text{instead of }%
x_{i}\text{, }y_{i}\text{ and }f\text{))}}}\\
&  =\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{n}b_{i,\sigma\left(  i\right)  }.
\end{align*}
Comparing this with%
\[
\det B=\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{n}b_{i,\sigma\left(  i\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{eq.det}), applied to }B\text{ and }b_{i,j}\\
\text{instead of }A\text{ and }a_{i,j}%
\end{array}
\right)  ,
\]
we obtain $\left[  t^{0}\right]  \left(  \det\left(  tA+B\right)  \right)
=\det B$. This proves Proposition \ref{prop.ta+b} \textbf{(b)}.

\textbf{(c)} We have%
\begin{align*}
\left[  t^{n}\right]  \underbrace{\left(  \det\left(  tA+B\right)  \right)
}_{=\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}f_{\sigma}}  &  =\left[
t^{n}\right]  \left(  \sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma
}f_{\sigma}\right)  =\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma
}\underbrace{\left[  t^{n}\right]  f_{\sigma}}_{\substack{=\prod_{i=1}%
^{n}a_{i,\sigma\left(  i\right)  }\\\text{(by Lemma \ref{lem.ta+b.prod}
\textbf{(b)} (applied to}\\a_{i,\sigma\left(  i\right)  }\text{, }%
b_{i,\sigma\left(  i\right)  }\text{ and }f_{\sigma}\\\text{instead of }%
x_{i}\text{, }y_{i}\text{ and }f\text{))}}}\\
&  =\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{n}a_{i,\sigma\left(  i\right)  }.
\end{align*}
Comparing this with (\ref{eq.det}), we obtain $\left[  t^{n}\right]  \left(
\det\left(  tA+B\right)  \right)  =\det A$. This proves Proposition
\ref{prop.ta+b} \textbf{(c)}.
\end{proof}

\begin{vershort}
\begin{proof}
[Proof of Corollary \ref{cor.chiA}.]The definition of $\chi_{A}$ yields
\newline$\chi_{A}=\det\left(  \underbrace{tI_{n}-A}_{=tI_{n}+\left(
-A\right)  }\right)  =\det\left(  tI_{n}+\left(  -A\right)  \right)  $. Hence,
Corollary \ref{cor.chiA} follows from Proposition \ref{prop.ta+b} (applied to
$I_{n}$ and $-A$ instead of $A$ and $B$). (For part \textbf{(b)}, we need the
additional observation that $\det\left(  -A\right)  =\left(  -1\right)
^{n}\det A$.)
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Corollary \ref{cor.chiA}.]The definition of $\chi_{A}$ yields%
\[
\chi_{A}=\det\left(  \underbrace{tI_{n}-A}_{=tI_{n}+\left(  -A\right)
}\right)  =\det\left(  tI_{n}+\left(  -A\right)  \right)  .
\]


\textbf{(a)} Proposition \ref{prop.ta+b} \textbf{(a)} (applied to $I_{n}$ and
$-A$ instead of $A$ and $B$) yields that $\det\left(  tI_{n}+\left(
-A\right)  \right)  \in\mathbb{K}\left[  t\right]  $ is a polynomial of degree
$\leq n$ in $t$. In other words, $\chi_{A}$ is a polynomial of degree $\leq n$
in $t$ (since $\chi_{A}=\det\left(  tI_{n}+\left(  -A\right)  \right)  $).
This proves Corollary \ref{cor.chiA} \textbf{(a)}.

\textbf{(b)} Since $A$ is an $n\times n$-matrix, we have $\det\left(
-A\right)  =\left(  -1\right)  ^{n}\det A$\ \ \ \ \footnote{This is a
well-known fact, but let me give a \textit{proof} for the sake of
completeness:
\par
From \cite[Proposition 6.12]{detnotes}, we obtain that $\det\left(  \lambda
A\right)  =\lambda^{n}\det A$ for every $\lambda\in\mathbb{K}$. Applying this
to $\lambda=-1$, we obtain $\det\left(  \left(  -1\right)  A\right)  =\left(
-1\right)  ^{n}\det A$. Since $\left(  -1\right)  A=-A$, this rewrites as
$\det\left(  -A\right)  =\left(  -1\right)  ^{n}\det A$. Qed.}.

Proposition \ref{prop.ta+b} \textbf{(b)} (applied to $I_{n}$ and $-A$ instead
of $A$ and $B$) yields that \newline$\left[  t^{0}\right]  \left(  \det\left(
tI_{n}+\left(  -A\right)  \right)  \right)  =\det\left(  -A\right)  $. In
other words, $\left[  t^{0}\right]  \chi_{A}=\left(  -1\right)  ^{n}\det A$
(since $\chi_{A}=\det\left(  tI_{n}+\left(  -A\right)  \right)  $). Comparing
this with $\det\left(  -A\right)  =\left(  -1\right)  ^{n}\det A$, we obtain
$\left[  t^{0}\right]  \chi_{A}=\left(  -1\right)  ^{n}\det A$. This proves
Corollary \ref{cor.chiA} \textbf{(b)}.

\textbf{(c)} Proposition \ref{prop.ta+b} \textbf{(c)} (applied to $I_{n}$ and
$-A$ instead of $A$ and $B$) yields that $\left[  t^{n}\right]  \left(
\det\left(  tI_{n}+\left(  -A\right)  \right)  \right)  =\det\left(
I_{n}\right)  =1$. In other words, $\left[  t^{n}\right]  \chi_{A}=1$ (since
$\chi_{A}=\det\left(  tI_{n}+\left(  -A\right)  \right)  $). This proves
Corollary \ref{cor.chiA} \textbf{(c)}.
\end{proof}
\end{verlong}

Let me state one more trivial observation as a corollary:

\begin{corollary}
\label{cor.CH.chiA}Let $n\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times n}$. For
every $j\in\mathbb{Z}$, define an element $c_{j}\in\mathbb{K}$ by
$c_{j}=\left[  t^{n-j}\right]  \chi_{A}$. Then, $\chi_{A}=\sum_{k=0}%
^{n}c_{n-k}t^{k}$.
\end{corollary}

\begin{vershort}
\begin{proof}
[Proof of Corollary \ref{cor.CH.chiA}.]For every $k\in\mathbb{Z}$, the
definition of $c_{n-k}$ yields%
\begin{equation}
c_{n-k}=\left[  t^{n-\left(  n-k\right)  }\right]  \chi_{A}=\left[
t^{k}\right]  \chi_{A}. \label{pf.cor.CH.chiA.1}%
\end{equation}


We know that $\chi_{A}\in\mathbb{K}\left[  t\right]  $ is a polynomial of
degree $\leq n$ in $t$ (by Corollary \ref{cor.chiA} \textbf{(a)}). Hence,%
\[
\chi_{A}=\sum_{k=0}^{n}\underbrace{\left(  \left[  t^{k}\right]  \chi
_{A}\right)  }_{\substack{=c_{n-k}\\\text{(by (\ref{pf.cor.CH.chiA.1}))}%
}}t^{k}=\sum_{k=0}^{n}c_{n-k}t^{k}.
\]
This proves Corollary \ref{cor.CH.chiA}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Corollary \ref{cor.CH.chiA}.]For every $k\in\mathbb{Z}$, we have%
\begin{align}
c_{n-k}  &  =\underbrace{\left[  t^{n-\left(  n-k\right)  }\right]
}_{=\left[  t^{k}\right]  }\chi_{A}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }c_{n-k}\right) \nonumber\\
&  =\left[  t^{k}\right]  \chi_{A}. \label{pf.lem.CH.D.a.cn-k}%
\end{align}


We know that $\chi_{A}\in\mathbb{K}\left[  t\right]  $ is a polynomial of
degree $\leq n$ in $t$ (by Corollary \ref{cor.chiA} \textbf{(a)}). Hence,
\begin{equation}
\left[  t^{k}\right]  \chi_{A}=0\ \ \ \ \ \ \ \ \ \ \text{for every integer
}k>n. \label{pf.lem.CH.D.a.1}%
\end{equation}


Now, recall that every polynomial $f\in\mathbb{K}\left[  t\right]  $ satisfies
$f=\sum_{k\in\mathbb{N}}\left(  \left[  t^{k}\right]  f\right)  t^{k}$.
Applying this to $f=\chi_{A}$, we obtain%
\begin{align*}
\chi_{A}  &  =\sum_{k\in\mathbb{N}}\left(  \left[  t^{k}\right]  \chi
_{A}\right)  t^{k}=\sum_{k=0}^{n}\left(  \left[  t^{k}\right]  \chi
_{A}\right)  t^{k}+\sum_{k=n+1}^{\infty}\underbrace{\left(  \left[
t^{k}\right]  \chi_{A}\right)  }_{\substack{=0\\\text{(by
(\ref{pf.lem.CH.D.a.1})}\\\text{(since }k\geq n+1>n\text{))}}}t^{k}\\
&  =\sum_{k=0}^{n}\left(  \left[  t^{k}\right]  \chi_{A}\right)
t^{k}+\underbrace{\sum_{k=n+1}^{\infty}0t^{k}}_{=0}=\sum_{k=0}^{n}%
\underbrace{\left(  \left[  t^{k}\right]  \chi_{A}\right)  }%
_{\substack{=c_{n-k}\\\text{(by (\ref{pf.lem.CH.D.a.cn-k}))}}}t^{k}=\sum
_{k=0}^{n}c_{n-k}t^{k}.
\end{align*}
This proves Corollary \ref{cor.CH.chiA}.
\end{proof}
\end{verlong}

\subsection{Reminders on the adjugate}

Let us now briefly introduce the adjugate of a matrix and state some of its properties.

We first recall the definitions (mostly quoting them from \cite[Chapter
6]{detnotes}):

\begin{definition}
\label{def.submatrix}Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $A=\left(
a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}$ be an $n\times m$-matrix.
Let $i_{1},i_{2},\ldots,i_{u}$ be some elements of $\left\{  1,2,\ldots
,n\right\}  $; let $j_{1},j_{2},\ldots,j_{v}$ be some elements of $\left\{
1,2,\ldots,m\right\}  $. Then, we define $\operatorname*{sub}\nolimits_{i_{1}%
,i_{2},\ldots,i_{u}}^{j_{1},j_{2},\ldots,j_{v}}A$ to be the $u\times v$-matrix
$\left(  a_{i_{x},j_{y}}\right)  _{1\leq x\leq u,\ 1\leq y\leq v}$.
\end{definition}

\begin{definition}
\label{def.hat-omit}Let $n\in\mathbb{N}$. Let $a_{1},a_{2},\ldots,a_{n}$ be
$n$ objects. Let $i\in\left\{  1,2,\ldots,n\right\}  $. Then, $\left(
a_{1},a_{2},\ldots,\widehat{a_{i}},\ldots,a_{n}\right)  $ shall mean the list
$\left(  a_{1},a_{2},\ldots,a_{i-1},a_{i+1},a_{i+2},\ldots,a_{n}\right)  $
(that is, the list $\left(  a_{1},a_{2},\ldots,a_{n}\right)  $ with its $i$-th
entry removed). (Thus, the \textquotedblleft hat\textquotedblright\ over the
$a_{i}$ means that this $a_{i}$ is being omitted from the list.)

For example, $\left(  1^{2},2^{2},\ldots,\widehat{5}^{2},\ldots,8^{2}\right)
=\left(  1^{2},2^{2},3^{2},4^{2},6^{2},7^{2},8^{2}\right)  $.
\end{definition}

\begin{definition}
\label{def.submatrix.minor}Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $A$
be an $n\times m$-matrix. For every $i\in\left\{  1,2,\ldots,n\right\}  $ and
$j\in\left\{  1,2,\ldots,m\right\}  $, we let $A_{\sim i,\sim j}$ be the
$\left(  n-1\right)  \times\left(  m-1\right)  $-matrix $\operatorname*{sub}%
\nolimits_{1,2,\ldots,\widehat{i},\ldots,n}^{1,2,\ldots,\widehat{j},\ldots
,m}A$. (Thus, $A_{\sim i,\sim j}$ is the matrix obtained from $A$ by crossing
out the $i$-th row and the $j$-th column.)
\end{definition}

\begin{definition}
\label{def.adj}Let $n\in\mathbb{N}$. Let $A$ be an $n\times n$-matrix. We
define a new $n\times n$-matrix $\operatorname*{adj}A$ by%
\[
\operatorname*{adj}A=\left(  \left(  -1\right)  ^{i+j}\det\left(  A_{\sim
j,\sim i}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq n}.
\]
This matrix $\operatorname*{adj}A$ is called the \textit{adjugate} of the
matrix $A$.
\end{definition}

The main property of the adjugate is the following fact:

\begin{theorem}
\label{thm.adj.inverse}Let $n\in\mathbb{N}$. Let $A$ be an $n\times n$-matrix.
Then,%
\[
A\cdot\operatorname*{adj}A=\operatorname*{adj}A\cdot A=\det A\cdot I_{n}.
\]
(Recall that $I_{n}$ denotes the $n\times n$ identity matrix. Expressions such
as $\operatorname*{adj}A\cdot A$ and $\det A\cdot I_{n}$ have to be understood
as $\left(  \operatorname*{adj}A\right)  \cdot A$ and $\left(  \det A\right)
\cdot I_{n}$, respectively.)
\end{theorem}

Theorem \ref{thm.adj.inverse} appears in almost any text on linear algebra
that considers the adjugate; for example, it appears in \cite[Chapter Four,
Section III, Theorem 1.9]{Heffer14}, in \cite[Proposition 2.38]{Knapp2016}, in
\cite[Theorem 4.11]{Willia89} and in \cite[Theorem 6.100]{detnotes}. (Again,
most of these sources only state it in the case when $\mathbb{K}$ is a field,
but the proofs given apply in all generality. Different texts use different
notations. The source that is closest to my notations here is \cite{detnotes},
since Theorem \ref{thm.adj.inverse} above is a verbatim copy of \cite[Theorem
6.100]{detnotes}.)

Let us state a simple fact:

\begin{lemma}
\label{lem.adj.linear}Let $n\in\mathbb{N}$. Let $u$ and $v$ be two elements of
$\left\{  1,2,\ldots,n\right\}  $. Let $\lambda$ and $\mu$ be two elements of
$\mathbb{K}$. Let $A$ and $B$ be two $n\times n$-matrices. Then,%
\[
\left(  \lambda A+\mu B\right)  _{\sim u,\sim v}=\lambda A_{\sim u,\sim v}+\mu
B_{\sim u,\sim v}.
\]

\end{lemma}

\begin{vershort}
\begin{proof}
[Proof of Lemma \ref{lem.adj.linear}.]Obvious.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.adj.linear}.]Let $\left(  p_{1},p_{2},\ldots
,p_{n-1}\right)  $ be the $\left(  n-1\right)  $-tuple $\left(  1,2,\ldots
,\widehat{u},\ldots,n\right)  $. Thus, $\left(  p_{1},p_{2},\ldots
,p_{n-1}\right)  =\left(  1,2,\ldots,\widehat{u},\ldots,n\right)  $.

Let $\left(  q_{1},q_{2},\ldots,q_{n-1}\right)  $ be the $\left(  n-1\right)
$-tuple $\left(  1,2,\ldots,\widehat{v},\ldots,n\right)  $. Thus, $\left(
q_{1},q_{2},\ldots,q_{n-1}\right)  =\left(  1,2,\ldots,\widehat{v}%
,\ldots,n\right)  $.

If $C=\left(  c_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$ is any
$n\times n$-matrix, then
\begin{equation}
C_{\sim u,\sim v}=\left(  c_{p_{i},q_{j}}\right)  _{1\leq i\leq n-1,\ 1\leq
j\leq n-1} \label{pf.lem.adj.linear.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.adj.linear.1}):} Let $C=\left(
c_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$ be any $n\times n$-matrix.
The definition of $C_{\sim u,\sim v}$ yields
\begin{align*}
C_{\sim u,\sim v}  &  =\operatorname*{sub}\nolimits_{1,2,\ldots,\widehat{u}%
,\ldots,n}^{1,2,\ldots,\widehat{v},\ldots,n}C=\operatorname*{sub}%
\nolimits_{p_{1},p_{2},\ldots,p_{n-1}}^{1,2,\ldots,\widehat{v},\ldots,n}C\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(  1,2,\ldots,\widehat{u}%
,\ldots,n\right)  =\left(  p_{1},p_{2},\ldots,p_{n-1}\right)  \right) \\
&  =\operatorname*{sub}\nolimits_{p_{1},p_{2},\ldots,p_{n-1}}^{q_{1}%
,q_{2},\ldots,q_{n-1}}C\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(
1,2,\ldots,\widehat{v},\ldots,n\right)  =\left(  q_{1},q_{2},\ldots
,q_{n-1}\right)  \right) \\
&  =\left(  c_{p_{x},q_{y}}\right)  _{1\leq x\leq n-1,\ 1\leq y\leq n-1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }\operatorname*{sub}%
\nolimits_{p_{1},p_{2},\ldots,p_{n-1}}^{q_{1},q_{2},\ldots,q_{n-1}}C\text{,
since }C=\left(  c_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}\right) \\
&  =\left(  c_{p_{i},q_{j}}\right)  _{1\leq i\leq n-1,\ 1\leq j\leq n-1}%
\end{align*}
(here, we have renamed the index $\left(  x,y\right)  $ as $\left(
i,j\right)  $). This proves (\ref{pf.lem.adj.linear.1}).}.

Write the $n\times n$-matrix $A$ in the form $A=\left(  a_{i,j}\right)
_{1\leq i\leq n,\ 1\leq j\leq n}$. Write the $n\times n$-matrix $B$ in the
form $B=\left(  b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$.

From (\ref{pf.lem.adj.linear.1}) (applied to $A$ and $a_{i,j}$ instead of $C$
and $c_{i,j}$), we obtain $A_{\sim u,\sim v}=\left(  a_{p_{i},q_{j}}\right)
_{1\leq i\leq n-1,\ 1\leq j\leq n-1}$. From (\ref{pf.lem.adj.linear.1})
(applied to $B$ and $b_{i,j}$ instead of $C$ and $c_{i,j}$), we obtain
$B_{\sim u,\sim v}=\left(  b_{p_{i},q_{j}}\right)  _{1\leq i\leq n-1,\ 1\leq
j\leq n-1}$.

But
\begin{align*}
&  \lambda\underbrace{A}_{=\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq
j\leq n}}+\mu\underbrace{B}_{=\left(  b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq
j\leq n}}\\
&  =\underbrace{\lambda\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n}}_{=\left(  \lambda a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}%
}+\underbrace{\mu\left(  b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}%
}_{=\left(  \mu b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}}=\left(
\lambda a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}+\left(  \mu
b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}\\
&  =\left(  \lambda a_{i,j}+\mu b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n}.
\end{align*}
Hence, (\ref{pf.lem.adj.linear.1}) (applied to $\lambda A+\mu B$ and $\lambda
a_{i,j}+\mu b_{i,j}$ instead of $C$ and $c_{i,j}$) shows that%
\[
\left(  \lambda A+\mu B\right)  _{\sim u,\sim v}=\left(  \lambda
a_{p_{i},q_{j}}+\mu b_{p_{i},q_{j}}\right)  _{1\leq i\leq n-1,\ 1\leq j\leq
n-1}.
\]
Comparing this with%
\begin{align*}
&  \lambda\underbrace{A_{\sim u,\sim v}}_{=\left(  a_{p_{i},q_{j}}\right)
_{1\leq i\leq n-1,\ 1\leq j\leq n-1}}+\mu\underbrace{B_{\sim u,\sim v}%
}_{=\left(  b_{p_{i},q_{j}}\right)  _{1\leq i\leq n-1,\ 1\leq j\leq n-1}}\\
&  =\underbrace{\lambda\left(  a_{p_{i},q_{j}}\right)  _{1\leq i\leq
n-1,\ 1\leq j\leq n-1}}_{=\left(  \lambda a_{p_{i},q_{j}}\right)  _{1\leq
i\leq n-1,\ 1\leq j\leq n-1}}+\underbrace{\mu\left(  b_{p_{i},q_{j}}\right)
_{1\leq i\leq n-1,\ 1\leq j\leq n-1}}_{=\left(  \mu b_{p_{i},q_{j}}\right)
_{1\leq i\leq n-1,\ 1\leq j\leq n-1}}\\
&  =\left(  \lambda a_{p_{i},q_{j}}\right)  _{1\leq i\leq n-1,\ 1\leq j\leq
n-1}+\left(  \mu b_{p_{i},q_{j}}\right)  _{1\leq i\leq n-1,\ 1\leq j\leq
n-1}\\
&  =\left(  \lambda a_{p_{i},q_{j}}+\mu b_{p_{i},q_{j}}\right)  _{1\leq i\leq
n-1,\ 1\leq j\leq n-1},
\end{align*}
we obtain $\left(  \lambda A+\mu B\right)  _{\sim u,\sim v}=\lambda A_{\sim
u,\sim v}+\mu B_{\sim u,\sim v}$. This proves Lemma \ref{lem.adj.linear}.
\end{proof}
\end{verlong}

Next, we prove a crucial, if simple, result:

\begin{proposition}
\label{prop.adj.poly}Let $n\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times n}$ be
an $n\times n$-matrix. Then, there exist $n$ matrices $D_{0},D_{1}%
,\ldots,D_{n-1}$ in $\mathbb{K}^{n\times n}$ such that%
\[
\operatorname*{adj}\left(  tI_{n}-A\right)  =\sum_{k=0}^{n-1}t^{k}%
D_{k}\ \ \ \ \ \ \ \ \ \ \text{in }\left(  \mathbb{K}\left[  t\right]
\right)  ^{n\times n}.
\]
(Here, of course, the matrix $D_{k}$ on the right hand side is understood as
an element of $\left(  \mathbb{K}\left[  t\right]  \right)  ^{n\times n}$.)
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.adj.poly}.]Fix $\left(  u,v\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}$. Then, Proposition \ref{prop.ta+b}
\textbf{(a)} (applied to $n-1$, $\left(  I_{n}\right)  _{\sim u,\sim v}$ and
$\left(  -A\right)  _{\sim u,\sim v}$ instead of $n$, $A$ and $B$) shows that
\newline$\det\left(  t\left(  I_{n}\right)  _{\sim u,\sim v}+\left(
-A\right)  _{\sim u,\sim v}\right)  \in\mathbb{K}\left[  t\right]  $ is a
polynomial of degree $\leq n-1$ in $t$. In other words, there exists an
$n$-tuple $\left(  d_{u,v,0},d_{u,v,1},\ldots,d_{u,v,n-1}\right)
\in\mathbb{K}^{n}$ such that
\[
\det\left(  t\left(  I_{n}\right)  _{\sim u,\sim v}+\left(  -A\right)  _{\sim
u,\sim v}\right)  =\sum_{k=0}^{n-1}d_{u,v,k}t^{k}.
\]
Consider this $\left(  d_{u,v,0},d_{u,v,1},\ldots,d_{u,v,n-1}\right)  $. But
Lemma \ref{lem.adj.linear} (applied to $\mathbb{K}\left[  t\right]  $, $t$,
$1$, $I_{n}$ and $-A$ instead of $\mathbb{K}$, $\lambda$, $\mu$, $A$ and $B$)
yields $\left(  tI_{n}-A\right)  _{\sim u,\sim v}=t\left(  I_{n}\right)
_{\sim u,\sim v}+\left(  -A\right)  _{\sim u,\sim v}$ (after some
simplifications). Thus,%
\begin{equation}
\det\left(  \left(  tI_{n}-A\right)  _{\sim u,\sim v}\right)  =\det\left(
t\left(  I_{n}\right)  _{\sim u,\sim v}+\left(  -A\right)  _{\sim u,\sim
v}\right)  =\sum_{k=0}^{n-1}d_{u,v,k}t^{k}. \label{pf.prop.adj.poly.short.2}%
\end{equation}


Now, forget that we fixed $\left(  u,v\right)  $. Thus, for every $\left(
u,v\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$, we have constructed an
$n$-tuple $\left(  d_{u,v,0},d_{u,v,1},\ldots,d_{u,v,n-1}\right)
\in\mathbb{K}^{n}$ satisfying (\ref{pf.prop.adj.poly.short.2}).

Now, the definition of $\operatorname*{adj}\left(  tI_{n}-A\right)  $ yields
\begin{align*}
\operatorname*{adj}\left(  tI_{n}-A\right)   &  =\left(  \left(  -1\right)
^{i+j}\underbrace{\det\left(  \left(  tI_{n}-A\right)  _{\sim j,\sim
i}\right)  }_{\substack{=\sum_{k=0}^{n-1}d_{j,i,k}t^{k}\\\text{(by
(\ref{pf.prop.adj.poly.short.2}), applied to }\left(  u,v\right)  =\left(
j,i\right)  \text{)}}}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}\\
&  =\left(  \underbrace{\left(  -1\right)  ^{i+j}\sum_{k=0}^{n-1}%
d_{j,i,k}t^{k}}_{=\sum_{k=0}^{n-1}t^{k}\left(  -1\right)  ^{i+j}d_{j,i,k}%
}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}=\left(  \sum_{k=0}^{n-1}%
t^{k}\left(  -1\right)  ^{i+j}d_{j,i,k}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n}.
\end{align*}
Comparing this with%
\[
\sum_{k=0}^{n-1}t^{k}\left(  \left(  -1\right)  ^{i+j}d_{j,i,k}\right)
_{1\leq i\leq n,\ 1\leq j\leq n}=\left(  \sum_{k=0}^{n-1}t^{k}\left(
-1\right)  ^{i+j}d_{j,i,k}\right)  _{1\leq i\leq n,\ 1\leq j\leq n},
\]
we obtain $\operatorname*{adj}\left(  tI_{n}-A\right)  =\sum_{k=0}^{n-1}%
t^{k}\left(  \left(  -1\right)  ^{i+j}d_{j,i,k}\right)  _{1\leq i\leq
n,\ 1\leq j\leq n}$. Hence, there exist $n$ matrices $D_{0},D_{1}%
,\ldots,D_{n-1}$ in $\mathbb{K}^{n\times n}$ such that%
\[
\operatorname*{adj}\left(  tI_{n}-A\right)  =\sum_{k=0}^{n-1}t^{k}%
D_{k}\ \ \ \ \ \ \ \ \ \ \text{in }\left(  \mathbb{K}\left[  t\right]
\right)  ^{n\times n}%
\]
(namely, $D_{k}=\left(  \left(  -1\right)  ^{i+j}d_{j,i,k}\right)  _{1\leq
i\leq n,\ 1\leq j\leq n}$ for every $k\in\left\{  0,1,\ldots,n-1\right\}  $).
This proves Proposition \ref{prop.adj.poly}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.adj.poly}.]Fix $\left(  u,v\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}$. We shall define an $n$-tuple
\newline$\left(  d_{u,v,0},d_{u,v,1},\ldots,d_{u,v,n-1}\right)  \in
\mathbb{K}^{n}$ satisfying%
\begin{equation}
\det\left(  \left(  tI_{n}-A\right)  _{\sim u,\sim v}\right)  =\sum
_{k=0}^{n-1}d_{u,v,k}t^{k}. \label{pf.prop.adj.poly.d-goal}%
\end{equation}


Indeed, $u$ and $v$ are two elements of $\left\{  1,2,\ldots,n\right\}  $
(since $\left(  u,v\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$). Hence,
Lemma \ref{lem.adj.linear} (applied to $\mathbb{K}\left[  t\right]  $, $t$,
$1$, $I_{n}$ and $-A$ instead of $\mathbb{K}$, $\lambda$, $\mu$, $A$ and $B$)
yields
\[
\left(  tI_{n}+1\left(  -A\right)  \right)  _{\sim u,\sim v}=t\left(
I_{n}\right)  _{\sim u,\sim v}+1\left(  -A\right)  _{\sim u,\sim v}=t\left(
I_{n}\right)  _{\sim u,\sim v}+\left(  -A\right)  _{\sim u,\sim v}.
\]
Since $tI_{n}+1\left(  -A\right)  =tI_{n}+\left(  -A\right)  =tI_{n}-A$, this
rewrites as
\begin{equation}
\left(  tI_{n}-A\right)  _{\sim u,\sim v}=t\left(  I_{n}\right)  _{\sim u,\sim
v}+\left(  -A\right)  _{\sim u,\sim v}. \label{pf.prop.adj.poly.3}%
\end{equation}


But $u\in\left\{  1,2,\ldots,n\right\}  $, so that $1\leq u\leq n$ and thus
$n\geq1$. Hence, $n-1\in\mathbb{N}$. Thus, Proposition \ref{prop.ta+b}
\textbf{(a)} (applied to $n-1$, $\left(  I_{n}\right)  _{\sim u,\sim v}$ and
$\left(  -A\right)  _{\sim u,\sim v}$ instead of $n$, $A$ and $B$) shows that
$\det\left(  t\left(  I_{n}\right)  _{\sim u,\sim v}+\left(  -A\right)  _{\sim
u,\sim v}\right)  \in\mathbb{K}\left[  t\right]  $ is a polynomial of degree
$\leq n-1$ in $t$. In other words, there exists some $n$-tuple $\left(
b_{0},b_{1},\ldots,b_{n-1}\right)  \in\mathbb{K}^{n}$ such that $\det\left(
t\left(  I_{n}\right)  _{\sim u,\sim v}+\left(  -A\right)  _{\sim u,\sim
v}\right)  =\sum_{k=0}^{n-1}b_{k}t^{k}$. Consider this $\left(  b_{0}%
,b_{1},\ldots,b_{n-1}\right)  $.

Taking determinants on both sides of (\ref{pf.prop.adj.poly.3}), we obtain%
\begin{equation}
\det\left(  \left(  tI_{n}-A\right)  _{\sim u,\sim v}\right)  =\det\left(
t\left(  I_{n}\right)  _{\sim u,\sim v}+\left(  -A\right)  _{\sim u,\sim
v}\right)  =\sum_{k=0}^{n-1}b_{k}t^{k}. \label{pf.prop.adj.poly.5}%
\end{equation}
Now, define an $n$-tuple $\left(  d_{u,v,0},d_{u,v,1},\ldots,d_{u,v,n-1}%
\right)  \in\mathbb{K}^{n}$ by $\left(  d_{u,v,0},d_{u,v,1},\ldots
,d_{u,v,n-1}\right)  =\left(  b_{0},b_{1},\ldots,b_{n-1}\right)  $. Thus,
$d_{u,v,k}=b_{k}$ for every $k\in\left\{  0,1,\ldots,n-1\right\}  $. Hence,
(\ref{pf.prop.adj.poly.5}) becomes%
\[
\det\left(  \left(  tI_{n}-A\right)  _{\sim u,\sim v}\right)  =\sum
_{k=0}^{n-1}\underbrace{b_{k}}_{=d_{u,v,k}}t^{k}=\sum_{k=0}^{n-1}%
d_{u,v,k}t^{k}.
\]
In other words, (\ref{pf.prop.adj.poly.d-goal}) is satisfied.

Now, forget that we have fixed $\left(  u,v\right)  $. Thus, for every
$\left(  u,v\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$, we have defined
an $n$-tuple $\left(  d_{u,v,0},d_{u,v,1},\ldots,d_{u,v,n-1}\right)
\in\mathbb{K}^{n}$ satisfying (\ref{pf.prop.adj.poly.d-goal}).

Now, the definition of $\operatorname*{adj}\left(  tI_{n}-A\right)  $ yields
\begin{align*}
\operatorname*{adj}\left(  tI_{n}-A\right)   &  =\left(  \left(  -1\right)
^{i+j}\underbrace{\det\left(  \left(  tI_{n}-A\right)  _{\sim j,\sim
i}\right)  }_{\substack{=\sum_{k=0}^{n-1}d_{j,i,k}t^{k}\\\text{(by
(\ref{pf.prop.adj.poly.d-goal}), applied}\\\text{to }\left(  u,v\right)
=\left(  j,i\right)  \text{)}}}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}\\
&  =\left(  \underbrace{\left(  -1\right)  ^{i+j}\sum_{k=0}^{n-1}%
d_{j,i,k}t^{k}}_{\substack{=\sum_{k=0}^{n-1}\left(  -1\right)  ^{i+j}%
d_{j,i,k}t^{k}\\=\sum_{k=0}^{n-1}t^{k}\left(  -1\right)  ^{i+j}d_{j,i,k}%
}}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}=\left(  \sum_{k=0}^{n-1}%
t^{k}\left(  -1\right)  ^{i+j}d_{j,i,k}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n}.
\end{align*}
Comparing this with%
\begin{align*}
\sum_{k=0}^{n-1}\underbrace{t^{k}\left(  \left(  -1\right)  ^{i+j}%
d_{j,i,k}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}}_{=\left(  t^{k}\left(
-1\right)  ^{i+j}d_{j,i,k}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}}  &
=\sum_{k=0}^{n-1}\left(  t^{k}\left(  -1\right)  ^{i+j}d_{j,i,k}\right)
_{1\leq i\leq n,\ 1\leq j\leq n}\\
&  =\left(  \sum_{k=0}^{n-1}t^{k}\left(  -1\right)  ^{i+j}d_{j,i,k}\right)
_{1\leq i\leq n,\ 1\leq j\leq n},
\end{align*}
we obtain $\operatorname*{adj}\left(  tI_{n}-A\right)  =\sum_{k=0}^{n-1}%
t^{k}\left(  \left(  -1\right)  ^{i+j}d_{j,i,k}\right)  _{1\leq i\leq
n,\ 1\leq j\leq n}$. Hence, there exist $n$ matrices $D_{0},D_{1}%
,\ldots,D_{n-1}$ in $\mathbb{K}^{n\times n}$ such that%
\[
\operatorname*{adj}\left(  tI_{n}-A\right)  =\sum_{k=0}^{n-1}t^{k}%
D_{k}\ \ \ \ \ \ \ \ \ \ \text{in }\left(  \mathbb{K}\left[  t\right]
\right)  ^{n\times n}%
\]
(namely, $D_{k}=\left(  \left(  -1\right)  ^{i+j}d_{j,i,k}\right)  _{1\leq
i\leq n,\ 1\leq j\leq n}$ for every $k\in\left\{  0,1,\ldots,n-1\right\}  $).
This proves Proposition \ref{prop.adj.poly}.
\end{proof}
\end{verlong}

\begin{noncompile}
For any two objects $i$ and $j$, define an element $\delta_{i,j}\in\mathbb{K}$
by $\delta_{i,j}=%
\begin{cases}
1, & \text{if }i=j;\\
0, & \text{if }i\neq j
\end{cases}
$. Then, $I_{n}=\left(  \delta_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$
(by the definition of the identity matrix $I_{n}$). Hence, $t\underbrace{I_{n}%
}_{=\left(  \delta_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}}=t\left(
\delta_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}=\left(  t\delta
_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$.
\end{noncompile}

\subsection{Polynomials with matrix entries: a trivial lemma}

\begin{lemma}
\label{lem.polymat.=}Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $\left(
B_{0},B_{1},\ldots,B_{m}\right)  \in\left(  \mathbb{K}^{n\times n}\right)
^{m+1}$ and $\left(  C_{0},C_{1},\ldots,C_{m}\right)  \in\left(
\mathbb{K}^{n\times n}\right)  ^{m+1}$ be two $\left(  m+1\right)  $-tuples of
matrices in $\mathbb{K}^{n\times n}$. Assume that%
\[
\sum_{k=0}^{m}t^{k}B_{k}=\sum_{k=0}^{m}t^{k}C_{k}\ \ \ \ \ \ \ \ \ \ \text{in
}\left(  \mathbb{K}\left[  t\right]  \right)  ^{n\times n}.
\]
Then, $B_{k}=C_{k}$ for every $k\in\left\{  0,1,\ldots,m\right\}  $.
\end{lemma}

\begin{vershort}
\begin{proof}
[Proof of Lemma \ref{lem.polymat.=}.]For every $k\in\left\{  0,1,\ldots
,m\right\}  $, write the matrix $B_{k}\in\mathbb{K}^{n\times n}$ in the form
$B_{k}=\left(  b_{k,i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$, and write
the matrix $C_{k}\in\mathbb{K}^{n\times n}$ in the form $C_{k}=\left(
c_{k,i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$.

Now, $\sum_{k=0}^{m}t^{k}B_{k}=\left(  \sum_{k=0}^{m}t^{k}b_{k,i,j}\right)
_{1\leq i\leq n,\ 1\leq j\leq n}$ (since $B_{k}=\left(  b_{k,i,j}\right)
_{1\leq i\leq n,\ 1\leq j\leq n}$ for every $k\in\left\{  0,1,\ldots
,m\right\}  $). Similarly, $\sum_{k=0}^{m}t^{k}C_{k}=\left(  \sum_{k=0}%
^{m}t^{k}c_{k,i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$. Thus,%
\[
\left(  \sum_{k=0}^{m}t^{k}b_{k,i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n}=\sum_{k=0}^{m}t^{k}B_{k}=\sum_{k=0}^{m}t^{k}C_{k}=\left(  \sum_{k=0}%
^{m}t^{k}c_{k,i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}.
\]
In other words,%
\[
\sum_{k=0}^{m}t^{k}b_{k,i,j}=\sum_{k=0}^{m}t^{k}c_{k,i,j}%
\]
for every $\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$.
Comparing coefficients on both sides of this equality, we obtain%
\[
b_{k,i,j}=c_{k,i,j}%
\]
for every $k\in\left\{  0,1,\ldots,m\right\}  $ for every $\left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}$. Now, every $k\in\left\{
0,1,\ldots,m\right\}  $ satisfies%
\[
B_{k}=\left(  \underbrace{b_{k,i,j}}_{=c_{k,i,j}}\right)  _{1\leq i\leq
n,\ 1\leq j\leq n}=\left(  c_{k,i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n}=C_{k}.
\]
This proves Lemma \ref{lem.polymat.=}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.polymat.=}.]For every $k\in\left\{  0,1,\ldots
,m\right\}  $, write the matrix $B_{k}\in\mathbb{K}^{n\times n}$ in the form
$B_{k}=\left(  b_{k,i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$. For every
$k\in\left\{  0,1,\ldots,m\right\}  $, write the matrix $C_{k}\in
\mathbb{K}^{n\times n}$ in the form $C_{k}=\left(  c_{k,i,j}\right)  _{1\leq
i\leq n,\ 1\leq j\leq n}$.

We have%
\begin{align*}
\sum_{k=0}^{m}t^{k}\underbrace{B_{k}}_{=\left(  b_{k,i,j}\right)  _{1\leq
i\leq n,\ 1\leq j\leq n}}  &  =\sum_{k=0}^{m}\underbrace{t^{k}\left(
b_{k,i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}}_{=\left(  t^{k}%
b_{k,i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}}=\sum_{k=0}^{m}\left(
t^{k}b_{k,i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}\\
&  =\left(  \sum_{k=0}^{m}t^{k}b_{k,i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n}.
\end{align*}
Comparing this with%
\begin{align*}
\sum_{k=0}^{m}t^{k}B_{k}  &  =\sum_{k=0}^{m}t^{k}\underbrace{C_{k}}_{=\left(
c_{k,i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}}=\sum_{k=0}^{m}%
\underbrace{t^{k}\left(  c_{k,i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}%
}_{=\left(  t^{k}c_{k,i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}}\\
&  =\sum_{k=0}^{m}\left(  t^{k}c_{k,i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n}=\left(  \sum_{k=0}^{m}t^{k}c_{k,i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n},
\end{align*}
we obtain%
\[
\left(  \sum_{k=0}^{m}t^{k}b_{k,i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n}=\left(  \sum_{k=0}^{m}t^{k}c_{k,i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n}.
\]
In other words,%
\begin{equation}
\sum_{k=0}^{m}t^{k}b_{k,i,j}=\sum_{k=0}^{m}t^{k}c_{k,i,j}
\label{pf.lem.polymat.=.1}%
\end{equation}
for every $\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$.

Now, fix $\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$. Then,%
\begin{align*}
\sum_{k=0}^{m}\underbrace{b_{k,i,j}t^{k}}_{=t^{k}b_{k,i,j}}  &  =\sum
_{k=0}^{m}t^{k}b_{k,i,j}=\sum_{k=0}^{m}\underbrace{t^{k}c_{k,i,j}}%
_{=c_{k,i,j}t^{k}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.polymat.=.1})}\right) \\
&  =\sum_{k=0}^{m}c_{k,i,j}t^{k}%
\end{align*}
in $\mathbb{K}\left[  t\right]  $. Comparing coefficients on both sides of
this equality, we obtain%
\begin{equation}
b_{k,i,j}=c_{k,i,j}\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\left\{
0,1,\ldots,m\right\}  . \label{pf.lem.polymat.=.2}%
\end{equation}


Now, forget that we fixed $\left(  i,j\right)  $. We thus have proven
(\ref{pf.lem.polymat.=.2}) for every $\left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}$ and every $k\in\left\{  0,1,\ldots,m\right\}  $.

Now, every $k\in\left\{  0,1,\ldots,m\right\}  $ satisfies%
\[
B_{k}=\left(  \underbrace{b_{k,i,j}}_{\substack{=c_{k,i,j}\\\text{(by
(\ref{pf.lem.polymat.=.2}))}}}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n}=\left(  c_{k,i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}=C_{k}%
\]
(since $C_{k}=\left(  c_{k,i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$).
This proves Lemma \ref{lem.polymat.=}.
\end{proof}
\end{verlong}

\subsection{Proof of the Cayley-Hamilton theorem}

We are now fully prepared for the proof of the Cayley-Hamilton theorem.
However, we are going to organize the crucial part of this proof as a lemma,
so that we can use it later in our proof of the trace Cayley-Hamilton theorem.

\begin{lemma}
\label{lem.CH.D}Let $n\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times n}$. For
every $j\in\mathbb{Z}$, define an element $c_{j}\in\mathbb{K}$ by
$c_{j}=\left[  t^{n-j}\right]  \chi_{A}$.

Let $D_{0},D_{1},\ldots,D_{n-1}$ be $n$ matrices in $\mathbb{K}^{n\times n}$
such that%
\begin{equation}
\operatorname*{adj}\left(  tI_{n}-A\right)  =\sum_{k=0}^{n-1}t^{k}%
D_{k}\ \ \ \ \ \ \ \ \ \ \text{in }\left(  \mathbb{K}\left[  t\right]
\right)  ^{n\times n}. \label{eq.lem.CH.D.adj}%
\end{equation}
Thus, an $n$-tuple $\left(  D_{0},D_{1},\ldots,D_{n-1}\right)  $ of matrices
in $\mathbb{K}^{n\times n}$ is defined. Extend this $n$-tuple to a family
$\left(  D_{k}\right)  _{k\in\mathbb{Z}}$ of matrices in $\mathbb{K}^{n\times
n}$ by setting%
\begin{equation}
\left(  D_{k}=0_{n\times n}\ \ \ \ \ \ \ \ \ \ \text{for every }k\in
\mathbb{Z}\setminus\left\{  0,1,\ldots,n-1\right\}  \right)  .
\label{eq.lem.CH.D.0}%
\end{equation}
Then:

\textbf{(a)} We have $\chi_{A}=\sum_{k=0}^{n}c_{n-k}t^{k}$.

\textbf{(b)} For every integer $k$, we have $c_{n-k}I_{n}=D_{k-1}-AD_{k}$.

\textbf{(c)} Every $k\in\mathbb{N}$ satisfies%
\[
\sum_{i=0}^{k}c_{k-i}A^{i}=D_{n-1-k}.
\]

\end{lemma}

\begin{vershort}
\begin{proof}
[Proof of Lemma \ref{lem.CH.D}.]\textbf{(a)} Lemma \ref{lem.CH.D} \textbf{(a)}
is just Corollary \ref{cor.CH.chiA}.

\textbf{(b)} We have%
\begin{align}
\sum_{k=0}^{n}t^{k}D_{k-1}  &  =t^{0}\underbrace{D_{0-1}}_{\substack{=D_{-1}%
=0_{n\times n}\\\text{(by (\ref{eq.lem.CH.D.0}))}}}+\sum_{k=1}^{n}t^{k}%
D_{k-1}=\sum_{k=1}^{n}t^{k}D_{k-1}=\sum_{k=0}^{n-1}\underbrace{t^{k+1}%
}_{=tt^{k}}\underbrace{D_{\left(  k+1\right)  -1}}_{=D_{k}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }k+1\text{ for
}k\text{ in the sum}\right) \nonumber\\
&  =\sum_{k=0}^{n-1}tt^{k}D_{k}=t\underbrace{\sum_{k=0}^{n-1}t^{k}D_{k}%
}_{\substack{=\operatorname*{adj}\left(  tI_{n}-A\right)  \\\text{(by
(\ref{eq.lem.CH.D.adj}))}}}=t\operatorname*{adj}\left(  tI_{n}-A\right)
\label{pf.lem.CH.D.short.1}%
\end{align}
and%
\begin{align}
\sum_{k=0}^{n}t^{k}D_{k}  &  =t^{n}\underbrace{D_{n}}_{\substack{=0_{n\times
n}\\\text{(by (\ref{eq.lem.CH.D.0}))}}}+\sum_{k=0}^{n-1}t^{k}D_{k}=\sum
_{k=0}^{n-1}t^{k}D_{k}\nonumber\\
&  =\operatorname*{adj}\left(  tI_{n}-A\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{eq.lem.CH.D.adj})}\right)  . \label{pf.lem.CH.D.short.2}%
\end{align}


But Theorem \ref{thm.adj.inverse} (applied to $\mathbb{K}\left[  t\right]  $
and $tI_{n}-A$ instead of $\mathbb{K}$ and $A$) shows that%
\[
\left(  tI_{n}-A\right)  \cdot\operatorname*{adj}\left(  tI_{n}-A\right)
=\operatorname*{adj}\left(  tI_{n}-A\right)  \cdot\left(  tI_{n}-A\right)
=\det\left(  tI_{n}-A\right)  \cdot I_{n}.
\]
Thus, in particular,%
\[
\left(  tI_{n}-A\right)  \cdot\operatorname*{adj}\left(  tI_{n}-A\right)
=\underbrace{\det\left(  tI_{n}-A\right)  }_{\substack{=\chi_{A}\\\text{(by
the definition of }\chi_{A}\text{)}}}\cdot I_{n}=\chi_{A}\cdot I_{n},
\]
so that%
\begin{align*}
\chi_{A}\cdot I_{n}  &  =\left(  tI_{n}-A\right)  \cdot\operatorname*{adj}%
\left(  tI_{n}-A\right) \\
&  =\underbrace{t\operatorname*{adj}\left(  tI_{n}-A\right)  }%
_{\substack{=\sum_{k=0}^{n}t^{k}D_{k-1}\\\text{(by (\ref{pf.lem.CH.D.short.1}%
))}}}-A\underbrace{\operatorname*{adj}\left(  tI_{n}-A\right)  }%
_{\substack{=\sum_{k=0}^{n}t^{k}D_{k}\\\text{(by (\ref{pf.lem.CH.D.short.2}%
))}}}=\sum_{k=0}^{n}t^{k}D_{k-1}-\underbrace{A\sum_{k=0}^{n}t^{k}D_{k}}%
_{=\sum_{k=0}^{n}t^{k}AD_{k}}\\
&  =\sum_{k=0}^{n}t^{k}D_{k-1}-\sum_{k=0}^{n}t^{k}AD_{k}=\sum_{k=0}^{n}%
t^{k}\left(  D_{k-1}-AD_{k}\right)  .
\end{align*}
Thus,%
\begin{align*}
\sum_{k=0}^{n}t^{k}\left(  D_{k-1}-AD_{k}\right)   &  =\underbrace{\chi_{A}%
}_{\substack{=\sum_{k=0}^{n}c_{n-k}t^{k}\\\text{(by Lemma \ref{lem.CH.D}
\textbf{(a)})}}}\cdot I_{n}=\left(  \sum_{k=0}^{n}c_{n-k}t^{k}\right)  \cdot
I_{n}\\
&  =\sum_{k=0}^{n}t^{k}c_{n-k}I_{n}.
\end{align*}
Lemma \ref{lem.polymat.=} (applied to $m=n$, $B_{k}=D_{k-1}-AD_{k}$ and
$C_{k}=c_{n-k}I_{n}$) thus shows that%
\begin{equation}
D_{k-1}-AD_{k}=c_{n-k}I_{n}\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\left\{
0,1,\ldots,n\right\}  . \label{pf.lem.CH.D.short.b.5}%
\end{equation}


Now, let $k$ be an integer. We must prove that $c_{n-k}I_{n}=D_{k-1}-AD_{k}$.

If $k\in\left\{  0,1,\ldots,n\right\}  $, then this follows from
(\ref{pf.lem.CH.D.short.b.5}). Thus, we WLOG assume that $k\notin\left\{
0,1,\ldots,n\right\}  $. Hence, $k-1\in\mathbb{Z}\setminus\left\{
0,1,\ldots,n-1\right\}  $, so that (\ref{eq.lem.CH.D.0}) (applied to $k-1$
instead of $k$) yields $D_{k-1}=0_{n\times n}$. Also, $k\notin\left\{
0,1,\ldots,n\right\}  $ leads to $k\in\mathbb{Z}\setminus\left\{
0,1,\ldots,n-1\right\}  $; therefore, (\ref{eq.lem.CH.D.0}) yields
$D_{k}=0_{n\times n}$. Now, $\underbrace{D_{k-1}}_{=0_{n\times n}%
}-A\underbrace{D_{k}}_{=0_{n\times n}}=0_{n\times n}-0_{n\times n}=0_{n\times
n}$.

On the other hand, $c_{n-k}=0$\ \ \ \ \footnote{\textit{Proof.} Recall that
$\chi_{A}$ is a polynomial of degree $\leq n$ (by Corollary \ref{cor.chiA}
\textbf{(a)}). Hence, $\left[  t^{k}\right]  \chi_{A}=0$ (since $k\notin%
\left\{  0,1,\ldots,n\right\}  $). Now, (\ref{pf.cor.CH.chiA.1}) yields
$c_{n-k}=\left[  t^{k}\right]  \chi_{A}=0$.}. Hence, $\underbrace{c_{n-k}%
}_{=0}I_{n}=0_{n\times n}$. Compared with $D_{k-1}-AD_{k}=0_{n\times n}$, this
yields $c_{n-k}I_{n}=D_{k-1}-AD_{k}$.

Hence, $c_{n-k}I_{n}=D_{k-1}-AD_{k}$ is proven. In other words, Lemma
\ref{lem.CH.D} \textbf{(b)} is proven.

\textbf{(c)} Let $k\in\mathbb{N}$. Then,%
\begin{align*}
\sum_{i=0}^{k}c_{k-i}A^{i}  &  =\sum_{i=n-k}^{n}\underbrace{c_{k-\left(
k-n+i\right)  }}_{=c_{n-i}}A^{k-n+i}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }k-n+i\text{
for }i\text{ in the sum}\right) \\
&  =\sum_{i=n-k}^{n}\underbrace{c_{n-i}A^{k-n+i}}_{=A^{k-n+i}c_{n-i}I_{n}%
}=\sum_{i=n-k}^{n}A^{k-n+i}\underbrace{c_{n-i}I_{n}}_{\substack{=D_{i-1}%
-AD_{i}\\\text{(by Lemma \ref{lem.CH.D} \textbf{(b)},}\\\text{applied to
}i\text{ instead of }k\text{)}}}\\
&  =\sum_{i=n-k}^{n}\underbrace{A^{k-n+i}\left(  D_{i-1}-AD_{i}\right)
}_{=A^{k-n+i}D_{i-1}-A^{k-n+i}AD_{i}}\\
&  =\sum_{i=n-k}^{n}\left(  A^{k-n+i}D_{i-1}-\underbrace{A^{k-n+i}%
A}_{=A^{k-n+i+1}=A^{k-n+\left(  i+1\right)  }}\ \ \underbrace{D_{i}%
}_{=D_{\left(  i+1\right)  -1}}\right) \\
&  =\sum_{i=n-k}^{n}\left(  A^{k-n+i}D_{i-1}-A^{k-n+\left(  i+1\right)
}D_{\left(  i+1\right)  -1}\right) \\
&  =\underbrace{A^{k-n+\left(  n-k\right)  }}_{=A^{0}=I_{n}}D_{n-k-1}%
-A^{k-n+\left(  n+1\right)  }\underbrace{D_{\left(  n+1\right)  -1}%
}_{\substack{=D_{n}=0_{n\times n}\\\text{(by (\ref{eq.lem.CH.D.0}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the telescope principle}\right) \\
&  =D_{n-k-1}=D_{n-1-k}.
\end{align*}
This proves Lemma \ref{lem.CH.D} \textbf{(c)}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.CH.D}.]Combining $n\in\mathbb{Z}$ with $n\notin%
\left\{  0,1,\ldots,n-1\right\}  $ (since $n>n-1$), we obtain $n\in
\mathbb{Z}\setminus\left\{  0,1,\ldots,n-1\right\}  $. Hence, $D_{n}%
=0_{n\times n}$ (by (\ref{eq.lem.CH.D.0}), applied to $k=n$).

Combining $-1\in\mathbb{Z}$ with $-1\notin\left\{  0,1,\ldots,n-1\right\}  $
(since $-1<0$), we obtain $-1\in\mathbb{Z}\setminus\left\{  0,1,\ldots
,n-1\right\}  $. Hence, $D_{-1}=0_{n\times n}$ (by (\ref{eq.lem.CH.D.0}),
applied to $k=-1$).

\textbf{(a)} Lemma \ref{lem.CH.D} \textbf{(a)} is identical with Corollary
\ref{cor.CH.chiA}, and therefore has already been proven.

\textbf{(b)} We have $n\in\mathbb{N}$, so that $0\in\left\{  0,1,\ldots
,n\right\}  $. Hence, we can split off the addend for $k=0$ from the sum
$\sum_{k=0}^{n}t^{k}D_{k-1}$. We thus obtain%
\begin{align}
\sum_{k=0}^{n}t^{k}D_{k-1}  &  =t^{0}\underbrace{D_{0-1}}_{=D_{-1}=0_{n\times
n}}+\sum_{k=1}^{n}t^{k}D_{k-1}=\underbrace{t^{0}0_{n\times n}}_{=0_{n\times
n}}+\sum_{k=1}^{n}t^{k}D_{k-1}\nonumber\\
&  =\sum_{k=1}^{n}t^{k}D_{k-1}=\sum_{k=0}^{n-1}\underbrace{t^{k+1}}_{=tt^{k}%
}\underbrace{D_{\left(  k+1\right)  -1}}_{=D_{k}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }k+1\text{ for
}k\text{ in the sum}\right) \nonumber\\
&  =\sum_{k=0}^{n-1}tt^{k}D_{k}=t\underbrace{\sum_{k=0}^{n-1}t^{k}D_{k}%
}_{\substack{=\operatorname*{adj}\left(  tI_{n}-A\right)  \\\text{(by
(\ref{eq.lem.CH.D.adj}))}}}=t\operatorname*{adj}\left(  tI_{n}-A\right)  .
\label{pf.lem.CH.D.1}%
\end{align}


Also, $n\in\mathbb{N}$, so that $n\in\left\{  0,1,\ldots,n\right\}  $. Thus,
we can split off the addend for $k=n$ from the sum $\sum_{k=0}^{n}t^{k}D_{k}$.
We thus obtain%
\begin{align}
\sum_{k=0}^{n}t^{k}D_{k}  &  =t^{n}\underbrace{D_{n}}_{=0_{n\times n}}%
+\sum_{k=0}^{n-1}t^{k}D_{k}=\underbrace{t^{0}0_{n\times n}}_{=0_{n\times n}%
}+\sum_{k=0}^{n-1}t^{k}D_{k}\nonumber\\
&  =\sum_{k=0}^{n-1}t^{k}D_{k}=\operatorname*{adj}\left(  tI_{n}-A\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{eq.lem.CH.D.adj})}\right)  .
\label{pf.lem.CH.D.2}%
\end{align}


But Theorem \ref{thm.adj.inverse} (applied to $\mathbb{K}\left[  t\right]  $
and $tI_{n}-A$ instead of $\mathbb{K}$ and $A$) shows that%
\[
\left(  tI_{n}-A\right)  \cdot\operatorname*{adj}\left(  tI_{n}-A\right)
=\operatorname*{adj}\left(  tI_{n}-A\right)  \cdot\left(  tI_{n}-A\right)
=\det\left(  tI_{n}-A\right)  \cdot I_{n}.
\]
Thus, in particular,%
\[
\left(  tI_{n}-A\right)  \cdot\operatorname*{adj}\left(  tI_{n}-A\right)
=\underbrace{\det\left(  tI_{n}-A\right)  }_{\substack{=\chi_{A}\\\text{(since
}\chi_{A}=\det\left(  tI_{n}-A\right)  \\\text{(by the definition of }\chi
_{A}\text{))}}}\cdot I_{n}=\chi_{A}\cdot I_{n},
\]
so that%
\begin{align*}
\chi_{A}\cdot I_{n}  &  =\left(  tI_{n}-A\right)  \cdot\operatorname*{adj}%
\left(  tI_{n}-A\right)  =t\underbrace{I_{n}\operatorname*{adj}\left(
tI_{n}-A\right)  }_{=\operatorname*{adj}\left(  tI_{n}-A\right)
}-A\operatorname*{adj}\left(  tI_{n}-A\right) \\
&  =\underbrace{t\operatorname*{adj}\left(  tI_{n}-A\right)  }%
_{\substack{=\sum_{k=0}^{n}t^{k}D_{k-1}\\\text{(by (\ref{pf.lem.CH.D.1}))}%
}}-A\underbrace{\operatorname*{adj}\left(  tI_{n}-A\right)  }_{\substack{=\sum
_{k=0}^{n}t^{k}D_{k}\\\text{(by (\ref{pf.lem.CH.D.2}))}}}\\
&  =\sum_{k=0}^{n}t^{k}D_{k-1}-\underbrace{A\sum_{k=0}^{n}t^{k}D_{k}}%
_{=\sum_{k=0}^{n}At^{k}D_{k}=\sum_{k=0}^{n}t^{k}AD_{k}}=\sum_{k=0}^{n}%
t^{k}D_{k-1}-\sum_{k=0}^{n}t^{k}AD_{k}\\
&  =\sum_{k=0}^{n}t^{k}\left(  D_{k-1}-AD_{k}\right)  .
\end{align*}
Thus,%
\begin{align*}
\sum_{k=0}^{n}t^{k}\left(  D_{k-1}-AD_{k}\right)   &  =\underbrace{\chi_{A}%
}_{\substack{=\sum_{k=0}^{n}c_{n-k}t^{k}\\\text{(by Lemma \ref{lem.CH.D}
\textbf{(a)})}}}\cdot I_{n}=\left(  \sum_{k=0}^{n}c_{n-k}t^{k}\right)  \cdot
I_{n}\\
&  =\sum_{k=0}^{n}\underbrace{c_{n-k}t^{k}}_{=t^{k}c_{n-k}}I_{n}=\sum
_{k=0}^{n}t^{k}c_{n-k}I_{n}.
\end{align*}
Lemma \ref{lem.polymat.=} (applied to $m=n$, $B_{k}=D_{k-1}-AD_{k}$ and
$C_{k}=c_{n-k}I_{n}$) thus shows that%
\begin{equation}
D_{k-1}-AD_{k}=c_{n-k}I_{n}\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\left\{
0,1,\ldots,n\right\}  . \label{pf.lem.CH.D.b.5}%
\end{equation}


Now, let $k$ be an integer. We must prove that $c_{n-k}I_{n}=D_{k-1}-AD_{k}$.

If $k\in\left\{  0,1,\ldots,n\right\}  $, then this follows from
(\ref{pf.lem.CH.D.b.5}). Thus, we can WLOG assume that we don't have
$k\in\left\{  0,1,\ldots,n\right\}  $. Assume this.

We have $k\notin\left\{  0,1,\ldots,n\right\}  $ (since we don't have
$k\in\left\{  0,1,\ldots,n\right\}  $). We have $k-1\notin\left\{
0,1,\ldots,n-1\right\}  $\ \ \ \ \footnote{\textit{Proof.} Assume the
contrary. Thus, $k-1\in\left\{  0,1,\ldots,n-1\right\}  $, so that $\left(
k-1\right)  +1\in\left\{  1,2,\ldots,n\right\}  $. Now, $k=\left(  k-1\right)
+1\in\left\{  1,2,\ldots,n\right\}  \subseteq\left\{  0,1,\ldots,n\right\}  $.
This contradicts $k\notin\left\{  0,1,\ldots,n\right\}  $. This contradiction
proves that our assumption was wrong, qed.}. Combining $k-1\in\mathbb{Z}$ with
$k-1\notin\left\{  0,1,\ldots,n-1\right\}  $, we obtain $k-1\in\mathbb{Z}%
\setminus\left\{  0,1,\ldots,n-1\right\}  $. Thus, (\ref{eq.lem.CH.D.0})
(applied to $k-1$ instead of $k$) yields $D_{k-1}=0_{n\times n}$.

We have $k\notin\left\{  0,1,\ldots,n-1\right\}  $%
\ \ \ \ \footnote{\textit{Proof.} Assume the contrary. Thus, $k\in\left\{
0,1,\ldots,n-1\right\}  $, so that $k\in\left\{  0,1,\ldots,n-1\right\}
\subseteq\left\{  0,1,\ldots,n\right\}  $. This contradicts $k\notin\left\{
0,1,\ldots,n\right\}  $. This contradiction proves that our assumption was
wrong, qed.}. Combining $k\in\mathbb{Z}$ with $k\notin\left\{  0,1,\ldots
,n-1\right\}  $, we obtain $k\in\mathbb{Z}\setminus\left\{  0,1,\ldots
,n-1\right\}  $. Thus, (\ref{eq.lem.CH.D.0}) yields $D_{k}=0_{n\times n}$.

Now, $\underbrace{D_{k-1}}_{=0_{n\times n}}-A\underbrace{D_{k}}_{=0_{n\times
n}}=0_{n\times n}-0_{n\times n}=0_{n\times n}$.

On the other hand, $c_{n-k}=0$\ \ \ \ \footnote{\textit{Proof.} We are in one
of the following two cases:
\par
\textit{Case 1:} We have $k\geq0$.
\par
\textit{Case 2:} We have $k<0$.
\par
Let us consider Case 1 first. In this case, we have $k\geq0$. Hence,
$k\in\mathbb{N}$. Combined with $k\notin\left\{  0,1,\ldots,n\right\}  $, this
yields $k\in\mathbb{N}\setminus\left\{  0,1,\ldots,n\right\}  =\left\{
n+1,n+2,n+3,\ldots\right\}  $, so that $k>n$. Thus, $\left[  t^{k}\right]
\chi_{A}=0$ (by (\ref{pf.lem.CH.D.a.1})). Now, (\ref{pf.lem.CH.D.a.cn-k})
yields $c_{n-k}=\left[  t^{k}\right]  \chi_{A}=0$. Thus, $c_{n-k}=0$ is proven
in Case 1.
\par
Let us now consider Case 2. In this case, we have $k<0$. Thus, $k$ is a
negative integer. Now, every $f\in\mathbb{K}\left[  t\right]  $ and every
negative integer $m$ satisfy $\left[  t^{m}\right]  f=0$ (since polynomials
have no monomials $t^{m}$ with negative $m$). Applying this to $f=\chi_{A}$
and $m=k$, we obtain $\left[  t^{k}\right]  \chi_{A}=0$. Now,
(\ref{pf.lem.CH.D.a.cn-k}) yields $c_{n-k}=\left[  t^{k}\right]  \chi_{A}=0$.
Thus, $c_{n-k}=0$ is proven in Case 2.
\par
Now, we have proven $c_{n-k}=0$ in each of the two Cases 1 and 2. Since these
two Cases cover all possibilities, this shows that $c_{n-k}=0$ always holds.
Qed.}. Hence, $\underbrace{c_{n-k}}_{=0}I_{n}=0I_{n}=0_{n\times n}$. Compared
with $D_{k-1}-AD_{k}=0_{n\times n}$, this yields $c_{n-k}I_{n}=D_{k-1}-AD_{k}$.

Hence, $c_{n-k}I_{n}=D_{k-1}-AD_{k}$ is proven. In other words, Lemma
\ref{lem.CH.D} \textbf{(b)} is proven.

\textbf{(c)} Let $k\in\mathbb{N}$. We have $k\geq0$ (since $k\in\mathbb{N}$)
and thus $n-\underbrace{k}_{\geq0}\leq n$. Hence, we can split off the addend
for $i=n$ from the sum $\sum_{i=n-k}^{n}A^{\left(  k-\left(  n-i\right)
\right)  +1}D_{i}$. We thus obtain%
\begin{align}
\sum_{i=n-k}^{n}A^{\left(  k-\left(  n-i\right)  \right)  +1}D_{i}  &
=A^{\left(  k-\left(  n-n\right)  \right)  +1}\underbrace{D_{n}}_{=0_{n\times
n}}+\sum_{i=n-k}^{n-1}A^{\left(  k-\left(  n-i\right)  \right)  +1}%
D_{i}\nonumber\\
&  =\underbrace{A^{\left(  k-\left(  n-n\right)  \right)  +1}0_{n\times n}%
}_{=0_{n\times n}}+\sum_{i=n-k}^{n-1}A^{\left(  k-\left(  n-i\right)  \right)
+1}D_{i}\nonumber\\
&  =\sum_{i=n-k}^{n-1}A^{\left(  k-\left(  n-i\right)  \right)  +1}D_{i}%
=\sum_{i=n-k+1}^{n}\underbrace{A^{\left(  k-\left(  n-\left(  i-1\right)
\right)  \right)  +1}}_{\substack{=A^{k-n+i}\\\text{(since }\left(  k-\left(
n-\left(  i-1\right)  \right)  \right)  +1=k-n+i\text{)}}}D_{i-1}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }i-1\text{ for
}i\text{ in the sum}\right) \nonumber\\
&  =\sum_{i=n-k+1}^{n}A^{k-n+i}D_{i-1}. \label{pf.lem.CH.D.c.1}%
\end{align}


Also, $n-k\leq n$. Hence, we can split off the addend for $i=n-k$ from the sum
$\sum_{i=n-k}^{n}A^{k-\left(  n-i\right)  }D_{i-1}$. We thus obtain%
\begin{align}
\sum_{i=n-k}^{n}A^{k-\left(  n-i\right)  }D_{i-1}  &
=\underbrace{A^{k-\left(  n-\left(  n-k\right)  \right)  }}_{\substack{=A^{0}%
\\\text{(since }k-\left(  n-\left(  n-k\right)  \right)  =0\text{)}}%
}D_{n-k-1}+\sum_{i=n-k+1}^{n}\underbrace{A^{k-\left(  n-i\right)  }%
}_{\substack{=A^{k-n+i}\\\text{(since }k-\left(  n-i\right)  =k-n+i\text{)}%
}}D_{i-1}\nonumber\\
&  =\underbrace{A^{0}}_{=I_{n}}D_{n-k-1}+\sum_{i=n-k+1}^{n}A^{k-n+i}%
D_{i-1}\nonumber\\
&  =D_{n-k-1}+\sum_{i=n-k+1}^{n}A^{k-n+i}D_{i-1}. \label{pf.lem.CH.D.c.2}%
\end{align}


Now,%
\begin{align*}
\sum_{i=0}^{k}c_{k-i}A^{i}  &  =\sum_{i=0}^{k}\underbrace{c_{k-\left(
k-i\right)  }}_{=c_{i}}A^{k-i}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have
substituted }k-i\text{ for }i\text{ in the sum}\right) \\
&  =\sum_{i=0}^{k}c_{i}A^{k-i}=\sum_{i=n-k}^{n}c_{n-i}\underbrace{A^{k-\left(
n-i\right)  }}_{=A^{k-\left(  n-i\right)  }I_{n}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }n-i\text{ for
}i\text{ in the sum}\right) \\
&  =\sum_{i=n-k}^{n}\underbrace{c_{n-i}A^{k-\left(  n-i\right)  }I_{n}%
}_{=A^{k-\left(  n-i\right)  }c_{n-i}I_{n}}=\sum_{i=n-k}^{n}A^{k-\left(
n-i\right)  }\underbrace{c_{n-i}I_{n}}_{\substack{=D_{i-1}-AD_{i}\\\text{(by
Lemma \ref{lem.CH.D} \textbf{(b)},}\\\text{applied to }i\text{ instead of
}k\text{)}}}\\
&  =\sum_{i=n-k}^{n}\underbrace{A^{k-\left(  n-i\right)  }\left(
D_{i-1}-AD_{i}\right)  }_{=A^{k-\left(  n-i\right)  }D_{i-1}-A^{k-\left(
n-i\right)  }AD_{i}}\\
&  =\sum_{i=n-k}^{n}\left(  A^{k-\left(  n-i\right)  }D_{i-1}-A^{k-\left(
n-i\right)  }AD_{i}\right) \\
&  =\sum_{i=n-k}^{n}A^{k-\left(  n-i\right)  }D_{i-1}-\sum_{i=n-k}%
^{n}\underbrace{A^{k-\left(  n-i\right)  }A}_{=A^{\left(  k-\left(
n-i\right)  \right)  +1}}D_{i}\\
&  =\underbrace{\sum_{i=n-k}^{n}A^{k-\left(  n-i\right)  }D_{i-1}%
}_{\substack{=D_{n-k-1}+\sum_{i=n-k+1}^{n}A^{k-n+i}D_{i-1}\\\text{(by
(\ref{pf.lem.CH.D.c.2}))}}}-\underbrace{\sum_{i=n-k}^{n}A^{\left(  k-\left(
n-i\right)  \right)  +1}D_{i}}_{\substack{=\sum_{i=n-k+1}^{n}A^{k-n+i}%
D_{i-1}\\\text{(by (\ref{pf.lem.CH.D.c.1}))}}}\\
&  =\left(  D_{n-k-1}+\sum_{i=n-k+1}^{n}A^{k-n+i}D_{i-1}\right)  -\left(
\sum_{i=n-k+1}^{n}A^{k-n+i}D_{i-1}\right) \\
&  =D_{n-k-1}=D_{n-1-k}%
\end{align*}
(since $n-k-1=n-1-k$). This proves Lemma \ref{lem.CH.D} \textbf{(c)}.
\end{proof}
\end{verlong}

\begin{vershort}
\begin{proof}
[Proof of Theorem \ref{thm.CH}.]For every $j\in\mathbb{Z}$, define an element
$c_{j}\in\mathbb{K}$ by $c_{j}=\left[  t^{n-j}\right]  \chi_{A}$.

Proposition \ref{prop.adj.poly} shows that there exist $n$ matrices
$D_{0},D_{1},\ldots,D_{n-1}$ in $\mathbb{K}^{n\times n}$ such that%
\[
\operatorname*{adj}\left(  tI_{n}-A\right)  =\sum_{k=0}^{n-1}t^{k}%
D_{k}\ \ \ \ \ \ \ \ \ \ \text{in }\left(  \mathbb{K}\left[  t\right]
\right)  ^{n\times n}.
\]
Consider these $D_{0},D_{1},\ldots,D_{n-1}$. Thus, an $n$-tuple $\left(
D_{0},D_{1},\ldots,D_{n-1}\right)  $ of matrices in $\mathbb{K}^{n\times n}$
is defined. Extend this $n$-tuple to a family $\left(  D_{k}\right)
_{k\in\mathbb{Z}}$ of matrices in $\mathbb{K}^{n\times n}$ by setting%
\[
D_{k}=0_{n\times n}\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\mathbb{Z}%
\setminus\left\{  0,1,\ldots,n-1\right\}  .
\]
Thus, in particular, $D_{-1}=0_{n\times n}$.

Lemma \ref{lem.CH.D} \textbf{(a)} shows that $\chi_{A}=\sum_{k=0}^{n}%
c_{n-k}t^{k}=\sum_{i=0}^{n}c_{n-i}t^{i}$. Substituting $A$ for $t$ in this
equality, we obtain%
\begin{align*}
\chi_{A}\left(  A\right)   &  =\sum_{i=0}^{n}c_{n-i}A^{i}=D_{n-1-n}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.CH.D} \textbf{(c)},
applied to }k=n\right) \\
&  =D_{-1}=0_{n\times n}.
\end{align*}
This proves Theorem \ref{thm.CH}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Theorem \ref{thm.CH}.]For every $j\in\mathbb{Z}$, define an element
$c_{j}\in\mathbb{K}$ by $c_{j}=\left[  t^{n-j}\right]  \chi_{A}$.

Proposition \ref{prop.adj.poly} shows that there exist $n$ matrices
$D_{0},D_{1},\ldots,D_{n-1}$ in $\mathbb{K}^{n\times n}$ such that%
\[
\operatorname*{adj}\left(  tI_{n}-A\right)  =\sum_{k=0}^{n-1}t^{k}%
D_{k}\ \ \ \ \ \ \ \ \ \ \text{in }\left(  \mathbb{K}\left[  t\right]
\right)  ^{n\times n}.
\]
Consider these $D_{0},D_{1},\ldots,D_{n-1}$. Thus, an $n$-tuple $\left(
D_{0},D_{1},\ldots,D_{n-1}\right)  $ of matrices in $\mathbb{K}^{n\times n}$
is defined. Extend this $n$-tuple to a family $\left(  D_{k}\right)
_{k\in\mathbb{Z}}$ of matrices in $\mathbb{K}^{n\times n}$ by setting%
\begin{equation}
\left(  D_{k}=0_{n\times n}\ \ \ \ \ \ \ \ \ \ \text{for every }k\in
\mathbb{Z}\setminus\left\{  0,1,\ldots,n-1\right\}  \right)  .
\label{pf.thm.CH.0}%
\end{equation}


Combining $-1\in\mathbb{Z}$ with $-1\notin\left\{  0,1,\ldots,n-1\right\}  $
(since $-1<0$), we obtain $-1\in\mathbb{Z}\setminus\left\{  0,1,\ldots
,n-1\right\}  $. Applying (\ref{pf.thm.CH.0}) to $k=-1$ therefore yields
$D_{-1}=0_{n\times n}$.

Lemma \ref{lem.CH.D} \textbf{(a)} shows that $\chi_{A}=\sum_{k=0}^{n}%
c_{n-k}t^{k}$. Substituting $A$ for $t$ in this equality, we obtain%
\begin{align*}
\chi_{A}\left(  A\right)   &  =\sum_{k=0}^{n}c_{n-k}A^{k}=\sum_{i=0}%
^{n}c_{n-i}A^{i}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have renamed the summation}\\
\text{index }k\text{ as }i
\end{array}
\right) \\
&  =D_{n-1-n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.CH.D}
\textbf{(c)}, applied to }k=n\right) \\
&  =D_{-1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n-1-n=-1\right) \\
&  =0_{n\times n}.
\end{align*}
This proves Theorem \ref{thm.CH}.
\end{proof}
\end{verlong}

\subsection{Derivations and determinants}

Now, let us make what seems to be a detour, and define $\mathbb{K}%
$-derivations of a $\mathbb{K}$-algebra\footnote{See \cite[Convention
1.1]{Grinbe16} for what we mean by a \textquotedblleft$\mathbb{K}%
$-algebra\textquotedblright. In a nutshell, we require $\mathbb{K}$-algebras
to be associative and unital, and we require the multiplication map on a
$\mathbb{K}$-algebra to be $\mathbb{K}$-bilinear.}:

\begin{definition}
\label{def.deriv}Let $\mathbb{L}$ be a $\mathbb{K}$-algebra. A $\mathbb{K}%
$-linear map $f:\mathbb{L}\rightarrow\mathbb{L}$ is said to be a $\mathbb{K}%
$\textit{-derivation} if it satisfies%
\begin{equation}
\left(  f\left(  ab\right)  =af\left(  b\right)  +f\left(  a\right)
b\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathbb{L}\text{ and }b\in
\mathbb{L}\right)  . \label{eq.def.deriv.axiom}%
\end{equation}

\end{definition}

The notion of a \textquotedblleft$\mathbb{K}$-derivation\textquotedblright\ is
a particular case of the notion of a \textquotedblleft$\mathbf{k}%
$-derivation\textquotedblright\ defined in \cite[Definition 1.5]{Grinbe16};
specifically, it is obtained from the latter when setting $\mathbf{k}%
=\mathbb{K}$, $A=\mathbb{L}$ and $M=\mathbb{L}$. This particular case will
suffice for us. Examples of $\mathbb{K}$-derivations abound (there are several
in \cite{Grinbe16}), but the only one we will need is the following:

\begin{proposition}
\label{prop.deriv.del}Let $\partial:\mathbb{K}\left[  t\right]  \rightarrow
\mathbb{K}\left[  t\right]  $ be the differentiation operator (i.e., the map
that sends every polynomial $f\in\mathbb{K}\left[  t\right]  $ to the
derivative of $f$). Then, $\partial:\mathbb{K}\left[  t\right]  \rightarrow
\mathbb{K}\left[  t\right]  $ is a $\mathbb{K}$-derivation.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.deriv.del}.]This follows from the fact that
$\partial\left(  ab\right)  =a\partial\left(  b\right)  +\partial\left(
a\right)  b$ for any two polynomials $a$ and $b$ (the well-known Leibniz law).
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.deriv.del}.]The well-known Leibniz law for the
derivative of a product yields that%
\begin{align*}
\partial\left(  ab\right)   &  =\partial\left(  a\right)  b+a\partial\left(
b\right) \\
&  =a\partial\left(  b\right)  +\partial\left(  a\right)
b\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathbb{K}\left[  t\right]  \text{
and }b\in\mathbb{K}\left[  t\right]  .
\end{align*}
But the map $\partial:\mathbb{K}\left[  t\right]  \rightarrow\mathbb{K}\left[
t\right]  $ is $\mathbb{K}$-linear. Hence, $\partial$ is a $\mathbb{K}%
$-derivation if and only if it satisfies%
\begin{equation}
\left(  \partial\left(  ab\right)  =a\partial\left(  b\right)  +\partial
\left(  a\right)  b\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathbb{K}\left[
t\right]  \text{ and }b\in\mathbb{K}\left[  t\right]  \right)  .
\label{pf.prop.deriv.del.1}%
\end{equation}
Hence, $\partial$ is a $\mathbb{K}$-derivation (since $\partial$ satisfies
(\ref{pf.prop.deriv.del.1})). This proves Proposition \ref{prop.deriv.del}.
\end{proof}
\end{verlong}

A fundamental fact about $\mathbb{K}$-derivations is the following:

\begin{proposition}
\label{prop.deriv.leibniz}Let $\mathbb{L}$ be a $\mathbb{K}$-algebra. Let
$f:\mathbb{L}\rightarrow\mathbb{L}$ be a $\mathbb{K}$-derivation. Let
$n\in\mathbb{N}$, and let $a_{1},a_{2},\ldots,a_{n}\in\mathbb{L}$. Then,%
\[
f\left(  a_{1}a_{2}\cdots a_{n}\right)  =\sum_{i=1}^{n}a_{1}a_{2}\cdots
a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{n}.
\]

\end{proposition}

This proposition is a particular case of \cite[Theorem 1.14]{Grinbe16}
(obtained by setting $\mathbf{k}=\mathbb{K}$, $A=\mathbb{L}$ and
$M=\mathbb{L}$); it is also easy to prove\footnote{First one should show that
$f\left(  1\right)  =0$ (by applying (\ref{eq.def.deriv.axiom}) to $a=1$ and
$b=1$). Then, one can prove Proposition \ref{prop.deriv.leibniz} by
straightforward induction on $n$.}.

What we are going to need is a formula for how a derivation acts on the
determinant of a matrix. We first introduce a notation:

\begin{definition}
\label{def.f-mat}Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $\mathbb{L}$
and $\mathbb{M}$ be rings. Let $f:\mathbb{L}\rightarrow\mathbb{M}$ be any map.
Then, $f^{n\times m}$ will denote the map from $\mathbb{L}^{n\times m}$ to
$\mathbb{M}^{n\times m}$ which sends every matrix $\left(  a_{i,j}\right)
_{1\leq i\leq n,\ 1\leq j\leq m}\in\mathbb{L}^{n\times m}$ to the matrix
$\left(  f\left(  a_{i,j}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq m}%
\in\mathbb{M}^{n\times m}$. (In other words, $f^{n\times m}$ is the map which
takes an $n\times m$-matrix in $\mathbb{L}^{n\times m}$, and applies $f$ to
each entry of this matrix.)
\end{definition}

\begin{theorem}
\label{thm.deriv.ddet}Let $\mathbb{L}$ be a commutative $\mathbb{K}$-algebra.
Let $f:\mathbb{L}\rightarrow\mathbb{L}$ be a $\mathbb{K}$-derivation. Let
$n\in\mathbb{N}$. Let $A\in\mathbb{L}^{n\times n}$. Then,%
\[
f\left(  \det A\right)  =\operatorname*{Tr}\left(  f^{n\times n}\left(
A\right)  \cdot\operatorname*{adj}A\right)  .
\]

\end{theorem}

Proving Theorem \ref{thm.deriv.ddet} will take us a while. Let us begin by
stating three lemmas:

\begin{lemma}
\label{lem.Tr(AB)}Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $A=\left(
a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}\in\mathbb{K}^{n\times m}$ and
$B=\left(  b_{i,j}\right)  _{1\leq i\leq m,\ 1\leq j\leq n}\in\mathbb{K}%
^{m\times n}$. Then,%
\[
\operatorname*{Tr}\left(  AB\right)  =\sum_{i=1}^{n}\sum_{j=1}^{m}%
a_{i,j}b_{j,i}.
\]

\end{lemma}

\begin{vershort}
\begin{proof}
[Proof of Lemma \ref{lem.Tr(AB)}.]The definition of $AB$ yields $AB=\left(
\sum_{k=1}^{m}a_{i,k}b_{k,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$ (since
$A=\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}$ and $B=\left(
b_{i,j}\right)  _{1\leq i\leq m,\ 1\leq j\leq n}$). Hence,%
\[
\operatorname*{Tr}\left(  AB\right)  =\sum_{i=1}^{n}\sum_{k=1}^{m}%
a_{i,k}b_{k,i}=\sum_{i=1}^{n}\sum_{j=1}^{m}a_{i,j}b_{j,i}%
\]
(here, we have renamed the summation index $k$ as $j$ in the second sum). This
proves Lemma \ref{lem.Tr(AB)}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.Tr(AB)}.]Recall that the trace $\operatorname*{Tr}C$
of an $n\times n$-matrix $C$ is the sum of all diagonal entries of $C$
(because this is how $\operatorname*{Tr}C$ is defined). In other words, if
$C=\left(  c_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$ is an $n\times
n$-matrix, then%
\begin{equation}
\operatorname*{Tr}C=\sum_{i=1}^{n}c_{i,i}. \label{pf.lem.Tr(AB).TrC=}%
\end{equation}


From $A=\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}$ and
$B=\left(  b_{i,j}\right)  _{1\leq i\leq m,\ 1\leq j\leq n}$, we obtain
\[
AB=\left(  \sum_{k=1}^{m}a_{i,k}b_{k,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n}%
\]
(by the definition of the matrix $AB$). Hence, (\ref{pf.lem.Tr(AB).TrC=})
(applied to $AB$ and $\sum_{k=1}^{m}a_{i,k}b_{k,j}$ instead of $C$ and
$c_{i,j}$) yields%
\[
\operatorname*{Tr}\left(  AB\right)  =\sum_{i=1}^{n}\underbrace{\sum_{k=1}%
^{m}a_{i,k}b_{k,i}}_{\substack{=\sum_{j=1}^{m}a_{i,j}b_{j,i}\\\text{(here, we
have renamed the}\\\text{summation index }k\text{ as }j\text{)}}}=\sum
_{i=1}^{n}\sum_{j=1}^{m}a_{i,j}b_{j,i}.
\]
This proves Lemma \ref{lem.Tr(AB)}.
\end{proof}
\end{verlong}

\begin{lemma}
\label{lem.deriv.leibniz-comm}Let $\mathbb{L}$ be a commutative $\mathbb{K}%
$-algebra. Let $f:\mathbb{L}\rightarrow\mathbb{L}$ be a $\mathbb{K}%
$-derivation. Let $n\in\mathbb{N}$, and let $a_{1},a_{2},\ldots,a_{n}%
\in\mathbb{L}$. Then,%
\[
f\left(  a_{1}a_{2}\cdots a_{n}\right)  =\sum_{k=1}^{n}f\left(  a_{k}\right)
\prod_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i\neq k}}a_{i}.
\]

\end{lemma}

\begin{vershort}
\begin{proof}
[Proof of Lemma \ref{lem.deriv.leibniz-comm}.]Proposition
\ref{prop.deriv.leibniz} yields%
\begin{align*}
f\left(  a_{1}a_{2}\cdots a_{n}\right)   &  =\sum_{i=1}^{n}a_{1}a_{2}\cdots
a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{n}\\
&  =\sum_{k=1}^{n}\underbrace{a_{1}a_{2}\cdots a_{k-1}f\left(  a_{k}\right)
a_{k+1}a_{k+2}\cdots a_{n}}_{\substack{=f\left(  a_{k}\right)  \left(
a_{1}a_{2}\cdots a_{k-1}\right)  \left(  a_{k+1}a_{k+2}\cdots a_{n}\right)
\\\text{(since }\mathbb{L}\text{ is commutative)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}i\text{ as }k\right) \\
&  =\sum_{k=1}^{n}f\left(  a_{k}\right)  \underbrace{\left(  a_{1}a_{2}\cdots
a_{k-1}\right)  \left(  a_{k+1}a_{k+2}\cdots a_{n}\right)  }_{=\prod
_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i\neq k}}a_{i}}=\sum
_{k=1}^{n}f\left(  a_{k}\right)  \prod_{\substack{i\in\left\{  1,2,\ldots
,n\right\}  ;\\i\neq k}}a_{i}.
\end{align*}
This proves Lemma \ref{lem.deriv.leibniz-comm}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.deriv.leibniz-comm}.]Let $k\in\left\{  1,2,\ldots
,n\right\}  $.

We recall a basic property of products (in a commutative ring): If $U$ and $V$
are two disjoint finite sets, and if $b_{i}$ is an element of $\mathbb{L}$ for
every $i\in U\cup V$, then%
\begin{equation}
\prod_{i\in U\cup V}b_{i}=\left(  \prod_{i\in U}b_{i}\right)  \left(
\prod_{i\in V}b_{i}\right)  . \label{pf.lem.deriv.leibniz-comm.1}%
\end{equation}
(This relies on the fact that the ring $\mathbb{L}$ is commutative.)

Now, the sets $\left\{  1,2,\ldots,k-1\right\}  $ and $\left\{  k+1,k+2,\ldots
,n\right\}  $ are disjoint\footnote{\textit{Proof.} Let $g\in\left\{
1,2,\ldots,k-1\right\}  \cap\left\{  k+1,k+2,\ldots,n\right\}  $. Then,
$g\in\left\{  1,2,\ldots,k-1\right\}  \cap\left\{  k+1,k+2,\ldots,n\right\}
\subseteq\left\{  1,2,\ldots,k-1\right\}  $, so that $g\leq k-1$. Also,
$g\in\left\{  1,2,\ldots,k-1\right\}  \cap\left\{  k+1,k+2,\ldots,n\right\}
\subseteq\left\{  k+1,k+2,\ldots,n\right\}  $, so that $g\geq k+1$. Combining
$g\leq k-1\leq k$ with $g\geq k+1>k$, we obtain a contradiction.
\par
Now, forget that we fixed $g$. We thus have found a contradiction for each
$g\in\left\{  1,2,\ldots,k-1\right\}  \cap\left\{  k+1,k+2,\ldots,n\right\}
$. Hence, there exists no $g\in\left\{  1,2,\ldots,k-1\right\}  \cap\left\{
k+1,k+2,\ldots,n\right\}  $. In other words, the set $\left\{  1,2,\ldots
,k-1\right\}  \cap\left\{  k+1,k+2,\ldots,n\right\}  $ is empty. In other
words, the sets $\left\{  1,2,\ldots,k-1\right\}  $ and $\left\{
k+1,k+2,\ldots,n\right\}  $ are disjoint. Qed.} and finite. Hence,
(\ref{pf.lem.deriv.leibniz-comm.1}) (applied to $U=\left\{  1,2,\ldots
,k-1\right\}  $, $V=\left\{  k+1,k+2,\ldots,n\right\}  $ and $b_{i}=a_{i}$)
yields%
\[
\prod_{i\in\left\{  1,2,\ldots,k-1\right\}  \cup\left\{  k+1,k+2,\ldots
,n\right\}  }a_{i}=\left(  \prod_{i\in\left\{  1,2,\ldots,k-1\right\}  }%
a_{i}\right)  \left(  \prod_{i\in\left\{  k+1,k+2,\ldots,n\right\}  }%
a_{i}\right)  .
\]


Now,%
\begin{align*}
\underbrace{\prod_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i\neq k}%
}}_{=\prod_{i\in\left\{  1,2,\ldots,n\right\}  \setminus\left\{  k\right\}  }%
}a_{i}  &  =\underbrace{\prod_{i\in\left\{  1,2,\ldots,n\right\}
\setminus\left\{  k\right\}  }}_{\substack{=\prod_{i\in\left\{  1,2,\ldots
,k-1\right\}  \cup\left\{  k+1,k+2,\ldots,n\right\}  }\\\text{(since }\left\{
1,2,\ldots,n\right\}  \setminus\left\{  k\right\}  \\=\left\{  1,2,\ldots
,k-1,k+1,k+2,\ldots,n\right\}  \\=\left\{  1,2,\ldots,k-1\right\}
\cup\left\{  k+1,k+2,\ldots,n\right\}  \text{)}}}a_{i}\\
&  =\prod_{i\in\left\{  1,2,\ldots,k-1\right\}  \cup\left\{  k+1,k+2,\ldots
,n\right\}  }a_{i}\\
&  =\underbrace{\left(  \prod_{i\in\left\{  1,2,\ldots,k-1\right\}  }%
a_{i}\right)  }_{=a_{1}a_{2}\cdots a_{k-1}}\underbrace{\left(  \prod
_{i\in\left\{  k+1,k+2,\ldots,n\right\}  }a_{i}\right)  }_{=a_{k+1}%
a_{k+2}\cdots a_{n}}\\
&  =\left(  a_{1}a_{2}\cdots a_{k-1}\right)  \left(  a_{k+1}a_{k+2}\cdots
a_{n}\right)  .
\end{align*}
Hence,%
\begin{align}
&  f\left(  a_{k}\right)  \underbrace{\prod_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\i\neq k}}a_{i}}_{=\left(  a_{1}a_{2}\cdots
a_{k-1}\right)  \left(  a_{k+1}a_{k+2}\cdots a_{n}\right)  }\nonumber\\
&  =\underbrace{f\left(  a_{k}\right)  \left(  a_{1}a_{2}\cdots a_{k-1}%
\right)  }_{\substack{=\left(  a_{1}a_{2}\cdots a_{k-1}\right)  f\left(
a_{k}\right)  \\\text{(since }\mathbb{L}\text{ is commutative)}}}\left(
a_{k+1}a_{k+2}\cdots a_{n}\right) \nonumber\\
&  =\left(  a_{1}a_{2}\cdots a_{k-1}\right)  f\left(  a_{k}\right)  \left(
a_{k+1}a_{k+2}\cdots a_{n}\right) \nonumber\\
&  =a_{1}a_{2}\cdots a_{k-1}f\left(  a_{k}\right)  a_{k+1}a_{k+2}\cdots a_{n}.
\label{pf.lem.deriv.leibniz-comm.3}%
\end{align}


Now, forget that we fixed $k$. We thus have shown that
(\ref{pf.lem.deriv.leibniz-comm.3}) holds for every $k\in\left\{
1,2,\ldots,n\right\}  $.

Now, Proposition \ref{prop.deriv.leibniz} yields%
\begin{align*}
f\left(  a_{1}a_{2}\cdots a_{n}\right)   &  =\sum_{i=1}^{n}a_{1}a_{2}\cdots
a_{i-1}f\left(  a_{i}\right)  a_{i+1}a_{i+2}\cdots a_{n}\\
&  =\sum_{k=1}^{n}\underbrace{a_{1}a_{2}\cdots a_{k-1}f\left(  a_{k}\right)
a_{k+1}a_{k+2}\cdots a_{n}}_{\substack{=f\left(  a_{k}\right)  \prod
_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i\neq k}}a_{i}\\\text{(by
(\ref{pf.lem.deriv.leibniz-comm.3}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have renamed the summation index
}i\text{ as }k\right) \\
&  =\sum_{k=1}^{n}f\left(  a_{k}\right)  \prod_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\i\neq k}}a_{i}.
\end{align*}
This proves Lemma \ref{lem.deriv.leibniz-comm}.
\end{proof}
\end{verlong}

\begin{lemma}
\label{lem.laplace.Apq}Let $n\in\mathbb{N}$. Let $A=\left(  a_{i,j}\right)
_{1\leq i\leq n,\ 1\leq j\leq n}$ be an $n\times n$-matrix. Let $p\in\left\{
1,2,\ldots,n\right\}  $ and $q\in\left\{  1,2,\ldots,n\right\}  $. Then,%
\[
\sum_{\substack{\sigma\in S_{n};\\\sigma\left(  p\right)  =q}}\left(
-1\right)  ^{\sigma}\prod_{\substack{i\in\left\{  1,2,\ldots,n\right\}
;\\i\neq p}}a_{i,\sigma\left(  i\right)  }=\left(  -1\right)  ^{p+q}%
\det\left(  A_{\sim p,\sim q}\right)  .
\]

\end{lemma}

Lemma \ref{lem.laplace.Apq} is \cite[Lemma 6.84]{detnotes}; it is also easy to
prove (it is the main step in the proof of the Laplace expansion formula for
the determinant).

\begin{vershort}
\begin{proof}
[Proof of Theorem \ref{thm.deriv.ddet}.]Write the matrix $A\in\mathbb{L}%
^{n\times n}$ in the form $A=\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq
j\leq n}$. Hence, $f^{n\times n}\left(  A\right)  =\left(  f\left(
a_{i,j}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq n}$ (by the definition
of $f^{n\times n}$). The definition of $\operatorname*{adj}A$ shows that
$\operatorname*{adj}A=\left(  \left(  -1\right)  ^{i+j}\det\left(  A_{\sim
j,\sim i}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq n}$. Hence, Lemma
\ref{lem.Tr(AB)} (applied to $\mathbb{L}$, $n$, $f^{n\times n}\left(
A\right)  $, $f\left(  a_{i,j}\right)  $, $\operatorname*{adj}A$ and $\left(
-1\right)  ^{i+j}\det\left(  A_{\sim j,\sim i}\right)  $ instead of
$\mathbb{K}$, $m$, $A$, $a_{i,j}$, $B$ and $b_{i,j}$) yields%
\begin{align}
\operatorname*{Tr}\left(  f^{n\times n}\left(  A\right)  \cdot
\operatorname*{adj}A\right)   &  =\sum_{i=1}^{n}\sum_{j=1}^{n}f\left(
a_{i,j}\right)  \left(  -1\right)  ^{j+i}\det\left(  A_{\sim i,\sim j}\right)
\nonumber\\
&  =\sum_{k=1}^{n}\sum_{j=1}^{n}f\left(  a_{k,j}\right)  \underbrace{\left(
-1\right)  ^{j+k}}_{=\left(  -1\right)  ^{k+j}}\det\left(  A_{\sim k,\sim
j}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have renamed the summation index }i\\
\text{as }k\text{ in the outer sum}%
\end{array}
\right) \nonumber\\
&  =\sum_{k=1}^{n}\sum_{j=1}^{n}f\left(  a_{k,j}\right)  \left(  -1\right)
^{k+j}\det\left(  A_{\sim k,\sim j}\right)  .
\label{pf.thm.deriv.ddet.short.Tr=}%
\end{align}


But the map $f$ is a $\mathbb{K}$-derivation, and thus is $\mathbb{K}$-linear.
Now, (\ref{eq.det}) (applied to $\mathbb{L}$ instead of $\mathbb{K}$) yields
$\det A=\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{n}a_{i,\sigma\left(  i\right)  }$. Applying $f$ to both sides of this
equality, we find%
\begin{align}
&  f\left(  \det A\right) \nonumber\\
&  =f\left(  \sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{n}a_{i,\sigma\left(  i\right)  }\right) \nonumber\\
&  =\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}f\left(
\underbrace{\prod_{i=1}^{n}a_{i,\sigma\left(  i\right)  }}_{=a_{1,\sigma
\left(  1\right)  }a_{2,\sigma\left(  2\right)  }\cdots a_{n,\sigma\left(
n\right)  }}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\text{
is }\mathbb{K}\text{-linear}\right) \nonumber\\
&  =\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\underbrace{f\left(
a_{1,\sigma\left(  1\right)  }a_{2,\sigma\left(  2\right)  }\cdots
a_{n,\sigma\left(  n\right)  }\right)  }_{\substack{=\sum_{k=1}^{n}f\left(
a_{k,\sigma\left(  k\right)  }\right)  \prod_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\i\neq k}}a_{i,\sigma\left(  i\right)  }\\\text{(by
Lemma \ref{lem.deriv.leibniz-comm}, applied to}\\a_{i,\sigma\left(  i\right)
}\text{ instead of }a_{i}\text{)}}}\nonumber\\
&  =\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\sum_{k=1}^{n}f\left(
a_{k,\sigma\left(  k\right)  }\right)  \prod_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\i\neq k}}a_{i,\sigma\left(  i\right)  }\nonumber\\
&  =\sum_{k=1}^{n}\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}f\left(
a_{k,\sigma\left(  k\right)  }\right)  \prod_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\i\neq k}}a_{i,\sigma\left(  i\right)  }.
\label{pf.thm.deriv.ddet.short.1}%
\end{align}
But every $k\in\left\{  1,2,\ldots,n\right\}  $ satisfies%
\begin{align*}
&  \underbrace{\sum_{\sigma\in S_{n}}}_{\substack{=\sum_{j\in\left\{
1,2,\ldots,n\right\}  }\sum_{\substack{\sigma\in S_{n};\\\sigma\left(
k\right)  =j}}\\\text{(since }\sigma\left(  k\right)  \in\left\{
1,2,\ldots,n\right\}  \text{ for each }\sigma\in S_{n}\text{)}}}\left(
-1\right)  ^{\sigma}f\left(  a_{k,\sigma\left(  k\right)  }\right)
\prod_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i\neq k}}a_{i,\sigma
\left(  i\right)  }\\
&  =\sum_{j\in\left\{  1,2,\ldots,n\right\}  }\sum_{\substack{\sigma\in
S_{n};\\\sigma\left(  k\right)  =j}}\left(  -1\right)  ^{\sigma}f\left(
\underbrace{a_{k,\sigma\left(  k\right)  }}_{\substack{=a_{k,j}\\\text{(since
}\sigma\left(  k\right)  =j\text{)}}}\right)  \prod_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\i\neq k}}a_{i,\sigma\left(  i\right)  }\\
&  =\sum_{j\in\left\{  1,2,\ldots,n\right\}  }\sum_{\substack{\sigma\in
S_{n};\\\sigma\left(  k\right)  =j}}\left(  -1\right)  ^{\sigma}f\left(
a_{k,j}\right)  \prod_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i\neq
k}}a_{i,\sigma\left(  i\right)  }\\
&  =\underbrace{\sum_{j\in\left\{  1,2,\ldots,n\right\}  }}_{=\sum_{j=1}^{n}%
}f\left(  a_{k,j}\right)  \underbrace{\sum_{\substack{\sigma\in S_{n}%
;\\\sigma\left(  k\right)  =j}}\left(  -1\right)  ^{\sigma}\prod
_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i\neq k}}a_{i,\sigma\left(
i\right)  }}_{\substack{=\left(  -1\right)  ^{k+j}\det\left(  A_{\sim k,\sim
j}\right)  \\\text{(by Lemma \ref{lem.laplace.Apq}, applied to}\\\mathbb{L}%
\text{, }k\text{ and }j\text{ instead of }\mathbb{K}\text{, }p\text{ and
}q\text{)}}}\\
&  =\sum_{j=1}^{n}f\left(  a_{k,j}\right)  \left(  -1\right)  ^{k+j}%
\det\left(  A_{\sim k,\sim j}\right)  .
\end{align*}
Hence, (\ref{pf.thm.deriv.ddet.short.1}) becomes%
\begin{align*}
f\left(  \det A\right)   &  =\sum_{k=1}^{n}\underbrace{\sum_{\sigma\in S_{n}%
}\left(  -1\right)  ^{\sigma}f\left(  a_{k,\sigma\left(  k\right)  }\right)
\prod_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i\neq k}}a_{i,\sigma
\left(  i\right)  }}_{=\sum_{j=1}^{n}f\left(  a_{k,j}\right)  \left(
-1\right)  ^{k+j}\det\left(  A_{\sim k,\sim j}\right)  }\\
&  =\sum_{k=1}^{n}\sum_{j=1}^{n}f\left(  a_{k,j}\right)  \left(  -1\right)
^{k+j}\det\left(  A_{\sim k,\sim j}\right)  =\operatorname*{Tr}\left(
f^{n\times n}\left(  A\right)  \cdot\operatorname*{adj}A\right)
\end{align*}
(by (\ref{pf.thm.deriv.ddet.short.Tr=})). This proves Theorem
\ref{thm.deriv.ddet}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Theorem \ref{thm.deriv.ddet}.]Write the matrix $A\in\mathbb{L}%
^{n\times n}$ in the form $A=\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq
j\leq n}$. Thus, $a_{i,j}\in\mathbb{L}$ for every $\left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}$.

From $A=\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$, we obtain
$f^{n\times n}\left(  A\right)  =\left(  f\left(  a_{i,j}\right)  \right)
_{1\leq i\leq n,\ 1\leq j\leq n}$ (by the definition of $f^{n\times n}$).
Recall that $\operatorname*{adj}A=\left(  \left(  -1\right)  ^{i+j}\det\left(
A_{\sim j,\sim i}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq n}$ (by the
definition of $\operatorname*{adj}A$). Hence, Lemma \ref{lem.Tr(AB)} (applied
to $\mathbb{L}$, $n$, $f^{n\times n}\left(  A\right)  $, $f\left(
a_{i,j}\right)  $, $\operatorname*{adj}A$ and $\left(  -1\right)  ^{i+j}%
\det\left(  A_{\sim j,\sim i}\right)  $ instead of $\mathbb{K}$, $m$, $A$,
$a_{i,j}$, $B$ and $b_{i,j}$) yields%
\begin{align}
\operatorname*{Tr}\left(  f^{n\times n}\left(  A\right)  \cdot
\operatorname*{adj}A\right)   &  =\sum_{i=1}^{n}\sum_{j=1}^{n}f\left(
a_{i,j}\right)  \left(  -1\right)  ^{j+i}\det\left(  A_{\sim i,\sim j}\right)
\nonumber\\
&  =\sum_{k=1}^{n}\sum_{j=1}^{n}f\left(  a_{k,j}\right)  \underbrace{\left(
-1\right)  ^{j+k}}_{\substack{=\left(  -1\right)  ^{k+j}\\\text{(since
}j+k=k+j\text{)}}}\det\left(  A_{\sim k,\sim j}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have renamed the summation index }i\\
\text{as }k\text{ in the outer sum}%
\end{array}
\right) \nonumber\\
&  =\sum_{k=1}^{n}\sum_{j=1}^{n}f\left(  a_{k,j}\right)  \left(  -1\right)
^{k+j}\det\left(  A_{\sim k,\sim j}\right)  . \label{pf.thm.deriv.ddet.Tr=}%
\end{align}


But the map $f$ is a $\mathbb{K}$-derivation, and thus is $\mathbb{K}$-linear.
Now, (\ref{eq.det}) (applied to $\mathbb{L}$ instead of $\mathbb{K}$) shows
that%
\[
\det A=\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{n}a_{i,\sigma\left(  i\right)  }.
\]
Applying the map $f$ to both sides of this equation, we obtain%
\begin{align}
&  f\left(  \det A\right) \nonumber\\
&  =f\left(  \sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{n}a_{i,\sigma\left(  i\right)  }\right) \nonumber\\
&  =\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}f\left(
\underbrace{\prod_{i=1}^{n}a_{i,\sigma\left(  i\right)  }}_{=a_{1,\sigma
\left(  1\right)  }a_{2,\sigma\left(  2\right)  }\cdots a_{n,\sigma\left(
n\right)  }}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }f\text{
is }\mathbb{K}\text{-linear}\right) \nonumber\\
&  =\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\underbrace{f\left(
a_{1,\sigma\left(  1\right)  }a_{2,\sigma\left(  2\right)  }\cdots
a_{n,\sigma\left(  n\right)  }\right)  }_{\substack{=\sum_{k=1}^{n}f\left(
a_{k,\sigma\left(  k\right)  }\right)  \prod_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\i\neq k}}a_{i,\sigma\left(  i\right)  }\\\text{(by
Lemma \ref{lem.deriv.leibniz-comm},}\\\text{applied to }a_{i,\sigma\left(
i\right)  }\text{ instead of }a_{i}\text{)}}}\nonumber\\
&  =\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\sum_{k=1}^{n}f\left(
a_{k,\sigma\left(  k\right)  }\right)  \prod_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\i\neq k}}a_{i,\sigma\left(  i\right)  }\nonumber\\
&  =\underbrace{\sum_{\sigma\in S_{n}}\sum_{k=1}^{n}}_{=\sum_{k=1}^{n}%
\sum_{\sigma\in S_{n}}}\left(  -1\right)  ^{\sigma}f\left(  a_{k,\sigma\left(
k\right)  }\right)  \prod_{\substack{i\in\left\{  1,2,\ldots,n\right\}
;\\i\neq k}}a_{i,\sigma\left(  i\right)  }\nonumber\\
&  =\sum_{k=1}^{n}\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}f\left(
a_{k,\sigma\left(  k\right)  }\right)  \prod_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\i\neq k}}a_{i,\sigma\left(  i\right)  }.
\label{pf.thm.deriv.ddet.1}%
\end{align}
But every $k\in\left\{  1,2,\ldots,n\right\}  $ satisfies%
\begin{align*}
&  \underbrace{\sum_{\sigma\in S_{n}}}_{\substack{=\sum_{j\in\left\{
1,2,\ldots,n\right\}  }\sum_{\substack{\sigma\in S_{n};\\\sigma\left(
k\right)  =j}}\\\text{(since }\sigma\left(  k\right)  \in\left\{
1,2,\ldots,n\right\}  \text{ for each }\sigma\in S_{n}\text{)}}}\left(
-1\right)  ^{\sigma}f\left(  a_{k,\sigma\left(  k\right)  }\right)
\prod_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i\neq k}}a_{i,\sigma
\left(  i\right)  }\\
&  =\sum_{j\in\left\{  1,2,\ldots,n\right\}  }\sum_{\substack{\sigma\in
S_{n};\\\sigma\left(  k\right)  =j}}\left(  -1\right)  ^{\sigma}f\left(
\underbrace{a_{k,\sigma\left(  k\right)  }}_{\substack{=a_{k,j}\\\text{(since
}\sigma\left(  k\right)  =j\text{)}}}\right)  \prod_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\i\neq k}}a_{i,\sigma\left(  i\right)  }\\
&  =\sum_{j\in\left\{  1,2,\ldots,n\right\}  }\underbrace{\sum
_{\substack{\sigma\in S_{n};\\\sigma\left(  k\right)  =j}}\left(  -1\right)
^{\sigma}f\left(  a_{k,j}\right)  \prod_{\substack{i\in\left\{  1,2,\ldots
,n\right\}  ;\\i\neq k}}a_{i,\sigma\left(  i\right)  }}_{=f\left(
a_{k,j}\right)  \sum_{\substack{\sigma\in S_{n};\\\sigma\left(  k\right)
=j}}\left(  -1\right)  ^{\sigma}\prod_{\substack{i\in\left\{  1,2,\ldots
,n\right\}  ;\\i\neq k}}a_{i,\sigma\left(  i\right)  }}\\
&  =\underbrace{\sum_{j\in\left\{  1,2,\ldots,n\right\}  }}_{=\sum_{j=1}^{n}%
}f\left(  a_{k,j}\right)  \underbrace{\sum_{\substack{\sigma\in S_{n}%
;\\\sigma\left(  k\right)  =j}}\left(  -1\right)  ^{\sigma}\prod
_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i\neq k}}a_{i,\sigma\left(
i\right)  }}_{\substack{=\left(  -1\right)  ^{k+j}\det\left(  A_{\sim k,\sim
j}\right)  \\\text{(by Lemma \ref{lem.laplace.Apq}, applied to}\\\mathbb{L}%
\text{, }k\text{ and }j\text{ instead of }\mathbb{K}\text{, }p\text{ and
}q\text{)}}}\\
&  =\sum_{j=1}^{n}f\left(  a_{k,j}\right)  \left(  -1\right)  ^{k+j}%
\det\left(  A_{\sim k,\sim j}\right)  .
\end{align*}
Hence, (\ref{pf.thm.deriv.ddet.1}) becomes%
\begin{align*}
&  f\left(  \det A\right) \\
&  =\sum_{k=1}^{n}\underbrace{\sum_{\sigma\in S_{n}}\left(  -1\right)
^{\sigma}f\left(  a_{k,\sigma\left(  k\right)  }\right)  \prod_{\substack{i\in
\left\{  1,2,\ldots,n\right\}  ;\\i\neq k}}a_{i,\sigma\left(  i\right)  }%
}_{=\sum_{j=1}^{n}f\left(  a_{k,j}\right)  \left(  -1\right)  ^{k+j}%
\det\left(  A_{\sim k,\sim j}\right)  }\\
&  =\sum_{k=1}^{n}\sum_{j=1}^{n}f\left(  a_{k,j}\right)  \left(  -1\right)
^{k+j}\det\left(  A_{\sim k,\sim j}\right)  =\operatorname*{Tr}\left(
f^{n\times n}\left(  A\right)  \cdot\operatorname*{adj}A\right)
\end{align*}
(by (\ref{pf.thm.deriv.ddet.Tr=})). This proves Theorem \ref{thm.deriv.ddet}.
\end{proof}
\end{verlong}

\subsection{The derivative of the characteristic polynomial}

The characteristic polynomial $\chi_{A}$ of a square matrix $A$ is, first of
all, a polynomial; and a polynomial has a derivative. We shall have need for a
formula for this derivative:

\begin{theorem}
\label{thm.ddet}Let $n\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times n}$. Let
$\partial:\mathbb{K}\left[  t\right]  \rightarrow\mathbb{K}\left[  t\right]  $
be the differentiation operator (i.e., the map that sends every polynomial
$f\in\mathbb{K}\left[  t\right]  $ to the derivative of $f$). Then,%
\[
\partial\chi_{A}=\operatorname*{Tr}\left(  \operatorname*{adj}\left(
tI_{n}-A\right)  \right)  .
\]

\end{theorem}

\begin{vershort}
\begin{proof}
[Proof of Theorem \ref{thm.ddet}.]Proposition \ref{prop.deriv.del} shows that
$\partial:\mathbb{K}\left[  t\right]  \rightarrow\mathbb{K}\left[  t\right]  $
is a $\mathbb{K}$-derivation. Now, consider the map $\partial^{n\times
n}:\left(  \mathbb{K}\left[  t\right]  \right)  ^{n\times n}\rightarrow\left(
\mathbb{K}\left[  t\right]  \right)  ^{n\times n}$ (defined according to
Definition \ref{def.f-mat}). It is easy to see that
\begin{equation}
\partial^{n\times n}\left(  tB-A\right)  =B \label{pf.thm.ddet.short.1}%
\end{equation}
for any $n\times n$-matrix $B\in\mathbb{K}^{n\times n}$%
\ \ \ \ \footnote{\textit{Proof.} Let $B\in\mathbb{K}^{n\times n}$ be an
$n\times n$-matrix. Write the matrix $B$ in the form $B=\left(  b_{i,j}%
\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$. Write the matrix $A$ in the form
$A=\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$. Both matrices
$A$ and $B$ belong to $\mathbb{K}^{n\times n}$; thus, every $\left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$ satisfies $a_{i,j}%
\in\mathbb{K}$ and $b_{i,j}\in\mathbb{K}$ and therefore $\partial\left(
tb_{i,j}-a_{i,j}\right)  =b_{i,j}$ (since $\partial$ is the differentiation
operator).
\par
Now,%
\begin{align*}
&  t\underbrace{B}_{=\left(  b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}%
}-\underbrace{A}_{=\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}}\\
&  =t\left(  b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}-\left(
a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}=\left(  tb_{i,j}%
-a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}.
\end{align*}
Hence, the definition of the map $\partial^{n\times n}$ yields%
\[
\partial^{n\times n}\left(  tB-A\right)  =\left(  \underbrace{\partial\left(
tb_{i,j}-a_{i,j}\right)  }_{=b_{i,j}}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n}=\left(  b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}=B,
\]
qed.}. Applying this to $B=I_{n}$, we obtain $\partial^{n\times n}\left(
tI_{n}-A\right)  =I_{n}$.

The definition of $\chi_{A}$ yields $\chi_{A}=\det\left(  tI_{n}-A\right)  $.
Applying the map $\partial$ to both sides of this equality, we obtain%
\begin{align*}
\partial\chi_{A}  &  =\partial\left(  \det\left(  tI_{n}-A\right)  \right)
=\operatorname*{Tr}\left(  \underbrace{\partial^{n\times n}\left(
tI_{n}-A\right)  }_{=I_{n}}\cdot\operatorname*{adj}\left(  tI_{n}-A\right)
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Theorem \ref{thm.deriv.ddet} (applied to }\mathbb{K}\left[  t\right]
\text{, }\partial\text{ and }tI_{n}-A\\
\text{instead of }\mathbb{L}\text{, }f\text{ and }A\text{)}%
\end{array}
\right) \\
&  =\operatorname*{Tr}\left(  \underbrace{I_{n}\cdot\operatorname*{adj}\left(
tI_{n}-A\right)  }_{=\operatorname*{adj}\left(  tI_{n}-A\right)  }\right)
=\operatorname*{Tr}\left(  \operatorname*{adj}\left(  tI_{n}-A\right)
\right)  .
\end{align*}
This proves Theorem \ref{thm.ddet}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Theorem \ref{thm.ddet}.]Proposition \ref{prop.deriv.del} shows that
$\partial:\mathbb{K}\left[  t\right]  \rightarrow\mathbb{K}\left[  t\right]  $
is a $\mathbb{K}$-derivation. A map $\partial^{n\times n}:\left(
\mathbb{K}\left[  t\right]  \right)  ^{n\times n}\rightarrow\left(
\mathbb{K}\left[  t\right]  \right)  ^{n\times n}$ is defined (according to
Definition \ref{def.f-mat}).

It is easy to see that $\partial^{n\times n}\left(  tB-A\right)  =B$ for any
$n\times n$-matrix $B\in\mathbb{K}^{n\times n}$%
\ \ \ \ \footnote{\textit{Proof.} Let $B\in\mathbb{K}^{n\times n}$ be an
$n\times n$-matrix. Write the matrix $B$ in the form $B=\left(  b_{i,j}%
\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$. Thus,
\begin{equation}
b_{i,j}\in\mathbb{K}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2} \label{pf.thm.ddet.fn1.1}%
\end{equation}
(since $B\in\mathbb{K}^{n\times n}$). Also, $t\underbrace{B}_{=\left(
b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}}=t\left(  b_{i,j}\right)
_{1\leq i\leq n,\ 1\leq j\leq n}=\left(  tb_{i,j}\right)  _{1\leq i\leq
n,\ 1\leq j\leq n}$.
\par
Write the matrix $A$ in the form $A=\left(  a_{i,j}\right)  _{1\leq i\leq
n,\ 1\leq j\leq n}$. Thus,%
\begin{equation}
a_{i,j}\in\mathbb{K}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2} \label{pf.thm.ddet.fn1.2}%
\end{equation}
(since $A\in\mathbb{K}^{n\times n}$).
\par
Fix $\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$. Then,
$b_{i,j}\in\mathbb{K}$ (by (\ref{pf.thm.ddet.fn1.1})) and $a_{i,j}%
\in\mathbb{K}$ (by (\ref{pf.thm.ddet.fn1.2})). Hence, the polynomial
$tb_{i,j}-a_{i,j}\in\mathbb{K}\left[  t\right]  $ has the coefficient
$b_{i,j}$ in front of $t^{1}$ and the coefficient $-a_{i,j}$ in front of
$t^{0}$, and all its other coefficients are $0$. Thus,
\begin{equation}
\partial\left(  tb_{i,j}-a_{i,j}\right)  =b_{i,j} \label{pf.thm.ddet.fn1.3}%
\end{equation}
(since $\partial$ is the differentiation operator on $\mathbb{K}\left[
t\right]  $).
\par
Now, forget that we fixed $\left(  i,j\right)  $. We thus have proven
(\ref{pf.thm.ddet.fn1.3}) for every $\left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}$.
\par
Now,%
\[
\underbrace{tB}_{=\left(  tb_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}%
}-\underbrace{A}_{=\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}%
}=\left(  tb_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}-\left(
a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}=\left(  tb_{i,j}%
-a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}.
\]
Hence, the definition of the map $\partial^{n\times n}$ yields%
\[
\partial^{n\times n}\left(  tB-A\right)  =\left(  \underbrace{\partial\left(
tb_{i,j}-a_{i,j}\right)  }_{\substack{=b_{i,j}\\\text{(by
(\ref{pf.thm.ddet.fn1.3}))}}}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}=\left(
b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}=B,
\]
qed.}. Applying this to $B=I_{n}$, we obtain $\partial^{n\times n}\left(
tI_{n}-A\right)  =I_{n}$.

The definition of $\chi_{A}$ yields $\chi_{A}=\det\left(  tI_{n}-A\right)  $.
Applying the map $\partial$ to both sides of this equality, we obtain%
\begin{align*}
\partial\chi_{A}  &  =\partial\left(  \det\left(  tI_{n}-A\right)  \right)
=\operatorname*{Tr}\left(  \underbrace{\partial^{n\times n}\left(
tI_{n}-A\right)  }_{=I_{n}}\cdot\operatorname*{adj}\left(  tI_{n}-A\right)
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Theorem \ref{thm.deriv.ddet} (applied to }\mathbb{K}\left[  t\right]
\text{, }\partial\text{ and }tI_{n}-A\\
\text{instead of }\mathbb{L}\text{, }f\text{ and }A\text{)}%
\end{array}
\right) \\
&  =\operatorname*{Tr}\left(  \underbrace{I_{n}\cdot\operatorname*{adj}\left(
tI_{n}-A\right)  }_{=\operatorname*{adj}\left(  tI_{n}-A\right)  }\right)
=\operatorname*{Tr}\left(  \operatorname*{adj}\left(  tI_{n}-A\right)
\right)  .
\end{align*}
This proves Theorem \ref{thm.ddet}.
\end{proof}
\end{verlong}

We can use Theorem \ref{thm.ddet} to obtain the following result:

\begin{proposition}
\label{prop.TrD}Let $n\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times n}$. For
every $j\in\mathbb{Z}$, define an element $c_{j}\in\mathbb{K}$ by
$c_{j}=\left[  t^{n-j}\right]  \chi_{A}$.

Let $D_{0},D_{1},\ldots,D_{n-1}$ be $n$ matrices in $\mathbb{K}^{n\times n}$
satisfying (\ref{eq.lem.CH.D.adj}). Thus, an $n$-tuple $\left(  D_{0}%
,D_{1},\ldots,D_{n-1}\right)  $ of matrices in $\mathbb{K}^{n\times n}$ is
defined. Extend this $n$-tuple to a family $\left(  D_{k}\right)
_{k\in\mathbb{Z}}$ of matrices in $\mathbb{K}^{n\times n}$ by setting
(\ref{eq.lem.CH.D.0}). Then, every $k\in\mathbb{Z}$ satisfies%
\begin{equation}
\operatorname*{Tr}\left(  D_{k}\right)  =\left(  k+1\right)  c_{n-\left(
k+1\right)  }. \label{eq.prop.TrD.claim}%
\end{equation}

\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.TrD}.]Let $\partial:\mathbb{K}\left[
t\right]  \rightarrow\mathbb{K}\left[  t\right]  $ be the differentiation
operator (i.e., the map that sends every polynomial $f\in\mathbb{K}\left[
t\right]  $ to the derivative of $f$).

Lemma \ref{lem.CH.D} \textbf{(a)} yields $\chi_{A}=\sum_{k=0}^{n}c_{n-k}t^{k}%
$. Applying the map $\partial$ to both sides of this equality, we obtain%
\begin{align*}
\partial\chi_{A}  &  =\partial\left(  \sum_{k=0}^{n}c_{n-k}t^{k}\right)
=\sum_{k=1}^{n}c_{n-k}kt^{k-1}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\partial\text{ is the differentiation}\\
\text{operator}%
\end{array}
\right) \\
&  =\sum_{k=1}^{n}kc_{n-k}t^{k-1}=\sum_{k=0}^{n-1}\left(  k+1\right)
c_{n-\left(  k+1\right)  }t^{k}%
\end{align*}
(here, we have substituted $k+1$ for $k$ in the sum). Comparing this with%
\begin{align*}
\partial\chi_{A}  &  =\operatorname*{Tr}\left(
\underbrace{\operatorname*{adj}\left(  tI_{n}-A\right)  }_{\substack{=\sum
_{k=0}^{n-1}t^{k}D_{k}\\\text{(by (\ref{eq.lem.CH.D.adj}))}}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem \ref{thm.ddet}}\right) \\
&  =\operatorname*{Tr}\left(  \sum_{k=0}^{n-1}t^{k}D_{k}\right)  =\sum
_{k=0}^{n-1}t^{k}\operatorname*{Tr}\left(  D_{k}\right)  =\sum_{k=0}%
^{n-1}\operatorname*{Tr}\left(  D_{k}\right)  t^{k},
\end{align*}
we obtain%
\[
\sum_{k=0}^{n-1}\operatorname*{Tr}\left(  D_{k}\right)  t^{k}=\sum_{k=0}%
^{n-1}\left(  k+1\right)  c_{n-\left(  k+1\right)  }t^{k}.
\]
This is an identity between two polynomials in $\mathbb{K}\left[  t\right]  $.
Comparing coefficients on both sides of this identity, we conclude that%
\begin{equation}
\operatorname*{Tr}\left(  D_{k}\right)  =\left(  k+1\right)  c_{n-\left(
k+1\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\left\{  0,1,\ldots
,n-1\right\}  . \label{pf.prop.TrD.short.c=T}%
\end{equation}


Now, let $k\in\mathbb{Z}$. We must prove (\ref{eq.prop.TrD.claim}).

If $k\in\left\{  0,1,\ldots,n-1\right\}  $, then (\ref{eq.prop.TrD.claim})
follows immediately from (\ref{pf.prop.TrD.short.c=T}). Hence, for the rest of
this proof, we WLOG assume that we don't have $k\in\left\{  0,1,\ldots
,n-1\right\}  $.

We don't have $k\in\left\{  0,1,\ldots,n-1\right\}  $. Thus, $k\in
\mathbb{Z}\setminus\left\{  0,1,\ldots,n-1\right\}  $. Hence,
(\ref{eq.lem.CH.D.0}) yields $D_{k}=0_{n\times n}$, so that
$\operatorname*{Tr}\left(  D_{k}\right)  =\operatorname*{Tr}\left(  0_{n\times
n}\right)  =0$.

Recall again that $k\in\mathbb{Z}\setminus\left\{  0,1,\ldots,n-1\right\}  $.
In other words, we have either $k<0$ or $k\geq n$. Thus, we are in one of the
following two cases:

\textit{Case 1:} We have $k<0$.

\textit{Case 2:} We have $k\geq n$.

Let us first consider Case 1. In this case, we have $k<0$. If $k=-1$, then
(\ref{eq.prop.TrD.claim}) holds\footnote{\textit{Proof.} Assume that $k=-1$.
Then, $k+1=0$, so that $\underbrace{\left(  k+1\right)  }_{=0}c_{n-\left(
k+1\right)  }=0$. Comparing this with $\operatorname*{Tr}\left(  D_{k}\right)
=0$, we obtain $\operatorname*{Tr}\left(  D_{k}\right)  =\left(  k+1\right)
c_{n-\left(  k+1\right)  }$; hence, (\ref{eq.prop.TrD.claim}) holds, qed.}.
Hence, for the rest of this proof, we WLOG assume that $k\neq-1$. Combining
$k<0$ with $k\neq-1$, we obtain $k<-1$. Hence, $k+1<0$.

The definition of $c_{n-\left(  k+1\right)  }$ yields $c_{n-\left(
k+1\right)  }=\left[  \underbrace{t^{n-\left(  n-\left(  k+1\right)  \right)
}}_{=t^{k+1}}\right]  \chi_{A}=\left[  t^{k+1}\right]  \chi_{A}=0$ (since
$k+1<0$, but $\chi_{A}$ is a polynomial). Hence, $\left(  k+1\right)
\underbrace{c_{n-\left(  k+1\right)  }}_{=0}=0$. Comparing this with
$\operatorname*{Tr}\left(  D_{k}\right)  =0$, we obtain $\operatorname*{Tr}%
\left(  D_{k}\right)  =\left(  k+1\right)  c_{n-\left(  k+1\right)  }$. Hence,
(\ref{eq.prop.TrD.claim}) is proven in Case 1.

Let us now consider Case 2. In this case, we have $k\geq n$. Thus, $k+1\geq
n+1>n$.

But $\chi_{A}$ is a polynomial of degree $\leq n$. Hence, $\left[
t^{m}\right]  \chi_{A}=0$ for every integer $m>n$. Applying this to $m=k+1$,
we obtain $\left[  t^{k+1}\right]  \chi_{A}=0$ (since $k+1>n$).

The definition of $c_{n-\left(  k+1\right)  }$ yields $c_{n-\left(
k+1\right)  }=\left[  \underbrace{t^{n-\left(  n-\left(  k+1\right)  \right)
}}_{=t^{k+1}}\right]  \chi_{A}=\left[  t^{k+1}\right]  \chi_{A}=0$. Hence,
$\left(  k+1\right)  \underbrace{c_{n-\left(  k+1\right)  }}_{=0}=0$.
Comparing this with $\operatorname*{Tr}\left(  D_{k}\right)  =0$, we obtain
$\operatorname*{Tr}\left(  D_{k}\right)  =\left(  k+1\right)  c_{n-\left(
k+1\right)  }$. Hence, (\ref{eq.prop.TrD.claim}) is proven in Case 2.

We have now proven (\ref{eq.prop.TrD.claim}) in each of the two Cases 1 and 2.
Thus, (\ref{eq.prop.TrD.claim}) always holds. Thus, Proposition \ref{prop.TrD}
is proven.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.TrD}.]For every commutative ring $\mathbb{L}$,
we consider the map $\operatorname*{Tr}:\mathbb{L}^{n\times n}\rightarrow
\mathbb{L}$ which sends every matrix $B\in\mathbb{L}^{n\times n}$ to
$\operatorname*{Tr}B\in\mathbb{L}$. This map $\operatorname*{Tr}%
:\mathbb{L}^{n\times n}\rightarrow\mathbb{L}$ is $\mathbb{L}$-linear. Applying
this to $\mathbb{L}=\mathbb{K}\left[  t\right]  $, we conclude that the map
$\operatorname*{Tr}:\left(  \mathbb{K}\left[  t\right]  \right)  ^{n\times
n}\rightarrow\mathbb{K}\left[  t\right]  $ is $\mathbb{K}\left[  t\right]  $-linear.

Let $\partial:\mathbb{K}\left[  t\right]  \rightarrow\mathbb{K}\left[
t\right]  $ be the differentiation operator (i.e., the map that sends every
polynomial $f\in\mathbb{K}\left[  t\right]  $ to the derivative of $f$).

Lemma \ref{lem.CH.D} \textbf{(a)} yields $\chi_{A}=\sum_{k=0}^{n}c_{n-k}t^{k}%
$. Applying the map $\partial$ to both sides of this equality, we obtain%
\begin{align*}
\partial\chi_{A}  &  =\partial\left(  \sum_{k=0}^{n}c_{n-k}t^{k}\right)
=\sum_{k=1}^{n}c_{n-k}kt^{k-1}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\partial\text{ is the differentiation}\\
\text{operator}%
\end{array}
\right) \\
&  =\sum_{k=0}^{n-1}\underbrace{c_{n-\left(  k+1\right)  }\left(  k+1\right)
}_{=\left(  k+1\right)  c_{n-\left(  k+1\right)  }}\underbrace{t^{\left(
k+1\right)  -1}}_{\substack{=t^{k}\\\text{(since }\left(  k+1\right)
-1=k\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have substituted }k+1\text{ for
}k\text{ in the sum}\right) \\
&  =\sum_{k=0}^{n-1}\left(  k+1\right)  c_{n-\left(  k+1\right)  }t^{k}.
\end{align*}
Thus,%
\begin{align*}
&  \sum_{k=0}^{n-1}\left(  k+1\right)  c_{n-\left(  k+1\right)  }t^{k}\\
&  =\partial\chi_{A}=\operatorname*{Tr}\left(  \underbrace{\operatorname*{adj}%
\left(  tI_{n}-A\right)  }_{\substack{=\sum_{k=0}^{n-1}t^{k}D_{k}\\\text{(by
(\ref{eq.lem.CH.D.adj}))}}}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
Theorem \ref{thm.ddet}}\right) \\
&  =\operatorname*{Tr}\left(  \sum_{k=0}^{n-1}t^{k}D_{k}\right)  =\sum
_{k=0}^{n-1}\underbrace{t^{k}\operatorname*{Tr}\left(  D_{k}\right)
}_{=\operatorname*{Tr}\left(  D_{k}\right)  t^{k}}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the map }\operatorname*{Tr}:\left(  \mathbb{K}\left[  t\right]
\right)  ^{n\times n}\rightarrow\mathbb{K}\left[  t\right] \\
\text{is }\mathbb{K}\left[  t\right]  \text{-linear}%
\end{array}
\right) \\
&  =\sum_{k=0}^{n-1}\operatorname*{Tr}\left(  D_{k}\right)  t^{k}.
\end{align*}
This is an identity between two polynomials in $\mathbb{K}\left[  t\right]  $.
Comparing coefficients on both sides of this identity, we conclude that%
\begin{equation}
\left(  k+1\right)  c_{n-\left(  k+1\right)  }=\operatorname*{Tr}\left(
D_{k}\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }k\in\left\{  0,1,\ldots
,n-1\right\}  . \label{pf.prop.TrD.c=T}%
\end{equation}


On the other hand, we have%
\begin{equation}
c_{n-k}=0\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\mathbb{Z}\setminus\left\{
0,1,\ldots,n\right\}  \label{pf.prop.TrD.c=0}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.TrD.c=0}):} We know that $\chi_{A}%
\in\mathbb{K}\left[  t\right]  $ is a polynomial of degree $\leq n$ in $t$ (by
Corollary \ref{cor.chiA} \textbf{(a)}). Hence,
\begin{equation}
\left[  t^{k}\right]  \chi_{A}=0\ \ \ \ \ \ \ \ \ \ \text{for every integer
}k>n. \label{pf.prop.TrD.c=0.pf.1}%
\end{equation}
Also, clearly, $\left[  t^{k}\right]  f=0$ for every negative integer $k$ and
every polynomial $f\in\mathbb{K}\left[  t\right]  $. Applying this to
$f=\chi_{A}$, we conclude that%
\begin{equation}
\left[  t^{k}\right]  \chi_{A}=0\ \ \ \ \ \ \ \ \ \ \text{for every negative
integer }k. \label{pf.prop.TrD.c=0.pf.2}%
\end{equation}
\par
Now, let $k\in\mathbb{Z}\setminus\left\{  0,1,\ldots,n\right\}  $. We must
prove that $c_{n-k}=0$. We have%
\begin{align*}
c_{n-k}  &  =\underbrace{\left[  t^{n-\left(  n-k\right)  }\right]
}_{=\left[  t^{k}\right]  }\chi_{A}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the
definition of }c_{n-k}\right) \\
&  =\left[  t^{k}\right]  \chi_{A}.
\end{align*}
\par
Now, we are in one of the following two cases:
\par
\textit{Case 1:} We have $k<0$.
\par
\textit{Case 2:} We have $k\geq0$.
\par
Let us first consider Case 1. In this case, we have $k<0$. Hence, $k$ is a
negative integer. Thus, (\ref{pf.prop.TrD.c=0.pf.2}) yields $\left[
t^{k}\right]  \chi_{A}=0$. Thus, $c_{n-k}=\left[  t^{k}\right]  \chi_{A}=0$.
Hence, $c_{n-k}=0$ is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $k\geq0$. Hence,
$k\in\mathbb{N}$. But $k\in\mathbb{Z}\setminus\left\{  0,1,\ldots,n\right\}
$; in other words, $k\in\mathbb{Z}$ and $k\notin\left\{  0,1,\ldots,n\right\}
$. Combining $k\in\mathbb{N}$ with $k\notin\left\{  0,1,\ldots,n\right\}  $,
we obtain $k\in\mathbb{N}\setminus\left\{  0,1,\ldots,n\right\}  =\left\{
n+1,n+2,n+3,\ldots\right\}  $. Hence, $k\geq n+1>n$, and thus $\left[
t^{k}\right]  \chi_{A}=0$ (by (\ref{pf.prop.TrD.c=0.pf.1})). Thus,
$c_{n-k}=\left[  t^{k}\right]  \chi_{A}=0$. Hence, $c_{n-k}=0$ is proven in
Case 2.
\par
Now, we have proven $c_{n-k}=0$ in each of the two Cases 1 and 2. Since these
two Cases cover all possibilities, this shows that $c_{n-k}=0$ always holds.
This proves (\ref{pf.prop.TrD.c=0}).}.

Now, let $k\in\mathbb{Z}$. We must prove (\ref{eq.prop.TrD.claim}).

If $k\in\left\{  0,1,\ldots,n-1\right\}  $, then (\ref{eq.prop.TrD.claim})
follows immediately from (\ref{pf.prop.TrD.c=T}). Hence, for the rest of this
proof, we can WLOG assume that we don't have $k\in\left\{  0,1,\ldots
,n-1\right\}  $. Assume this.

We are in one of the following three cases:

\textit{Case 1:} We have $k+1\in\left\{  0,1,\ldots,n\right\}  $.

\textit{Case 2:} We have $k+1\notin\left\{  0,1,\ldots,n\right\}  $.

Let us first consider Case 1. In this case, we have $k+1\in\left\{
0,1,\ldots,n\right\}  $. Hence, $\left(  k+1\right)  -1\in\left\{
-1,0,\ldots,n-1\right\}  $. Thus, $k=\left(  k+1\right)  -1\in\left\{
-1,0,\ldots,n-1\right\}  $.

But $k\notin\left\{  0,1,\ldots,n-1\right\}  $ (since we don't have
$k\in\left\{  0,1,\ldots,n-1\right\}  $). Combining $k\in\left\{
-1,0,\ldots,n-1\right\}  $ with $k\notin\left\{  0,1,\ldots,n-1\right\}  $, we
obtain%
\[
k\in\left\{  -1,0,\ldots,n-1\right\}  \setminus\left\{  0,1,\ldots
,n-1\right\}  \subseteq\left\{  -1\right\}  .
\]
In other words, $k=-1$.

Now, $k=-1<0$, so that $k\notin\left\{  0,1,\ldots,n-1\right\}  $. Combining
$k\in\mathbb{Z}$ with $k\notin\left\{  0,1,\ldots,n-1\right\}  $, we obtain
$k\in\mathbb{Z}\setminus\left\{  0,1,\ldots,n-1\right\}  $. Thus,
(\ref{eq.lem.CH.D.0}) yields $D_{k}=0_{n\times n}$, so that
$\operatorname*{Tr}\left(  \underbrace{D_{k}}_{=0_{n\times n}}\right)
=\operatorname*{Tr}\left(  0_{n\times n}\right)  =0$. Comparing this with
$\underbrace{\left(  k+1\right)  }_{\substack{=0\\\text{(since }k=-1\text{)}%
}}c_{n-\left(  k+1\right)  }=0$, we obtain $\operatorname*{Tr}\left(
D_{k}\right)  =\left(  k+1\right)  c_{n-\left(  k+1\right)  }$. Thus,
(\ref{eq.prop.TrD.claim}) is proven in Case 1.

Let us now consider Case 2. In this case, we have $k+1\notin\left\{
0,1,\ldots,n\right\}  $. Combining $k+1\in\mathbb{Z}$ with $k+1\notin\left\{
0,1,\ldots,n\right\}  $, we obtain $k+1\in\mathbb{Z}\setminus\left\{
0,1,\ldots,n\right\}  $. Thus, (\ref{pf.prop.TrD.c=0}) (applied to $k+1$
instead of $k$) yields $c_{n-\left(  k+1\right)  }=0$. On the other hand,
$k\notin\left\{  0,1,\ldots,n-1\right\}  $\ \ \ \ \footnote{\textit{Proof.}
Assume the contrary. Thus, $k\in\left\{  0,1,\ldots,n-1\right\}  $. Hence,
$k+1\in\left\{  1,2,\ldots,n\right\}  \subseteq\left\{  0,1,\ldots,n\right\}
$. This contradicts $k+1\notin\left\{  0,1,\ldots,n\right\}  $. This
contradiction proves that our assumption was wrong, qed.}. Combining
$k\in\mathbb{Z}$ with $k\notin\left\{  0,1,\ldots,n-1\right\}  $, we obtain
$k\in\mathbb{Z}\setminus\left\{  0,1,\ldots,n-1\right\}  $. Hence,
(\ref{eq.lem.CH.D.0}) yields $D_{k}=0_{n\times n}$. Thus, $\operatorname*{Tr}%
\left(  \underbrace{D_{k}}_{=0_{n\times n}}\right)  =\operatorname*{Tr}\left(
0_{n\times n}\right)  =0$. Comparing this with $\left(  k+1\right)
\underbrace{c_{n-\left(  k+1\right)  }}_{=0}=0$, we obtain $\operatorname*{Tr}%
\left(  D_{k}\right)  =\left(  k+1\right)  c_{n-\left(  k+1\right)  }$. Thus,
(\ref{eq.prop.TrD.claim}) is proven in Case 2.

We have now proven (\ref{eq.prop.TrD.claim}) in each of the two Cases 1 and 2.
Since these two Cases cover all possibilities, this shows that
(\ref{eq.prop.TrD.claim}) always holds. Thus, Proposition \ref{prop.TrD} is proven.
\end{proof}
\end{verlong}

\subsection{Proof of the trace Cayley-Hamilton theorem}

Now, we can finally prove the trace Cayley-Hamilton theorem itself:

\begin{vershort}
\begin{proof}
[Proof of Theorem \ref{thm.TCH}.]Proposition \ref{prop.adj.poly} shows that
there exist $n$ matrices $D_{0},D_{1},\ldots,D_{n-1}$ in $\mathbb{K}^{n\times
n}$ such that%
\[
\operatorname*{adj}\left(  tI_{n}-A\right)  =\sum_{k=0}^{n-1}t^{k}%
D_{k}\ \ \ \ \ \ \ \ \ \ \text{in }\left(  \mathbb{K}\left[  t\right]
\right)  ^{n\times n}.
\]
Consider these $D_{0},D_{1},\ldots,D_{n-1}$. Thus, an $n$-tuple $\left(
D_{0},D_{1},\ldots,D_{n-1}\right)  $ of matrices in $\mathbb{K}^{n\times n}$
is defined. Extend this $n$-tuple to a family $\left(  D_{k}\right)
_{k\in\mathbb{Z}}$ of matrices in $\mathbb{K}^{n\times n}$ by setting%
\[
\left(  D_{k}=0_{n\times n}\ \ \ \ \ \ \ \ \ \ \text{for every }k\in
\mathbb{Z}\setminus\left\{  0,1,\ldots,n-1\right\}  \right)  .
\]


Now, let $k\in\mathbb{N}$. Then, Proposition \ref{prop.TrD} (applied to
$n-1-k$ instead of $k$) yields
\[
\operatorname*{Tr}\left(  D_{n-1-k}\right)  =\underbrace{\left(  \left(
n-1-k\right)  +1\right)  }_{=n-k}\underbrace{c_{n-\left(  \left(
n-1-k\right)  +1\right)  }}_{\substack{=c_{k}\\\text{(since }n-\left(  \left(
n-1-k\right)  +1\right)  =k\text{)}}}=\left(  n-k\right)  c_{k}.
\]
Thus,%
\begin{align*}
\left(  n-k\right)  c_{k}  &  =\operatorname*{Tr}\left(  \underbrace{D_{n-1-k}%
}_{\substack{=\sum_{i=0}^{k}c_{k-i}A^{i}\\\text{(by Lemma \ref{lem.CH.D}
\textbf{(c)})}}}\right)  =\operatorname*{Tr}\left(  \sum_{i=0}^{k}c_{k-i}%
A^{i}\right) \\
&  =\sum_{i=0}^{k}c_{k-i}\operatorname*{Tr}\left(  A^{i}\right)
=\underbrace{c_{k-0}}_{=c_{k}}\operatorname*{Tr}\left(  \underbrace{A^{0}%
}_{=I_{n}}\right)  +\sum_{i=1}^{k}c_{k-i}\operatorname*{Tr}\left(
A^{i}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}i=0\text{ from the sum}\right) \\
&  =c_{k}\underbrace{\operatorname*{Tr}\left(  I_{n}\right)  }_{=n}+\sum
_{i=1}^{k}\underbrace{c_{k-i}\operatorname*{Tr}\left(  A^{i}\right)
}_{=\operatorname*{Tr}\left(  A^{i}\right)  c_{k-i}}=c_{k}n+\sum_{i=1}%
^{k}\operatorname*{Tr}\left(  A^{i}\right)  c_{k-i}.
\end{align*}
Solving this equation for $\sum_{i=1}^{k}\operatorname*{Tr}\left(
A^{i}\right)  c_{k-i}$, we obtain%
\[
\sum_{i=1}^{k}\operatorname*{Tr}\left(  A^{i}\right)  c_{k-i}%
=\underbrace{\left(  n-k\right)  c_{k}}_{=nc_{k}-kc_{k}}-\underbrace{c_{k}%
n}_{=nc_{k}}=nc_{k}-kc_{k}-nc_{k}=-kc_{k}.
\]
Adding $kc_{k}$ to both sides of this equation, we obtain $kc_{k}+\sum
_{i=1}^{k}\operatorname*{Tr}\left(  A^{i}\right)  c_{k-i}=0$. This proves
Theorem \ref{thm.TCH}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Theorem \ref{thm.TCH}.]Consider the map $\operatorname*{Tr}%
:\mathbb{K}^{n\times n}\rightarrow\mathbb{K}$ which sends every matrix
$B\in\mathbb{K}^{n\times n}$ to $\operatorname*{Tr}B\in\mathbb{K}$. This map
$\operatorname*{Tr}:\mathbb{K}^{n\times n}\rightarrow\mathbb{K}$ is
$\mathbb{K}$-linear.

The identity matrix $I_{n}$ has diagonal entries $1,1,\ldots,1$. Hence, its
trace is%
\[
\operatorname*{Tr}\left(  I_{n}\right)  =\underbrace{1+1+\cdots+1}_{n\text{
times}}=n\cdot1=n.
\]


Proposition \ref{prop.adj.poly} shows that there exist $n$ matrices
$D_{0},D_{1},\ldots,D_{n-1}$ in $\mathbb{K}^{n\times n}$ such that%
\[
\operatorname*{adj}\left(  tI_{n}-A\right)  =\sum_{k=0}^{n-1}t^{k}%
D_{k}\ \ \ \ \ \ \ \ \ \ \text{in }\left(  \mathbb{K}\left[  t\right]
\right)  ^{n\times n}.
\]
Consider these $D_{0},D_{1},\ldots,D_{n-1}$. Thus, an $n$-tuple $\left(
D_{0},D_{1},\ldots,D_{n-1}\right)  $ of matrices in $\mathbb{K}^{n\times n}$
is defined. Extend this $n$-tuple to a family $\left(  D_{k}\right)
_{k\in\mathbb{Z}}$ of matrices in $\mathbb{K}^{n\times n}$ by setting%
\[
\left(  D_{k}=0_{n\times n}\ \ \ \ \ \ \ \ \ \ \text{for every }k\in
\mathbb{Z}\setminus\left\{  0,1,\ldots,n-1\right\}  \right)  .
\]


Now, let $k\in\mathbb{N}$. Then, Proposition \ref{prop.TrD} (applied to
$n-1-k$ instead of $k$) yields
\[
\operatorname*{Tr}\left(  D_{n-1-k}\right)  =\underbrace{\left(  \left(
n-1-k\right)  +1\right)  }_{=n-k}\underbrace{c_{n-\left(  \left(
n-1-k\right)  +1\right)  }}_{\substack{=c_{k}\\\text{(since }n-\left(  \left(
n-1-k\right)  +1\right)  =k\text{)}}}=\left(  n-k\right)  c_{k}.
\]
Thus,%
\begin{align*}
\left(  n-k\right)  c_{k}  &  =\operatorname*{Tr}\left(  \underbrace{D_{n-1-k}%
}_{\substack{=\sum_{i=0}^{k}c_{k-i}A^{i}\\\text{(by Lemma \ref{lem.CH.D}
\textbf{(c)})}}}\right)  =\operatorname*{Tr}\left(  \sum_{i=0}^{k}c_{k-i}%
A^{i}\right) \\
&  =\sum_{i=0}^{k}c_{k-i}\operatorname*{Tr}\left(  A^{i}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since the map }\operatorname*{Tr}%
:\mathbb{K}^{n\times n}\rightarrow\mathbb{K}\text{ is }\mathbb{K}%
\text{-linear}\right) \\
&  =\underbrace{c_{k-0}}_{=c_{k}}\operatorname*{Tr}\left(  \underbrace{A^{0}%
}_{=I_{n}}\right)  +\sum_{i=1}^{k}c_{k-i}\operatorname*{Tr}\left(
A^{i}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }i=0\text{ from the sum}\\
\text{(since }0\in\left\{  0,1,\ldots,k\right\}  \text{ (since }k\in
\mathbb{N}\text{))}%
\end{array}
\right) \\
&  =c_{k}\underbrace{\operatorname*{Tr}\left(  I_{n}\right)  }_{=n}+\sum
_{i=1}^{k}\underbrace{c_{k-i}\operatorname*{Tr}\left(  A^{i}\right)
}_{=\operatorname*{Tr}\left(  A^{i}\right)  c_{k-i}}=c_{k}n+\sum_{i=1}%
^{k}\operatorname*{Tr}\left(  A^{i}\right)  c_{k-i}.
\end{align*}
Solving this equation for $\sum_{i=1}^{k}\operatorname*{Tr}\left(
A^{i}\right)  c_{k-i}$, we obtain%
\[
\sum_{i=1}^{k}\operatorname*{Tr}\left(  A^{i}\right)  c_{k-i}%
=\underbrace{\left(  n-k\right)  c_{k}}_{=nc_{k}-kc_{k}}-\underbrace{c_{k}%
n}_{=nc_{k}}=nc_{k}-kc_{k}-nc_{k}=-kc_{k}.
\]
Adding $kc_{k}$ to both sides of this equation, we obtain $kc_{k}+\sum
_{i=1}^{k}\operatorname*{Tr}\left(  A^{i}\right)  c_{k-i}=0$. This proves
Theorem \ref{thm.TCH}.
\end{proof}
\end{verlong}

\subsection{A corollary}

The following fact (which can also be easily proven by other means) follows
readily from Theorem \ref{thm.TCH}:

\begin{corollary}
\label{cor.TCH.c1}Let $n\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times n}$.
Then, $\left[  t^{n-1}\right]  \chi_{A}=-\operatorname*{Tr}A$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.TCH.c1}.]For every $j\in\mathbb{Z}$, define an
element $c_{j}\in\mathbb{K}$ by $c_{j}=\left[  t^{n-j}\right]  \chi_{A}$. The
definition of $c_{1}$ yields $c_{1}=\left[  t^{n-1}\right]  \chi_{A}$. The
definition of $c_{0}$ yields $c_{0}=\underbrace{\left[  t^{n-0}\right]
}_{=\left[  t^{n}\right]  }\chi_{A}=\left[  t^{n}\right]  \chi_{A}=1$ (by
Corollary \ref{cor.chiA} \textbf{(c)}).

Theorem \ref{thm.TCH} (applied to $k=1$) yields $1c_{1}+\sum_{i=1}%
^{1}\operatorname*{Tr}\left(  A^{i}\right)  c_{1-i}=0$. Thus,%
\[
1c_{1}=-\underbrace{\sum_{i=1}^{1}\operatorname*{Tr}\left(  A^{i}\right)
c_{1-i}}_{=\operatorname*{Tr}\left(  A^{1}\right)  c_{1-1}}%
=-\operatorname*{Tr}\left(  \underbrace{A^{1}}_{=A}\right)
\underbrace{c_{1-1}}_{=c_{0}=1}=-\operatorname*{Tr}A.
\]
Comparing this with $1c_{1}=c_{1}=\left[  t^{n-1}\right]  \chi_{A}$, we obtain
$\left[  t^{n-1}\right]  \chi_{A}=-\operatorname*{Tr}A$. This proves Corollary
\ref{cor.TCH.c1}.
\end{proof}

\section{Application: Nilpotency and traces}

\subsection{A nilpotency criterion}

As an application of Theorem \ref{thm.TCH}, let us now prove the following
fact (generalizing \cite{m.se1798703}):

\begin{corollary}
\label{cor.nilpcrit}Let $n\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times n}$.
Assume that
\begin{equation}
\operatorname*{Tr}\left(  A^{i}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every
}i\in\left\{  1,2,\ldots,n\right\}  . \label{eq.cor.nilpcrit.cond}%
\end{equation}


\textbf{(a)} Then, $n!A^{n}=0_{n\times n}$.

\textbf{(b)} If $\mathbb{K}$ is a commutative $\mathbb{Q}$-algebra, then
$A^{n}=0_{n\times n}$.

\textbf{(c)} We have $n!\chi_{A}=n!t^{n}$.

\textbf{(d)} If $\mathbb{K}$ is a commutative $\mathbb{Q}$-algebra, then
$\chi_{A}=t^{n}$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.nilpcrit}.]For every $j\in\mathbb{Z}$, define an
element $c_{j}\in\mathbb{K}$ by $c_{j}=\left[  t^{n-j}\right]  \chi_{A}$. The
definition of $c_{0}$ yields $c_{0}=\underbrace{\left[  t^{n-0}\right]
}_{=\left[  t^{n}\right]  }\chi_{A}=\left[  t^{n}\right]  \chi_{A}=1$ (by
Corollary \ref{cor.chiA} \textbf{(c)}).

We now claim that%
\begin{equation}
kc_{k}=0\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\left\{  1,2,\ldots
,n\right\}  . \label{pf.cor.nilpcrit.main}%
\end{equation}


\begin{vershort}
[\textit{Proof of (\ref{pf.cor.nilpcrit.main}):} Let $k\in\left\{
1,2,\ldots,n\right\}  $. Then, every $i\in\left\{  1,2,\ldots,k\right\}  $
satisfies $i\in\left\{  1,2,\ldots,n\right\}  $ and therefore also
\begin{equation}
\operatorname*{Tr}\left(  A^{i}\right)  =0
\label{pf.cor.nilpcrit.main.pf.short.2}%
\end{equation}
(by (\ref{eq.cor.nilpcrit.cond})). Now, Theorem \ref{thm.TCH} yields%
\[
kc_{k}+\sum_{i=1}^{k}\operatorname*{Tr}\left(  A^{i}\right)  c_{k-i}=0.
\]
Solving this equation for $kc_{k}$, we obtain%
\[
kc_{k}=-\sum_{i=1}^{k}\underbrace{\operatorname*{Tr}\left(  A^{i}\right)
}_{\substack{=0\\\text{(by (\ref{pf.cor.nilpcrit.main.pf.short.2}))}}%
}c_{k-i}=-\underbrace{\sum_{i=1}^{k}0c_{k-i}}_{=0}=-0=0.
\]
This proves (\ref{pf.cor.nilpcrit.main}).]
\end{vershort}

\begin{verlong}
[\textit{Proof of (\ref{pf.cor.nilpcrit.main}):} Let $k\in\left\{
1,2,\ldots,n\right\}  $. Then, $k\leq n$, so that $\left\{  1,2,\ldots
,k\right\}  \subseteq\left\{  1,2,\ldots,n\right\}  $. Thus, every
$i\in\left\{  1,2,\ldots,k\right\}  $ satisfies $i\in\left\{  1,2,\ldots
,k\right\}  \subseteq\left\{  1,2,\ldots,n\right\}  $ and therefore also
\begin{equation}
\operatorname*{Tr}\left(  A^{i}\right)  =0 \label{pf.cor.nilpcrit.main.pf.2}%
\end{equation}
(by (\ref{eq.cor.nilpcrit.cond})). Now, Theorem \ref{thm.TCH} yields%
\[
kc_{k}+\sum_{i=1}^{k}\operatorname*{Tr}\left(  A^{i}\right)  c_{k-i}=0.
\]
Solving this equation for $kc_{k}$, we obtain%
\[
kc_{k}=-\sum_{i=1}^{k}\underbrace{\operatorname*{Tr}\left(  A^{i}\right)
}_{\substack{=0\\\text{(by (\ref{pf.cor.nilpcrit.main.pf.2}))}}}c_{k-i}%
=-\underbrace{\sum_{i=1}^{k}0c_{k-i}}_{=0}=-0=0.
\]
This proves (\ref{pf.cor.nilpcrit.main}).]
\end{verlong}

Now, we claim that%
\begin{equation}
n!c_{k}=0\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\left\{  1,2,\ldots
,n\right\}  . \label{pf.cor.nilpcrit.main-n}%
\end{equation}


\begin{vershort}
[\textit{Proof of (\ref{pf.cor.nilpcrit.main-n}):} Let $k\in\left\{
1,2,\ldots,n\right\}  $. The product $1\cdot2\cdot\cdots\cdot n$ contains $k$
as a factor, and thus is a multiple of $k$; in other words, $n!$ is a multiple
of $k$ (since $n!=1\cdot2\cdot\cdots\cdot n$). Hence, $n!c_{k}$ is a multiple
of $kc_{k}$. Thus, (\ref{pf.cor.nilpcrit.main-n}) follows from
(\ref{pf.cor.nilpcrit.main}).]
\end{vershort}

\begin{verlong}
[\textit{Proof of (\ref{pf.cor.nilpcrit.main-n}):} Let $k\in\left\{
1,2,\ldots,n\right\}  $. Recall that
\[
n!=1\cdot2\cdot\cdots\cdot n=\prod_{i\in\left\{  1,2,\ldots,n\right\}
}i=k\cdot\prod_{\substack{i\in\left\{  1,2,\ldots,n\right\}  ;\\i\neq k}}i
\]
(here, we have split off the factor for $i=k$ from the product (since
$k\in\left\{  1,2,\ldots,n\right\}  $)). Thus,%
\[
\underbrace{n!}_{=k\cdot\prod_{\substack{i\in\left\{  1,2,\ldots,n\right\}
;\\i\neq k}}i}c_{k}=k\cdot\left(  \prod_{\substack{i\in\left\{  1,2,\ldots
,n\right\}  ;\\i\neq k}}i\right)  c_{k}=\left(  \prod_{\substack{i\in\left\{
1,2,\ldots,n\right\}  ;\\i\neq k}}i\right)  \underbrace{kc_{k}}%
_{\substack{=0\\\text{(by (\ref{pf.cor.nilpcrit.main}))}}}=0.
\]
This proves (\ref{pf.cor.nilpcrit.main-n}).]
\end{verlong}

Finally, we observe that%
\begin{equation}
n!c_{n-k}=0\ \ \ \ \ \ \ \ \ \ \text{for every }k\in\left\{  0,1,\ldots
,n-1\right\}  . \label{pf.cor.nilpcrit.main-n2}%
\end{equation}


[\textit{Proof of (\ref{pf.cor.nilpcrit.main-n2}):} Let $k\in\left\{
0,1,\ldots,n-1\right\}  $. Then, $n-k\in\left\{  1,2,\ldots,n\right\}  $.
Hence, (\ref{pf.cor.nilpcrit.main-n}) (applied to $n-k$ instead of $k$) yields
$n!c_{n-k}=0$. This proves (\ref{pf.cor.nilpcrit.main-n2}).]

\begin{vershort}
Now, Corollary \ref{cor.CH.chiA} yields $\chi_{A}=\sum_{k=0}^{n}c_{n-k}t^{k}$.
Substituting $A$ for $t$ in this equality, we obtain $\chi_{A}\left(
A\right)  =\sum_{k=0}^{n}c_{n-k}A^{k}$. Multiplying both sides of the latter
equality by $n!$, we obtain%
\begin{align*}
n!\chi_{A}\left(  A\right)   &  =n!\sum_{k=0}^{n}c_{n-k}A^{k}=\sum_{k=0}%
^{n}n!c_{n-k}A^{k}=\sum_{k=0}^{n-1}\underbrace{n!c_{n-k}}%
_{\substack{=0\\\text{(by (\ref{pf.cor.nilpcrit.main-n2}))}}}A^{k}%
+n!\underbrace{c_{n-n}}_{=c_{0}=1}A^{n}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}k=n\text{ from the sum}\right) \\
&  =\underbrace{\sum_{k=0}^{n-1}0A^{k}}_{=0}+n!A^{n}=n!A^{n}.
\end{align*}
Hence,
\[
n!A^{n}=n!\underbrace{\chi_{A}\left(  A\right)  }_{\substack{=0_{n\times
n}\\\text{(by Theorem \ref{thm.CH})}}}=0_{n\times n}.
\]
This proves Corollary \ref{cor.nilpcrit} \textbf{(a)}.
\end{vershort}

\begin{verlong}
Now, Corollary \ref{cor.CH.chiA} yields $\chi_{A}=\sum_{k=0}^{n}c_{n-k}t^{k}$.
Substituting $A$ for $t$ in this equality, we obtain $\chi_{A}\left(
A\right)  =\sum_{k=0}^{n}c_{n-k}A^{k}$. Multiplying both sides of the latter
equality by $n!$, we obtain%
\begin{align*}
n!\chi_{A}\left(  A\right)   &  =n!\sum_{k=0}^{n}c_{n-k}A^{k}=\sum_{k=0}%
^{n}n!c_{n-k}A^{k}=\sum_{k=0}^{n-1}\underbrace{n!c_{n-k}}%
_{\substack{=0\\\text{(by (\ref{pf.cor.nilpcrit.main-n2}))}}}A^{k}%
+n!\underbrace{c_{n-n}}_{=c_{0}=1}A^{n}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }k=n\text{ from the sum}\\
\text{(since }n\in\left\{  0,1,\ldots,n\right\}  \text{ (since }n\in
\mathbb{N}\text{))}%
\end{array}
\right) \\
&  =\underbrace{\sum_{k=0}^{n-1}0A^{k}}_{=0}+n!A^{n}=n!A^{n}.
\end{align*}
Hence,
\[
n!A^{n}=n!\underbrace{\chi_{A}\left(  A\right)  }_{\substack{=0_{n\times
n}\\\text{(by Theorem \ref{thm.CH})}}}=0_{n\times n}.
\]
This proves Corollary \ref{cor.nilpcrit} \textbf{(a)}.
\end{verlong}

\textbf{(b)} Assume that $\mathbb{K}$ is a commutative $\mathbb{Q}$-algebra.
Corollary \ref{cor.nilpcrit} \textbf{(a)} yields $n!A^{n}=0_{n\times n}$. Now,
$\dfrac{1}{n!}\in\mathbb{Q}$, so that we can multiply an $n\times n$-matrix in
$\mathbb{K}^{n\times n}$ by $\dfrac{1}{n!}$ (since $\mathbb{K}$ is a
$\mathbb{Q}$-algebra). We have $\underbrace{\dfrac{1}{n!}n!}_{=1}A^{n}=A^{n}$.
Hence, $A^{n}=\dfrac{1}{n!}\underbrace{n!A^{n}}_{=0_{n\times n}}=\dfrac{1}%
{n!}0_{n\times n}=0_{n\times n}$. This proves Corollary \ref{cor.nilpcrit}
\textbf{(b)}.

\begin{vershort}
\textbf{(c)} Multiplying the equality $\chi_{A}=\sum_{k=0}^{n}c_{n-k}t^{k}$ by
$n!$, we obtain%
\begin{align*}
n!\chi_{A}  &  =n!\sum_{k=0}^{n}c_{n-k}t^{k}=\sum_{k=0}^{n}n!c_{n-k}t^{k}%
=\sum_{k=0}^{n-1}\underbrace{n!c_{n-k}}_{\substack{=0\\\text{(by
(\ref{pf.cor.nilpcrit.main-n2}))}}}t^{k}+n!\underbrace{c_{n-n}}_{=c_{0}%
=1}t^{n}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}k=n\text{ from the sum}\right) \\
&  =\underbrace{\sum_{k=0}^{n-1}0t^{k}}_{=0}+n!t^{n}=n!t^{n}.
\end{align*}
This proves Corollary \ref{cor.nilpcrit} \textbf{(c)}.
\end{vershort}

\begin{verlong}
\textbf{(c)} Multiplying both sides of the equality $\chi_{A}=\sum_{k=0}%
^{n}c_{n-k}t^{k}$ by $n!$, we obtain%
\begin{align*}
n!\chi_{A}  &  =n!\sum_{k=0}^{n}c_{n-k}t^{k}=\sum_{k=0}^{n}n!c_{n-k}t^{k}%
=\sum_{k=0}^{n-1}\underbrace{n!c_{n-k}}_{\substack{=0\\\text{(by
(\ref{pf.cor.nilpcrit.main-n2}))}}}t^{k}+n!\underbrace{c_{n-n}}_{=c_{0}%
=1}t^{n}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }k=n\text{ from the sum}\\
\text{(since }n\in\left\{  0,1,\ldots,n\right\}  \text{ (since }n\in
\mathbb{N}\text{))}%
\end{array}
\right) \\
&  =\underbrace{\sum_{k=0}^{n-1}0t^{k}}_{=0}+n!t^{n}=n!t^{n}.
\end{align*}
This proves Corollary \ref{cor.nilpcrit} \textbf{(c)}.
\end{verlong}

\textbf{(d)} Assume that $\mathbb{K}$ is a commutative $\mathbb{Q}$-algebra.
Corollary \ref{cor.nilpcrit} \textbf{(c)} yields $n!\chi_{A}=n!t^{n}$.

Now, $\dfrac{1}{n!}\in\mathbb{Q}$, so that we can multiply any polynomial in
$\mathbb{K}\left[  t\right]  $ by $\dfrac{1}{n!}$ (since $\mathbb{K}$ is a
$\mathbb{Q}$-algebra). We have $\underbrace{\dfrac{1}{n!}n!}_{=1}\chi_{A}%
=\chi_{A}$. Hence, $\chi_{A}=\dfrac{1}{n!}\underbrace{n!\chi_{A}}_{=n!t^{n}%
}=\dfrac{1}{n!}n!t^{n}=t^{n}$. This proves Corollary \ref{cor.nilpcrit}
\textbf{(d)}.
\end{proof}

\subsection{A converse direction}

The following result -- in a sense, a converse of Corollary \ref{cor.nilpcrit}
\textbf{(d)} -- also follows from Theorem \ref{thm.TCH}:

\begin{corollary}
\label{cor.nilpcrit-conv}Let $n\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times
n}$. Assume that $\chi_{A}=t^{n}$. Then, $\operatorname*{Tr}\left(
A^{i}\right)  =0$ for every positive integer $i$.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.nilpcrit-conv}.]For every $j\in\mathbb{Z}$,
define an element $c_{j}\in\mathbb{K}$ by $c_{j}=\left[  t^{n-j}\right]
\chi_{A}$. Then, each positive integer $j$ satisfies%
\begin{equation}
c_{j}=0. \label{pf.cor.nilpcrit-conv.1}%
\end{equation}


\begin{vershort}
[\textit{Proof of (\ref{pf.cor.nilpcrit-conv.1}):} Let $j$ be a positive
integer. Thus, $j\neq0$, so that $n-j\neq n$ and thus $\left[  t^{n-j}\right]
\left(  t^{n}\right)  =0$. In view of $\chi_{A}=t^{n}$, this rewrites as
$\left[  t^{n-j}\right]  \chi_{A}=0$. But the definition of $c_{j}$ yields
$c_{j}=\left[  t^{n-j}\right]  \chi_{A}=0$. This proves
(\ref{pf.cor.nilpcrit-conv.1}).]
\end{vershort}

\begin{verlong}
[\textit{Proof of (\ref{pf.cor.nilpcrit-conv.1}):} Let $j$ be a positive
integer. Thus, $j\neq0$. Hence, $n-j\neq n$.

It is thus easy to see that $\left[  t^{n-j}\right]  \left(  t^{n}\right)
=0$\ \ \ \ \footnote{\textit{Proof.} We are in one of the following two cases:
\par
\textit{Case 1:} We have $n-j\geq0$.
\par
\textit{Case 2:} We have $n-j<0$.
\par
Let us first consider Case 1. In this case, we have $n-j\geq0$. Thus, $\left[
t^{n-j}\right]  \left(  t^{n}\right)  $ is the coefficient of $t^{n-j}$ in
$t^{n}$ (by the definition of $\left[  t^{n-j}\right]  \left(  t^{n}\right)
$). Thus,%
\[
\left[  t^{n-j}\right]  \left(  t^{n}\right)  =\left(  \text{the coefficient
of }t^{n-j}\text{ in }t^{n}\right)  =0
\]
(since $t^{n-j}$ and $t^{j}$ are two distinct monomials). Thus, $\left[
t^{n-j}\right]  \left(  t^{n}\right)  =0$ is proven in Case 1.
\par
Let us now consider Case 2. In this case, we have $n-j<0$. Thus, $n-j$ is a
negative integer. Now, every $f\in\mathbb{K}\left[  t\right]  $ and every
negative integer $m$ satisfy $\left[  t^{m}\right]  f=0$ (since polynomials
have no monomials $t^{m}$ with negative $m$). Applying this to $f=t^{n}$ and
$m=n-j$, we obtain $\left[  t^{n-j}\right]  \left(  t^{n}\right)  =0$. Hence,
$\left[  t^{n-j}\right]  \left(  t^{n}\right)  =0$ is proven in Case 2.
\par
Now, we have proven that $\left[  t^{n-j}\right]  \left(  t^{n}\right)  =0$ in
each of the two Cases 1 and 2. Since these two Cases cover all possibilities,
this shows that $\left[  t^{n-j}\right]  \left(  t^{n}\right)  =0$ always
holds. Qed.}. But the definition of $c_{j}$ yields $c_{j}=\left[
t^{n-j}\right]  \underbrace{\chi_{A}}_{=t^{n}}=\left[  t^{n-j}\right]  \left(
t^{n}\right)  =0$. This proves (\ref{pf.cor.nilpcrit-conv.1}).]
\end{verlong}

The definition of $c_{0}$ yields $c_{0}=\underbrace{\left[  t^{n-0}\right]
}_{=\left[  t^{n}\right]  }\chi_{A}=\left[  t^{n}\right]  \chi_{A}=1$ (by
Corollary \ref{cor.chiA} \textbf{(c)}).

Now, we claim that%
\begin{equation}
\operatorname*{Tr}\left(  A^{p}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every
positive integer }p. \label{pf.cor.nilpcrit-conv.2}%
\end{equation}


[\textit{Proof of (\ref{pf.cor.nilpcrit-conv.2}):} We shall prove
(\ref{pf.cor.nilpcrit-conv.2}) by strong induction on $p$:

\textit{Induction step:} Fix a positive integer $k$. Assume (as the induction
hypothesis) that (\ref{pf.cor.nilpcrit-conv.2}) holds whenever $p<k$. We must
now prove that (\ref{pf.cor.nilpcrit-conv.2}) holds for $p=k$.

From (\ref{pf.cor.nilpcrit-conv.1}) (applied to $j=k$), we obtain $c_{k}=0$.

\begin{vershort}
We have assumed that (\ref{pf.cor.nilpcrit-conv.2}) holds whenever $p<k$. In
other words,%
\begin{equation}
\operatorname*{Tr}\left(  A^{p}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every
positive integer }p<k. \label{pf.cor.nilpcrit-conv.2.pf.short.IH}%
\end{equation}


Now, Theorem \ref{thm.TCH} yields%
\[
kc_{k}+\sum_{i=1}^{k}\operatorname*{Tr}\left(  A^{i}\right)  c_{k-i}=0.
\]
Hence,%
\begin{align*}
0  &  =k\underbrace{c_{k}}_{=0}+\sum_{i=1}^{k}\operatorname*{Tr}\left(
A^{i}\right)  c_{k-i}=\sum_{i=1}^{k}\operatorname*{Tr}\left(  A^{i}\right)
c_{k-i}=\sum_{i=1}^{k-1}\underbrace{\operatorname*{Tr}\left(  A^{i}\right)
}_{\substack{=0\\\text{(by (\ref{pf.cor.nilpcrit-conv.2.pf.short.IH}%
),}\\\text{applied}\\\text{to }p=i\text{)}}}c_{k-i}+\operatorname*{Tr}\left(
A^{k}\right)  \underbrace{c_{k-k}}_{=c_{0}=1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we have split off the addend for
}i=k\text{ from the sum}\right) \\
&  =\underbrace{\sum_{i=1}^{k-1}0c_{k-i}}_{=0}+\operatorname*{Tr}\left(
A^{k}\right)  =\operatorname*{Tr}\left(  A^{k}\right)  .
\end{align*}

\end{vershort}

\begin{verlong}
We have assumed that (\ref{pf.cor.nilpcrit-conv.2}) holds whenever $p<k$. In
other words,%
\begin{equation}
\operatorname*{Tr}\left(  A^{p}\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every
positive integer }p<k. \label{pf.cor.nilpcrit-conv.2.pf.IH}%
\end{equation}
Thus, for each $i\in\left\{  1,2,\ldots,k-1\right\}  $, we have%
\begin{equation}
\operatorname*{Tr}\left(  A^{i}\right)  =0 \label{pf.cor.nilpcrit-conv.2.pf.3}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.cor.nilpcrit-conv.2.pf.3}):} Let
$i\in\left\{  1,2,\ldots,k-1\right\}  $. Thus, $1\leq i\leq k-1$. Hence, $i$
is a positive integer (since $i\in\left\{  1,2,\ldots,k-1\right\}  $) and
satisfies $i\leq k-1<k$. Therefore, (\ref{pf.cor.nilpcrit-conv.2.pf.IH})
(applied to $p=i$) yields $\operatorname*{Tr}\left(  A^{i}\right)  =0$. Qed.}.

Now, Theorem \ref{thm.TCH} yields%
\[
kc_{k}+\sum_{i=1}^{k}\operatorname*{Tr}\left(  A^{i}\right)  c_{k-i}=0.
\]
Hence,%
\[
\sum_{i=1}^{k}\operatorname*{Tr}\left(  A^{i}\right)  c_{k-i}%
=-\underbrace{c_{k}}_{=0}=-0=0.
\]
Thus,%
\begin{align*}
0  &  =\sum_{i=1}^{k}\operatorname*{Tr}\left(  A^{i}\right)  c_{k-i}%
=\sum_{i=1}^{k-1}\underbrace{\operatorname*{Tr}\left(  A^{i}\right)
}_{\substack{=0\\\text{(by (\ref{pf.cor.nilpcrit-conv.2.pf.3}))}}%
}c_{k-i}+\operatorname*{Tr}\left(  A^{k}\right)  \underbrace{c_{k-k}}%
_{=c_{0}=1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }i=k\text{ from the sum}\\
\text{(since }k\in\left\{  1,2,\ldots,k\right\}  \text{ (since }k\text{ is a
positive integer))}%
\end{array}
\right) \\
&  =\underbrace{\sum_{i=1}^{k-1}0c_{k-i}}_{=0}+\operatorname*{Tr}\left(
A^{k}\right)  =\operatorname*{Tr}\left(  A^{k}\right)  .
\end{align*}

\end{verlong}

Thus, $\operatorname*{Tr}\left(  A^{k}\right)  =0$. In other words,
(\ref{pf.cor.nilpcrit-conv.2}) holds for $p=k$. This completes the induction
step. Thus, (\ref{pf.cor.nilpcrit-conv.2}) is proven by strong induction.]

We have thus proven that $\operatorname*{Tr}\left(  A^{p}\right)  =0$ for
every positive integer $p$. Renaming the variable $p$ as $i$ in this
statement, we conclude that $\operatorname*{Tr}\left(  A^{i}\right)  =0$ for
every positive integer $i$. This proves Corollary \ref{cor.nilpcrit-conv}.
\end{proof}

\section{More on the adjugate}

I shall now discuss various other properties of the adjugate
$\operatorname*{adj}A$ of a square matrix $A$.

\subsection{Functoriality}

For any $n\in\mathbb{N}$ and $m\in\mathbb{N}$, a homomorphism $f:\mathbb{L}%
\rightarrow\mathbb{M}$ between two rings $\mathbb{L}$ and $\mathbb{M}$ gives
rise to a map $f^{n\times m}:\mathbb{L}^{n\times m}\rightarrow\mathbb{M}%
^{n\times m}$ (as defined in Definition \ref{def.f-mat}). We recall some
classical properties of these maps $f^{n\times m}$:

\begin{proposition}
\label{prop.functor.clas}Let $\mathbb{L}$ and $\mathbb{M}$ be two commutative
rings. Let $f:\mathbb{L}\rightarrow\mathbb{M}$ be a ring homomorphism.

\textbf{(a)} For every $n\in\mathbb{N}$ and $m\in\mathbb{N}$, the map
$f^{n\times m}:\mathbb{L}^{n\times m}\rightarrow\mathbb{M}^{n\times m}$ is a
homomorphism of additive groups.

\textbf{(b)} Every $n\in\mathbb{N}$ satisfies $f^{n\times n}\left(
I_{n}\right)  =I_{n}$.

\textbf{(c)} For every $n\in\mathbb{N}$, $m\in\mathbb{N}$, $p\in\mathbb{N}$,
$A\in\mathbb{L}^{n\times m}$ and $B\in\mathbb{L}^{m\times p}$, we have
$f^{n\times p}\left(  AB\right)  =f^{n\times m}\left(  A\right)  \cdot
f^{m\times p}\left(  B\right)  $.

\textbf{(d)} For every $n\in\mathbb{N}$ and $m\in\mathbb{N}$ and every
$A\in\mathbb{L}^{n\times m}$ and $\lambda\in\mathbb{L}$, we have $f^{n\times
m}\left(  \lambda A\right)  =f\left(  \lambda\right)  f^{n\times m}\left(
A\right)  $.
\end{proposition}

Now, let me state the classical (and simple) fact which is often (somewhat
incompletely) subsumed under the slogan \textquotedblleft ring homomorphisms
preserve determinants and adjugates\textquotedblright:

\begin{proposition}
\label{prop.functor.det}Let $\mathbb{L}$ and $\mathbb{M}$ be two commutative
rings. Let $f:\mathbb{L}\rightarrow\mathbb{M}$ be a ring homomorphism. Let
$n\in\mathbb{N}$. Let $A\in\mathbb{L}^{n\times n}$.

\textbf{(a)} We have $f\left(  \det A\right)  =\det\left(  f^{n\times
n}\left(  A\right)  \right)  $.

\textbf{(b)} Any two elements $u$ and $v$ of $\left\{  1,2,\ldots,n\right\}  $
satisfy $f^{\left(  n-1\right)  \times\left(  n-1\right)  }\left(  A_{\sim
u,\sim v}\right)  =\left(  f^{n\times n}\left(  A\right)  \right)  _{\sim
u,\sim v}$.

\textbf{(c)} We have $f^{n\times n}\left(  \operatorname*{adj}A\right)
=\operatorname*{adj}\left(  f^{n\times n}\left(  A\right)  \right)  $.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.functor.det}.]Proving Proposition
\ref{prop.functor.det} is completely straightforward, and left to the reader.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.functor.det}.]Write the $n\times n$-matrix $A$
in the form $A=\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$.
Then,
\[
f^{n\times n}\left(  \underbrace{A}_{=\left(  a_{i,j}\right)  _{1\leq i\leq
n,\ 1\leq j\leq n}}\right)  =f^{n\times n}\left(  \left(  a_{i,j}\right)
_{1\leq i\leq n,\ 1\leq j\leq n}\right)  =\left(  f\left(  a_{i,j}\right)
\right)  _{1\leq i\leq n,\ 1\leq j\leq n}%
\]
(by the definition of $f^{n\times n}$). Hence, (\ref{eq.det}) (applied to
$\mathbb{M}$, $f^{n\times n}\left(  A\right)  $ and $f\left(  a_{i,j}\right)
$ instead of $\mathbb{K}$, $A$ and $a_{i,j}$) yields%
\begin{equation}
\det\left(  f^{n\times n}\left(  A\right)  \right)  =\sum_{\sigma\in S_{n}%
}\left(  -1\right)  ^{\sigma}\prod_{i=1}^{n}f\left(  a_{i,\sigma\left(
i\right)  }\right)  . \label{pf.prop.functor.det.a.1}%
\end{equation}


\textbf{(a)} The equality (\ref{eq.det}) (applied to $\mathbb{L}$ instead of
$\mathbb{K}$) yields%
\[
\det A=\sum_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\prod_{i=1}%
^{n}a_{i,\sigma\left(  i\right)  }.
\]
Applying the map $f$ to both sides of this equality, we obtain%
\begin{align*}
f\left(  \det A\right)   &  =f\left(  \sum_{\sigma\in S_{n}}\left(  -1\right)
^{\sigma}\prod_{i=1}^{n}a_{i,\sigma\left(  i\right)  }\right)  =\sum
_{\sigma\in S_{n}}\left(  -1\right)  ^{\sigma}\prod_{i=1}^{n}f\left(
a_{i,\sigma\left(  i\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is a ring homomorphism}%
\right) \\
&  =\det\left(  f^{n\times n}\left(  A\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.functor.det.a.1})}\right)
.
\end{align*}
This proves Proposition \ref{prop.functor.det} \textbf{(a)}.

\textbf{(b)} Let $u$ and $v$ be two elements of $\left\{  1,2,\ldots
,n\right\}  $.

Let $\left(  p_{1},p_{2},\ldots,p_{n-1}\right)  $ be the $\left(  n-1\right)
$-tuple $\left(  1,2,\ldots,\widehat{u},\ldots,n\right)  $. Thus, $\left(
p_{1},p_{2},\ldots,p_{n-1}\right)  =\left(  1,2,\ldots,\widehat{u}%
,\ldots,n\right)  $.

Let $\left(  q_{1},q_{2},\ldots,q_{n-1}\right)  $ be the $\left(  n-1\right)
$-tuple $\left(  1,2,\ldots,\widehat{v},\ldots,n\right)  $. Thus, $\left(
q_{1},q_{2},\ldots,q_{n-1}\right)  =\left(  1,2,\ldots,\widehat{v}%
,\ldots,n\right)  $.

If $C=\left(  c_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$ is any
$n\times n$-matrix over a commutative ring $\mathbb{K}$, then
\begin{equation}
C_{\sim u,\sim v}=\left(  c_{p_{i},q_{j}}\right)  _{1\leq i\leq n-1,\ 1\leq
j\leq n-1} \label{pf.prop.functor.det.b.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.functor.det.b.1}):} The equality
(\ref{pf.prop.functor.det.b.1}) is precisely the equality
(\ref{pf.lem.adj.linear.1}), and thus has already been proven.}.

From (\ref{pf.prop.functor.det.b.1}) (applied to $\mathbb{L}$, $A$ and
$a_{i,j}$ instead of $\mathbb{K}$, $C$ and $c_{i,j}$), we obtain $A_{\sim
u,\sim v}=\left(  a_{p_{i},q_{j}}\right)  _{1\leq i\leq n-1,\ 1\leq j\leq
n-1}$. Hence, the definition of the map $f^{\left(  n-1\right)  \times\left(
n-1\right)  }$ yields%
\begin{equation}
f^{\left(  n-1\right)  \times\left(  n-1\right)  }\left(  A_{\sim u,\sim
v}\right)  =\left(  f\left(  a_{p_{i},q_{j}}\right)  \right)  _{1\leq i\leq
n-1,\ 1\leq j\leq n-1}. \label{pf.prop.functor.det.b.2}%
\end{equation}


From (\ref{pf.prop.functor.det.b.1}) (applied to $\mathbb{M}$, $f^{n\times
n}\left(  A\right)  $ and $f\left(  a_{i,j}\right)  $ instead of $\mathbb{K}$,
$C$ and $c_{i,j}$), we obtain $\left(  f^{n\times n}\left(  A\right)  \right)
_{\sim u,\sim v}=\left(  f\left(  a_{p_{i},q_{j}}\right)  \right)  _{1\leq
i\leq n-1,\ 1\leq j\leq n-1}$ (since $f^{n\times n}\left(  A\right)  =\left(
f\left(  a_{i,j}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq n}$).
Comparing this with (\ref{pf.prop.functor.det.b.2}), we obtain $f^{\left(
n-1\right)  \times\left(  n-1\right)  }\left(  A_{\sim u,\sim v}\right)
=\left(  f^{n\times n}\left(  A\right)  \right)  _{\sim u,\sim v}$. This
proves Proposition \ref{prop.functor.det} \textbf{(b)}.

\textbf{(c)} Let $\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$.
Then, $A_{\sim j,\sim i}\in\mathbb{L}^{\left(  n-1\right)  \times\left(
n-1\right)  }$. Hence, Proposition \ref{prop.functor.det} \textbf{(a)}
(applied to $n-1$ and $A_{\sim j,\sim i}$ instead of $n$ and $A$) yields%
\[
f\left(  \det\left(  A_{\sim j,\sim i}\right)  \right)  =\det\left(
\underbrace{f^{\left(  n-1\right)  \times\left(  n-1\right)  }\left(  A_{\sim
j,\sim i}\right)  }_{\substack{=\left(  f^{n\times n}\left(  A\right)
\right)  _{\sim j,\sim i}\\\text{(by Proposition \ref{prop.functor.det}
\textbf{(b)}}\\\text{(applied to }u=j\text{ and }v=i\text{))}}}\right)
=\det\left(  \left(  f^{n\times n}\left(  A\right)  \right)  _{\sim j,\sim
i}\right)  .
\]
But $f$ is a ring homomorphism; hence,%
\begin{align}
&  f\left(  \left(  -1\right)  ^{i+j}\det\left(  A_{\sim j,\sim i}\right)
\right) \nonumber\\
&  =\left(  -1\right)  ^{i+j}\underbrace{f\left(  \det\left(  A_{\sim j,\sim
i}\right)  \right)  }_{=\det\left(  \left(  f^{n\times n}\left(  A\right)
\right)  _{\sim j,\sim i}\right)  }=\left(  -1\right)  ^{i+j}\det\left(
\left(  f^{n\times n}\left(  A\right)  \right)  _{\sim j,\sim i}\right)  .
\label{pf.prop.functor.det.c.2}%
\end{align}


Now, forget that we fixed $\left(  i,j\right)  $. We thus have shown that
(\ref{pf.prop.functor.det.c.2}) holds for every $\left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}$.

The definition of $\operatorname*{adj}A$ yields $\operatorname*{adj}A=\left(
\left(  -1\right)  ^{i+j}\det\left(  A_{\sim j,\sim i}\right)  \right)
_{1\leq i\leq n,\ 1\leq j\leq n}$. Hence, the definition of the map
$f^{n\times n}$ yields%
\begin{align}
f^{n\times n}\left(  \operatorname*{adj}A\right)   &  =\left(
\underbrace{f\left(  \left(  -1\right)  ^{i+j}\det\left(  A_{\sim j,\sim
i}\right)  \right)  }_{\substack{=\left(  -1\right)  ^{i+j}\det\left(  \left(
f^{n\times n}\left(  A\right)  \right)  _{\sim j,\sim i}\right)  \\\text{(by
(\ref{pf.prop.functor.det.c.2}))}}}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n}\nonumber\\
&  =\left(  \left(  -1\right)  ^{i+j}\det\left(  \left(  f^{n\times n}\left(
A\right)  \right)  _{\sim j,\sim i}\right)  \right)  _{1\leq i\leq n,\ 1\leq
j\leq n}. \label{pf.prop.functor.det.c.3}%
\end{align}


But the definition of $\operatorname*{adj}\left(  f^{n\times n}\left(
A\right)  \right)  $ yields%
\begin{align*}
\operatorname*{adj}\left(  f^{n\times n}\left(  A\right)  \right)   &
=\left(  \left(  -1\right)  ^{i+j}\det\left(  \left(  f^{n\times n}\left(
A\right)  \right)  _{\sim j,\sim i}\right)  \right)  _{1\leq i\leq n,\ 1\leq
j\leq n}\\
&  =f^{n\times n}\left(  \operatorname*{adj}A\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.prop.functor.det.c.3})}\right)
.
\end{align*}
This proves Proposition \ref{prop.functor.det} \textbf{(c)}.
\end{proof}
\end{verlong}

\subsection{The evaluation homomorphism}

We shall apply the above to relate the determinant and the adjugate of a
matrix $A$ with those of the matrix $tI_{n}+A$:

\begin{proposition}
\label{prop.tInA.ev}Let $\varepsilon:\mathbb{K}\left[  t\right]
\rightarrow\mathbb{K}$ be the map which sends every polynomial $p\in
\mathbb{K}\left[  t\right]  $ to its value $p\left(  0\right)  $. It is
well-known that $\varepsilon$ is a $\mathbb{K}$-algebra homomorphism.

Let $n\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times n}$. Consider the matrix
$tI_{n}+A\in\left(  \mathbb{K}\left[  t\right]  \right)  ^{n\times n}$. Then:

\textbf{(a)} We have $\varepsilon\left(  \det\left(  tI_{n}+A\right)  \right)
=\det A$.

\textbf{(b)} We have $\varepsilon^{n\times n}\left(  \operatorname*{adj}%
\left(  tI_{n}+A\right)  \right)  =\operatorname*{adj}A$.

\textbf{(c)} We have $\varepsilon^{n\times n}\left(  tI_{n}+A\right)  =A$.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.tInA.ev}.]We have%
\[
\varepsilon^{n\times n}\left(  tB+A\right)  =A
\]
for every $B\in\mathbb{K}^{n\times n}$\ \ \ \ \footnote{\textit{Proof.} This
equality is similar to (\ref{pf.thm.ddet.short.1}), and is proven
analogously.}. Applying this to $B=I_{n}$, we obtain $\varepsilon^{n\times
n}\left(  tI_{n}+A\right)  =A$. This proves Proposition \ref{prop.tInA.ev}
\textbf{(c)}.

\textbf{(a)} Proposition \ref{prop.functor.det} \textbf{(a)} (applied to
$\mathbb{K}\left[  t\right]  $, $\mathbb{K}$, $\varepsilon$ and $tI_{n}+A$
instead of $\mathbb{L}$, $\mathbb{M}$, $f$ and $A$) yields%
\[
\varepsilon\left(  \det\left(  tI_{n}+A\right)  \right)  =\det\left(
\underbrace{\varepsilon^{n\times n}\left(  tI_{n}+A\right)  }_{=A}\right)
=\det A.
\]
This proves Proposition \ref{prop.tInA.ev} \textbf{(a)}.

\textbf{(b)} Proposition \ref{prop.functor.det} \textbf{(c)} (applied to
$\mathbb{K}\left[  t\right]  $, $\mathbb{K}$, $\varepsilon$ and $tI_{n}+A$
instead of $\mathbb{L}$, $\mathbb{M}$, $f$ and $A$) yields%
\[
\varepsilon^{n\times n}\left(  \operatorname*{adj}\left(  tI_{n}+A\right)
\right)  =\operatorname*{adj}\left(  \underbrace{\varepsilon^{n\times
n}\left(  tI_{n}+A\right)  }_{=A}\right)  =\operatorname*{adj}A.
\]
This proves Proposition \ref{prop.tInA.ev} \textbf{(b)}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.tInA.ev}.]We have%
\begin{equation}
\varepsilon^{n\times n}\left(  tB+A\right)  =A \label{pf.prop.tInA.ev.1}%
\end{equation}
for every $B\in\mathbb{K}^{n\times n}$\ \ \ \ \footnote{\textit{Proof of
(\ref{pf.prop.tInA.ev.1}):} Let $B\in\mathbb{K}^{n\times n}$. Write the
$n\times n$-matrix $B\in\mathbb{K}^{n\times n}$ in the form $B=\left(
b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$. Thus,
\begin{equation}
b_{i,j}\in\mathbb{K}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2} \label{pf.prop.tInA.ev.1.pf.1}%
\end{equation}
(since $B\in\mathbb{K}^{n\times n}$). Also, $t\underbrace{B}_{=\left(
b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}}=t\left(  b_{i,j}\right)
_{1\leq i\leq n,\ 1\leq j\leq n}=\left(  tb_{i,j}\right)  _{1\leq i\leq
n,\ 1\leq j\leq n}$.
\par
Write the $n\times n$-matrix $A\in\mathbb{K}^{n\times n}$ in the form
$A=\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$. Thus,%
\begin{equation}
a_{i,j}\in\mathbb{K}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2} \label{pf.prop.tInA.ev.1.pf.2}%
\end{equation}
(since $A\in\mathbb{K}^{n\times n}$).
\par
Fix $\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$. Then,
$b_{i,j}\in\mathbb{K}$ (by (\ref{pf.prop.tInA.ev.1.pf.1})) and $a_{i,j}%
\in\mathbb{K}$ (by (\ref{pf.prop.tInA.ev.1.pf.2})). Hence, the value of the
polynomial $tb_{i,j}-a_{i,j}\in\mathbb{K}\left[  t\right]  $ at $0$ is
$\left(  tb_{i,j}+a_{i,j}\right)  \left(  0\right)  =a_{i,j}$. But the
definition of $\varepsilon$ yields%
\begin{equation}
\varepsilon\left(  tb_{i,j}+a_{i,j}\right)  =\left(  tb_{i,j}+a_{i,j}\right)
\left(  0\right)  =a_{i,j}. \label{pf.prop.tInA.ev.1.pf.4}%
\end{equation}
\par
Now, forget that we fixed $\left(  i,j\right)  $. We thus have proven
(\ref{pf.prop.tInA.ev.1.pf.4}) for every $\left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}.$
\par
Now,%
\[
\underbrace{tB}_{=\left(  tb_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}%
}+\underbrace{A}_{=\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}%
}=\left(  tb_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}+\left(
a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}=\left(  tb_{i,j}%
+a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}.
\]
Hence, the definition of the map $\varepsilon^{n\times n}$ yields%
\[
\varepsilon^{n\times n}\left(  tB+A\right)  =\left(  \underbrace{\varepsilon
\left(  tb_{i,j}+a_{i,j}\right)  }_{\substack{=a_{i,j}\\\text{(by
(\ref{pf.prop.tInA.ev.1.pf.4}))}}}\right)  _{1\leq i\leq n,\ 1\leq j\leq
n}=\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}=A.
\]
Hence, (\ref{pf.prop.tInA.ev.1}) is proven.}. Applying this to $B=I_{n}$, we
obtain $\varepsilon^{n\times n}\left(  tI_{n}+A\right)  =A$. This proves
Proposition \ref{prop.tInA.ev} \textbf{(c)}.

The map $\varepsilon$ is a $\mathbb{K}$-algebra homomorphism, thus a ring homomorphism.

\textbf{(a)} Proposition \ref{prop.functor.det} \textbf{(a)} (applied to
$\mathbb{K}\left[  t\right]  $, $\mathbb{K}$, $\varepsilon$ and $tI_{n}+A$
instead of $\mathbb{L}$, $\mathbb{M}$, $f$ and $A$) yields%
\[
\varepsilon\left(  \det\left(  tI_{n}+A\right)  \right)  =\det\left(
\underbrace{\varepsilon^{n\times n}\left(  tI_{n}+A\right)  }_{=A}\right)
=\det A.
\]
This proves Proposition \ref{prop.tInA.ev} \textbf{(a)}.

\textbf{(b)} Proposition \ref{prop.functor.det} \textbf{(c)} (applied to
$\mathbb{K}\left[  t\right]  $, $\mathbb{K}$, $\varepsilon$ and $tI_{n}+A$
instead of $\mathbb{L}$, $\mathbb{M}$, $f$ and $A$) yields%
\[
\varepsilon^{n\times n}\left(  \operatorname*{adj}\left(  tI_{n}+A\right)
\right)  =\operatorname*{adj}\left(  \underbrace{\varepsilon^{n\times
n}\left(  tI_{n}+A\right)  }_{=A}\right)  =\operatorname*{adj}A.
\]
This proves Proposition \ref{prop.tInA.ev} \textbf{(b)}.
\end{proof}
\end{verlong}

If $A\in\mathbb{K}^{n\times n}$ is a square matrix, then the matrix
$tI_{n}+A\in\left(  \mathbb{K}\left[  t\right]  \right)  ^{n\times n}$ has a
property which the matrix $A$ might not have: namely, its determinant is
regular. Let us first define what this means:

\begin{definition}
\label{def.regular}Let $\mathbb{A}$ be a commutative ring. Let $a\in
\mathbb{A}$. The element $a$ of $\mathbb{A}$ is said to be \textit{regular} if
and only if every $x\in\mathbb{A}$ satisfying $ax=0$ satisfies $x=0$.

Instead of saying that $a$ is regular, one can also say that \textquotedblleft%
$a$ is \textit{cancellable}\textquotedblright, or that \textquotedblleft$a$ is
a \textit{non-zero-divisor}\textquotedblright.
\end{definition}

A basic property of regular elements is the following:

\begin{lemma}
\label{lem.regular.cancel}Let $\mathbb{A}$ be a commutative ring. Let $a$ be a
regular element of $\mathbb{A}$. Let $b$ and $c$ be two elements of
$\mathbb{A}$ such that $ab=ac$. Then, $b=c$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.regular.cancel}.]We have $a\left(  b-c\right)
=\underbrace{ab}_{=ac}-ac=ac-ac=0$.

Now, recall that the element $a$ of $\mathbb{A}$ is regular if and only if
every $x\in\mathbb{A}$ satisfying $ax=0$ satisfies $x=0$ (by the definition of
\textquotedblleft regular\textquotedblright). Hence, every $x\in\mathbb{A}$
satisfying $ax=0$ satisfies $x=0$ (because the element $a$ of $\mathbb{A}$ is
regular). Applying this to $x=b-c$, we obtain $b-c=0$ (since $a\left(
b-c\right)  =0$). Thus, $b=c$. This proves Lemma \ref{lem.regular.cancel}.
\end{proof}

Regular elements, of course, can also be cancelled from matrix equations:

\begin{lemma}
\label{lem.regular.mat-cancel}Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let
$a$ be a regular element of $\mathbb{K}$. Let $B\in\mathbb{K}^{n\times m}$ and
$C\in\mathbb{K}^{n\times m}$ be such that $aB=aC$. Then, $B=C$.
\end{lemma}

\begin{vershort}
\begin{proof}
[Proof of Lemma \ref{lem.regular.mat-cancel}.]Write the $n\times m$-matrices
$B$ and $C$ in the forms $B=\left(  b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq
j\leq m}$ and $C=\left(  c_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}$.
Then, $aB=\left(  ab_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}$ and
$aC=\left(  ac_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}$. Hence,
\[
\left(  ab_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}=aB=aC=\left(
ac_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}.
\]
In other words,
\[
ab_{i,j}=ac_{i,j}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  \times\left\{  1,2,\ldots,m\right\}  .
\]
Thus,
\[
b_{i,j}=c_{i,j}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  \times\left\{  1,2,\ldots,m\right\}
\]
(by Lemma \ref{lem.regular.cancel}, applied to $b=b_{i,j}$ and $c=c_{i,j}$).
Hence, $\left(  b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}=\left(
c_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}$. Thus, $B=\left(
b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}=\left(  c_{i,j}\right)
_{1\leq i\leq n,\ 1\leq j\leq m}=C$. Lemma \ref{lem.regular.mat-cancel} is proven.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.regular.mat-cancel}.]Write the $n\times m$-matrix
$B\in\mathbb{K}^{n\times m}$ in the form $B=\left(  b_{i,j}\right)  _{1\leq
i\leq n,\ 1\leq j\leq m}$.

Write the $n\times m$-matrix $C\in\mathbb{K}^{n\times m}$ in the form
$C=\left(  c_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}$.

Now,%
\[
a\underbrace{B}_{=\left(  b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}%
}=a\left(  b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}=\left(
ab_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m},
\]
so that%
\[
\left(  ab_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}=aB=a\underbrace{C}%
_{=\left(  c_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}}=a\left(
c_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}=\left(  ac_{i,j}\right)
_{1\leq i\leq n,\ 1\leq j\leq m}.
\]
In other words,
\begin{equation}
ab_{i,j}=ac_{i,j}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  \times\left\{  1,2,\ldots,m\right\}  .
\label{pf.lem.regular.det.cancel.a.1}%
\end{equation}


Now,
\begin{equation}
b_{i,j}=c_{i,j}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  \times\left\{  1,2,\ldots,m\right\}
\label{pf.lem.regular.det.cancel.a.2}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.regular.det.cancel.a.2}):} Let
$\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  \times\left\{
1,2,\ldots,m\right\}  $. From (\ref{pf.lem.regular.det.cancel.a.1}), we obtain
$ab_{i,j}=ac_{i,j}$. Lemma \ref{lem.regular.cancel} (applied to $\mathbb{K}$,
$b_{i,j}$ and $c_{i,j}$ instead of $\mathbb{A}$, $b$ and $c$) thus yields
$b_{i,j}=c_{i,j}$ (since the element $a$ of $\mathbb{K}$ is regular). This
proves (\ref{pf.lem.regular.det.cancel.a.2}).}. Hence, $\left(  b_{i,j}%
\right)  _{1\leq i\leq n,\ 1\leq j\leq m}=\left(  c_{i,j}\right)  _{1\leq
i\leq n,\ 1\leq j\leq m}$. Thus, $B=\left(  b_{i,j}\right)  _{1\leq i\leq
n,\ 1\leq j\leq m}=\left(  c_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}%
=C$. This proves Lemma \ref{lem.regular.mat-cancel}.
\end{proof}
\end{verlong}

One important way to construct regular elements is the following fact:

\begin{proposition}
\label{prop.pol.monic-reg}Let $n\in\mathbb{N}$. Let $p\in\mathbb{K}\left[
t\right]  $ be a monic polynomial of degree $n$. Then, the element $p$ of
$\mathbb{K}\left[  t\right]  $ is regular.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.pol.monic-reg}.]Proposition
\ref{prop.pol.monic-reg} is precisely \cite[Corollary 3.15]{Grinbe16b}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.pol.monic-reg}.]Proposition
\ref{prop.pol.monic-reg} is precisely \cite[Corollary 3.15]{Grinbe16b}. (The
notations used in \cite{Grinbe16b} differ from ours as follows: Our ring
$\mathbb{K}$ and our indeterminate $t$ correspond to the ring $A$ and the
indeterminate $X$ in \cite{Grinbe16b}.)
\end{proof}
\end{verlong}

\begin{corollary}
\label{cor.tInA.reg}Let $n\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times n}$.
Consider the matrix $tI_{n}+A\in\left(  \mathbb{K}\left[  t\right]  \right)
^{n\times n}$.

Then, the element $\det\left(  tI_{n}+A\right)  $ of $\mathbb{K}\left[
t\right]  $ is regular.
\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.tInA.reg}.]Proposition \ref{prop.ta+b}
\textbf{(a)} (applied to $I_{n}$ and $A$ instead of $A$ and $B$) yields that
$\det\left(  tI_{n}+A\right)  \in\mathbb{K}\left[  t\right]  $ is a polynomial
of degree $\leq n$ in $t$. Proposition \ref{prop.ta+b} \textbf{(c)} (applied
to $I_{n}$ and $A$ instead of $A$ and $B$) yields that $\left[  t^{n}\right]
\left(  \det\left(  tI_{n}+A\right)  \right)  =\det\left(  I_{n}\right)  =1$.

So we know that the polynomial $\det\left(  tI_{n}+A\right)  \in
\mathbb{K}\left[  t\right]  $ is a polynomial of degree $\leq n$, and that the
coefficient of $t^{n}$ in this polynomial is $\left[  t^{n}\right]  \left(
\det\left(  tI_{n}+A\right)  \right)  =1$. In other words, the polynomial
$\det\left(  tI_{n}+A\right)  \in\mathbb{K}\left[  t\right]  $ is monic of
degree $n$. Thus, Proposition \ref{prop.pol.monic-reg} (applied to
$p=\det\left(  tI_{n}+A\right)  $) shows that the element $\det\left(
tI_{n}+A\right)  $ of $\mathbb{K}\left[  t\right]  $ is regular. This proves
Corollary \ref{cor.tInA.reg}.
\end{proof}

A square matrix whose determinant is regular can be cancelled from equations,
as the following lemma shows:

\begin{lemma}
\label{lem.regular.det.cancel}Let $n\in\mathbb{N}$. Let $A\in\mathbb{K}%
^{n\times n}$. Assume that the element $\det A$ of $\mathbb{K}$ is regular.
Let $m\in\mathbb{N}$.

\textbf{(a)} If $B\in\mathbb{K}^{n\times m}$ and $C\in\mathbb{K}^{n\times m}$
are such that $AB=AC$, then $B=C$.

\textbf{(b)} If $B\in\mathbb{K}^{m\times n}$ and $C\in\mathbb{K}^{m\times n}$
are such that $BA=CA$, then $B=C$.
\end{lemma}

\begin{vershort}
\begin{proof}
[Proof of Lemma \ref{lem.regular.det.cancel}.]Define an element $a$ of
$\mathbb{K}$ by $a=\det A$. Recall that the element $\det A$ of $\mathbb{K}$
is regular. In other words, the element $a$ of $\mathbb{K}$ is regular (since
$a=\det A$). Theorem \ref{thm.adj.inverse} yields $A\cdot\operatorname*{adj}%
A=\operatorname*{adj}A\cdot A=\det A\cdot I_{n}$.

\textbf{(a)} Let $B\in\mathbb{K}^{n\times m}$ and $C\in\mathbb{K}^{n\times m}$
be such that $AB=AC$. We must prove that $B=C$.

We have
\[
\underbrace{\operatorname*{adj}A\cdot A}_{=\det A\cdot I_{n}}%
B=\underbrace{\det A}_{=a}\cdot\underbrace{I_{n}B}_{=B}=aB.
\]
Thus,
\[
aB=\operatorname*{adj}A\cdot\underbrace{AB}_{=AC}%
=\underbrace{\operatorname*{adj}A\cdot A}_{=\det A\cdot I_{n}}%
C=\underbrace{\det A}_{=a}\cdot\underbrace{I_{n}C}_{=C}=aC.
\]


Lemma \ref{lem.regular.mat-cancel} thus yields $B=C$. This proves Lemma
\ref{lem.regular.det.cancel} \textbf{(a)}.

\textbf{(b)} The proof of Lemma \ref{lem.regular.det.cancel} \textbf{(b)} is
similar to the proof of Lemma \ref{lem.regular.det.cancel} \textbf{(a)} (but
now we need to work with $BA\cdot\operatorname*{adj}A$ and $CA\cdot
\operatorname*{adj}A$ instead of $\operatorname*{adj}A\cdot AB$ and
$\operatorname*{adj}A\cdot AC$). The details are left to the reader.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.regular.det.cancel}.]Define an element $a$ of
$\mathbb{K}$ by $a=\det A$. Recall that the element $\det A$ of $\mathbb{K}$
is regular. In other words, the element $a$ of $\mathbb{K}$ is regular (since
$a=\det A$).

Theorem \ref{thm.adj.inverse} yields $A\cdot\operatorname*{adj}%
A=\operatorname*{adj}A\cdot A=\det A\cdot I_{n}$.

\textbf{(a)} Let $B\in\mathbb{K}^{n\times m}$ and $C\in\mathbb{K}^{n\times m}$
be such that $AB=AC$. We must prove that $B=C$.

We have
\[
\underbrace{\operatorname*{adj}A\cdot A}_{=\det A\cdot I_{n}}%
B=\underbrace{\det A}_{=a}\cdot\underbrace{I_{n}B}_{=B}=aB.
\]
Hence,%
\[
aB=\operatorname*{adj}A\cdot\underbrace{AB}_{=AC}%
=\underbrace{\operatorname*{adj}A\cdot A}_{=\det A\cdot I_{n}}%
C=\underbrace{\det A}_{=a}\cdot\underbrace{I_{n}C}_{=C}=aC.
\]


Lemma \ref{lem.regular.mat-cancel} thus yields $B=C$ (since the element $a$ of
$\mathbb{K}$ is regular). This proves Lemma \ref{lem.regular.det.cancel}
\textbf{(a)}.

\textbf{(b)} Let $B\in\mathbb{K}^{m\times n}$ and $C\in\mathbb{K}^{m\times n}$
be such that $BA=CA$. We must prove that $B=C$.

We have
\[
B\underbrace{A\cdot\operatorname*{adj}A}_{=\det A\cdot I_{n}}=B\cdot\det
A\cdot I_{n}=\underbrace{\det A}_{=a}\cdot B=aB.
\]
Hence,%
\[
aB=\underbrace{BA}_{=CA}\cdot\operatorname*{adj}A=C\underbrace{A\cdot
\operatorname*{adj}A}_{=\det A\cdot I_{n}}=C\cdot\det A\cdot I_{n}%
=\underbrace{\det A}_{=a}\cdot C=aC.
\]
Lemma \ref{lem.regular.mat-cancel} (applied to $m$ and $n$ instead of $n$ and
$m$) thus yields $B=C$ (since the element $a$ of $\mathbb{K}$ is regular).
This proves Lemma \ref{lem.regular.det.cancel} \textbf{(b)}.
\end{proof}
\end{verlong}

\subsection{The adjugate of a product}

Corollary \ref{cor.tInA.reg} can be put to use in several circumstances. Here
is a simple example:

\begin{theorem}
\label{thm.adj.adj(AB)}Let $n\in\mathbb{N}$. Let $A$ and $B$ be two $n\times
n$-matrices. Then,%
\[
\operatorname*{adj}\left(  AB\right)  =\operatorname*{adj}B\cdot
\operatorname*{adj}A.
\]

\end{theorem}

Theorem \ref{thm.adj.adj(AB)} is the statement of \cite[Exercise
6.33]{detnotes}; see \cite[solution of Exercise 6.33]{detnotes} for a proof of
this theorem. We shall show a different proof of it now.

We begin by showing a particular case of Theorem \ref{thm.adj.adj(AB)}:

\begin{lemma}
\label{lem.adj.adj(AB).reg}Let $n\in\mathbb{N}$. Let $A$ and $B$ be two
$n\times n$-matrices. Assume that the elements $\det A$ and $\det B$ of
$\mathbb{K}$ are regular. Then, $\operatorname*{adj}\left(  AB\right)
=\operatorname*{adj}B\cdot\operatorname*{adj}A$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.adj.adj(AB).reg}.]Theorem \ref{thm.adj.inverse}
yields%
\[
A\cdot\operatorname*{adj}A=\operatorname*{adj}A\cdot A=\det A\cdot I_{n}.
\]
Theorem \ref{thm.adj.inverse} (applied to $B$ instead of $A$) yields%
\[
B\cdot\operatorname*{adj}B=\operatorname*{adj}B\cdot B=\det B\cdot I_{n}.
\]
Theorem \ref{thm.adj.inverse} (applied to $AB$ instead of $A$) yields%
\[
AB\cdot\operatorname*{adj}\left(  AB\right)  =\operatorname*{adj}\left(
AB\right)  \cdot AB=\det\left(  AB\right)  \cdot I_{n}.
\]


Now,
\begin{align*}
A\underbrace{B\cdot\operatorname*{adj}B}_{=\det B\cdot I_{n}}\cdot
\operatorname*{adj}A  &  =\underbrace{A\cdot\det B\cdot I_{n}}_{=\det B\cdot
A}\cdot\operatorname*{adj}A\\
&  =\det B\cdot\underbrace{A\cdot\operatorname*{adj}A}_{=\det A\cdot I_{n}%
}=\det B\cdot\det A\cdot I_{n}\\
&  =\det A\cdot\det B\cdot I_{n}.
\end{align*}
Comparing this with%
\[
AB\cdot\operatorname*{adj}\left(  AB\right)  =\underbrace{\det\left(
AB\right)  }_{=\det A\cdot\det B}\cdot I_{n}=\det A\cdot\det B\cdot I_{n},
\]
we obtain $AB\cdot\operatorname*{adj}B\cdot\operatorname*{adj}A=AB\cdot
\operatorname*{adj}\left(  AB\right)  $. Lemma \ref{lem.regular.det.cancel}
\textbf{(a)} (applied to $n$, $B\cdot\operatorname*{adj}B\cdot
\operatorname*{adj}A$ and $B\cdot\operatorname*{adj}\left(  AB\right)  $
instead of $m$, $B$ and $C$) therefore yields $B\cdot\operatorname*{adj}%
B\cdot\operatorname*{adj}A=B\cdot\operatorname*{adj}\left(  AB\right)  $
(since the element $\det A$ of $\mathbb{K}$ is regular). Thus, Lemma
\ref{lem.regular.det.cancel} \textbf{(a)} (applied to $n$, $B$,
$\operatorname*{adj}B\cdot\operatorname*{adj}A$ and $\operatorname*{adj}%
\left(  AB\right)  $ instead of $m$, $A$, $B$ and $C$) yields
$\operatorname*{adj}B\cdot\operatorname*{adj}A=\operatorname*{adj}\left(
AB\right)  $ (since the element $\det B$ of $\mathbb{K}$ is regular). This
proves Lemma \ref{lem.adj.adj(AB).reg}.
\end{proof}

We now derive Theorem \ref{thm.adj.adj(AB)} from this lemma:

\begin{vershort}
\begin{proof}
[Proof of Theorem \ref{thm.adj.adj(AB)}.]Define the $\mathbb{K}$-algebra
homomorphism $\varepsilon:\mathbb{K}\left[  t\right]  \rightarrow\mathbb{K}$
as in Proposition \ref{prop.tInA.ev}.

Define two matrices $\widetilde{A}$ and $\widetilde{B}$ in $\left(
\mathbb{K}\left[  t\right]  \right)  ^{n\times n}$ by $\widetilde{A}=tI_{n}+A$
and $\widetilde{B}=tI_{n}+B$.

From $\widetilde{A}=tI_{n}+A$, we obtain $\varepsilon^{n\times n}\left(
\operatorname*{adj}\widetilde{A}\right)  =\varepsilon^{n\times n}\left(
\operatorname*{adj}\left(  tI_{n}+A\right)  \right)  =\operatorname*{adj}A$
(by Proposition \ref{prop.tInA.ev} \textbf{(b)}). Similarly, $\varepsilon
^{n\times n}\left(  \operatorname*{adj}\widetilde{B}\right)
=\operatorname*{adj}B$.

From $\widetilde{A}=tI_{n}+A$, we obtain $\varepsilon^{n\times n}\left(
\widetilde{A}\right)  =\varepsilon^{n\times n}\left(  tI_{n}+A\right)  =A$ (by
Proposition \ref{prop.tInA.ev} \textbf{(c)}). Similarly, $\varepsilon^{n\times
n}\left(  \widetilde{B}\right)  =B$.

Corollary \ref{cor.tInA.reg} shows that the element $\det\left(
tI_{n}+A\right)  $ of $\mathbb{K}\left[  t\right]  $ is regular. In other
words, the element $\det\widetilde{A}$ of $\mathbb{K}\left[  t\right]  $ is
regular (since $\widetilde{A}=tI_{n}+A$). Similarly, the element
$\det\widetilde{B}$ of $\mathbb{K}\left[  t\right]  $ is regular. Lemma
\ref{lem.adj.adj(AB).reg} (applied to $\mathbb{K}\left[  t\right]  $,
$\widetilde{A}$ and $\widetilde{B}$ instead of $\mathbb{K}$, $A$ and $B$) thus
yields%
\[
\operatorname*{adj}\left(  \widetilde{A}\widetilde{B}\right)
=\operatorname*{adj}\widetilde{B}\cdot\operatorname*{adj}\widetilde{A}.
\]
Applying the map $\varepsilon^{n\times n}$ to both sides of this equality, we
obtain%
\begin{align*}
\varepsilon^{n\times n}\left(  \operatorname*{adj}\left(  \widetilde{A}%
\widetilde{B}\right)  \right)   &  =\varepsilon^{n\times n}\left(
\operatorname*{adj}\widetilde{B}\cdot\operatorname*{adj}\widetilde{A}\right)
=\underbrace{\varepsilon^{n\times n}\left(  \operatorname*{adj}\widetilde{B}%
\right)  }_{\substack{=\operatorname*{adj}B}}\cdot\underbrace{\varepsilon
^{n\times n}\left(  \operatorname*{adj}\widetilde{A}\right)  }%
_{=\operatorname*{adj}A}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Proposition \ref{prop.functor.clas} \textbf{(c)}, applied to}\\
\mathbb{K}\left[  t\right]  \text{, }\mathbb{K}\text{, }\varepsilon\text{,
}n\text{, }n\text{, }\operatorname*{adj}\widetilde{B}\text{ and }%
\operatorname*{adj}\widetilde{A}\\
\text{instead of }\mathbb{L}\text{, }\mathbb{M}\text{, }f\text{, }m\text{,
}p\text{, }A\text{ and }B
\end{array}
\right) \\
&  =\operatorname*{adj}B\cdot\operatorname*{adj}A.
\end{align*}
Hence,%
\begin{equation}
\operatorname*{adj}B\cdot\operatorname*{adj}A=\varepsilon^{n\times n}\left(
\operatorname*{adj}\left(  \widetilde{A}\widetilde{B}\right)  \right)
=\operatorname*{adj}\left(  \varepsilon^{n\times n}\left(  \widetilde{A}%
\widetilde{B}\right)  \right)  \label{pf.thm.adj.adj(AB).short.1}%
\end{equation}
(by Proposition \ref{prop.functor.det} \textbf{(c)}, applied to $\mathbb{K}%
\left[  t\right]  $, $\mathbb{K}$, $\varepsilon$ and $\widetilde{A}%
\widetilde{B}$ instead of $\mathbb{L}$, $\mathbb{M}$, $f$ and $A$).

But Proposition \ref{prop.functor.clas} \textbf{(c)} (applied to
$\mathbb{K}\left[  t\right]  $, $\mathbb{K}$, $\varepsilon$, $n$, $n$,
$\widetilde{A}$ and $\widetilde{B}$ instead of $\mathbb{L}$, $\mathbb{M}$,
$f$, $m$, $p$, $A$ and $B$) shows that%
\[
\varepsilon^{n\times n}\left(  \widetilde{A}\widetilde{B}\right)
=\underbrace{\varepsilon^{n\times n}\left(  \widetilde{A}\right)  }_{=A}%
\cdot\underbrace{\varepsilon^{n\times n}\left(  \widetilde{B}\right)  }%
_{=B}=AB.
\]
Hence, (\ref{pf.thm.adj.adj(AB).short.1}) becomes%
\[
\operatorname*{adj}B\cdot\operatorname*{adj}A=\operatorname*{adj}\left(
\underbrace{\varepsilon^{n\times n}\left(  \widetilde{A}\widetilde{B}\right)
}_{=AB}\right)  =\operatorname*{adj}\left(  AB\right)  .
\]
This proves Theorem \ref{thm.adj.adj(AB)}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Theorem \ref{thm.adj.adj(AB)}.]Define the $\mathbb{K}$-algebra
homomorphism $\varepsilon:\mathbb{K}\left[  t\right]  \rightarrow\mathbb{K}$
as in Proposition \ref{prop.tInA.ev}.

Consider the matrices $tI_{n}+A$ and $tI_{n}+B$ in $\left(  \mathbb{K}\left[
t\right]  \right)  ^{n\times n}$. Corollary \ref{cor.tInA.reg} shows that the
element $\det\left(  tI_{n}+A\right)  $ of $\mathbb{K}\left[  t\right]  $ is
regular. Corollary \ref{cor.tInA.reg} (applied to $B$ instead of $A$) shows
that the element $\det\left(  tI_{n}+B\right)  $ of $\mathbb{K}\left[
t\right]  $ is regular. Lemma \ref{lem.adj.adj(AB).reg} (applied to
$\mathbb{K}\left[  t\right]  $, $tI_{n}+A$ and $tI_{n}+B$ instead of
$\mathbb{K}$, $A$ and $B$) thus yields%
\[
\operatorname*{adj}\left(  \left(  tI_{n}+A\right)  \left(  tI_{n}+B\right)
\right)  =\operatorname*{adj}\left(  tI_{n}+B\right)  \cdot\operatorname*{adj}%
\left(  tI_{n}+A\right)  .
\]
Applying the map $\varepsilon^{n\times n}$ to both sides of this equality, we
obtain%
\begin{align*}
&  \varepsilon^{n\times n}\left(  \operatorname*{adj}\left(  \left(
tI_{n}+A\right)  \left(  tI_{n}+B\right)  \right)  \right) \\
&  =\varepsilon^{n\times n}\left(  \operatorname*{adj}\left(  tI_{n}+B\right)
\cdot\operatorname*{adj}\left(  tI_{n}+A\right)  \right) \\
&  =\underbrace{\varepsilon^{n\times n}\left(  \operatorname*{adj}\left(
tI_{n}+B\right)  \right)  }_{\substack{=\operatorname*{adj}B\\\text{(by
Proposition \ref{prop.tInA.ev} \textbf{(b)}}\\\text{(applied to }B\text{
instead of }A\text{))}}}\cdot\underbrace{\varepsilon^{n\times n}\left(
\operatorname*{adj}\left(  tI_{n}+A\right)  \right)  }%
_{\substack{=\operatorname*{adj}A\\\text{(by Proposition \ref{prop.tInA.ev}
\textbf{(b)})}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Proposition \ref{prop.functor.clas} \textbf{(c)} (applied to}\\
\mathbb{K}\left[  t\right]  \text{, }\mathbb{K}\text{, }\varepsilon\text{,
}n\text{, }n\text{, }\operatorname*{adj}\left(  tI_{n}+B\right)  \text{ and
}\operatorname*{adj}\left(  tI_{n}+A\right) \\
\text{instead of }\mathbb{L}\text{, }\mathbb{M}\text{, }f\text{, }m\text{,
}p\text{, }A\text{ and }B\text{)}%
\end{array}
\right) \\
&  =\operatorname*{adj}B\cdot\operatorname*{adj}A.
\end{align*}
Hence,%
\begin{align}
\operatorname*{adj}B\cdot\operatorname*{adj}A  &  =\varepsilon^{n\times
n}\left(  \operatorname*{adj}\left(  \left(  tI_{n}+A\right)  \left(
tI_{n}+B\right)  \right)  \right) \nonumber\\
&  =\operatorname*{adj}\left(  \varepsilon^{n\times n}\left(  \left(
tI_{n}+A\right)  \left(  tI_{n}+B\right)  \right)  \right)
\label{pf.thm.adj.adj(AB).1}%
\end{align}
(by Proposition \ref{prop.functor.det} \textbf{(c)} (applied to $\mathbb{K}%
\left[  t\right]  $, $\mathbb{K}$, $\varepsilon$ and $\left(  tI_{n}+A\right)
\left(  tI_{n}+B\right)  $ instead of $\mathbb{L}$, $\mathbb{M}$, $f$ and $A$)).

But Proposition \ref{prop.functor.clas} \textbf{(c)} (applied to
$\mathbb{K}\left[  t\right]  $, $\mathbb{K}$, $\varepsilon$, $n$, $n$,
$tI_{n}+A$ and $tI_{n}+B$ instead of $\mathbb{L}$, $\mathbb{M}$, $f$, $m$,
$p$, $A$ and $B$) shows that%
\[
\varepsilon^{n\times n}\left(  \left(  tI_{n}+A\right)  \left(  tI_{n}%
+B\right)  \right)  =\underbrace{\varepsilon^{n\times n}\left(  tI_{n}%
+A\right)  }_{\substack{=A\\\text{(by Proposition \ref{prop.tInA.ev}
\textbf{(c)})}}}\cdot\underbrace{\varepsilon^{n\times n}\left(  tI_{n}%
+B\right)  }_{\substack{=B\\\text{(by Proposition \ref{prop.tInA.ev}
\textbf{(c)}}\\\text{(applied to }B\text{ instead of }A\text{))}}}=AB.
\]
Hence, (\ref{pf.thm.adj.adj(AB).1}) becomes%
\[
\operatorname*{adj}B\cdot\operatorname*{adj}A=\operatorname*{adj}\left(
\underbrace{\varepsilon^{n\times n}\left(  \left(  tI_{n}+A\right)  \left(
tI_{n}+B\right)  \right)  }_{=AB}\right)  =\operatorname*{adj}\left(
AB\right)  .
\]
This proves Theorem \ref{thm.adj.adj(AB)}.
\end{proof}
\end{verlong}

\subsection{Determinant and adjugate of an adjugate}

Our next target is the following result:

\begin{theorem}
\label{thm.adj.adjadj}Let $n\in\mathbb{N}$. Let $A$ be an $n\times n$-matrix.

\textbf{(a)} If $n\geq1$, then $\det\left(  \operatorname*{adj}A\right)
=\left(  \det A\right)  ^{n-1}$.

\textbf{(b)} If $n\geq2$, then $\operatorname*{adj}\left(  \operatorname*{adj}%
A\right)  =\left(  \det A\right)  ^{n-2}A$.
\end{theorem}

Again, we shall first prove it in a particular case:

\begin{lemma}
\label{lem.adj.adjadj.reg}Let $n\in\mathbb{N}$. Let $A$ be an $n\times
n$-matrix. Assume that the element $\det A$ of $\mathbb{K}$ is regular.

\textbf{(a)} If $n\geq1$, then $\det\left(  \operatorname*{adj}A\right)
=\left(  \det A\right)  ^{n-1}$.

\textbf{(b)} If $n\geq2$, then $\operatorname*{adj}\left(  \operatorname*{adj}%
A\right)  =\left(  \det A\right)  ^{n-2}A$.
\end{lemma}

Before we start proving Lemma \ref{lem.adj.adjadj.reg}, let us first recall
the following fact: If $n\in\mathbb{N}$, $\lambda\in\mathbb{K}$ and
$C\in\mathbb{K}^{n\times n}$, then%
\begin{equation}
\det\left(  \lambda C\right)  =\lambda^{n}\det C.
\label{pf.lem.adj.adjadj.reg.lam1}%
\end{equation}
(In fact, this is precisely \cite[Proposition 6.12]{detnotes} (applied to $C$
instead of $A$).)

\begin{proof}
[Proof of Lemma \ref{lem.adj.adjadj.reg}.]Theorem \ref{thm.adj.inverse} yields%
\[
A\cdot\operatorname*{adj}A=\operatorname*{adj}A\cdot A=\det A\cdot I_{n}.
\]


\textbf{(a)} Assume that $n\geq1$. Now,%
\begin{align*}
\det\left(  \underbrace{A\cdot\operatorname*{adj}A}_{=\det A\cdot I_{n}%
}\right)   &  =\det\left(  \det A\cdot I_{n}\right)  =\left(  \det A\right)
^{n}\underbrace{\det\left(  I_{n}\right)  }_{=1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.lem.adj.adjadj.reg.lam1})
(applied to }\det A\text{ and }I_{n}\text{ instead of }\lambda\text{ and
}C\text{)}\right) \\
&  =\left(  \det A\right)  ^{n}=\det A\cdot\left(  \det A\right)  ^{n-1}.
\end{align*}
Thus,%
\[
\det A\cdot\left(  \det A\right)  ^{n-1}=\det\left(  A\cdot\operatorname*{adj}%
A\right)  =\det A\cdot\det\left(  \operatorname*{adj}A\right)  .
\]
Hence, Lemma \ref{lem.regular.cancel} (applied to $\mathbb{A}=\mathbb{K}$,
$a=\det A$, $b=\left(  \det A\right)  ^{n-1}$ and $c=\det\left(
\operatorname*{adj}A\right)  $) yields $\left(  \det A\right)  ^{n-1}%
=\det\left(  \operatorname*{adj}A\right)  $ (since $\det A$ is a regular
element of $\mathbb{K}$). This proves Lemma \ref{lem.adj.adjadj.reg}
\textbf{(a)}.

\textbf{(b)} Assume that $n\geq2$. Thus, $n-1\geq1$ and $n\geq2\geq1$. Now,
Lemma \ref{lem.adj.adjadj.reg} \textbf{(a)} yields
\begin{align*}
\det\left(  \operatorname*{adj}A\right)   &  =\left(  \det A\right)
^{n-1}=\det A\cdot\underbrace{\left(  \det A\right)  ^{\left(  n-1\right)
-1}}_{=\left(  \det A\right)  ^{n-2}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}n-1\geq1\right) \\
&  =\det A\cdot\left(  \det A\right)  ^{n-2}.
\end{align*}


But Theorem \ref{thm.adj.inverse} (applied to $\operatorname*{adj}A$ instead
of $A$) yields%
\[
\operatorname*{adj}A\cdot\operatorname*{adj}\left(  \operatorname*{adj}%
A\right)  =\operatorname*{adj}\left(  \operatorname*{adj}A\right)
\cdot\operatorname*{adj}A=\det\left(  \operatorname*{adj}A\right)  \cdot
I_{n}.
\]
Now,%
\begin{align*}
A\cdot\underbrace{\operatorname*{adj}A\cdot\operatorname*{adj}\left(
\operatorname*{adj}A\right)  }_{=\det\left(  \operatorname*{adj}A\right)
\cdot I_{n}}  &  =A\cdot\det\left(  \operatorname*{adj}A\right)  \cdot
I_{n}=\underbrace{\det\left(  \operatorname*{adj}A\right)  }_{=\det
A\cdot\left(  \det A\right)  ^{n-2}}A\\
&  =\det A\cdot\left(  \det A\right)  ^{n-2}A.
\end{align*}
Hence,%
\begin{align*}
\det A\cdot\left(  \det A\right)  ^{n-2}A  &  =\underbrace{A\cdot
\operatorname*{adj}A}_{=\det A\cdot I_{n}}\cdot\operatorname*{adj}\left(
\operatorname*{adj}A\right)  =\det A\cdot I_{n}\cdot\operatorname*{adj}\left(
\operatorname*{adj}A\right) \\
&  =\det A\cdot\operatorname*{adj}\left(  \operatorname*{adj}A\right)  .
\end{align*}
Hence, Lemma \ref{lem.regular.mat-cancel} (applied to $n$, $\det A$, $\left(
\det A\right)  ^{n-2}A$ and $\operatorname*{adj}\left(  \operatorname*{adj}%
A\right)  $ instead of $m$, $a$, $B$ and $C$) yields $\left(  \det A\right)
^{n-2}A=\operatorname*{adj}\left(  \operatorname*{adj}A\right)  $ (since $\det
A$ is a regular element of $\mathbb{K}$). This proves Lemma
\ref{lem.adj.adjadj.reg} \textbf{(b)}.
\end{proof}

Let us now derive Theorem \ref{thm.adj.adjadj} from this lemma:

\begin{proof}
[Proof of Theorem \ref{thm.adj.adjadj}.]Define the $\mathbb{K}$-algebra
homomorphism $\varepsilon:\mathbb{K}\left[  t\right]  \rightarrow\mathbb{K}$
as in Proposition \ref{prop.tInA.ev}.

Define a matrix $\widetilde{A}\in\left(  \mathbb{K}\left[  t\right]  \right)
^{n\times n}$ by $\widetilde{A}=tI_{n}+A$. Corollary \ref{cor.tInA.reg} shows
that the element $\det\left(  tI_{n}+A\right)  $ of $\mathbb{K}\left[
t\right]  $ is regular. In other words, the element $\det\widetilde{A}$ of
$\mathbb{K}\left[  t\right]  $ is regular (since $\widetilde{A}=tI_{n}+A$).

We have $\varepsilon^{n\times n}\left(  \operatorname*{adj}%
\underbrace{\widetilde{A}}_{=tI_{n}+A}\right)  =\varepsilon^{n\times n}\left(
\operatorname*{adj}\left(  tI_{n}+A\right)  \right)  =\operatorname*{adj}A$
(by Proposition \ref{prop.tInA.ev} \textbf{(b)}). Also, $\varepsilon\left(
\det\underbrace{\widetilde{A}}_{=tI_{n}+A}\right)  =\varepsilon\left(
\det\left(  tI_{n}+A\right)  \right)  =\det A$ (by Proposition
\ref{prop.tInA.ev} \textbf{(a)}).

\textbf{(a)} Assume that $n\geq1$. Lemma \ref{lem.adj.adjadj.reg} \textbf{(a)}
(applied to $\mathbb{K}\left[  t\right]  $ and $\widetilde{A}$ instead of
$\mathbb{K}$ and $A$) yields $\det\left(  \operatorname*{adj}\widetilde{A}%
\right)  =\left(  \det\widetilde{A}\right)  ^{n-1}$.

Now, Proposition \ref{prop.functor.det} \textbf{(a)} (applied to
$\mathbb{K}\left[  t\right]  $, $\mathbb{K}$, $\varepsilon$ and
$\operatorname*{adj}\widetilde{A}$ instead of $\mathbb{L}$, $\mathbb{M}$, $f$
and $A$) yields%
\[
\varepsilon\left(  \det\left(  \operatorname*{adj}\widetilde{A}\right)
\right)  =\det\left(  \underbrace{\varepsilon^{n\times n}\left(
\operatorname*{adj}\widetilde{A}\right)  }_{=\operatorname*{adj}A}\right)
=\det\left(  \operatorname*{adj}A\right)  .
\]
Hence,%
\begin{align*}
\det\left(  \operatorname*{adj}A\right)   &  =\varepsilon\left(
\underbrace{\det\left(  \operatorname*{adj}\widetilde{A}\right)  }_{=\left(
\det\widetilde{A}\right)  ^{n-1}}\right)  =\varepsilon\left(  \left(
\det\widetilde{A}\right)  ^{n-1}\right) \\
&  =\left(  \underbrace{\varepsilon\left(  \det\widetilde{A}\right)  }_{=\det
A}\right)  ^{n-1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varepsilon\text{ is
a }\mathbb{K}\text{-algebra homomorphism}\right) \\
&  =\left(  \det A\right)  ^{n-1}.
\end{align*}
This proves Theorem \ref{thm.adj.adjadj} \textbf{(a)}.

\textbf{(b)} Assume that $n\geq2$. Lemma \ref{lem.adj.adjadj.reg} \textbf{(b)}
(applied to $\mathbb{K}\left[  t\right]  $ and $\widetilde{A}$ instead of
$\mathbb{K}$ and $A$) yields $\operatorname*{adj}\left(  \operatorname*{adj}%
\widetilde{A}\right)  =\left(  \det\widetilde{A}\right)  ^{n-2}\widetilde{A}$.
We have $\varepsilon^{n\times n}\left(  \underbrace{\widetilde{A}}_{=tI_{n}%
+A}\right)  =\varepsilon^{n\times n}\left(  tI_{n}+A\right)  =A$ (by
Proposition \ref{prop.tInA.ev} \textbf{(c)}). Proposition
\ref{prop.functor.det} \textbf{(c)} (applied to $\mathbb{K}\left[  t\right]
$, $\mathbb{K}$, $\varepsilon$ and $\operatorname*{adj}\widetilde{A}$ instead
of $\mathbb{L}$, $\mathbb{M}$, $f$ and $A$) yields%
\[
\varepsilon^{n\times n}\left(  \operatorname*{adj}\left(  \operatorname*{adj}%
\widetilde{A}\right)  \right)  =\operatorname*{adj}\left(
\underbrace{\varepsilon^{n\times n}\left(  \operatorname*{adj}\widetilde{A}%
\right)  }_{=\operatorname*{adj}A}\right)  =\operatorname*{adj}\left(
\operatorname*{adj}A\right)  .
\]
Thus,%
\begin{align*}
\operatorname*{adj}\left(  \operatorname*{adj}A\right)   &  =\varepsilon
^{n\times n}\left(  \underbrace{\operatorname*{adj}\left(  \operatorname*{adj}%
\widetilde{A}\right)  }_{=\left(  \det\widetilde{A}\right)  ^{n-2}%
\widetilde{A}}\right)  =\varepsilon^{n\times n}\left(  \left(  \det
\widetilde{A}\right)  ^{n-2}\widetilde{A}\right) \\
&  =\underbrace{\varepsilon\left(  \left(  \det\widetilde{A}\right)
^{n-2}\right)  }_{\substack{=\left(  \varepsilon\left(  \det\widetilde{A}%
\right)  \right)  ^{n-2}\\\text{(since }\varepsilon\text{ is a }%
\mathbb{K}\text{-algebra}\\\text{homomorphism)}}}\underbrace{\varepsilon
^{n\times n}\left(  \widetilde{A}\right)  }_{=A}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Proposition \ref{prop.functor.clas} (applied to }\mathbb{K}\left[
t\right]  \text{, }\mathbb{K}\text{,}\\
\varepsilon\text{, }n\text{, }\widetilde{A}\text{ and }\left(  \det
\widetilde{A}\right)  ^{n-2}\text{ instead of }\mathbb{L}\text{, }%
\mathbb{M}\text{, }f\text{, }m\text{, }A\text{ and }\lambda\text{)}%
\end{array}
\right) \\
&  =\left(  \underbrace{\varepsilon\left(  \det\widetilde{A}\right)  }_{=\det
A}\right)  ^{n-2}A=\left(  \det A\right)  ^{n-2}A.
\end{align*}
This proves Theorem \ref{thm.adj.adjadj} \textbf{(b)}.
\end{proof}

\subsection{The adjugate of $A$ as a polynomial in $A$}

Next, let us show that the adjugate of a square matrix $A$ is a polynomial in
$A$ (with coefficients that depend on $A$, but are scalars -- not matrices):

\begin{theorem}
\label{thm.adj.poly}Let $n\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times n}$.
For every $j\in\mathbb{Z}$, define an element $c_{j}\in\mathbb{K}$ by
$c_{j}=\left[  t^{n-j}\right]  \chi_{A}$. Then,%
\[
\operatorname*{adj}A=\left(  -1\right)  ^{n-1}\sum_{i=0}^{n-1}c_{n-1-i}A^{i}.
\]

\end{theorem}

One consequence of Theorem \ref{thm.adj.poly} is that every $n\times n$-matrix
which commutes with a given $n\times n$-matrix $A$ must also commute with
$\operatorname*{adj}A$.

We prepare for the proof of Theorem \ref{thm.adj.poly} with two really simple facts:

\begin{lemma}
\label{lem.adj.lam}Let $n\in\mathbb{N}$. Let $u$ and $v$ be two elements of
$\left\{  1,2,\ldots,n\right\}  $. Let $\lambda\in\mathbb{K}$. Let $A$ be an
$n\times n$-matrix. Then,%
\[
\left(  \lambda A\right)  _{\sim u,\sim v}=\lambda A_{\sim u,\sim v}.
\]

\end{lemma}

\begin{vershort}
\begin{proof}
[Proof of Lemma \ref{lem.adj.lam}.]This follows from Lemma
\ref{lem.adj.linear} (applied to $\mu=0$ and $B=A$).
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.adj.lam}.]Lemma \ref{lem.adj.linear} (applied to
$\mu=0$ and $B=A$) yields%
\[
\left(  \lambda A+0A\right)  _{\sim u,\sim v}=\lambda A_{\sim u,\sim
v}+\underbrace{0A_{\sim u,\sim v}}_{=0_{\left(  n-1\right)  \times\left(
n-1\right)  }}=\lambda A_{\sim u,\sim v}.
\]
Since $\lambda A+\underbrace{0A}_{=0_{n\times n}}=\lambda A$, this rewrites as
$\left(  \lambda A\right)  _{\sim u,\sim v}=\lambda A_{\sim u,\sim v}$. This
proves Lemma \ref{lem.adj.lam}.
\end{proof}
\end{verlong}

\begin{proposition}
\label{prop.adj.lamA}Let $n$ be a positive integer. Let $A\in\mathbb{K}%
^{n\times n}$ and $\lambda\in\mathbb{K}$. Then, $\operatorname*{adj}\left(
\lambda A\right)  =\lambda^{n-1}\operatorname*{adj}A$.
\end{proposition}

\begin{vershort}
\begin{proof}
[Proof of Proposition \ref{prop.adj.lamA}.]Recalling the definitions of
$\operatorname*{adj}\left(  \lambda A\right)  $ and $\operatorname*{adj}A$
(and using Lemma \ref{lem.adj.lam}), the reader can easily reduce Proposition
\ref{prop.adj.lamA} to (\ref{pf.lem.adj.adjadj.reg.lam1}) (applied to $n-1$
and $A_{\sim j,\sim i}$ instead of $n$ and $C$).
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Proposition \ref{prop.adj.lamA}.]Every $\left(  i,j\right)
\in\left\{  1,2,\ldots,n\right\}  ^{2}$ satisfies
\begin{equation}
\det\left(  \left(  \lambda A\right)  _{\sim j,\sim i}\right)  =\lambda
^{n-1}\det\left(  A_{\sim j,\sim i}\right)  \label{pf.prop.adj.lamA.1}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.prop.adj.lamA.1}):} Let $\left(
i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}$. Thus, $i$ and $j$ are two
elements of $\left\{  1,2,\ldots,n\right\}  $. Hence, Lemma \ref{lem.adj.lam}
(applied to $u=j$ and $v=i$) yields $\left(  \lambda A\right)  _{\sim j,\sim
i}=\lambda A_{\sim j,\sim i}$. But $i\in\left\{  1,2,\ldots,n\right\}  $, so
that $1\leq i\leq n$ and thus $n\geq1$. Hence, $n-1\in\mathbb{N}$. Also,
$A_{\sim j,\sim i}\in\mathbb{K}^{\left(  n-1\right)  \times\left(  n-1\right)
}$. Hence, (\ref{pf.lem.adj.adjadj.reg.lam1}) (applied to $n-1$ and $A_{\sim
j,\sim i}$ instead of $n$ and $C$) yields $\det\left(  \lambda A_{\sim j,\sim
i}\right)  =\lambda^{n-1}\det\left(  A_{\sim j,\sim i}\right)  $. Now,%
\[
\det\left(  \underbrace{\left(  \lambda A\right)  _{\sim j,\sim i}}_{=\lambda
A_{\sim j,\sim i}}\right)  =\det\left(  \lambda A_{\sim j,\sim i}\right)
=\lambda^{n-1}\det\left(  A_{\sim j,\sim i}\right)  .
\]
This proves (\ref{pf.prop.adj.lamA.1}).}. Now, the definition of
$\operatorname*{adj}A$ yields $\operatorname*{adj}A=\left(  \left(  -1\right)
^{i+j}\det\left(  A_{\sim j,\sim i}\right)  \right)  _{1\leq i\leq n,\ 1\leq
j\leq n}$. Multiplying both sides of this equality by $\lambda^{n-1}$, we
obtain%
\begin{align}
\lambda^{n-1}\operatorname*{adj}A  &  =\lambda^{n-1}\left(  \left(  -1\right)
^{i+j}\det\left(  A_{\sim j,\sim i}\right)  \right)  _{1\leq i\leq n,\ 1\leq
j\leq n}\nonumber\\
&  =\left(  \underbrace{\lambda^{n-1}\left(  -1\right)  ^{i+j}}_{=\left(
-1\right)  ^{i+j}\lambda^{n-1}}\det\left(  A_{\sim j,\sim i}\right)  \right)
_{1\leq i\leq n,\ 1\leq j\leq n}\nonumber\\
&  =\left(  \left(  -1\right)  ^{i+j}\lambda^{n-1}\det\left(  A_{\sim j,\sim
i}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq n}.
\label{pf.prop.adj.lamA.4}%
\end{align}


On the other hand, the definition of $\operatorname*{adj}\left(  \lambda
A\right)  $ yields%
\begin{align*}
\operatorname*{adj}\left(  \lambda A\right)   &  =\left(  \left(  -1\right)
^{i+j}\underbrace{\det\left(  \left(  \lambda A\right)  _{\sim j,\sim
i}\right)  }_{\substack{=\lambda^{n-1}\det\left(  A_{\sim j,\sim i}\right)
\\\text{(by (\ref{pf.prop.adj.lamA.1}))}}}\right)  _{1\leq i\leq n,\ 1\leq
j\leq n}\\
&  =\left(  \left(  -1\right)  ^{i+j}\lambda^{n-1}\det\left(  A_{\sim j,\sim
i}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq n}=\lambda^{n-1}%
\operatorname*{adj}A
\end{align*}
(by (\ref{pf.prop.adj.lamA.4})). This proves Proposition \ref{prop.adj.lamA}.
\end{proof}
\end{verlong}

Now, let me show a slightly simpler variant of Theorem \ref{thm.adj.poly}:

\begin{lemma}
\label{lem.adj.poly.-A}Let $n$ be a positive integer. Let $A\in\mathbb{K}%
^{n\times n}$. For every $j\in\mathbb{Z}$, define an element $c_{j}%
\in\mathbb{K}$ by $c_{j}=\left[  t^{n-j}\right]  \chi_{A}$. Then,%
\[
\operatorname*{adj}\left(  -A\right)  =\sum_{i=0}^{n-1}c_{n-1-i}A^{i}.
\]

\end{lemma}

\begin{vershort}
\begin{proof}
[Proof of Lemma \ref{lem.adj.poly.-A}.]For every $j\in\mathbb{Z}$, define an
element $c_{j}\in\mathbb{K}$ by $c_{j}=\left[  t^{n-j}\right]  \chi_{A}$.

Proposition \ref{prop.adj.poly} shows that there exist $n$ matrices
$D_{0},D_{1},\ldots,D_{n-1}$ in $\mathbb{K}^{n\times n}$ such that%
\begin{equation}
\operatorname*{adj}\left(  tI_{n}-A\right)  =\sum_{k=0}^{n-1}t^{k}%
D_{k}\ \ \ \ \ \ \ \ \ \ \text{in }\left(  \mathbb{K}\left[  t\right]
\right)  ^{n\times n}. \label{pf.lem.adj.poly.-A.short.1}%
\end{equation}
Consider these $D_{0},D_{1},\ldots,D_{n-1}$. Thus, an $n$-tuple $\left(
D_{0},D_{1},\ldots,D_{n-1}\right)  $ of matrices in $\mathbb{K}^{n\times n}$
is defined. Extend this $n$-tuple to a family $\left(  D_{k}\right)
_{k\in\mathbb{Z}}$ of matrices in $\mathbb{K}^{n\times n}$ by setting
(\ref{eq.lem.CH.D.0}). Lemma \ref{lem.CH.D} \textbf{(c)} (applied to $k=n-1$)
yields%
\begin{equation}
\sum_{i=0}^{n-1}c_{n-1-i}A^{i}=D_{n-1-\left(  n-1\right)  }=D_{0}.
\label{pf.lem.adj.poly.-A.short.2}%
\end{equation}


On the other hand, define the $\mathbb{K}$-algebra homomorphism $\varepsilon
:\mathbb{K}\left[  t\right]  \rightarrow\mathbb{K}$ as in Proposition
\ref{prop.tInA.ev}. This homomorphism $\varepsilon$ satisfies $\varepsilon
\left(  t\right)  =0$. Also, it satisfies $\varepsilon\left(  u\right)  =u$
for every $u\in\mathbb{K}$. Hence, the map $\varepsilon^{n\times n}:\left(
\mathbb{K}\left[  t\right]  \right)  ^{n\times n}\rightarrow\mathbb{K}%
^{n\times n}$ (defined as in Definition \ref{def.f-mat}) satisfies%
\begin{equation}
\varepsilon^{n\times n}\left(  F\right)  =F\ \ \ \ \ \ \ \ \ \ \text{for every
}F\in\mathbb{K}^{n\times n}. \label{pf.lem.adj.poly.-A.short.eF}%
\end{equation}


But Proposition \ref{prop.functor.clas} \textbf{(a)} (applied to
$\mathbb{L}=\mathbb{K}\left[  t\right]  $, $\mathbb{M}=\mathbb{K}$,
$f=\varepsilon$ and $m=n$) yields that the map $\varepsilon^{n\times
n}:\left(  \mathbb{K}\left[  t\right]  \right)  ^{n\times n}\rightarrow
\mathbb{K}^{n\times n}$ is a homomorphism of additive groups. Hence,%
\begin{align}
\varepsilon^{n\times n}\left(  \sum_{k=0}^{n-1}t^{k}D_{k}\right)   &
=\sum_{k=0}^{n-1}\underbrace{\varepsilon^{n\times n}\left(  t^{k}D_{k}\right)
}_{\substack{=\varepsilon\left(  t^{k}\right)  \varepsilon^{n\times n}\left(
D_{k}\right)  \\\text{(by Proposition \ref{prop.functor.clas} \textbf{(d)}
(applied to}\\\mathbb{K}\left[  t\right]  \text{, }\mathbb{K}\text{,
}\varepsilon\text{, }n\text{, }D_{k}\text{ and }t^{k}\\\text{instead of
}\mathbb{L}\text{, }\mathbb{M}\text{, }f\text{, }m\text{, }A\text{ and
}\lambda\text{))}}}\nonumber\\
&  =\sum_{k=0}^{n-1}\underbrace{\varepsilon\left(  t^{k}\right)
}_{\substack{=\left(  \varepsilon\left(  t\right)  \right)  ^{k}\\\text{(since
}\varepsilon\text{ is a ring}\\\text{homomorphism)}}}\underbrace{\varepsilon
^{n\times n}\left(  D_{k}\right)  }_{\substack{=D_{k}\\\text{(by
(\ref{pf.lem.adj.poly.-A.short.eF})}\\\text{(applied to }F=D_{k}\text{))}%
}}\nonumber\\
&  =\sum_{k=0}^{n-1}\left(  \underbrace{\varepsilon\left(  t\right)  }%
_{=0}\right)  ^{k}D_{k}=\sum_{k=0}^{n-1}0^{k}D_{k}=\underbrace{0^{0}}%
_{=1}D_{0}+\sum_{k=1}^{n-1}\underbrace{0^{k}}_{\substack{=0\\\text{(since
}k\geq1\text{)}}}D_{k}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }k=0\text{ from the sum}\\
\text{(since }0\in\left\{  0,1,\ldots,n-1\right\}  \text{)}%
\end{array}
\right) \nonumber\\
&  =D_{0}+\underbrace{\sum_{k=1}^{n-1}0D_{k}}_{=0_{n\times n}}=D_{0}.
\label{pf.lem.adj.poly.-A.short.7}%
\end{align}
But applying the map $\varepsilon^{n\times n}$ to both sides of the equality
(\ref{pf.lem.adj.poly.-A.short.1}), we obtain%
\[
\varepsilon^{n\times n}\left(  \operatorname*{adj}\left(  tI_{n}-A\right)
\right)  =\varepsilon^{n\times n}\left(  \sum_{k=0}^{n-1}t^{k}D_{k}\right)
=D_{0}%
\]
(by (\ref{pf.lem.adj.poly.-A.short.7})). Thus,%
\[
D_{0}=\varepsilon^{n\times n}\left(  \operatorname*{adj}\left(
\underbrace{tI_{n}-A}_{=tI_{n}+\left(  -A\right)  }\right)  \right)
=\varepsilon^{n\times n}\left(  \operatorname*{adj}\left(  tI_{n}+\left(
-A\right)  \right)  \right)  =\operatorname*{adj}\left(  -A\right)
\]
(by Proposition \ref{prop.tInA.ev} \textbf{(b)}, applied to $-A$ instead of
$A$). Hence, (\ref{pf.lem.adj.poly.-A.short.2}) becomes%
\[
\sum_{i=0}^{n-1}c_{n-1-i}A^{i}=D_{0}=\operatorname*{adj}\left(  -A\right)  .
\]
This proves Lemma \ref{lem.adj.poly.-A}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.adj.poly.-A}.]For every $j\in\mathbb{Z}$, define an
element $c_{j}\in\mathbb{K}$ by $c_{j}=\left[  t^{n-j}\right]  \chi_{A}$.

Proposition \ref{prop.adj.poly} shows that there exist $n$ matrices
$D_{0},D_{1},\ldots,D_{n-1}$ in $\mathbb{K}^{n\times n}$ such that%
\begin{equation}
\operatorname*{adj}\left(  tI_{n}-A\right)  =\sum_{k=0}^{n-1}t^{k}%
D_{k}\ \ \ \ \ \ \ \ \ \ \text{in }\left(  \mathbb{K}\left[  t\right]
\right)  ^{n\times n}. \label{pf.lem.adj.poly.-A.1}%
\end{equation}
Consider these $D_{0},D_{1},\ldots,D_{n-1}$. Thus, an $n$-tuple $\left(
D_{0},D_{1},\ldots,D_{n-1}\right)  $ of matrices in $\mathbb{K}^{n\times n}$
is defined. Extend this $n$-tuple to a family $\left(  D_{k}\right)
_{k\in\mathbb{Z}}$ of matrices in $\mathbb{K}^{n\times n}$ by setting
(\ref{eq.lem.CH.D.0}). Lemma \ref{lem.CH.D} \textbf{(c)} (applied to $k=n-1$)
yields%
\begin{equation}
\sum_{i=0}^{n-1}c_{n-1-i}A^{i}=D_{n-1-\left(  n-1\right)  }=D_{0}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }n-1-\left(  n-1\right)  =0\right)  .
\label{pf.lem.adj.poly.-A.2}%
\end{equation}


On the other hand, define the $\mathbb{K}$-algebra homomorphism $\varepsilon
:\mathbb{K}\left[  t\right]  \rightarrow\mathbb{K}$ as in Proposition
\ref{prop.tInA.ev}. This homomorphism $\varepsilon$ satisfies $\varepsilon
\left(  t\right)  =0$. Also, it satisfies%
\begin{equation}
\varepsilon\left(  u\right)  =u\ \ \ \ \ \ \ \ \ \ \text{for every }%
u\in\mathbb{K}. \label{pf.lem.adj.poly.-A.eu}%
\end{equation}
Hence, the map $\varepsilon^{n\times n}:\left(  \mathbb{K}\left[  t\right]
\right)  ^{n\times n}\rightarrow\mathbb{K}^{n\times n}$ (defined as in
Definition \ref{def.f-mat}) satisfies%
\begin{equation}
\varepsilon^{n\times n}\left(  F\right)  =F\ \ \ \ \ \ \ \ \ \ \text{for every
}F\in\mathbb{K}^{n\times n} \label{pf.lem.adj.poly.-A.eF}%
\end{equation}
\footnote{\textit{Proof of (\ref{pf.lem.adj.poly.-A.eF}):} Let $F\in
\mathbb{K}^{n\times n}$. Write the $n\times n$-matrix $F$ in the form
$F=\left(  f_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$. Then,
$f_{i,j}\in\mathbb{K}$ for every $\left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}$ (since $F\in\mathbb{K}^{n\times n}$). Hence,%
\begin{equation}
\varepsilon\left(  f_{i,j}\right)  =f_{i,j}\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}
\label{pf.lem.adj.poly.-A.eF.pf.1}%
\end{equation}
(by (\ref{pf.lem.adj.poly.-A.eu}) (applied to $u=f_{i,j}$)).
\par
Applying the map $\varepsilon^{n\times n}$ to both sides of the equality
$F=\left(  f_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}$, we obtain%
\begin{align*}
\varepsilon^{n\times n}\left(  F\right)   &  =\varepsilon^{n\times n}\left(
\left(  f_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}\right)  =\left(
\underbrace{\varepsilon\left(  f_{i,j}\right)  }_{\substack{=f_{i,j}%
\\\text{(by (\ref{pf.lem.adj.poly.-A.eF.pf.1}))}}}\right)  _{1\leq i\leq
n,\ 1\leq j\leq n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the
map }\varepsilon^{n\times n}\right) \\
&  =\left(  f_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq n}=F.
\end{align*}
This proves (\ref{pf.lem.adj.poly.-A.eF}).}.

The integer $n$ is positive. Thus, $n-1\in\mathbb{N}$, so that $0\in\left\{
0,1,\ldots,n-1\right\}  $.

But $\varepsilon$ is a ring homomorphism (since $\varepsilon$ is a
$\mathbb{K}$-algebra homomorphism). Hence, Proposition \ref{prop.functor.clas}
\textbf{(a)} (applied to $\mathbb{L}=\mathbb{K}\left[  t\right]  $,
$\mathbb{M}=\mathbb{K}$, $f=\varepsilon$ and $m=n$) yields that the map
$\varepsilon^{n\times n}:\left(  \mathbb{K}\left[  t\right]  \right)
^{n\times n}\rightarrow\mathbb{K}^{n\times n}$ is a homomorphism of additive
groups. Hence,%
\begin{align}
\varepsilon^{n\times n}\left(  \sum_{k=0}^{n-1}t^{k}D_{k}\right)   &
=\sum_{k=0}^{n-1}\underbrace{\varepsilon^{n\times n}\left(  t^{k}D_{k}\right)
}_{\substack{=\varepsilon\left(  t^{k}\right)  \varepsilon^{n\times n}\left(
D_{k}\right)  \\\text{(by Proposition \ref{prop.functor.clas} \textbf{(d)}%
}\\\text{(applied to }\mathbb{K}\left[  t\right]  \text{, }\mathbb{K}\text{,
}\varepsilon\text{, }n\text{, }D_{k}\text{ and }t^{k}\\\text{instead of
}\mathbb{L}\text{, }\mathbb{M}\text{, }f\text{, }m\text{, }A\text{ and
}\lambda\text{))}}}\nonumber\\
&  =\sum_{k=0}^{n-1}\underbrace{\varepsilon\left(  t^{k}\right)
}_{\substack{=\left(  \varepsilon\left(  t\right)  \right)  ^{k}\\\text{(since
}\varepsilon\text{ is a ring}\\\text{homomorphism)}}}\underbrace{\varepsilon
^{n\times n}\left(  D_{k}\right)  }_{\substack{=D_{k}\\\text{(by
(\ref{pf.lem.adj.poly.-A.eF})}\\\text{(applied to }F=D_{k}\text{))}%
}}\nonumber\\
&  =\sum_{k=0}^{n-1}\left(  \underbrace{\varepsilon\left(  t\right)  }%
_{=0}\right)  ^{k}D_{k}=\sum_{k=0}^{n-1}0^{k}D_{k}=\underbrace{0^{0}}%
_{=1}D_{0}+\sum_{k=1}^{n-1}\underbrace{0^{k}}_{\substack{=0\\\text{(since
}k\geq1\text{)}}}D_{k}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we have split off the addend for }k=0\text{ from the sum}\\
\text{(since }0\in\left\{  0,1,\ldots,n-1\right\}  \text{)}%
\end{array}
\right) \nonumber\\
&  =D_{0}+\underbrace{\sum_{k=1}^{n-1}0D_{k}}_{=0_{n\times n}}=D_{0}.
\label{pf.lem.adj.poly.-A.7}%
\end{align}
But applying the map $\varepsilon^{n\times n}$ to both sides of the equality
(\ref{pf.lem.adj.poly.-A.1}), we obtain%
\[
\varepsilon^{n\times n}\left(  \operatorname*{adj}\left(  tI_{n}-A\right)
\right)  =\varepsilon^{n\times n}\left(  \sum_{k=0}^{n-1}t^{k}D_{k}\right)
=D_{0}%
\]
(by (\ref{pf.lem.adj.poly.-A.7})). Thus,%
\[
D_{0}=\varepsilon^{n\times n}\left(  \operatorname*{adj}\left(
\underbrace{tI_{n}-A}_{=tI_{n}+\left(  -A\right)  }\right)  \right)
=\varepsilon^{n\times n}\left(  \operatorname*{adj}\left(  tI_{n}+\left(
-A\right)  \right)  \right)  =\operatorname*{adj}\left(  -A\right)
\]
(by Proposition \ref{prop.tInA.ev} \textbf{(b)} (applied to $-A$ instead of
$A$)). Hence, (\ref{pf.lem.adj.poly.-A.2}) becomes%
\[
\sum_{i=0}^{n-1}c_{n-1-i}A^{i}=D_{0}=\operatorname*{adj}\left(  -A\right)  .
\]
This proves Lemma \ref{lem.adj.poly.-A}.
\end{proof}
\end{verlong}

Finally, we are ready to prove Theorem \ref{thm.adj.poly}:

\begin{vershort}
\begin{proof}
[Proof of Theorem \ref{thm.adj.poly}.]We must prove the equality
$\operatorname*{adj}A=\left(  -1\right)  ^{n-1}\sum_{i=0}^{n-1}c_{n-1-i}A^{i}%
$. This is an equality between two $n\times n$-matrices, and thus obviously
holds if $n=0$. Hence, we WLOG assume that $n\neq0$. Thus, $n$ is a positive
integer. Hence, Proposition \ref{prop.adj.lamA} (applied to $\lambda=-1$)
yields
\[
\operatorname*{adj}\left(  -A\right)  =\left(  -1\right)  ^{n-1}%
\operatorname*{adj}A.
\]
Therefore,%
\[
\operatorname*{adj}A=\left(  -1\right)  ^{n-1}\underbrace{\operatorname*{adj}%
\left(  -A\right)  }_{\substack{=\sum_{i=0}^{n-1}c_{n-1-i}A^{i}\\\text{(by
Lemma \ref{lem.adj.poly.-A})}}}=\left(  -1\right)  ^{n-1}\sum_{i=0}%
^{n-1}c_{n-1-i}A^{i}.
\]
This proves Theorem \ref{thm.adj.poly}.
\end{proof}
\end{vershort}

\begin{verlong}
\begin{proof}
[Proof of Theorem \ref{thm.adj.poly}.]We must prove the equality
$\operatorname*{adj}A=\left(  -1\right)  ^{n-1}\sum_{i=0}^{n-1}c_{n-1-i}A^{i}%
$. If $n=0$, then this equality is obviously true\footnote{\textit{Proof.}
This equality is an equality between two $n\times n$-matrices. Thus, if $n=0$,
then this equality is an equality between two $0\times0$-matrices, and
therefore is true (because any equality between two $0\times0$-matrices is
true (since there exists only one $0\times0$-matrix)). Qed.}. Hence, for the
rest of this proof, we can WLOG assume that we don't have $n=0$. Assume this.

We have $n\neq0$ (since we don't have $n=0$). Thus, $n$ is a positive integer
(since $n\in\mathbb{N}$). Hence, Proposition \ref{prop.adj.lamA} (applied to
$\lambda=-1$) yields
\[
\operatorname*{adj}\left(  \left(  -1\right)  A\right)  =\left(  -1\right)
^{n-1}\operatorname*{adj}A.
\]
Now,%
\[
\left(  -1\right)  ^{n-1}\operatorname*{adj}\left(  \underbrace{-A}_{=\left(
-1\right)  A}\right)  =\left(  -1\right)  ^{n-1}%
\underbrace{\operatorname*{adj}\left(  \left(  -1\right)  A\right)
}_{=\left(  -1\right)  ^{n-1}\operatorname*{adj}A}=\underbrace{\left(
-1\right)  ^{n-1}\left(  -1\right)  ^{n-1}}_{\substack{=\left(  -1\right)
^{\left(  n-1\right)  +\left(  n-1\right)  }=1\\\text{(since }\left(
n-1\right)  +\left(  n-1\right)  =2\left(  n-1\right)  \\\text{is even)}%
}}\operatorname*{adj}A=\operatorname*{adj}A,
\]
so that%
\[
\operatorname*{adj}A=\left(  -1\right)  ^{n-1}\underbrace{\operatorname*{adj}%
\left(  -A\right)  }_{\substack{=\sum_{i=0}^{n-1}c_{n-1-i}A^{i}\\\text{(by
Lemma \ref{lem.adj.poly.-A})}}}=\left(  -1\right)  ^{n-1}\sum_{i=0}%
^{n-1}c_{n-1-i}A^{i}.
\]
This proves Theorem \ref{thm.adj.poly}.
\end{proof}
\end{verlong}

\subsection{Minors of the adjugate: Jacobi's theorem}

A \textit{minor} of a matrix $A$ is defined to be a determinant of a square
submatrix of $A$. A theorem due to Jacobi connects the minors of
$\operatorname*{adj}A$ (for a square matrix $A$) with the minors of $A$.
Before we can state this theorem, let us introduce some notations:

\begin{definition}
\label{def.submatrix2}Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let
$A=\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}$ be an $n\times
m$-matrix. Let $i_{1},i_{2},\ldots,i_{u}$ be some elements of $\left\{
1,2,\ldots,n\right\}  $; let $j_{1},j_{2},\ldots,j_{v}$ be some elements of
$\left\{  1,2,\ldots,m\right\}  $. Then, we shall use $\operatorname*{sub}%
\nolimits_{\left(  i_{1},i_{2},\ldots,i_{u}\right)  }^{\left(  j_{1}%
,j_{2},\ldots,j_{v}\right)  }A$ as a synonym for the $u\times v$-matrix
$\operatorname*{sub}\nolimits_{i_{1},i_{2},\ldots,i_{u}}^{j_{1},j_{2}%
,\ldots,j_{v}}A$. Thus, for every $\mathbf{i}\in\left\{  1,2,\ldots,n\right\}
^{u}$ and $\mathbf{j}\in\left\{  1,2,\ldots,m\right\}  ^{v}$, a $u\times
v$-matrix $\operatorname*{sub}\nolimits_{\mathbf{i}}^{\mathbf{j}}A$ is defined.
\end{definition}

\begin{definition}
If $I$ is a finite set of integers, then $\sum I$ shall denote the sum of all
elements of $I$. (Thus, $\sum I=\sum_{i\in I}i$.)
\end{definition}

\begin{definition}
If $I$ is a finite set of integers, then $w\left(  I\right)  $ shall denote
the list of all elements of $I$ in increasing order (with no repetitions).
(For example, $w\left(  \left\{  3,4,8\right\}  \right)  =\left(
3,4,8\right)  $.)
\end{definition}

The following fact is obvious:

\begin{remark}
\label{rmk.w(I).in}Let $n\in\mathbb{N}$. Let $I$ be a subset of $\left\{
1,2,\ldots,n\right\}  $. Then, $w\left(  I\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{\left\vert I\right\vert }$.
\end{remark}

\begin{verlong}
\begin{proof}
[Proof of Remark \ref{rmk.w(I).in}.]We know that $w\left(  I\right)  $ is the
list of all elements of $I$ in increasing order (with no repetitions) (by the
definition of $w\left(  I\right)  $). Thus, $w\left(  I\right)  $ is a list of
$\left\vert I\right\vert $ elements.

Also, $w\left(  I\right)  $ is a list of elements of $I$ (since $w\left(
I\right)  $ is the list of all elements of $I$ in increasing order (with no
repetitions)). Thus, $w\left(  I\right)  $ is a list of $\left\vert
I\right\vert $ elements of $I$. In other words, $w\left(  I\right)  \in
I^{\left\vert I\right\vert }$. But $I\subseteq\left\{  1,2,\ldots,n\right\}  $
(since $I$ is a subset of $\left\{  1,2,\ldots,n\right\}  $), and thus
$I^{\left\vert I\right\vert }\subseteq\left\{  1,2,\ldots,n\right\}
^{\left\vert I\right\vert }$. Hence, $w\left(  I\right)  \in I^{\left\vert
I\right\vert }\subseteq\left\{  1,2,\ldots,n\right\}  ^{\left\vert
I\right\vert }$. This proves Remark \ref{rmk.w(I).in}.
\end{proof}
\end{verlong}

Now, we can state Jacobi's theorem\footnote{This is \cite[Corollary
7.256]{detnotes}. It also appears in \cite[Theorem 2.5.2]{Prasolov} (in a
different form).}:

\begin{theorem}
\label{thm.jacobi-complement}Let $n\in\mathbb{N}$. For any subset $I$ of
$\left\{  1,2,\ldots,n\right\}  $, we let $\widetilde{I}$ denote the
complement $\left\{  1,2,\ldots,n\right\}  \setminus I$ of $I$.

Let $A$ be an $n\times n$-matrix.

Let $P$ and $Q$ be two subsets of $\left\{  1,2,\ldots,n\right\}  $ such that
$\left\vert P\right\vert =\left\vert Q\right\vert \geq1$. Then,%
\[
\det\left(  \operatorname*{sub}\nolimits_{w\left(  P\right)  }^{w\left(
Q\right)  }\left(  \operatorname*{adj}A\right)  \right)  =\left(  -1\right)
^{\sum P+\sum Q}\left(  \det A\right)  ^{\left\vert Q\right\vert -1}%
\det\left(  \operatorname*{sub}\nolimits_{w\left(  \widetilde{Q}\right)
}^{w\left(  \widetilde{P}\right)  }A\right)  .
\]

\end{theorem}

We shall not give a standalone proof of this theorem; instead, we will merely
derive it from results proven in \cite{detnotes}. Namely, in \cite[Corollary
7.255]{detnotes}, the following was proven:

\begin{lemma}
\label{lem.sol.addexe.jacobi-complement.adjform}Let $n\in\mathbb{N}$. For any
subset $I$ of $\left\{  1,2,\ldots,n\right\}  $, we let $\widetilde{I}$ denote
the complement $\left\{  1,2,\ldots,n\right\}  \setminus I$ of $I$.

Let $A$ be an $n\times n$-matrix.

Let $P$ and $Q$ be two subsets of $\left\{  1,2,\ldots,n\right\}  $ such that
$\left\vert P\right\vert =\left\vert Q\right\vert $. Then,%
\[
\det A\cdot\det\left(  \operatorname*{sub}\nolimits_{w\left(  P\right)
}^{w\left(  Q\right)  }\left(  \operatorname*{adj}A\right)  \right)  =\left(
-1\right)  ^{\sum P+\sum Q}\left(  \det A\right)  ^{\left\vert Q\right\vert
}\det\left(  \operatorname*{sub}\nolimits_{w\left(  \widetilde{Q}\right)
}^{w\left(  \widetilde{P}\right)  }A\right)  .
\]

\end{lemma}

We shall also use the following obvious lemma:

\begin{lemma}
\label{lem.functor.subij}Let $\mathbb{L}$ and $\mathbb{M}$ be two commutative
rings. Let $f:\mathbb{L}\rightarrow\mathbb{M}$ be any map. Let $n\in
\mathbb{N}$ and $m\in\mathbb{N}$. Let $A\in\mathbb{L}^{n\times m}$.

Let $u\in\mathbb{N}$ and $v\in\mathbb{N}$. Let $\mathbf{i}\in\left\{
1,2,\ldots,n\right\}  ^{u}$ and $\mathbf{j}\in\left\{  1,2,\ldots,m\right\}
^{v}$. Then,%
\[
f^{u\times v}\left(  \operatorname*{sub}\nolimits_{\mathbf{i}}^{\mathbf{j}%
}A\right)  =\operatorname*{sub}\nolimits_{\mathbf{i}}^{\mathbf{j}}\left(
f^{n\times m}\left(  A\right)  \right)  .
\]

\end{lemma}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.functor.subij}.]We have $\mathbf{i}\in\left\{
1,2,\ldots,n\right\}  ^{u}$. Thus, we can write $\mathbf{i}$ in the form
$\mathbf{i}=\left(  i_{1},i_{2},\ldots,i_{u}\right)  $ for some $u$ elements
$i_{1},i_{2},\ldots,i_{u}$ of $\left\{  1,2,\ldots,n\right\}  $. Consider
these $i_{1},i_{2},\ldots,i_{u}$.

We have $\mathbf{j}\in\left\{  1,2,\ldots,m\right\}  ^{v}$. Thus, we can write
$\mathbf{j}$ in the form $\mathbf{j}=\left(  j_{1},j_{2},\ldots,j_{v}\right)
$ for some $v$ elements $j_{1},j_{2},\ldots,j_{v}$ of $\left\{  1,2,\ldots
,m\right\}  $. Consider these $j_{1},j_{2},\ldots,j_{v}$.

Write the $n\times m$-matrix $A$ in the form $A=\left(  a_{i,j}\right)
_{1\leq i\leq n,\ 1\leq j\leq m}$. Then, $f^{n\times m}\left(  A\right)
=\left(  f\left(  a_{i,j}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq m}$
(by the definition of the map $f^{n\times m}$).

But%
\begin{align*}
\operatorname*{sub}\nolimits_{\mathbf{i}}^{\mathbf{j}}A  &
=\operatorname*{sub}\nolimits_{\left(  i_{1},i_{2},\ldots,i_{u}\right)
}^{\left(  j_{1},j_{2},\ldots,j_{v}\right)  }A\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\mathbf{i}=\left(  i_{1},i_{2},\ldots,i_{u}\right)  \text{ and
}\mathbf{j}=\left(  j_{1},j_{2},\ldots,j_{v}\right)  \right) \\
&  =\operatorname*{sub}\nolimits_{i_{1},i_{2},\ldots,i_{u}}^{j_{1}%
,j_{2},\ldots,j_{v}}A=\left(  a_{i_{x},j_{y}}\right)  _{1\leq x\leq u,\ 1\leq
y\leq v}%
\end{align*}
(by the definition of $\operatorname*{sub}\nolimits_{i_{1},i_{2},\ldots,i_{u}%
}^{j_{1},j_{2},\ldots,j_{v}}A$, since $A=\left(  a_{i,j}\right)  _{1\leq i\leq
n,\ 1\leq j\leq m}$). Applying the map $f^{u\times v}$ to both sides of this
equality, we find%
\[
f^{u\times v}\left(  \operatorname*{sub}\nolimits_{\mathbf{i}}^{\mathbf{j}%
}A\right)  =f^{u\times v}\left(  \left(  a_{i_{x},j_{y}}\right)  _{1\leq x\leq
u,\ 1\leq y\leq v}\right)  =\left(  f\left(  a_{i_{x},j_{y}}\right)  \right)
_{1\leq x\leq u,\ 1\leq y\leq v}%
\]
(by the definition of $f^{u\times v}$). Comparing this with%
\begin{align*}
\operatorname*{sub}\nolimits_{\mathbf{i}}^{\mathbf{j}}\left(  f^{n\times
m}\left(  A\right)  \right)   &  =\operatorname*{sub}\nolimits_{\left(
i_{1},i_{2},\ldots,i_{u}\right)  }^{\left(  j_{1},j_{2},\ldots,j_{v}\right)
}\left(  f^{n\times m}\left(  A\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathbf{i}=\left(  i_{1}%
,i_{2},\ldots,i_{u}\right)  \text{ and }\mathbf{j}=\left(  j_{1},j_{2}%
,\ldots,j_{v}\right)  \right) \\
&  =\operatorname*{sub}\nolimits_{i_{1},i_{2},\ldots,i_{u}}^{j_{1}%
,j_{2},\ldots,j_{v}}\left(  f^{n\times m}\left(  A\right)  \right)  =\left(
f\left(  a_{i_{x},j_{y}}\right)  \right)  _{1\leq x\leq u,\ 1\leq y\leq v}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by the definition of }\operatorname*{sub}\nolimits_{i_{1},i_{2}%
,\ldots,i_{u}}^{j_{1},j_{2},\ldots,j_{v}}\left(  f^{n\times m}\left(
A\right)  \right)  \text{,}\\
\text{since }f^{n\times m}\left(  A\right)  =\left(  f\left(  a_{i,j}\right)
\right)  _{1\leq i\leq n,\ 1\leq j\leq m}%
\end{array}
\right)  ,
\end{align*}
we obtain $f^{u\times v}\left(  \operatorname*{sub}\nolimits_{\mathbf{i}%
}^{\mathbf{j}}A\right)  =\operatorname*{sub}\nolimits_{\mathbf{i}}%
^{\mathbf{j}}\left(  f^{n\times m}\left(  A\right)  \right)  $. This proves
Lemma \ref{lem.functor.subij}.
\end{proof}
\end{verlong}

\begin{proof}
[Proof of Theorem \ref{thm.jacobi-complement}.]Define the $\mathbb{K}$-algebra
homomorphism $\varepsilon:\mathbb{K}\left[  t\right]  \rightarrow\mathbb{K}$
as in Proposition \ref{prop.tInA.ev}.

Define a matrix $\widetilde{A}\in\left(  \mathbb{K}\left[  t\right]  \right)
^{n\times n}$ by $\widetilde{A}=tI_{n}+A$. Corollary \ref{cor.tInA.reg} shows
that the element $\det\left(  tI_{n}+A\right)  $ of $\mathbb{K}\left[
t\right]  $ is regular. In other words, the element $\det\widetilde{A}$ of
$\mathbb{K}\left[  t\right]  $ is regular (since $\widetilde{A}=tI_{n}+A$).

We have $\left\vert Q\right\vert -1\in\mathbb{N}$ (since $\left\vert
Q\right\vert \geq1$). Lemma \ref{lem.sol.addexe.jacobi-complement.adjform}
(applied to $\mathbb{K}\left[  t\right]  $ and $\widetilde{A}$ instead of
$\mathbb{K}$ and $A$) yields%
\begin{align*}
\det\widetilde{A}\cdot\det\left(  \operatorname*{sub}\nolimits_{w\left(
P\right)  }^{w\left(  Q\right)  }\left(  \operatorname*{adj}\widetilde{A}%
\right)  \right)   &  =\left(  -1\right)  ^{\sum P+\sum Q}\underbrace{\left(
\det\widetilde{A}\right)  ^{\left\vert Q\right\vert }}_{\substack{=\left(
\det\widetilde{A}\right)  \left(  \det\widetilde{A}\right)  ^{\left\vert
Q\right\vert -1}\\\text{(since }\left\vert Q\right\vert \geq1\text{)}}%
}\det\left(  \operatorname*{sub}\nolimits_{w\left(  \widetilde{Q}\right)
}^{w\left(  \widetilde{P}\right)  }\widetilde{A}\right) \\
&  =\left(  -1\right)  ^{\sum P+\sum Q}\left(  \det\widetilde{A}\right)
\left(  \det\widetilde{A}\right)  ^{\left\vert Q\right\vert -1}\det\left(
\operatorname*{sub}\nolimits_{w\left(  \widetilde{Q}\right)  }^{w\left(
\widetilde{P}\right)  }\widetilde{A}\right) \\
&  =\det\widetilde{A}\cdot\left(  -1\right)  ^{\sum P+\sum Q}\left(
\det\widetilde{A}\right)  ^{\left\vert Q\right\vert -1}\det\left(
\operatorname*{sub}\nolimits_{w\left(  \widetilde{Q}\right)  }^{w\left(
\widetilde{P}\right)  }\widetilde{A}\right)  .
\end{align*}
Hence, Lemma \ref{lem.regular.cancel} (applied to $\mathbb{A}=\mathbb{K}%
\left[  t\right]  $, $a=\det\widetilde{A}$, $b=\det\left(  \operatorname*{sub}%
\nolimits_{w\left(  P\right)  }^{w\left(  Q\right)  }\left(
\operatorname*{adj}\widetilde{A}\right)  \right)  $ and $c=\left(  -1\right)
^{\sum P+\sum Q}\left(  \det\widetilde{A}\right)  ^{\left\vert Q\right\vert
-1}\det\left(  \operatorname*{sub}\nolimits_{w\left(  \widetilde{Q}\right)
}^{w\left(  \widetilde{P}\right)  }\widetilde{A}\right)  $) yields%
\[
\det\left(  \operatorname*{sub}\nolimits_{w\left(  P\right)  }^{w\left(
Q\right)  }\left(  \operatorname*{adj}\widetilde{A}\right)  \right)  =\left(
-1\right)  ^{\sum P+\sum Q}\left(  \det\widetilde{A}\right)  ^{\left\vert
Q\right\vert -1}\det\left(  \operatorname*{sub}\nolimits_{w\left(
\widetilde{Q}\right)  }^{w\left(  \widetilde{P}\right)  }\widetilde{A}\right)
\]
(since the element $\det\widetilde{A}$ of $\mathbb{K}\left[  t\right]  $ is
regular). Applying the map $\varepsilon$ to both sides of this equality, we
obtain%
\begin{align}
&  \varepsilon\left(  \det\left(  \operatorname*{sub}\nolimits_{w\left(
P\right)  }^{w\left(  Q\right)  }\left(  \operatorname*{adj}\widetilde{A}%
\right)  \right)  \right) \nonumber\\
&  =\varepsilon\left(  \left(  -1\right)  ^{\sum P+\sum Q}\left(
\det\widetilde{A}\right)  ^{\left\vert Q\right\vert -1}\det\left(
\operatorname*{sub}\nolimits_{w\left(  \widetilde{Q}\right)  }^{w\left(
\widetilde{P}\right)  }\widetilde{A}\right)  \right) \nonumber\\
&  =\left(  -1\right)  ^{\sum P+\sum Q}\left(  \varepsilon\left(
\det\widetilde{A}\right)  \right)  ^{\left\vert Q\right\vert -1}%
\varepsilon\left(  \det\left(  \operatorname*{sub}\nolimits_{w\left(
\widetilde{Q}\right)  }^{w\left(  \widetilde{P}\right)  }\widetilde{A}\right)
\right)  \label{pf.thm.jacobi-complement.1}%
\end{align}
(since $\varepsilon$ is a $\mathbb{K}$-algebra homomorphism).

\begin{vershort}
The definition of $\widetilde{P}$ yields $\widetilde{P}=\left\{
1,2,\ldots,n\right\}  \setminus P$. Hence,
\begin{align*}
\left\vert \widetilde{P}\right\vert  &  =\underbrace{\left\vert \left\{
1,2,\ldots,n\right\}  \right\vert }_{=n}-\left\vert P\right\vert
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }P\subseteq\left\{  1,2,\ldots
,n\right\}  \right) \\
&  =n-\left\vert P\right\vert .
\end{align*}
Similarly, $\left\vert \widetilde{Q}\right\vert =n-\left\vert Q\right\vert $.
Notice that $\left\vert \widetilde{P}\right\vert =n-\underbrace{\left\vert
P\right\vert }_{=\left\vert Q\right\vert }=n-\left\vert Q\right\vert $.
\end{vershort}

\begin{verlong}
The definition of $\widetilde{P}$ yields $\widetilde{P}=\left\{
1,2,\ldots,n\right\}  \setminus P\subseteq\left\{  1,2,\ldots,n\right\}  $. In
other words, $\widetilde{P}$ is a subset of $\left\{  1,2,\ldots,n\right\}  $.
The same argument (applied to $Q$ instead of $P$) yields that $\widetilde{Q}$
is a subset of $\left\{  1,2,\ldots,n\right\}  $.

The set $P$ is a subset of $\left\{  1,2,\ldots,n\right\}  $, and thus is
finite (since $\left\{  1,2,\ldots,n\right\}  $ is finite). The set
$\widetilde{P}$ is a subset of $\left\{  1,2,\ldots,n\right\}  $, and thus is
finite (since $\left\{  1,2,\ldots,n\right\}  $ is finite).

Furthermore,%
\begin{align*}
\left\vert \underbrace{\widetilde{P}}_{=\left\{  1,2,\ldots,n\right\}
\setminus P}\right\vert  &  =\left\vert \left\{  1,2,\ldots,n\right\}
\setminus P\right\vert =\underbrace{\left\vert \left\{  1,2,\ldots,n\right\}
\right\vert }_{=n}-\left\vert P\right\vert \ \ \ \ \ \ \ \ \ \ \left(
\text{since }P\subseteq\left\{  1,2,\ldots,n\right\}  \right) \\
&  =n-\left\vert P\right\vert .
\end{align*}
The same argument (applied to $Q$ instead of $P$) yields $\left\vert
\widetilde{Q}\right\vert =n-\left\vert Q\right\vert $. Notice that $\left\vert
\widetilde{P}\right\vert =n-\underbrace{\left\vert P\right\vert }_{=\left\vert
Q\right\vert }=n-\left\vert Q\right\vert $. Thus, $n-\left\vert Q\right\vert
=\left\vert \widetilde{P}\right\vert \in\mathbb{N}$ (since $\widetilde{P}$ is
a finite set).
\end{verlong}

But Remark \ref{rmk.w(I).in} (applied to $I=P$) yields $w\left(  P\right)
\in\left\{  1,2,\ldots,n\right\}  ^{\left\vert P\right\vert }=\left\{
1,2,\ldots,n\right\}  ^{\left\vert Q\right\vert }$ (since $\left\vert
P\right\vert =\left\vert Q\right\vert $). Also, Remark \ref{rmk.w(I).in}
(applied to $I=Q$) yields $w\left(  Q\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{\left\vert Q\right\vert }$. Furthermore, Remark
\ref{rmk.w(I).in} (applied to $I=\widetilde{P}$) yields $w\left(
\widetilde{P}\right)  \in\left\{  1,2,\ldots,n\right\}  ^{\left\vert
\widetilde{P}\right\vert }=\left\{  1,2,\ldots,n\right\}  ^{n-\left\vert
Q\right\vert }$ (since $\left\vert \widetilde{P}\right\vert =n-\left\vert
Q\right\vert $). Finally, Remark \ref{rmk.w(I).in} (applied to
$I=\widetilde{Q}$) yields $w\left(  \widetilde{Q}\right)  \in\left\{
1,2,\ldots,n\right\}  ^{\left\vert \widetilde{Q}\right\vert }=\left\{
1,2,\ldots,n\right\}  ^{n-\left\vert Q\right\vert }$ (since $\left\vert
\widetilde{Q}\right\vert =n-\left\vert Q\right\vert $).

Recall that $\operatorname*{adj}\widetilde{A}\in\left(  \mathbb{K}\left[
t\right]  \right)  ^{n\times n}$. Furthermore,%
\[
\varepsilon^{n\times n}\left(  \operatorname*{adj}\underbrace{\widetilde{A}%
}_{=tI_{n}+A}\right)  =\varepsilon^{n\times n}\left(  \operatorname*{adj}%
\left(  tI_{n}+A\right)  \right)  =\operatorname*{adj}A
\]
(by Proposition \ref{prop.tInA.ev} \textbf{(b)}).

We have $w\left(  P\right)  \in\left\{  1,2,\ldots,n\right\}  ^{\left\vert
Q\right\vert }$ and $w\left(  Q\right)  \in\left\{  1,2,\ldots,n\right\}
^{\left\vert Q\right\vert }$. Hence,%
\[
\operatorname*{sub}\nolimits_{w\left(  P\right)  }^{w\left(  Q\right)
}\left(  \operatorname*{adj}\widetilde{A}\right)  \in\left(  \mathbb{K}\left[
t\right]  \right)  ^{\left\vert Q\right\vert \times\left\vert Q\right\vert }.
\]
Thus, Proposition \ref{prop.functor.det} \textbf{(a)} (applied to
$\mathbb{K}\left[  t\right]  $, $\mathbb{K}$, $\varepsilon$, $\left\vert
Q\right\vert $ and $\operatorname*{sub}\nolimits_{w\left(  P\right)
}^{w\left(  Q\right)  }\left(  \operatorname*{adj}\widetilde{A}\right)  $
instead of $\mathbb{L}$, $\mathbb{M}$, $f$, $n$ and $A$) yields%
\begin{align}
\varepsilon\left(  \det\left(  \operatorname*{sub}\nolimits_{w\left(
P\right)  }^{w\left(  Q\right)  }\left(  \operatorname*{adj}\widetilde{A}%
\right)  \right)  \right)   &  =\det\left(  \underbrace{\varepsilon
^{\left\vert Q\right\vert \times\left\vert Q\right\vert }\left(
\operatorname*{sub}\nolimits_{w\left(  P\right)  }^{w\left(  Q\right)
}\left(  \operatorname*{adj}\widetilde{A}\right)  \right)  }%
_{\substack{=\operatorname*{sub}\nolimits_{w\left(  P\right)  }^{w\left(
Q\right)  }\left(  \varepsilon^{n\times n}\left(  \operatorname*{adj}%
\widetilde{A}\right)  \right)  \\\text{(by Lemma \ref{lem.functor.subij}
(applied to }\mathbb{K}\left[  t\right]  \text{, }\mathbb{K}\text{,
}\varepsilon\text{, }n\text{, }\operatorname*{adj}\widetilde{A}\text{,
}\left\vert Q\right\vert \text{, }\left\vert Q\right\vert \text{,}\\w\left(
P\right)  \text{ and }w\left(  Q\right)  \text{ instead of }\mathbb{L}\text{,
}\mathbb{M}\text{, }f\text{, }m\text{, }A\text{, }u\text{, }v\text{,
}\mathbf{i}\text{ and }\mathbf{j}\text{))}}}\right) \nonumber\\
&  =\det\left(  \operatorname*{sub}\nolimits_{w\left(  P\right)  }^{w\left(
Q\right)  }\left(  \underbrace{\varepsilon^{n\times n}\left(
\operatorname*{adj}\widetilde{A}\right)  }_{=\operatorname*{adj}A}\right)
\right) \nonumber\\
&  =\det\left(  \operatorname*{sub}\nolimits_{w\left(  P\right)  }^{w\left(
Q\right)  }\left(  \operatorname*{adj}A\right)  \right)  .
\label{pf.thm.jacobi-complement.3}%
\end{align}
Comparing this with (\ref{pf.thm.jacobi-complement.1}), we obtain%
\begin{align}
&  \det\left(  \operatorname*{sub}\nolimits_{w\left(  P\right)  }^{w\left(
Q\right)  }\left(  \operatorname*{adj}A\right)  \right) \nonumber\\
&  =\left(  -1\right)  ^{\sum P+\sum Q}\left(  \varepsilon\left(
\det\widetilde{A}\right)  \right)  ^{\left\vert Q\right\vert -1}%
\varepsilon\left(  \det\left(  \operatorname*{sub}\nolimits_{w\left(
\widetilde{Q}\right)  }^{w\left(  \widetilde{P}\right)  }\widetilde{A}\right)
\right)  . \label{pf.thm.jacobi-complement.4}%
\end{align}


Recall that $\varepsilon^{n\times n}\left(  \underbrace{\widetilde{A}%
}_{=tI_{n}+A}\right)  =\varepsilon^{n\times n}\left(  tI_{n}+A\right)  =A$ (by
Proposition \ref{prop.tInA.ev} \textbf{(c)}).

On the other hand, $w\left(  \widetilde{P}\right)  \in\left\{  1,2,\ldots
,n\right\}  ^{n-\left\vert Q\right\vert }$ and $w\left(  \widetilde{Q}\right)
\in\left\{  1,2,\ldots,n\right\}  ^{n-\left\vert Q\right\vert }$. Hence,
\[
\operatorname*{sub}\nolimits_{w\left(  \widetilde{Q}\right)  }^{w\left(
\widetilde{P}\right)  }\widetilde{A}\in\left(  \mathbb{K}\left[  t\right]
\right)  ^{\left(  n-\left\vert Q\right\vert \right)  \times\left(
n-\left\vert Q\right\vert \right)  }.
\]
Hence, Proposition \ref{prop.functor.det} \textbf{(a)} (applied to
$\mathbb{K}\left[  t\right]  $, $\mathbb{K}$, $\varepsilon$, $n-\left\vert
Q\right\vert $ and $\operatorname*{sub}\nolimits_{w\left(  \widetilde{Q}%
\right)  }^{w\left(  \widetilde{P}\right)  }\widetilde{A}$ instead of
$\mathbb{L}$, $\mathbb{M}$, $f$, $n$ and $A$) yields%
\begin{align}
\varepsilon\left(  \det\left(  \operatorname*{sub}\nolimits_{w\left(
\widetilde{Q}\right)  }^{w\left(  \widetilde{P}\right)  }\widetilde{A}\right)
\right)   &  =\det\left(  \underbrace{\varepsilon^{\left(  n-\left\vert
Q\right\vert \right)  \times\left(  n-\left\vert Q\right\vert \right)
}\left(  \operatorname*{sub}\nolimits_{w\left(  \widetilde{Q}\right)
}^{w\left(  \widetilde{P}\right)  }\widetilde{A}\right)  }%
_{\substack{=\operatorname*{sub}\nolimits_{w\left(  \widetilde{Q}\right)
}^{w\left(  \widetilde{P}\right)  }\left(  \varepsilon^{n\times n}\left(
\widetilde{A}\right)  \right)  \\\text{(by Lemma \ref{lem.functor.subij}
(applied to }\mathbb{K}\left[  t\right]  \text{, }\mathbb{K}\text{,
}\varepsilon\text{, }n\text{, }\widetilde{A}\text{, }n-\left\vert Q\right\vert
\text{, }n-\left\vert Q\right\vert \text{,}\\w\left(  \widetilde{Q}\right)
\text{ and }w\left(  \widetilde{P}\right)  \text{ instead of }\mathbb{L}%
\text{, }\mathbb{M}\text{, }f\text{, }m\text{, }A\text{, }u\text{, }v\text{,
}\mathbf{i}\text{ and }\mathbf{j}\text{))}}}\right) \nonumber\\
&  =\det\left(  \operatorname*{sub}\nolimits_{w\left(  \widetilde{Q}\right)
}^{w\left(  \widetilde{P}\right)  }\left(  \underbrace{\varepsilon^{n\times
n}\left(  \widetilde{A}\right)  }_{=A}\right)  \right) \nonumber\\
&  =\det\left(  \operatorname*{sub}\nolimits_{w\left(  \widetilde{Q}\right)
}^{w\left(  \widetilde{P}\right)  }A\right)  .
\label{pf.thm.jacobi-complement.6}%
\end{align}
Also, $\varepsilon\left(  \det\underbrace{\widetilde{A}}_{=tI_{n}+A}\right)
=\varepsilon\left(  \det\left(  tI_{n}+A\right)  \right)  =\det A$ (by
Proposition \ref{prop.tInA.ev} \textbf{(a)}).

Now, (\ref{pf.thm.jacobi-complement.4}) becomes%
\begin{align*}
&  \det\left(  \operatorname*{sub}\nolimits_{w\left(  P\right)  }^{w\left(
Q\right)  }\left(  \operatorname*{adj}A\right)  \right) \\
&  =\left(  -1\right)  ^{\sum P+\sum Q}\left(  \underbrace{\varepsilon\left(
\det\widetilde{A}\right)  }_{=\det A}\right)  ^{\left\vert Q\right\vert
-1}\underbrace{\varepsilon\left(  \det\left(  \operatorname*{sub}%
\nolimits_{w\left(  \widetilde{Q}\right)  }^{w\left(  \widetilde{P}\right)
}\widetilde{A}\right)  \right)  }_{\substack{=\det\left(  \operatorname*{sub}%
\nolimits_{w\left(  \widetilde{Q}\right)  }^{w\left(  \widetilde{P}\right)
}A\right)  \\\text{(by (\ref{pf.thm.jacobi-complement.6}))}}}\\
&  =\left(  -1\right)  ^{\sum P+\sum Q}\left(  \det A\right)  ^{\left\vert
Q\right\vert -1}\det\left(  \operatorname*{sub}\nolimits_{w\left(
\widetilde{Q}\right)  }^{w\left(  \widetilde{P}\right)  }A\right)  .
\end{align*}
This proves Theorem \ref{thm.jacobi-complement}.
\end{proof}

\subsection{Another application of the $tI_{n}+A$ strategy}

The strategy that we have used to prove Theorem \ref{thm.adj.adj(AB)}, Theorem
\ref{thm.adj.adjadj} and Theorem \ref{thm.jacobi-complement} (namely,
replacing a matrix $A\in\mathbb{K}^{n\times n}$ by the matrix $tI_{n}%
+A\in\left(  \mathbb{K}\left[  t\right]  \right)  ^{n\times n}$, whose
determinant is a regular element of $\mathbb{K}\left[  t\right]  $; and then
applying the homomorphism $\varepsilon$ to get back to $A$) has many
applications; not all of them concern the adjugate of a matrix. As an example
of such an application, let us prove a neat property of commuting matrices:

\begin{theorem}
\label{thm.det.AS+B}Let $n\in\mathbb{N}$. Let $A$, $B$ and $S$ be three
$n\times n$-matrices such that $AB=BA$. Then,%
\[
\det\left(  AS+B\right)  =\det\left(  SA+B\right)  .
\]

\end{theorem}

Again, we start by showing a particular case of this theorem:

\begin{lemma}
\label{lem.det.AS+B.reg}Let $n\in\mathbb{N}$. Let $A$, $B$ and $S$ be three
$n\times n$-matrices such that $AB=BA$. Assume that the element $\det A$ of
$\mathbb{K}$ is regular. Then,%
\[
\det\left(  AS+B\right)  =\det\left(  SA+B\right)  .
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.det.AS+B.reg}.]Define two $n\times n$-matrices $X$
and $Y$ by $X=AS+B$ and $Y=SA+B$. Comparing%
\[
\underbrace{X}_{=AS+B}A=\left(  AS+B\right)  A=ASA+BA
\]
with%
\[
A\underbrace{Y}_{=SA+B}=A\left(  SA+B\right)  =ASA+\underbrace{AB}%
_{=BA}=ASA+BA,
\]
we obtain $XA=AY$. Now, comparing%
\[
\det\left(  \underbrace{XA}_{=AY}\right)  =\det\left(  AY\right)  =\det
A\cdot\det Y
\]
with%
\[
\det\left(  XA\right)  =\det X\cdot\det A=\det A\cdot\det X,
\]
we obtain $\det A\cdot\det X=\det A\cdot\det Y$. Lemma
\ref{lem.regular.cancel} (applied to $\mathbb{K}$, $\det A$, $\det X$ and
$\det Y$ instead of $\mathbb{A}$, $a$, $b$ and $c$) thus yields $\det X=\det
Y$ (since the element $\det A$ of $\mathbb{K}$ is regular). In view of
$X=AS+B$ and $Y=SA+B$, this rewrites as $\det\left(  AS+B\right)  =\det\left(
SA+B\right)  $. This proves Lemma \ref{lem.det.AS+B.reg}.
\end{proof}

\begin{proof}
[Proof of Theorem \ref{thm.det.AS+B}.]Define the $\mathbb{K}$-algebra
homomorphism $\varepsilon:\mathbb{K}\left[  t\right]  \rightarrow\mathbb{K}$
as in Proposition \ref{prop.tInA.ev}. Thus, $\varepsilon$ is a ring
homomorphism. Hence, Proposition \ref{prop.functor.clas} \textbf{(a)} (applied
to $\mathbb{L}=\mathbb{K}\left[  t\right]  $, $\mathbb{M}=\mathbb{K}$ and
$m=n$) shows that the map $\varepsilon^{n\times n}:\left(  \mathbb{K}\left[
t\right]  \right)  ^{n\times n}\rightarrow\mathbb{K}^{n\times n}$ is a
homomorphism of additive groups.

Recall that every $n\times n$-matrix in $\mathbb{K}^{n\times n}$ can be
considered as a matrix in $\left(  \mathbb{K}\left[  t\right]  \right)
^{n\times n}$. In other words, for each $F\in\mathbb{K}^{n\times n}$, we can
consider $F$ as a matrix in $\left(  \mathbb{K}\left[  t\right]  \right)
^{n\times n}$; therefore, $\varepsilon^{n\times n}\left(  F\right)  $ is
well-defined. We have
\begin{equation}
\varepsilon^{n\times n}\left(  F\right)  =F\ \ \ \ \ \ \ \ \ \ \text{for every
}F\in\mathbb{K}^{n\times n}. \label{pf.thm.det.AS+B.FF}%
\end{equation}


\begin{vershort}
\noindent(In fact, the proof of (\ref{pf.thm.det.AS+B.FF}) is identical with
the proof of (\ref{pf.lem.adj.poly.-A.short.eF}) we gave above.)
\end{vershort}

\begin{verlong}
\noindent(In fact, the proof of (\ref{pf.thm.det.AS+B.FF}) is identical with
the proof of (\ref{pf.lem.adj.poly.-A.eF}) we gave above.)
\end{verlong}

Let $\widetilde{A}$ be the matrix $tI_{n}+A\in\left(  \mathbb{K}\left[
t\right]  \right)  ^{n\times n}$. Thus, $\widetilde{A}=tI_{n}+A$. Applying the
map $\varepsilon^{n\times n}$ to both sides of this equality, we find
$\varepsilon^{n\times n}\left(  \widetilde{A}\right)  =\varepsilon^{n\times
n}\left(  tI_{n}+A\right)  =A$ (by Proposition \ref{prop.tInA.ev} \textbf{(c)}).

Corollary \ref{cor.tInA.reg} shows that the element $\det\left(
tI_{n}+A\right)  $ of $\mathbb{K}\left[  t\right]  $ is regular. In other
words, the element $\det\widetilde{A}$ of $\mathbb{K}\left[  t\right]  $ is
regular (since $\widetilde{A}=tI_{n}+A$).

Let us consider the matrix $S\in\mathbb{K}^{n\times n}$ as a matrix in
$\left(  \mathbb{K}\left[  t\right]  \right)  ^{n\times n}$ (since every
$n\times n$-matrix in $\mathbb{K}^{n\times n}$ can be considered as a matrix
in $\left(  \mathbb{K}\left[  t\right]  \right)  ^{n\times n}$).

Similarly, let us consider the matrix $B\in\mathbb{K}^{n\times n}$ as a matrix
in $\left(  \mathbb{K}\left[  t\right]  \right)  ^{n\times n}$. Then,%
\begin{align*}
\underbrace{\widetilde{A}}_{=tI_{n}+A}B  &  =\left(  tI_{n}+A\right)
B=t\underbrace{I_{n}B}_{=B=BI_{n}}+\underbrace{AB}_{=BA}=\underbrace{tBI_{n}%
}_{=B\cdot tI_{n}}+BA\\
&  =B\cdot tI_{n}+BA=B\underbrace{\left(  tI_{n}+A\right)  }_{=\widetilde{A}%
}=B\widetilde{A}.
\end{align*}
Hence, Lemma \ref{lem.det.AS+B.reg} (applied to $\mathbb{K}\left[  t\right]  $
and $\widetilde{A}$ instead of $\mathbb{K}$ and $A$) yields
\begin{equation}
\det\left(  \widetilde{A}S+B\right)  =\det\left(  S\widetilde{A}+B\right)  .
\label{pf.thm.det.AS+B.1}%
\end{equation}


Proposition \ref{prop.functor.det} \textbf{(a)} (applied to $\mathbb{K}\left[
t\right]  $, $\mathbb{K}$, $\varepsilon$ and $\widetilde{A}S+B$ instead of
$\mathbb{L}$, $\mathbb{M}$, $f$ and $A$) yields
\[
\varepsilon\left(  \det\left(  \widetilde{A}S+B\right)  \right)  =\det\left(
\varepsilon^{n\times n}\left(  \widetilde{A}S+B\right)  \right)  .
\]
In view of%
\begin{align*}
\varepsilon^{n\times n}\left(  \widetilde{A}S+B\right)   &
=\underbrace{\varepsilon^{n\times n}\left(  \widetilde{A}S\right)
}_{\substack{=\varepsilon^{n\times n}\left(  \widetilde{A}\right)
\cdot\varepsilon^{n\times n}\left(  S\right)  \\\text{(by Proposition
\ref{prop.functor.clas} \textbf{(b)}}\\\text{(applied to }\mathbb{K}\left[
t\right]  \text{, }\mathbb{K}\text{, }\varepsilon\text{, }n\text{, }n\text{,
}\widetilde{A}\text{ and }S\\\text{instead of }\mathbb{L}\text{, }%
\mathbb{M}\text{, }f\text{, }m\text{, }p\text{, }A\text{ and }B\text{))}%
}}+\varepsilon^{n\times n}\left(  B\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the map }\varepsilon^{n\times n}\text{ is a homomorphism}\\
\text{of additive groups}%
\end{array}
\right) \\
&  =\underbrace{\varepsilon^{n\times n}\left(  \widetilde{A}\right)  }%
_{=A}\cdot\underbrace{\varepsilon^{n\times n}\left(  S\right)  }%
_{\substack{=S\\\text{(by (\ref{pf.thm.det.AS+B.FF})}\\\text{(applied to
}F=S\text{))}}}+\underbrace{\varepsilon^{n\times n}\left(  B\right)
}_{\substack{=B\\\text{(by (\ref{pf.thm.det.AS+B.FF})}\\\text{(applied to
}F=B\text{))}}}\\
&  =AS+B,
\end{align*}
this becomes%
\begin{equation}
\varepsilon\left(  \det\left(  \widetilde{A}S+B\right)  \right)  =\det\left(
\underbrace{\varepsilon^{n\times n}\left(  \widetilde{A}S+B\right)  }%
_{=AS+B}\right)  =\det\left(  AS+B\right)  . \label{pf.thm.det.AS+B.2}%
\end{equation}


\begin{vershort}
Similarly,%
\[
\varepsilon\left(  \det\left(  S\widetilde{A}+B\right)  \right)  =\det\left(
SA+B\right)  .
\]

\end{vershort}

\begin{verlong}
Proposition \ref{prop.functor.det} \textbf{(a)} (applied to $\mathbb{K}\left[
t\right]  $, $\mathbb{K}$, $\varepsilon$ and $S\widetilde{A}+B$ instead of
$\mathbb{L}$, $\mathbb{M}$, $f$ and $A$) yields
\[
\varepsilon\left(  \det\left(  S\widetilde{A}+B\right)  \right)  =\det\left(
\varepsilon^{n\times n}\left(  S\widetilde{A}+B\right)  \right)  .
\]
In view of%
\begin{align*}
\varepsilon^{n\times n}\left(  S\widetilde{A}+B\right)   &
=\underbrace{\varepsilon^{n\times n}\left(  S\widetilde{A}\right)
}_{\substack{=\varepsilon^{n\times n}\left(  S\right)  \cdot\varepsilon
^{n\times n}\left(  \widetilde{A}\right)  \\\text{(by Proposition
\ref{prop.functor.clas} \textbf{(b)}}\\\text{(applied to }\mathbb{K}\left[
t\right]  \text{, }\mathbb{K}\text{, }\varepsilon\text{, }n\text{, }n\text{,
}S\text{ and }\widetilde{A}\\\text{instead of }\mathbb{L}\text{, }%
\mathbb{M}\text{, }f\text{, }m\text{, }p\text{, }A\text{ and }B\text{))}%
}}+\varepsilon^{n\times n}\left(  B\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the map }\varepsilon^{n\times n}\text{ is a homomorphism}\\
\text{of additive groups}%
\end{array}
\right) \\
&  =\underbrace{\varepsilon^{n\times n}\left(  S\right)  }%
_{\substack{=S\\\text{(by (\ref{pf.thm.det.AS+B.FF})}\\\text{(applied to
}F=S\text{))}}}\cdot\underbrace{\varepsilon^{n\times n}\left(  \widetilde{A}%
\right)  }_{=A}+\underbrace{\varepsilon^{n\times n}\left(  B\right)
}_{\substack{=B\\\text{(by (\ref{pf.thm.det.AS+B.FF})}\\\text{(applied to
}F=B\text{))}}}\\
&  =SA+B,
\end{align*}
this becomes%
\[
\varepsilon\left(  \det\left(  S\widetilde{A}+B\right)  \right)  =\det\left(
\underbrace{\varepsilon^{n\times n}\left(  S\widetilde{A}+B\right)  }%
_{=SA+B}\right)  =\det\left(  SA+B\right)  .
\]

\end{verlong}

Comparing this with%
\[
\varepsilon\left(  \underbrace{\det\left(  S\widetilde{A}+B\right)
}_{\substack{=\det\left(  \widetilde{A}S+B\right)  \\\text{(by
(\ref{pf.thm.det.AS+B.1}))}}}\right)  =\varepsilon\left(  \det\left(
\widetilde{A}S+B\right)  \right)  =\det\left(  AS+B\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.det.AS+B.2})}\right)  ,
\]
we obtain $\det\left(  AS+B\right)  =\det\left(  SA+B\right)  $. This proves
Theorem \ref{thm.det.AS+B}.
\end{proof}

\subsection{Another application of the strategy: block matrices}

The same strategy (replacing $A\in\mathbb{K}^{n\times n}$ by $tI_{n}%
+A\in\left(  \mathbb{K}\left[  t\right]  \right)  ^{n\times n}$) turns out to
be useful in proving a formula for determinants of block matrices with a
certain property.

We will use \cite[Definition 6.89]{detnotes} in this section. Roughly
speaking, this definition says that if $n$, $n^{\prime}$, $m$ and $m^{\prime}$
are four nonnegative integers, and if $A\in\mathbb{K}^{n\times m}$,
$B\in\mathbb{K}^{n\times m^{\prime}}$, $C\in\mathbb{K}^{n^{\prime}\times m}$
and $D\in\mathbb{K}^{n^{\prime}\times m^{\prime}}$ are four matrices, then
$\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  $ shall denote the $\left(  n+n^{\prime}\right)  \times\left(
m+m^{\prime}\right)  $-matrix obtained by \textquotedblleft gluing the
matrices $A$, $B$, $C$ and $D$ together\textquotedblright\ in the way the
notation suggests (i.e., the matrix $B$ is glued to the right edge of $A$, and
then the matrices $C$ and $D$ are glued to the bottom edges of $A$ and $B$,
respectively). For example, if $n=2$, $n^{\prime}=2$, $m=2$ and $m^{\prime}%
=2$, and if%
\begin{align}
A  &  =\left(
\begin{array}
[c]{cc}%
a_{1,1} & a_{1,2}\\
a_{2,1} & a_{2,2}%
\end{array}
\right)  ,\ \ \ \ \ \ \ \ \ \ B=\left(
\begin{array}
[c]{cc}%
b_{1,1} & b_{1,2}\\
b_{2,1} & b_{2,2}%
\end{array}
\right)  ,\label{eq.block2x2.ex2x2.1}\\
C  &  =\left(
\begin{array}
[c]{cc}%
c_{1,1} & c_{1,2}\\
c_{2,1} & c_{2,2}%
\end{array}
\right)  ,\ \ \ \ \ \ \ \ \ \ \text{and }D=\left(
\begin{array}
[c]{cc}%
d_{1,1} & d_{1,2}\\
d_{2,1} & d_{2,2}%
\end{array}
\right)  , \label{eq.block2x2.ex2x2.2}%
\end{align}
then%
\[
\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  =\left(
\begin{array}
[c]{cccc}%
a_{1,1} & a_{1,2} & b_{1,1} & b_{1,2}\\
a_{2,1} & a_{2,2} & b_{2,1} & b_{2,2}\\
c_{1,1} & c_{1,2} & d_{1,1} & d_{1,2}\\
c_{2,1} & c_{2,2} & d_{2,1} & d_{2,2}%
\end{array}
\right)  .
\]
There are more general versions of this \textquotedblleft gluing
operation\textquotedblright\ that allow for more than four matrices; but we
will only concern ourselves with the case of four matrices.

We are aiming to prove the following theorem:

\begin{theorem}
\label{thm.block2x2.detAD-CB}Let $n\in\mathbb{N}$. Let $A$, $B$, $C$ and $D$
be four $n\times n$-matrices such that $AC=CA$. Then, the $\left(  2n\right)
\times\left(  2n\right)  $-matrix $\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  $ satisfies%
\[
\det\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  =\det\left(  AD-CB\right)  .
\]

\end{theorem}

Theorem \ref{thm.block2x2.detAD-CB} appears, e.g., in \cite[(14)]{Silvest}.
Our proof of this theorem will closely follow \cite[proof of Lemma 2]%
{Silvest}. We will use the following obvious lemma:

\begin{lemma}
\label{lem.functor.block2x2}Let $\mathbb{L}$ and $\mathbb{M}$ be two
commutative rings. Let $f:\mathbb{L}\rightarrow\mathbb{M}$ be any map. Let
$n$, $n^{\prime}$, $m$ and $m^{\prime}$ be four nonnegative integers. Let
$A\in\mathbb{L}^{n\times m}$, $B\in\mathbb{L}^{n\times m^{\prime}}$,
$C\in\mathbb{L}^{n^{\prime}\times m}$ and $D\in\mathbb{L}^{n^{\prime}\times
m^{\prime}}$ be four matrices. Then,
\[
f^{\left(  n+n^{\prime}\right)  \times\left(  m+m^{\prime}\right)  }\left(
\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  \right)  =\left(
\begin{array}
[c]{cc}%
f^{n\times m}\left(  A\right)  & f^{n\times m^{\prime}}\left(  B\right) \\
f^{n^{\prime}\times m}\left(  C\right)  & f^{n^{\prime}\times m^{\prime}%
}\left(  D\right)
\end{array}
\right)  .
\]

\end{lemma}

\begin{example}
For this example, set $n=2$ and $n^{\prime}=2$ and $m=2$ and $m^{\prime}=2$,
and let the $2\times2$-matrices $A$, $B$, $C$ and $D$ be given by
(\ref{eq.block2x2.ex2x2.1}) and (\ref{eq.block2x2.ex2x2.2}). Then, Lemma
\ref{lem.functor.block2x2} says that%
\[
f^{4\times4}\left(  \left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  \right)  =\left(
\begin{array}
[c]{cc}%
f^{2\times2}\left(  A\right)  & f^{2\times2}\left(  B\right) \\
f^{2\times2}\left(  C\right)  & f^{2\times2}\left(  D\right)
\end{array}
\right)  .
\]
Both the left and the right hand side of this equality are easily seen to
equal
\[
\left(
\begin{array}
[c]{cccc}%
f\left(  a_{1,1}\right)  & f\left(  a_{1,2}\right)  & f\left(  b_{1,1}\right)
& f\left(  b_{1,2}\right) \\
f\left(  a_{2,1}\right)  & f\left(  a_{2,2}\right)  & f\left(  b_{2,1}\right)
& f\left(  b_{2,2}\right) \\
f\left(  c_{1,1}\right)  & f\left(  c_{1,2}\right)  & f\left(  d_{1,1}\right)
& f\left(  d_{1,2}\right) \\
f\left(  c_{2,1}\right)  & f\left(  c_{2,2}\right)  & f\left(  d_{2,1}\right)
& f\left(  d_{2,2}\right)
\end{array}
\right)  .
\]

\end{example}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.functor.block2x2}.]Write the $n\times m$-matrix $A$
in the form $A=\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}$.

Write the $n\times m^{\prime}$-matrix $B$ in the form $B=\left(
b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m^{\prime}}$.

Write the $n^{\prime}\times m$-matrix $C$ in the form $C=\left(
c_{i,j}\right)  _{1\leq i\leq n^{\prime},\ 1\leq j\leq m}$.

Write the $n^{\prime}\times m^{\prime}$-matrix $D$ in the form $D=\left(
d_{i,j}\right)  _{1\leq i\leq n^{\prime},\ 1\leq j\leq m^{\prime}}$.

Now, we have $A=\left(  a_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}$ and
$B=\left(  b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m^{\prime}}$ and
\newline$C=\left(  c_{i,j}\right)  _{1\leq i\leq n^{\prime},\ 1\leq j\leq m}$
and $D=\left(  d_{i,j}\right)  _{1\leq i\leq n^{\prime},\ 1\leq j\leq
m^{\prime}}$. Thus, the definition of the block matrix $\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  $\ \ \ \ \footnote{or, more precisely, the equality \cite[(435)]%
{detnotes} (applied to $\mathbb{L}$ instead of $\mathbb{K}$)} yields%
\[
\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  =\left(
\begin{cases}
a_{i,j}, & \text{if }i\leq n\text{ and }j\leq m;\\
b_{i,j-m}, & \text{if }i\leq n\text{ and }j>m;\\
c_{i-n,j}, & \text{if }i>n\text{ and }j\leq m;\\
d_{i-n,j-m}, & \text{if }i>n\text{ and }j>m
\end{cases}
\right)  _{1\leq i\leq n+n^{\prime},\ 1\leq j\leq m+m^{\prime}}.
\]
Hence, the definition of the map $f^{\left(  n+n^{\prime}\right)
\times\left(  m+m^{\prime}\right)  }$ yields%
\begin{align}
&  f^{\left(  n+n^{\prime}\right)  \times\left(  m+m^{\prime}\right)  }\left(
\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  \right) \nonumber\\
&  =\left(  f\left(
\begin{cases}
a_{i,j}, & \text{if }i\leq n\text{ and }j\leq m;\\
b_{i,j-m}, & \text{if }i\leq n\text{ and }j>m;\\
c_{i-n,j}, & \text{if }i>n\text{ and }j\leq m;\\
d_{i-n,j-m}, & \text{if }i>n\text{ and }j>m
\end{cases}
\right)  \right)  _{1\leq i\leq n+n^{\prime},\ 1\leq j\leq m+m^{\prime}%
}\nonumber\\
&  =\left(
\begin{cases}
f\left(  a_{i,j}\right)  , & \text{if }i\leq n\text{ and }j\leq m;\\
f\left(  b_{i,j-m}\right)  , & \text{if }i\leq n\text{ and }j>m;\\
f\left(  c_{i-n,j}\right)  , & \text{if }i>n\text{ and }j\leq m;\\
f\left(  d_{i-n,j-m}\right)  , & \text{if }i>n\text{ and }j>m
\end{cases}
\right)  _{1\leq i\leq n+n^{\prime},\ 1\leq j\leq m+m^{\prime}}
\label{pf.lem.functor.block2x2.L=}%
\end{align}
(since each $i\in\left\{  1,2,\ldots,n+n^{\prime}\right\}  $ and $j\in\left\{
1,2,\ldots,m+m^{\prime}\right\}  $ satisfy \newline$f\left(
\begin{cases}
a_{i,j}, & \text{if }i\leq n\text{ and }j\leq m;\\
b_{i,j-m}, & \text{if }i\leq n\text{ and }j>m;\\
c_{i-n,j}, & \text{if }i>n\text{ and }j\leq m;\\
d_{i-n,j-m}, & \text{if }i>n\text{ and }j>m
\end{cases}
\right)  =%
\begin{cases}
f\left(  a_{i,j}\right)  , & \text{if }i\leq n\text{ and }j\leq m;\\
f\left(  b_{i,j-m}\right)  , & \text{if }i\leq n\text{ and }j>m;\\
f\left(  c_{i-n,j}\right)  , & \text{if }i>n\text{ and }j\leq m;\\
f\left(  d_{i-n,j-m}\right)  , & \text{if }i>n\text{ and }j>m
\end{cases}
$).

On the other hand, the definition of the map $f^{n\times m}$ yields
$f^{n\times m}\left(  A\right)  =\left(  f\left(  a_{i,j}\right)  \right)
_{1\leq i\leq n,\ 1\leq j\leq m}$ (since $A=\left(  a_{i,j}\right)  _{1\leq
i\leq n,\ 1\leq j\leq m}$). The definition of the map $f^{n\times m^{\prime}}$
yields $f^{n\times m^{\prime}}\left(  B\right)  =\left(  f\left(
b_{i,j}\right)  \right)  _{1\leq i\leq n,\ 1\leq j\leq m^{\prime}}$ (since
$B=\left(  b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m^{\prime}}$). The
definition of the map $f^{n^{\prime}\times m}$ yields $f^{n^{\prime}\times
m}\left(  C\right)  =\left(  f\left(  c_{i,j}\right)  \right)  _{1\leq i\leq
n^{\prime},\ 1\leq j\leq m}$ (since $C=\left(  c_{i,j}\right)  _{1\leq i\leq
n^{\prime},\ 1\leq j\leq m}$). The definition of the map $f^{n^{\prime}\times
m^{\prime}}$ yields $f^{n^{\prime}\times m^{\prime}}\left(  D\right)  =\left(
f\left(  d_{i,j}\right)  \right)  _{1\leq i\leq n^{\prime},\ 1\leq j\leq
m^{\prime}}$ (since $D=\left(  d_{i,j}\right)  _{1\leq i\leq n^{\prime
},\ 1\leq j\leq m^{\prime}}$).

Now, we have%
\begin{align*}
f^{n\times m}\left(  A\right)   &  =\left(  f\left(  a_{i,j}\right)  \right)
_{1\leq i\leq n,\ 1\leq j\leq m}\ \ \ \ \ \ \ \ \ \ \text{and}\\
f^{n\times m^{\prime}}\left(  B\right)   &  =\left(  f\left(  b_{i,j}\right)
\right)  _{1\leq i\leq n,\ 1\leq j\leq m^{\prime}}%
\ \ \ \ \ \ \ \ \ \ \text{and}\\
f^{n^{\prime}\times m}\left(  C\right)   &  =\left(  f\left(  c_{i,j}\right)
\right)  _{1\leq i\leq n^{\prime},\ 1\leq j\leq m}%
\ \ \ \ \ \ \ \ \ \ \text{and}\\
f^{n^{\prime}\times m^{\prime}}\left(  D\right)   &  =\left(  f\left(
d_{i,j}\right)  \right)  _{1\leq i\leq n^{\prime},\ 1\leq j\leq m^{\prime}}.
\end{align*}
Hence, the definition of the block matrix $\left(
\begin{array}
[c]{cc}%
f^{n\times m}\left(  A\right)  & f^{n\times m^{\prime}}\left(  B\right) \\
f^{n^{\prime}\times m}\left(  C\right)  & f^{n^{\prime}\times m^{\prime}%
}\left(  D\right)
\end{array}
\right)  $\ \ \ \ \footnote{or, more precisely, the equality \cite[(435)]%
{detnotes} (applied to $\mathbb{M}$, $f^{n\times m}\left(  A\right)  $,
$f\left(  a_{i,j}\right)  $, $f^{n\times m^{\prime}}\left(  B\right)  $,
$f\left(  b_{i,j}\right)  $, $f^{n^{\prime}\times m}\left(  C\right)  $,
$f\left(  c_{i,j}\right)  $, $f^{n^{\prime}\times m^{\prime}}\left(  D\right)
$ and $f\left(  d_{i,j}\right)  $ instead of $\mathbb{K}$, $A$, $a_{i,j}$,
$B$, $b_{i,j}$, $C$, $c_{i,j}$, $D$ and $d_{i,j}$)} yields%
\begin{align*}
&  \left(
\begin{array}
[c]{cc}%
f^{n\times m}\left(  A\right)  & f^{n\times m^{\prime}}\left(  B\right) \\
f^{n^{\prime}\times m}\left(  C\right)  & f^{n^{\prime}\times m^{\prime}%
}\left(  D\right)
\end{array}
\right) \\
&  =\left(
\begin{cases}
f\left(  a_{i,j}\right)  , & \text{if }i\leq n\text{ and }j\leq m;\\
f\left(  b_{i,j-m}\right)  , & \text{if }i\leq n\text{ and }j>m;\\
f\left(  c_{i-n,j}\right)  , & \text{if }i>n\text{ and }j\leq m;\\
f\left(  d_{i-n,j-m}\right)  , & \text{if }i>n\text{ and }j>m
\end{cases}
\right)  _{1\leq i\leq n+n^{\prime},\ 1\leq j\leq m+m^{\prime}}.
\end{align*}
Comparing this with (\ref{pf.lem.functor.block2x2.L=}), we obtain%
\[
f^{\left(  n+n^{\prime}\right)  \times\left(  m+m^{\prime}\right)  }\left(
\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  \right)  =\left(
\begin{array}
[c]{cc}%
f^{n\times m}\left(  A\right)  & f^{n\times m^{\prime}}\left(  B\right) \\
f^{n^{\prime}\times m}\left(  C\right)  & f^{n^{\prime}\times m^{\prime}%
}\left(  D\right)
\end{array}
\right)  .
\]
This proves Lemma \ref{lem.functor.block2x2}.
\end{proof}
\end{verlong}

Next, let us recall a result from \cite{detnotes} (a version of the Schur
complement theorem):

\begin{proposition}
\label{prop.block2x2.VB+WD}Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let
$A\in\mathbb{K}^{n\times n}$, $B\in\mathbb{K}^{n\times m}$, $C\in
\mathbb{K}^{m\times n}$ and $D\in\mathbb{K}^{m\times m}$. Furthermore, let
$W\in\mathbb{K}^{m\times m}$ and $V\in\mathbb{K}^{m\times n}$ be such that
$VA=-WC$. Then,%
\[
\det W\cdot\det\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  =\det A\cdot\det\left(  VB+WD\right)  .
\]

\end{proposition}

Proposition \ref{prop.block2x2.VB+WD} appears (with proof) in \cite[Exercise
6.35]{detnotes}, so we will not prove it here.

Let us next prove the particular case of Theorem \ref{thm.block2x2.detAD-CB}
in which we assume $\det A$ to be regular:

\begin{lemma}
\label{lem.block2x2.detAD-CB.reg}Let $n\in\mathbb{N}$. Let $A$, $B$, $C$ and
$D$ be four $n\times n$-matrices such that $AC=CA$. Assume that the element
$\det A$ of $\mathbb{K}$ is regular. Then, the $\left(  2n\right)
\times\left(  2n\right)  $-matrix $\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  $ satisfies%
\[
\det\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  =\det\left(  AD-CB\right)  .
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.block2x2.detAD-CB.reg}.]The matrix $\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  $ is an $\left(  n+n\right)  \times\left(  n+n\right)  $-matrix (by
its definition), i.e., a $\left(  2n\right)  \times\left(  2n\right)  $-matrix
(since $n+n=2n$).

We have $CA=-\left(  -A\right)  C$ (since $-\left(  -A\right)  C=AC=CA$).
Thus, Proposition \ref{prop.block2x2.VB+WD} (applied to $V=C$ and $W=-A$)
yields%
\begin{align}
\det\left(  -A\right)  \cdot\det\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)   &  =\det A\cdot\det\left(  \underbrace{CB+\left(  -A\right)
D}_{=CB-AD}\right) \nonumber\\
&  =\det A\cdot\det\left(  CB-AD\right)  .
\label{pf.lem.block2x2.detAD-CB.reg.1}%
\end{align}
But (\ref{pf.lem.adj.adjadj.reg.lam1}) (applied to $-1$ and $CB-AD$ instead of
$\lambda$ and $C$) yields
\begin{equation}
\det\left(  \left(  -1\right)  \left(  CB-AD\right)  \right)  =\left(
-1\right)  ^{n}\det\left(  CB-AD\right)  .
\label{pf.lem.block2x2.detAD-CB.reg.2}%
\end{equation}
Also, (\ref{pf.lem.adj.adjadj.reg.lam1}) (applied to $-1$ and $-A$ instead of
$\lambda$ and $C$) yields $\det\left(  \left(  -1\right)  \left(  -A\right)
\right)  =\left(  -1\right)  ^{n}\det\left(  -A\right)  $. In view of $\left(
-1\right)  \left(  -A\right)  =A$, this rewrites as $\det A=\left(  -1\right)
^{n}\det\left(  -A\right)  $. Hence,%
\begin{align*}
&  \underbrace{\det A}_{=\left(  -1\right)  ^{n}\det\left(  -A\right)  }%
\cdot\det\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right) \\
&  =\left(  -1\right)  ^{n}\underbrace{\det\left(  -A\right)  \cdot\det\left(
%
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  }_{\substack{=\det A\cdot\det\left(  CB-AD\right)  \\\text{(by
(\ref{pf.lem.block2x2.detAD-CB.reg.1}))}}}\\
&  =\left(  -1\right)  ^{n}\det A\cdot\det\left(  CB-AD\right)  =\det
A\cdot\underbrace{\left(  -1\right)  ^{n}\det\left(  CB-AD\right)
}_{\substack{=\det\left(  \left(  -1\right)  \left(  CB-AD\right)  \right)
\\\text{(by (\ref{pf.lem.block2x2.detAD-CB.reg.2}))}}}\\
&  =\det A\cdot\det\left(  \underbrace{\left(  -1\right)  \left(
CB-AD\right)  }_{=AD-CB}\right)  =\det A\cdot\det\left(  AD-CB\right)  .
\end{align*}
Lemma \ref{lem.regular.cancel} (applied to $\mathbb{K}$, $\det A$,
$\det\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  $ and $\det\left(  AD-CB\right)  $ instead of $\mathbb{A}$, $a$, $b$
and $c$) thus yields $\det\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  =\det\left(  AD-CB\right)  $ (since the element $\det A$ of
$\mathbb{K}$ is regular). This proves Lemma \ref{lem.block2x2.detAD-CB.reg}.
\end{proof}

We are now ready to prove Theorem \ref{thm.block2x2.detAD-CB}:

\begin{proof}
[Proof of Theorem \ref{thm.block2x2.detAD-CB}.]The matrix $\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  $ is a $\left(  2n\right)  \times\left(  2n\right)  $-matrix. (This
is proven in the same way as in our proof of Lemma
\ref{lem.block2x2.detAD-CB.reg}.)

Define the $\mathbb{K}$-algebra homomorphism $\varepsilon:\mathbb{K}\left[
t\right]  \rightarrow\mathbb{K}$ as in Proposition \ref{prop.tInA.ev}. Thus,
$\varepsilon$ is a ring homomorphism. Hence, Proposition
\ref{prop.functor.clas} \textbf{(a)} (applied to $\mathbb{L}=\mathbb{K}\left[
t\right]  $, $\mathbb{M}=\mathbb{K}$ and $m=n$) shows that the map
$\varepsilon^{n\times n}:\left(  \mathbb{K}\left[  t\right]  \right)
^{n\times n}\rightarrow\mathbb{K}^{n\times n}$ is a homomorphism of additive groups.

Recall that every $n\times n$-matrix in $\mathbb{K}^{n\times n}$ can be
considered as a matrix in $\left(  \mathbb{K}\left[  t\right]  \right)
^{n\times n}$. In other words, for each $F\in\mathbb{K}^{n\times n}$, we can
consider $F$ as a matrix in $\left(  \mathbb{K}\left[  t\right]  \right)
^{n\times n}$; therefore, $\varepsilon^{n\times n}\left(  F\right)  $ is
well-defined. We have
\begin{equation}
\varepsilon^{n\times n}\left(  F\right)  =F\ \ \ \ \ \ \ \ \ \ \text{for every
}F\in\mathbb{K}^{n\times n}. \label{pf.thm.block2x2.detAD-CB.FF}%
\end{equation}


\begin{vershort}
\noindent(In fact, the proof of (\ref{pf.thm.block2x2.detAD-CB.FF}) is
identical with the proof of (\ref{pf.lem.adj.poly.-A.short.eF}) we gave above.)
\end{vershort}

\begin{verlong}
\noindent(In fact, the proof of (\ref{pf.thm.block2x2.detAD-CB.FF}) is
identical with the proof of (\ref{pf.lem.adj.poly.-A.eF}) we gave above.)
\end{verlong}

Let $\widetilde{A}$ be the matrix $tI_{n}+A\in\left(  \mathbb{K}\left[
t\right]  \right)  ^{n\times n}$. Thus, $\widetilde{A}=tI_{n}+A$. Applying the
map $\varepsilon^{n\times n}$ to both sides of this equality, we find
$\varepsilon^{n\times n}\left(  \widetilde{A}\right)  =\varepsilon^{n\times
n}\left(  tI_{n}+A\right)  =A$ (by Proposition \ref{prop.tInA.ev} \textbf{(c)}).

Corollary \ref{cor.tInA.reg} shows that the element $\det\left(
tI_{n}+A\right)  $ of $\mathbb{K}\left[  t\right]  $ is regular. In other
words, the element $\det\widetilde{A}$ of $\mathbb{K}\left[  t\right]  $ is
regular (since $\widetilde{A}=tI_{n}+A$).

Let us consider the matrix $B\in\mathbb{K}^{n\times n}$ as a matrix in
$\left(  \mathbb{K}\left[  t\right]  \right)  ^{n\times n}$ (since every
$n\times n$-matrix in $\mathbb{K}^{n\times n}$ can be considered as a matrix
in $\left(  \mathbb{K}\left[  t\right]  \right)  ^{n\times n}$). Similarly,
let us consider the matrices $C$ and $D$ as matrices in $\left(
\mathbb{K}\left[  t\right]  \right)  ^{n\times n}$.

\begin{vershort}
Notice that (\ref{pf.thm.block2x2.detAD-CB.FF}) (applied to $F=B$) yields
$\varepsilon^{n\times n}\left(  B\right)  =B$. Similarly, $\varepsilon
^{n\times n}\left(  C\right)  =C$ and $\varepsilon^{n\times n}\left(
D\right)  =D$.
\end{vershort}

\begin{verlong}
Notice that (\ref{pf.thm.block2x2.detAD-CB.FF}) (applied to $F=B$) yields
$\varepsilon^{n\times n}\left(  B\right)  =B$. Also,
(\ref{pf.thm.block2x2.detAD-CB.FF}) (applied to $F=C$) yields $\varepsilon
^{n\times n}\left(  C\right)  =C$. Also, (\ref{pf.thm.block2x2.detAD-CB.FF})
(applied to $F=D$) yields $\varepsilon^{n\times n}\left(  D\right)  =D$.
\end{verlong}

Now,%
\begin{align*}
\underbrace{\widetilde{A}}_{=tI_{n}+A}C  &  =\left(  tI_{n}+A\right)
C=t\underbrace{I_{n}C}_{=C=CI_{n}}+\underbrace{AC}_{=CA}=\underbrace{tCI_{n}%
}_{=C\cdot tI_{n}}+CA\\
&  =C\cdot tI_{n}+CA=C\underbrace{\left(  tI_{n}+A\right)  }_{=\widetilde{A}%
}=C\widetilde{A}.
\end{align*}
Thus, Lemma \ref{lem.block2x2.detAD-CB.reg} (applied to $\mathbb{K}\left[
t\right]  $ and $\widetilde{A}$ instead of $\mathbb{K}$ and $A$) yields
\[
\det\left(
\begin{array}
[c]{cc}%
\widetilde{A} & B\\
C & D
\end{array}
\right)  =\det\left(  \widetilde{A}D-CB\right)
\]
(since the element $\det\widetilde{A}$ of $\mathbb{K}\left[  t\right]  $ is
regular). Applying the map $\varepsilon$ to both sides of this equality, we
find%
\begin{equation}
\varepsilon\left(  \det\left(
\begin{array}
[c]{cc}%
\widetilde{A} & B\\
C & D
\end{array}
\right)  \right)  =\varepsilon\left(  \det\left(  \widetilde{A}D-CB\right)
\right)  . \label{pf.thm.block2x2.detAD-CB.2}%
\end{equation}
But Proposition \ref{prop.functor.det} \textbf{(a)} (applied to $\mathbb{K}%
\left[  t\right]  $, $\mathbb{K}$, $\varepsilon$ and $\widetilde{A}D-CB$
instead of $\mathbb{L}$, $\mathbb{M}$, $f$ and $A$) yields
\[
\varepsilon\left(  \det\left(  \widetilde{A}D-CB\right)  \right)  =\det\left(
\varepsilon^{n\times n}\left(  \widetilde{A}D-CB\right)  \right)  .
\]
In view of%
\begin{align*}
\varepsilon^{n\times n}\left(  \widetilde{A}D-CB\right)   &
=\underbrace{\varepsilon^{n\times n}\left(  \widetilde{A}D\right)
}_{\substack{=\varepsilon^{n\times n}\left(  \widetilde{A}\right)
\cdot\varepsilon^{n\times n}\left(  D\right)  \\\text{(by Proposition
\ref{prop.functor.clas} \textbf{(b)}}\\\text{(applied to }\mathbb{K}\left[
t\right]  \text{, }\mathbb{K}\text{, }\varepsilon\text{, }n\text{, }n\text{,
}\widetilde{A}\text{ and }D\\\text{instead of }\mathbb{L}\text{, }%
\mathbb{M}\text{, }f\text{, }m\text{, }p\text{, }A\text{ and }B\text{))}%
}}-\underbrace{\varepsilon^{n\times n}\left(  CB\right)  }%
_{\substack{=\varepsilon^{n\times n}\left(  C\right)  \cdot\varepsilon
^{n\times n}\left(  B\right)  \\\text{(by Proposition \ref{prop.functor.clas}
\textbf{(b)}}\\\text{(applied to }\mathbb{K}\left[  t\right]  \text{,
}\mathbb{K}\text{, }\varepsilon\text{, }n\text{, }n\text{, }C\text{ and
}B\\\text{instead of }\mathbb{L}\text{, }\mathbb{M}\text{, }f\text{, }m\text{,
}p\text{, }A\text{ and }B\text{))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the map }\varepsilon^{n\times n}\text{ is a homomorphism}\\
\text{of additive groups}%
\end{array}
\right) \\
&  =\underbrace{\varepsilon^{n\times n}\left(  \widetilde{A}\right)  }%
_{=A}\cdot\underbrace{\varepsilon^{n\times n}\left(  D\right)  }%
_{=D}-\underbrace{\varepsilon^{n\times n}\left(  C\right)  }_{=C}%
\cdot\underbrace{\varepsilon^{n\times n}\left(  B\right)  }_{\substack{=B}}\\
&  =AD-CB,
\end{align*}
this becomes%
\begin{align}
\varepsilon\left(  \det\left(  \widetilde{A}D-CB\right)  \right)   &
=\det\left(  \underbrace{\varepsilon^{n\times n}\left(  \widetilde{A}%
D-CB\right)  }_{=AD-CB}\right) \nonumber\\
&  =\det\left(  AD-CB\right)  . \label{pf.thm.block2x2.detAD-CB.3}%
\end{align}


But Proposition \ref{prop.functor.det} \textbf{(a)} (applied to $\mathbb{K}%
\left[  t\right]  $, $\mathbb{K}$, $\varepsilon$, $n+n$ and $\left(
\begin{array}
[c]{cc}%
\widetilde{A} & B\\
C & D
\end{array}
\right)  $ instead of $\mathbb{L}$, $\mathbb{M}$, $f$, $n$ and $A$) yields%
\begin{equation}
\varepsilon\left(  \det\left(
\begin{array}
[c]{cc}%
\widetilde{A} & B\\
C & D
\end{array}
\right)  \right)  =\det\left(  \varepsilon^{\left(  n+n\right)  \times\left(
n+n\right)  }\left(  \left(
\begin{array}
[c]{cc}%
\widetilde{A} & B\\
C & D
\end{array}
\right)  \right)  \right)  . \label{pf.thm.block2x2.detAD-CB.4}%
\end{equation}


On the other hand, Lemma \ref{lem.functor.block2x2} (applied to $\mathbb{K}%
\left[  t\right]  $, $\mathbb{K}$, $\varepsilon$, $n$, $n$, $n$ and
$\widetilde{A}$ instead of $\mathbb{L}$, $\mathbb{M}$, $f$, $n^{\prime}$, $m$,
$m^{\prime}$ and $A$) yields
\[
\varepsilon^{\left(  n+n\right)  \times\left(  n+n\right)  }\left(  \left(
\begin{array}
[c]{cc}%
\widetilde{A} & B\\
C & D
\end{array}
\right)  \right)  =\left(
\begin{array}
[c]{cc}%
\varepsilon^{n\times n}\left(  \widetilde{A}\right)  & \varepsilon^{n\times
n}\left(  B\right) \\
\varepsilon^{n\times n}\left(  C\right)  & \varepsilon^{n\times n}\left(
D\right)
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)
\]
(since $\varepsilon^{n\times n}\left(  \widetilde{A}\right)  =A$ and
$\varepsilon^{n\times n}\left(  B\right)  =B$ and $\varepsilon^{n\times
n}\left(  C\right)  =C$ and $\varepsilon^{n\times n}\left(  D\right)  =D$).
Taking determinants on both sides of this equality, we find%
\[
\det\left(  \varepsilon^{\left(  n+n\right)  \times\left(  n+n\right)
}\left(  \left(
\begin{array}
[c]{cc}%
\widetilde{A} & B\\
C & D
\end{array}
\right)  \right)  \right)  =\det\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  .
\]
Hence,%
\begin{align*}
\det\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)   &  =\det\left(  \varepsilon^{\left(  n+n\right)  \times\left(
n+n\right)  }\left(  \left(
\begin{array}
[c]{cc}%
\widetilde{A} & B\\
C & D
\end{array}
\right)  \right)  \right) \\
&  =\varepsilon\left(  \det\left(
\begin{array}
[c]{cc}%
\widetilde{A} & B\\
C & D
\end{array}
\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.block2x2.detAD-CB.4})}\right) \\
&  =\varepsilon\left(  \det\left(  \widetilde{A}D-CB\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.thm.block2x2.detAD-CB.2}%
)}\right) \\
&  =\det\left(  AD-CB\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.block2x2.detAD-CB.3})}\right)  .
\end{align*}
This completes the proof of Theorem \ref{thm.block2x2.detAD-CB}.
\end{proof}

\subsection{The trace of the adjugate}

The following neat result follows so easily from Theorem \ref{thm.adj.poly}
and Theorem \ref{thm.TCH} that it would be strange not to mention it:

\begin{theorem}
\label{thm.adj.Tr}Let $n\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times n}$. For
every $j\in\mathbb{Z}$, define an element $c_{j}\in\mathbb{K}$ by
$c_{j}=\left[  t^{n-j}\right]  \chi_{A}$. Then,
\[
\operatorname*{Tr}\left(  \operatorname*{adj}A\right)  =\left(  -1\right)
^{n-1}c_{n-1}=\left(  -1\right)  ^{n-1}\left[  t^{1}\right]  \chi_{A}.
\]

\end{theorem}

In other words, the trace of the adjugate $\operatorname*{adj}A$ of an
$n\times n$-matrix is the coefficient of $t$ in the characteristic polynomial
$\chi_{A}$.

\begin{proof}
[Proof of Theorem \ref{thm.adj.Tr}.]The definition of $c_{n-1}$ yields
$c_{n-1}=\left[  t^{n-\left(  n-1\right)  }\right]  \chi_{A}=\left[
t^{1}\right]  \chi_{A}$ (since $n-\left(  n-1\right)  =1$).

It is easy to see that Theorem \ref{thm.adj.Tr} holds for $n=0$%
\ \ \ \ \footnote{\textit{Proof.} Assume that $n=0$. Thus, $1>0=n$. But
Corollary \ref{cor.chiA} \textbf{(a)} yields that $\chi_{A}\in\mathbb{K}%
\left[  t\right]  $ is a polynomial of degree $\leq n$ in $t$. Hence, $\left[
t^{m}\right]  \chi_{A}=0$ for every integer $m>n$. Applying this to $m=1$, we
obtain $\left[  t^{1}\right]  \chi_{A}=0$ (since $1>n$). Also,
$\operatorname*{adj}A$ is an $n\times n$-matrix, and thus a $0\times0$-matrix
(since $n=0$). Hence, $\operatorname*{Tr}\left(  \operatorname*{adj}A\right)
=0$ (since the trace of a $0\times0$-matrix is $0$). Comparing this with
$\left(  -1\right)  ^{n-1}\underbrace{c_{n-1}}_{=\left[  t^{1}\right]
\chi_{A}=0}=0$, we obtain $\operatorname*{Tr}\left(  \operatorname*{adj}%
A\right)  =\left(  -1\right)  ^{n-1}\underbrace{c_{n-1}}_{=\left[
t^{1}\right]  \chi_{A}}=\left(  -1\right)  ^{n-1}\left[  t^{1}\right]
\chi_{A}$. Hence, we have proven Theorem \ref{thm.adj.Tr} under the assumption
that $n=0$.}. Thus, for the rest of this proof, we can WLOG assume that we
don't have $n=0$. Assume this. Hence, $n\neq0$, so that $n\geq1$ (since
$n\in\mathbb{N}$). Therefore, $n-1\in\mathbb{N}$.

Thus, Theorem \ref{thm.TCH} (applied to $k=n-1$) yields%
\[
\left(  n-1\right)  c_{n-1}+\sum_{i=1}^{n-1}\operatorname*{Tr}\left(
A^{i}\right)  c_{n-1-i}=0.
\]
Subtracting $\left(  n-1\right)  c_{n-1}$ from both sides of this equation, we
obtain%
\begin{equation}
\sum_{i=1}^{n-1}\operatorname*{Tr}\left(  A^{i}\right)  c_{n-1-i}=-\left(
n-1\right)  c_{n-1}. \label{pf.thm.adj.Tr.1}%
\end{equation}


But Theorem \ref{thm.adj.poly} yields%
\[
\operatorname*{adj}A=\left(  -1\right)  ^{n-1}\sum_{i=0}^{n-1}c_{n-1-i}A^{i}.
\]
Applying the map $\operatorname*{Tr}:\mathbb{K}^{n\times n}\rightarrow
\mathbb{K}$ to both sides of this equality, we obtain%
\begin{align}
\operatorname*{Tr}\left(  \operatorname*{adj}A\right)   &  =\operatorname*{Tr}%
\left(  \left(  -1\right)  ^{n-1}\sum_{i=0}^{n-1}c_{n-1-i}A^{i}\right)
\nonumber\\
&  =\left(  -1\right)  ^{n-1}\sum_{i=0}^{n-1}c_{n-1-i}\operatorname*{Tr}%
\left(  A^{i}\right)  \label{pf.thm.adj.Tr.2}%
\end{align}
(since the map $\operatorname*{Tr}:\mathbb{K}^{n\times n}\rightarrow
\mathbb{K}$ is $\mathbb{K}$-linear). But $n-1\geq0$ (since $n\geq1$);
therefore, $0\in\left\{  0,1,\ldots,n-1\right\}  $. Hence, we can split off
the addend for $i=0$ from the sum $\sum_{i=0}^{n-1}c_{n-1-i}\operatorname*{Tr}%
\left(  A^{i}\right)  $. We thus obtain
\begin{align*}
\sum_{i=0}^{n-1}c_{n-1-i}\operatorname*{Tr}\left(  A^{i}\right)   &
=\underbrace{c_{n-1-0}}_{=c_{n-1}}\operatorname*{Tr}\left(  \underbrace{A^{0}%
}_{=I_{n}}\right)  +\sum_{i=1}^{n-1}\underbrace{c_{n-1-i}\operatorname*{Tr}%
\left(  A^{i}\right)  }_{=\operatorname*{Tr}\left(  A^{i}\right)  c_{n-1-i}}\\
&  =c_{n-1}\underbrace{\operatorname*{Tr}\left(  I_{n}\right)  }%
_{=n}+\underbrace{\sum_{i=1}^{n-1}\operatorname*{Tr}\left(  A^{i}\right)
c_{n-1-i}}_{\substack{=-\left(  n-1\right)  c_{n-1}\\\text{(by
(\ref{pf.thm.adj.Tr.1}))}}}=c_{n-1}n+\left(  -\left(  n-1\right)
c_{n-1}\right) \\
&  =\underbrace{\left(  n-\left(  n-1\right)  \right)  }_{=1}c_{n-1}=c_{n-1}.
\end{align*}
Hence, (\ref{pf.thm.adj.Tr.2}) becomes%
\[
\operatorname*{Tr}\left(  \operatorname*{adj}A\right)  =\left(  -1\right)
^{n-1}\underbrace{\sum_{i=0}^{n-1}c_{n-1-i}\operatorname*{Tr}\left(
A^{i}\right)  }_{=c_{n-1}}=\left(  -1\right)  ^{n-1}\underbrace{c_{n-1}%
}_{=\left[  t^{1}\right]  \chi_{A}}=\left(  -1\right)  ^{n-1}\left[
t^{1}\right]  \chi_{A}.
\]
This proves Theorem \ref{thm.adj.Tr}.
\end{proof}

\subsection{Yet another application to block matrices}

Let us show one further formula for determinants of certain block matrices
that can be proved using our \textquotedblleft replace $A$ by $tI_{n}%
+A$\textquotedblright\ strategy.

We will again use \cite[Definition 6.89]{detnotes} in this section. We shall
furthermore use the following notation:

\begin{definition}
\label{def.sol.det.rk1upd.scalar}If $B$ is any $1\times1$-matrix, then
$\operatorname*{ent}B$ will denote the $\left(  1,1\right)  $-th entry of $B$.
(This entry is, of course, the only entry of $B$. Thus, the $1\times1$-matrix
$B$ satisfies $B=\left(
\begin{array}
[c]{c}%
\operatorname*{ent}B
\end{array}
\right)  $.)
\end{definition}

We now claim the following:

\begin{theorem}
\label{thm.block.puqv}Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let
$A\in\mathbb{K}^{n\times n}$ and $D\in\mathbb{K}^{m\times m}$ be two square
matrices. Let $p\in\mathbb{K}^{n\times1}$ and $q\in\mathbb{K}^{m\times1}$ be
two column vectors. Let $v\in\mathbb{K}^{1\times m}$ and $u\in\mathbb{K}%
^{1\times n}$ be two row vectors. Then,%
\[
\det\left(
\begin{array}
[c]{cc}%
A & pv\\
qu & D
\end{array}
\right)  =\det A\cdot\det D-\operatorname*{ent}\left(  u\left(
\operatorname*{adj}A\right)  p\right)  \cdot\operatorname*{ent}\left(
v\left(  \operatorname*{adj}D\right)  q\right)  .
\]

\end{theorem}

\begin{example}
Let us see what Theorem \ref{thm.block.puqv} says in the case when $n=2$ and
$m=2$. Indeed, let $n=2$ and $m=2$ and%
\begin{align*}
A  &  =\left(
\begin{array}
[c]{cc}%
a_{1,1} & a_{1,2}\\
a_{2,1} & a_{2,2}%
\end{array}
\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ D=\left(
\begin{array}
[c]{cc}%
d_{1,1} & d_{1,2}\\
d_{2,1} & d_{2,2}%
\end{array}
\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\\
p  &  =\left(
\begin{array}
[c]{c}%
p_{1}\\
p_{2}%
\end{array}
\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ q=\left(
\begin{array}
[c]{c}%
q_{1}\\
q_{2}%
\end{array}
\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\\
v  &  =\left(
\begin{array}
[c]{cc}%
v_{1} & v_{2}%
\end{array}
\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ u=\left(
\begin{array}
[c]{cc}%
u_{1} & u_{2}%
\end{array}
\right)  .
\end{align*}
Then,
\[
pv=\left(
\begin{array}
[c]{cc}%
p_{1}v_{1} & p_{1}v_{2}\\
p_{2}v_{1} & p_{2}v_{2}%
\end{array}
\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ qu=\left(
\begin{array}
[c]{cc}%
q_{1}u_{1} & q_{1}u_{2}\\
q_{2}u_{1} & q_{2}u_{2}%
\end{array}
\right)  .
\]
Hence,%
\[
\left(
\begin{array}
[c]{cc}%
A & pv\\
qu & D
\end{array}
\right)  =\left(
\begin{array}
[c]{cccc}%
a_{1,1} & a_{1,2} & p_{1}v_{1} & p_{1}v_{2}\\
a_{2,1} & a_{2,2} & p_{2}v_{1} & p_{2}v_{2}\\
q_{1}u_{1} & q_{1}u_{2} & d_{1,1} & d_{1,2}\\
q_{2}u_{1} & q_{2}u_{2} & d_{2,1} & d_{2,2}%
\end{array}
\right)  .
\]
Hence, the claim of Theorem \ref{thm.block.puqv} rewrites as follows in our
case:
\begin{align*}
&  \det\left(
\begin{array}
[c]{cccc}%
a_{1,1} & a_{1,2} & p_{1}v_{1} & p_{1}v_{2}\\
a_{2,1} & a_{2,2} & p_{2}v_{1} & p_{2}v_{2}\\
q_{1}u_{1} & q_{1}u_{2} & d_{1,1} & d_{1,2}\\
q_{2}u_{1} & q_{2}u_{2} & d_{2,1} & d_{2,2}%
\end{array}
\right) \\
&  =\det\left(
\begin{array}
[c]{cc}%
a_{1,1} & a_{1,2}\\
a_{2,1} & a_{2,2}%
\end{array}
\right)  \cdot\det\left(
\begin{array}
[c]{cc}%
d_{1,1} & d_{1,2}\\
d_{2,1} & d_{2,2}%
\end{array}
\right) \\
&  \ \ \ \ \ \ \ \ \ \ -\operatorname*{ent}\left(  \left(
\begin{array}
[c]{cc}%
u_{1} & u_{2}%
\end{array}
\right)  \left(  \operatorname*{adj}\left(
\begin{array}
[c]{cc}%
a_{1,1} & a_{1,2}\\
a_{2,1} & a_{2,2}%
\end{array}
\right)  \right)  \left(
\begin{array}
[c]{c}%
p_{1}\\
p_{2}%
\end{array}
\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \cdot\operatorname*{ent}\left(
\left(
\begin{array}
[c]{cc}%
v_{1} & v_{2}%
\end{array}
\right)  \left(  \operatorname*{adj}\left(
\begin{array}
[c]{cc}%
d_{1,1} & d_{1,2}\\
d_{2,1} & d_{2,2}%
\end{array}
\right)  \right)  \left(
\begin{array}
[c]{c}%
q_{1}\\
q_{2}%
\end{array}
\right)  \right)  .
\end{align*}

\end{example}

In order to prove Theorem \ref{thm.block.puqv}, we will need one standard
result (known as the \textit{matrix determinant lemma}):

\begin{theorem}
\label{thm.sol.det.rk1upd.claim}Let $n\in\mathbb{N}$. Let $u$ be a column
vector with $n$ entries, and let $v$ be a row vector with $n$ entries. (Thus,
$uv$ is an $n\times n$-matrix, whereas $vu$ is a $1\times1$-matrix.) Let $A$
be an $n\times n$-matrix. Then,%
\[
\det\left(  A+uv\right)  =\det A+\operatorname*{ent}\left(  v\left(
\operatorname*{adj}A\right)  u\right)  .
\]

\end{theorem}

See \cite[Theorem 7.262]{detnotes} for a proof of Theorem
\ref{thm.sol.det.rk1upd.claim}.

We will furthermore need the following three trivial lemmas:

\begin{lemma}
\label{lem.ent.scalp}Let $C\in\mathbb{K}^{1\times1}$ be an $1\times1$-matrix.
Let $\lambda\in\mathbb{K}$. Then, $\operatorname*{ent}\left(  \lambda
C\right)  =\lambda\operatorname*{ent}C$.
\end{lemma}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.ent.scalp}.]Since $C$ is a $1\times1$-matrix, we have
$C=\left(
\begin{array}
[c]{c}%
\operatorname*{ent}C
\end{array}
\right)  $ (since $\operatorname*{ent}C$ denotes the $\left(  1,1\right)  $-th
entry of $C$). Hence, $\lambda C=\lambda\left(
\begin{array}
[c]{c}%
\operatorname*{ent}C
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
\lambda\operatorname*{ent}C
\end{array}
\right)  $. Now, the definition of $\operatorname*{ent}\left(  \lambda
C\right)  $ yields
\begin{align*}
\operatorname*{ent}\left(  \lambda C\right)   &  =\left(  \text{the }\left(
1,1\right)  \text{-th entry of }\lambda C\right) \\
&  =\lambda\operatorname*{ent}C\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\lambda C=\left(
\begin{array}
[c]{c}%
\lambda\operatorname*{ent}C
\end{array}
\right)  \right)  .
\end{align*}
This proves Lemma \ref{lem.ent.scalp}.
\end{proof}
\end{verlong}

\begin{lemma}
\label{lem.ent.prod}Let $B$ and $C$ be two $1\times1$-matrices. Then,
$\operatorname*{ent}\left(  BC\right)  =\operatorname*{ent}B\cdot
\operatorname*{ent}C$.
\end{lemma}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.ent.prod}.]Since $B$ is a $1\times1$-matrix, we have
$B=\left(
\begin{array}
[c]{c}%
\operatorname*{ent}B
\end{array}
\right)  $ (since $\operatorname*{ent}B$ denotes the $\left(  1,1\right)  $-th
entry of $B$). Similarly, $C=\left(
\begin{array}
[c]{c}%
\operatorname*{ent}C
\end{array}
\right)  $. Thus,
\[
BC=\left(
\begin{array}
[c]{c}%
\operatorname*{ent}B
\end{array}
\right)  \left(
\begin{array}
[c]{c}%
\operatorname*{ent}C
\end{array}
\right)  =\left(
\begin{array}
[c]{c}%
\sum_{k=1}^{1}\operatorname*{ent}B\cdot\operatorname*{ent}C
\end{array}
\right)
\]
(by the definition of the product of two matrices). Now, the definition of
$\operatorname*{ent}\left(  BC\right)  $ yields%
\begin{align*}
\operatorname*{ent}\left(  BC\right)   &  =\left(  \text{the }\left(
1,1\right)  \text{-th entry of }BC\right) \\
&  =\sum_{k=1}^{1}\operatorname*{ent}B\cdot\operatorname*{ent}%
C\ \ \ \ \ \ \ \ \ \ \left(  \text{since }BC=\left(
\begin{array}
[c]{c}%
\sum_{k=1}^{1}\operatorname*{ent}B\cdot\operatorname*{ent}C
\end{array}
\right)  \right) \\
&  =\operatorname*{ent}B\cdot\operatorname*{ent}C.
\end{align*}
This proves Lemma \ref{lem.ent.prod}.
\end{proof}
\end{verlong}

\begin{lemma}
\label{lem.ent.map}Let $\mathbb{L}$ and $\mathbb{M}$ be rings. Let
$f:\mathbb{L}\rightarrow\mathbb{M}$ be any map. Let $B\in\mathbb{L}^{1\times
1}$. Then, $\operatorname*{ent}\left(  f^{1\times1}\left(  B\right)  \right)
=f\left(  \operatorname*{ent}B\right)  $.
\end{lemma}

\begin{verlong}
\begin{proof}
[Proof of Lemma \ref{lem.ent.prod}.]Since $B$ is a $1\times1$-matrix, we have
$B=\left(
\begin{array}
[c]{c}%
\operatorname*{ent}B
\end{array}
\right)  $ (since $\operatorname*{ent}B$ denotes the $\left(  1,1\right)  $-th
entry of $B$). Thus, $f^{1\times1}\left(  B\right)  =f^{1\times1}\left(
\left(
\begin{array}
[c]{c}%
\operatorname*{ent}B
\end{array}
\right)  \right)  =\left(
\begin{array}
[c]{c}%
f\left(  \operatorname*{ent}B\right)
\end{array}
\right)  $ (by the definition of $f^{1\times1}$). Now, the definition of
$\operatorname*{ent}\left(  f^{1\times1}\left(  B\right)  \right)  $ yields%
\begin{align*}
\operatorname*{ent}\left(  f^{1\times1}\left(  B\right)  \right)   &  =\left(
\text{the }\left(  1,1\right)  \text{-th entry of }f^{1\times1}\left(
B\right)  \right) \\
&  =f\left(  \operatorname*{ent}B\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }f^{1\times1}\left(  B\right)  =\left(
\begin{array}
[c]{c}%
f\left(  \operatorname*{ent}B\right)
\end{array}
\right)  \right)  .
\end{align*}
This proves Lemma \ref{lem.ent.prod}.
\end{proof}
\end{verlong}

Finally, we will need a simple property of regular elements in a commutative ring:

\begin{lemma}
\label{lem.reg.power}Let $\mathbb{A}$ be a commutative ring. Let $a$ be a
regular element of $\mathbb{A}$. Let $m\in\mathbb{N}$. Then, $a^{m}$ is a
regular element of $\mathbb{A}$.
\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.reg.power}.]The element $a$ is regular. In other
words,%
\begin{equation}
\text{every }x\in\mathbb{A}\text{ satisfying }ax=0\text{ satisfies }x=0
\label{pf.lem.reg.power.ass}%
\end{equation}
(by the definition of \textquotedblleft regular\textquotedblright).

Now, let $x\in\mathbb{A}$ satisfy $a^{m}x=0$. We shall show that $x=0$.
Indeed, we shall first prove that%
\begin{equation}
a^{m-i}x=0\ \ \ \ \ \ \ \ \ \ \text{for each }i\in\left\{  0,1,\ldots
,m\right\}  \text{.} \label{pf.lem.reg.power.1}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.reg.power.1}):} We proceed by induction on $i$:

\textit{Induction base:} We have $a^{m-0}x=a^{m}x=0$ (by assumption). Hence,
(\ref{pf.lem.reg.power.1}) holds for $i=0$.

\textit{Induction step:} Let $j\in\left\{  1,2,\ldots,m\right\}  $. Assume
that (\ref{pf.lem.reg.power.1}) holds for $i=j-1$. We must show that
(\ref{pf.lem.reg.power.1}) holds for $i=j$ as well.

We have assumed that (\ref{pf.lem.reg.power.1}) holds for $i=j-1$. In other
words, $a^{m-\left(  j-1\right)  }x=0$. In other words, $a^{m-j+1}x=0$ (since
$m-\left(  j-1\right)  =m-j+1$). In other words, $aa^{m-j}x=0$ (since
$a^{m-j+1}=aa^{m-j}$). Hence, (\ref{pf.lem.reg.power.ass}) (applied to
$a^{m-j}x$ instead of $x$) yields $a^{m-j}x=0$. In other words,
(\ref{pf.lem.reg.power.1}) holds for $i=j$. This completes the induction step.
Thus, (\ref{pf.lem.reg.power.1}) is proved by induction.] \medskip

Now, (\ref{pf.lem.reg.power.1}) (applied to $i=m$) yields $a^{m-m}x=0$. Since
$\underbrace{a^{m-m}}_{=a^{0}=1}x=x$, this rewrites as $x=0$.

Forget that we fixed $x$. We thus have shown that every $x\in\mathbb{A}$
satisfying $a^{m}x=0$ satisfies $x=0$. In other words, the element $a^{m}$ of
$\mathbb{A}$ is regular (by the definition of \textquotedblleft
regular\textquotedblright). This proves Lemma \ref{lem.reg.power}.
\end{proof}

Now, we can approach the proof of Theorem \ref{thm.block.puqv} using the same
technique as various theorems proved above. We begin by proving it in the case
when $\det A$ is regular:

\begin{lemma}
\label{lem.block.puqv-reg}Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let
$A\in\mathbb{K}^{n\times n}$ and $D\in\mathbb{K}^{m\times m}$ be two square
matrices. Assume that the element $\det A$ of $\mathbb{K}$ is regular. Let
$p\in\mathbb{K}^{n\times1}$ and $q\in\mathbb{K}^{m\times1}$ be two column
vectors. Let $v\in\mathbb{K}^{1\times m}$ and $u\in\mathbb{K}^{1\times n}$ be
two row vectors. Then,%
\[
\det\left(
\begin{array}
[c]{cc}%
A & pv\\
qu & D
\end{array}
\right)  =\det A\cdot\det D-\operatorname*{ent}\left(  u\left(
\operatorname*{adj}A\right)  p\right)  \cdot\operatorname*{ent}\left(
v\left(  \operatorname*{adj}D\right)  q\right)  .
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.block.puqv-reg}.]Set $\lambda=\det A$. Thus, the
element $\lambda$ of $\mathbb{K}$ is regular (since the element $\det A$ of
$\mathbb{K}$ is regular). Hence, Lemma \ref{lem.reg.power} (applied to
$\mathbb{A}=\mathbb{K}$ and $a=\lambda$) shows that $\lambda^{m}$ is a regular
element of $\mathbb{K}$.

It is furthermore easy to see that%
\begin{equation}
\lambda\operatorname*{adj}\left(  \lambda D\right)  =\lambda^{m}%
\operatorname*{adj}D. \label{pf.lem.block.puqv-reg.ladjlD}%
\end{equation}


[\textit{Proof of (\ref{pf.lem.block.puqv-reg.ladjlD}):} If $m=0$, then
(\ref{pf.lem.block.puqv-reg.ladjlD}) holds for trivial
reasons\footnote{\textit{Proof.} Assume that $m=0$. The matrices
$\lambda\operatorname*{adj}\left(  \lambda D\right)  $ and $\lambda
^{m}\operatorname*{adj}D$ are $m\times m$-matrices (since $D$ is an $m\times
m$-matrix). In other words, the matrices $\lambda\operatorname*{adj}\left(
\lambda D\right)  $ and $\lambda^{m}\operatorname*{adj}D$ are $0\times
0$-matrices (since $m=0$). Hence, these two matrices $\lambda
\operatorname*{adj}\left(  \lambda D\right)  $ and $\lambda^{m}%
\operatorname*{adj}D$ are equal (since there exists only one $0\times
0$-matrix, and therefore any two $0\times0$-matrices are equal). In other
words, $\lambda\operatorname*{adj}\left(  \lambda D\right)  =\lambda
^{m}\operatorname*{adj}D$. Thus, we have proved
(\ref{pf.lem.block.puqv-reg.ladjlD}) under the assumption that $m=0$.}. Thus,
for the rest of this proof, we WLOG assume that we don't have $m=0$. Hence,
$m$ is a positive integer (since $m\in\mathbb{N}$). Therefore, Proposition
\ref{prop.adj.lamA} (applied to $m$ and $D$ instead of $n$ and $A$) yields
$\operatorname*{adj}\left(  \lambda D\right)  =\lambda^{m-1}%
\operatorname*{adj}D$. Hence, $\lambda\underbrace{\operatorname*{adj}\left(
\lambda D\right)  }_{=\lambda^{m-1}\operatorname*{adj}D}=\underbrace{\lambda
\lambda^{m-1}}_{=\lambda^{m}}\operatorname*{adj}D=\lambda^{m}%
\operatorname*{adj}D$. This proves (\ref{pf.lem.block.puqv-reg.ladjlD}).]
\medskip

Define two matrices $W\in\mathbb{K}^{m\times m}$ and $V\in\mathbb{K}^{m\times
n}$ by%
\[
W=\lambda I_{m}\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ V=-qu\operatorname*{adj}A.
\]
Then,%
\begin{align*}
\underbrace{V}_{=-qu\operatorname*{adj}A}A  &
=-qu\underbrace{\operatorname*{adj}A\cdot A}_{\substack{=\det A\cdot
I_{n}\\\text{(by Theorem \ref{thm.adj.inverse})}}}=-qu\underbrace{\det
A}_{\substack{=\lambda\\\text{(since }\lambda=\det A\text{)}}}\cdot
I_{n}=-qu\lambda I_{n}\\
&  =-\lambda quI_{n}=-\lambda qu=-Wqu
\end{align*}
(since $-\underbrace{W}_{=\lambda I_{m}}qu=-\lambda I_{m}qu=-\lambda qu$).
Hence, Proposition \ref{prop.block2x2.VB+WD} (applied to $B=pv$ and $C=qu$)
yields
\[
\det W\cdot\det\left(
\begin{array}
[c]{cc}%
A & pv\\
qu & D
\end{array}
\right)  =\det A\cdot\det\left(  Vpv+WD\right)  .
\]
In view of%
\begin{align*}
\det\underbrace{W}_{=\lambda I_{m}}  &  =\det\left(  \lambda I_{m}\right)
=\lambda^{m}\underbrace{\det\left(  I_{m}\right)  }_{=1}%
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{pf.lem.adj.adjadj.reg.lam1}), applied to }m\text{ and }I_{m}\\
\text{instead of }n\text{ and }C
\end{array}
\right) \\
&  =\lambda^{m}%
\end{align*}
and $\det A=\lambda$, we can rewrite this as%
\begin{equation}
\lambda^{m}\cdot\det\left(
\begin{array}
[c]{cc}%
A & pv\\
qu & D
\end{array}
\right)  =\lambda\cdot\det\left(  Vpv+WD\right)  .
\label{pf.lem.block.puqv-reg.3}%
\end{equation}


Let us define two $1\times1$-matrices $B$ and $C$ by
\[
B=u\left(  \operatorname*{adj}A\right)  p\in\mathbb{K}^{1\times1}%
\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ C=v\cdot\operatorname*{adj}%
\left(  \lambda D\right)  \cdot\left(  -q\right)  \in\mathbb{K}^{1\times1}.
\]


Next, we observe that%
\begin{align}
&  \underbrace{V}_{=-qu\operatorname*{adj}A}pv+\underbrace{W}_{=\lambda I_{m}%
}D\nonumber\\
&  =\underbrace{\left(  -qu\operatorname*{adj}A\right)  pv}_{=\left(
-q\right)  u\left(  \operatorname*{adj}A\right)  pv}+\lambda\underbrace{I_{m}%
D}_{=D}=\left(  -q\right)  \underbrace{u\left(  \operatorname*{adj}A\right)
p}_{\substack{=B\\\text{(since }B=u\left(  \operatorname*{adj}A\right)
p\text{)}}}v+\lambda D=\left(  -q\right)  Bv+\lambda D\nonumber\\
&  =\lambda D+\left(  -q\right)  Bv=\lambda D+\left(  -q\right)  \left(
Bv\right)  . \label{pf.lem.block.puqv-reg.4}%
\end{align}
Note that $\lambda D\in\mathbb{K}^{m\times m}$ and $-q\in\mathbb{K}^{m\times
1}$ and $Bv\in\mathbb{K}^{1\times m}$. Hence, Theorem
\ref{thm.sol.det.rk1upd.claim} (applied to $m$, $-q$, $Bv$ and $D$ instead of
$n$, $u$, $v$ and $A$) yields%
\begin{align*}
&  \det\left(  \lambda D+\left(  -q\right)  \left(  Bv\right)  \right) \\
&  =\underbrace{\det\left(  \lambda D\right)  }_{\substack{=\lambda^{m}\det
D\\\text{(by (\ref{pf.lem.adj.adjadj.reg.lam1}), applied to }m\text{ and
}D\\\text{instead of }n\text{ and }A\text{)}}}+\operatorname*{ent}\left(
\underbrace{\left(  Bv\right)  \cdot\operatorname*{adj}\left(  \lambda
D\right)  \cdot\left(  -q\right)  }_{=B\left(  v\cdot\operatorname*{adj}%
\left(  \lambda D\right)  \cdot\left(  -q\right)  \right)  }\right) \\
&  =\lambda^{m}\det D+\operatorname*{ent}\left(  B\underbrace{\left(
v\cdot\operatorname*{adj}\left(  \lambda D\right)  \cdot\left(  -q\right)
\right)  }_{\substack{=C\\\text{(since }C=v\cdot\operatorname*{adj}\left(
\lambda D\right)  \cdot\left(  -q\right)  \text{)}}}\right)  =\lambda^{m}\det
D+\underbrace{\operatorname*{ent}\left(  BC\right)  }%
_{\substack{=\operatorname*{ent}B\cdot\operatorname*{ent}C\\\text{(by Lemma
\ref{lem.ent.prod})}}}\\
&  =\lambda^{m}\det D+\operatorname*{ent}B\cdot\operatorname*{ent}C.
\end{align*}
In view of (\ref{pf.lem.block.puqv-reg.4}), we can rewrite this as%
\[
\det\left(  Vpv+WD\right)  =\lambda^{m}\det D+\operatorname*{ent}%
B\cdot\operatorname*{ent}C.
\]
Thus, (\ref{pf.lem.block.puqv-reg.3}) becomes%
\begin{align}
\lambda^{m}\cdot\det\left(
\begin{array}
[c]{cc}%
A & pv\\
qu & D
\end{array}
\right)   &  =\lambda\cdot\underbrace{\det\left(  Vpv+WD\right)  }%
_{=\lambda^{m}\det D+\operatorname*{ent}B\cdot\operatorname*{ent}C}\nonumber\\
&  =\lambda\cdot\left(  \lambda^{m}\det D+\operatorname*{ent}B\cdot
\operatorname*{ent}C\right) \nonumber\\
&  =\lambda\lambda^{m}\det D+\underbrace{\lambda\operatorname*{ent}%
B\cdot\operatorname*{ent}C}_{=\lambda\operatorname*{ent}C\cdot
\operatorname*{ent}B}\nonumber\\
&  =\lambda\lambda^{m}\det D+\lambda\operatorname*{ent}C\cdot
\operatorname*{ent}B\nonumber\\
&  =\lambda\lambda^{m}\det D-\left(  -\lambda\right)  \operatorname*{ent}%
C\cdot\operatorname*{ent}B. \label{pf.lem.block.puqv-reg.6}%
\end{align}
However, Lemma \ref{lem.ent.scalp} (applied to $-\lambda$ instead of $\lambda
$) yields $\operatorname*{ent}\left(  \left(  -\lambda\right)  C\right)
=\left(  -\lambda\right)  \operatorname*{ent}C$. Thus,%
\begin{align*}
\left(  -\lambda\right)  \operatorname*{ent}C  &  =\operatorname*{ent}\left(
\left(  -\lambda\right)  \underbrace{C}_{=v\cdot\operatorname*{adj}\left(
\lambda D\right)  \cdot\left(  -q\right)  }\right)  =\operatorname*{ent}%
\left(  \underbrace{\left(  -\lambda\right)  v\cdot\operatorname*{adj}\left(
\lambda D\right)  \cdot\left(  -q\right)  }_{\substack{=\lambda v\cdot
\operatorname*{adj}\left(  \lambda D\right)  \cdot q\\=v\cdot\lambda
\operatorname*{adj}\left(  \lambda D\right)  \cdot q}}\right) \\
&  =\operatorname*{ent}\left(  v\cdot\underbrace{\lambda\operatorname*{adj}%
\left(  \lambda D\right)  }_{\substack{=\lambda^{m}\operatorname*{adj}%
D\\\text{(by (\ref{pf.lem.block.puqv-reg.ladjlD}))}}}\cdot q\right)
=\operatorname*{ent}\left(  \underbrace{v\cdot\left(  \lambda^{m}%
\operatorname*{adj}D\right)  \cdot q}_{=\lambda^{m}\cdot v\left(
\operatorname*{adj}D\right)  q}\right) \\
&  =\operatorname*{ent}\left(  \lambda^{m}\cdot v\left(  \operatorname*{adj}%
D\right)  q\right)  =\lambda^{m}\operatorname*{ent}\left(  v\left(
\operatorname*{adj}D\right)  q\right)
\end{align*}
(by Lemma \ref{lem.ent.scalp}, applied to $\lambda^{m}$ and $v\left(
\operatorname*{adj}D\right)  q$ instead of $\lambda$ and $C$). Thus,
(\ref{pf.lem.block.puqv-reg.6}) becomes%
\begin{align*}
\lambda^{m}\cdot\det\left(
\begin{array}
[c]{cc}%
A & pv\\
qu & D
\end{array}
\right)   &  =\lambda\lambda^{m}\det D-\underbrace{\left(  -\lambda\right)
\operatorname*{ent}C}_{=\lambda^{m}\operatorname*{ent}\left(  v\left(
\operatorname*{adj}D\right)  q\right)  }\cdot\operatorname*{ent}B\\
&  =\lambda\lambda^{m}\det D-\lambda^{m}\operatorname*{ent}\left(  v\left(
\operatorname*{adj}D\right)  q\right)  \cdot\operatorname*{ent}B\\
&  =\lambda^{m}\cdot\left(  \lambda\det D-\operatorname*{ent}\left(  v\left(
\operatorname*{adj}D\right)  q\right)  \cdot\operatorname*{ent}B\right)  .
\end{align*}
Since $\lambda^{m}$ is a regular element of $\mathbb{K}$, we can thus conclude
that%
\[
\det\left(
\begin{array}
[c]{cc}%
A & pv\\
qu & D
\end{array}
\right)  =\lambda\det D-\operatorname*{ent}\left(  v\left(
\operatorname*{adj}D\right)  q\right)  \cdot\operatorname*{ent}B
\]
(by Lemma \ref{lem.regular.cancel}, applied to $\mathbb{A}=\mathbb{K}$ and
$a=\lambda^{m}$ and $b=\det\left(
\begin{array}
[c]{cc}%
A & pv\\
qu & D
\end{array}
\right)  $ and $c=\lambda\det D-\operatorname*{ent}\left(  v\left(
\operatorname*{adj}D\right)  q\right)  \cdot\operatorname*{ent}B$). In view of
$\lambda=\det A$ and $B=u\left(  \operatorname*{adj}A\right)  p$, we can
rewrite this as
\begin{align*}
\det\left(
\begin{array}
[c]{cc}%
A & pv\\
qu & D
\end{array}
\right)   &  =\det A\cdot\det D-\operatorname*{ent}\left(  v\left(
\operatorname*{adj}D\right)  q\right)  \cdot\operatorname*{ent}\left(
u\left(  \operatorname*{adj}A\right)  p\right) \\
&  =\det A\cdot\det D-\operatorname*{ent}\left(  u\left(  \operatorname*{adj}%
A\right)  p\right)  \cdot\operatorname*{ent}\left(  v\left(
\operatorname*{adj}D\right)  q\right)  .
\end{align*}
This proves Lemma \ref{lem.block.puqv-reg}.
\end{proof}

We can now derive Theorem \ref{thm.block.puqv} from Lemma
\ref{lem.block.puqv-reg} by the same recipe as before:

\begin{proof}
[Proof of Theorem \ref{thm.block.puqv}.]Define the $\mathbb{K}$-algebra
homomorphism $\varepsilon:\mathbb{K}\left[  t\right]  \rightarrow\mathbb{K}$
as in Proposition \ref{prop.tInA.ev}. Thus, $\varepsilon$ is a ring homomorphism.

Recall that every $n\times n$-matrix in $\mathbb{K}^{n\times n}$ can be
considered as a matrix in $\left(  \mathbb{K}\left[  t\right]  \right)
^{n\times n}$. More generally, every $k\times\ell$-matrix in $\mathbb{K}%
^{k\times\ell}$ (for any nonnegative integers $k$ and $\ell$) can be
considered as a matrix in $\left(  \mathbb{K}\left[  t\right]  \right)
^{k\times\ell}$. In other words, for each $F\in\mathbb{K}^{k\times\ell}$, we
can consider $F$ as a matrix in $\left(  \mathbb{K}\left[  t\right]  \right)
^{k\times\ell}$; therefore, $\varepsilon^{k\times\ell}\left(  F\right)  $ is
well-defined. We have
\begin{equation}
\varepsilon^{k\times\ell}\left(  F\right)  =F\ \ \ \ \ \ \ \ \ \ \text{for
every }F\in\mathbb{K}^{k\times\ell}. \label{pf.thm.block.puqv.FF}%
\end{equation}


\begin{vershort}
\noindent(In fact, the proof of (\ref{pf.thm.block.puqv.FF}) is analogous to
the proof of (\ref{pf.lem.adj.poly.-A.short.eF}) we gave above.)
\end{vershort}

\begin{verlong}
\noindent\footnote{\textit{Proof of (\ref{pf.thm.block.puqv.FF}):} Let
$F\in\mathbb{K}^{k\times\ell}$. Write the $k\times\ell$-matrix $F$ in the form
$F=\left(  f_{i,j}\right)  _{1\leq i\leq k,\ 1\leq j\leq\ell}$. Then,
$f_{i,j}\in\mathbb{K}$ for every $\left(  i,j\right)  \in\left\{
1,2,\ldots,n\right\}  ^{2}$ (since $F\in\mathbb{K}^{k\times\ell}$). Hence,%
\begin{equation}
\varepsilon\left(  f_{i,j}\right)  =f_{i,j}\ \ \ \ \ \ \ \ \ \ \text{for every
}\left(  i,j\right)  \in\left\{  1,2,\ldots,n\right\}  ^{2}%
\end{equation}
(by (\ref{pf.lem.adj.poly.-A.eu}) (applied to $u=f_{i,j}$)).
\par
Applying the map $\varepsilon^{k\times\ell}$ to both sides of the equality
$F=\left(  f_{i,j}\right)  _{1\leq i\leq k,\ 1\leq j\leq\ell}$, we obtain%
\begin{align*}
\varepsilon^{k\times\ell}\left(  F\right)   &  =\varepsilon^{k\times\ell
}\left(  \left(  f_{i,j}\right)  _{1\leq i\leq k,\ 1\leq j\leq\ell}\right)
=\left(  \underbrace{\varepsilon\left(  f_{i,j}\right)  }_{\substack{=f_{i,j}%
\\\text{(by (\ref{pf.lem.adj.poly.-A.eF.pf.1}))}}}\right)  _{1\leq i\leq
k,\ 1\leq j\leq\ell}\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of the
map }\varepsilon^{k\times\ell}\right) \\
&  =\left(  f_{i,j}\right)  _{1\leq i\leq k,\ 1\leq j\leq\ell}=F.
\end{align*}
This proves (\ref{pf.thm.block.puqv.FF}).}
\end{verlong}

Let $\widetilde{A}$ be the matrix $tI_{n}+A\in\left(  \mathbb{K}\left[
t\right]  \right)  ^{n\times n}$. Thus, $\widetilde{A}=tI_{n}+A$. Applying the
map $\varepsilon^{n\times n}$ to both sides of this equality, we find
$\varepsilon^{n\times n}\left(  \widetilde{A}\right)  =\varepsilon^{n\times
n}\left(  tI_{n}+A\right)  =A$ (by Proposition \ref{prop.tInA.ev} \textbf{(c)}).

Corollary \ref{cor.tInA.reg} shows that the element $\det\left(
tI_{n}+A\right)  $ of $\mathbb{K}\left[  t\right]  $ is regular. In other
words, the element $\det\widetilde{A}$ of $\mathbb{K}\left[  t\right]  $ is
regular (since $\widetilde{A}=tI_{n}+A$).

Let us consider the matrix $A\in\mathbb{K}^{n\times n}$ as a matrix in
$\left(  \mathbb{K}\left[  t\right]  \right)  ^{n\times n}$ (since every
$n\times n$-matrix in $\mathbb{K}^{n\times n}$ can be considered as a matrix
in $\left(  \mathbb{K}\left[  t\right]  \right)  ^{n\times n}$). Similarly,
let us consider the matrices $D\in\mathbb{K}^{m\times m}$, $p\in
\mathbb{K}^{n\times1}$, $q\in\mathbb{K}^{m\times1}$, $v\in\mathbb{K}^{1\times
m}$ and $u\in\mathbb{K}^{1\times n}$ as matrices in $\left(  \mathbb{K}\left[
t\right]  \right)  ^{m\times m}$, $\left(  \mathbb{K}\left[  t\right]
\right)  ^{n\times1}$, $\left(  \mathbb{K}\left[  t\right]  \right)
^{m\times1}$, $\left(  \mathbb{K}\left[  t\right]  \right)  ^{1\times m}$ and
$\left(  \mathbb{K}\left[  t\right]  \right)  ^{1\times n}$, respectively.

\begin{vershort}
Notice that (\ref{pf.thm.block.puqv.FF}) (applied to $k=n$ and $\ell=n$ and
$F=A$) yields $\varepsilon^{n\times n}\left(  A\right)  =A$. Similarly,
$\varepsilon^{m\times m}\left(  D\right)  =D$ and $\varepsilon^{n\times
1}\left(  p\right)  =p$ and $\varepsilon^{m\times1}\left(  q\right)  =q$ and
$\varepsilon^{1\times m}\left(  v\right)  =v$ and $\varepsilon^{1\times
n}\left(  u\right)  =u$ and $\varepsilon^{n\times m}\left(  pv\right)  =pv$
and $\varepsilon^{m\times n}\left(  qu\right)  =qu$.
\end{vershort}

\begin{verlong}
Notice that (\ref{pf.thm.block.puqv.FF}) (applied to $F=A$) yields
$\varepsilon^{n\times n}\left(  A\right)  =A$. Also,
(\ref{pf.thm.block.puqv.FF}) (applied to $k=m$ and $\ell=m$ and $F=D$) yields
$\varepsilon^{m\times m}\left(  D\right)  =D$. Also,
(\ref{pf.thm.block.puqv.FF}) (applied to $k=n$ and $\ell=1$ and $F=p$) yields
$\varepsilon^{n\times1}\left(  p\right)  =p$. Also,
(\ref{pf.thm.block.puqv.FF}) (applied to $k=m$ and $\ell=1$ and $F=q$) yields
$\varepsilon^{m\times1}\left(  q\right)  =q$. Also,
(\ref{pf.thm.block.puqv.FF}) (applied to $k=1$ and $\ell=m$ and $F=v$) yields
$\varepsilon^{1\times m}\left(  v\right)  =v$. Also, Also,
(\ref{pf.thm.block.puqv.FF}) (applied to $k=1$ and $\ell=n$ and $F=u$) yields
$\varepsilon^{1\times n}\left(  u\right)  =u$. Also,
(\ref{pf.thm.block.puqv.FF}) (applied to $k=n$ and $\ell=m$ and $F=pv$) yields
$\varepsilon^{n\times m}\left(  pv\right)  =pv$. Also,
(\ref{pf.thm.block.puqv.FF}) (applied to $k=m$ and $\ell=n$ and $F=qu$) yields
$\varepsilon^{m\times n}\left(  qu\right)  =qu$.
\end{verlong}

In the proof of Theorem \ref{thm.adj.adjadj}, we have already shown that
\[
\varepsilon^{n\times n}\left(  \operatorname*{adj}\widetilde{A}\right)
=\operatorname*{adj}A\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \varepsilon\left(  \det\widetilde{A}\right)  =\det A.
\]


Next, we claim that%
\begin{equation}
\varepsilon\left(  \operatorname*{ent}\left(  u\left(  \operatorname*{adj}%
\widetilde{A}\right)  p\right)  \right)  =\operatorname*{ent}\left(  u\left(
\operatorname*{adj}A\right)  p\right)  . \label{pf.thm.block.puqv.ent1}%
\end{equation}


\begin{vershort}
[\textit{Proof of (\ref{pf.thm.block.puqv.ent1}):} Define the $1\times
1$-matrix $B=u\left(  \operatorname*{adj}\widetilde{A}\right)  p\in\left(
\mathbb{K}\left[  t\right]  \right)  ^{1\times1}$. Then, Lemma
\ref{lem.ent.map} (applied to $\mathbb{L}=\mathbb{K}\left[  t\right]  $ and
$\mathbb{M}=\mathbb{K}$ and $f=\varepsilon$) yields $\operatorname*{ent}%
\left(  \varepsilon^{1\times1}\left(  B\right)  \right)  =\varepsilon\left(
\operatorname*{ent}B\right)  $.

However, from $B=u\left(  \operatorname*{adj}\widetilde{A}\right)  p=u\left(
\left(  \operatorname*{adj}\widetilde{A}\right)  p\right)  $, we obtain%
\begin{align*}
\varepsilon^{1\times1}\left(  B\right)   &  =\varepsilon^{1\times1}\left(
u\left(  \left(  \operatorname*{adj}\widetilde{A}\right)  p\right)  \right) \\
&  =\underbrace{\varepsilon^{1\times n}\left(  u\right)  }_{=u}\cdot
\underbrace{\varepsilon^{n\times1}\left(  \left(  \operatorname*{adj}%
\widetilde{A}\right)  p\right)  }_{\substack{=\varepsilon^{n\times n}\left(
\operatorname*{adj}\widetilde{A}\right)  \cdot\varepsilon^{n\times1}\left(
p\right)  \\\text{(by Theorem \ref{prop.functor.clas} \textbf{(c)})}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem \ref{prop.functor.clas}
\textbf{(c)}}\right) \\
&  =u\cdot\underbrace{\varepsilon^{n\times n}\left(  \operatorname*{adj}%
\widetilde{A}\right)  }_{=\operatorname*{adj}A}\cdot\underbrace{\varepsilon
^{n\times1}\left(  p\right)  }_{=p}=u\left(  \operatorname*{adj}A\right)  p.
\end{align*}
In view of this, we can rewrite the equality $\operatorname*{ent}\left(
\varepsilon^{1\times1}\left(  B\right)  \right)  =\varepsilon\left(
\operatorname*{ent}B\right)  $ (which we have proved in the previous
paragraph) as
\[
\operatorname*{ent}\left(  u\left(  \operatorname*{adj}A\right)  p\right)
=\varepsilon\left(  \operatorname*{ent}B\right)  =\varepsilon\left(
\operatorname*{ent}\left(  u\left(  \operatorname*{adj}\widetilde{A}\right)
p\right)  \right)
\]
(since $B=u\left(  \operatorname*{adj}\widetilde{A}\right)  p$). This proves
(\ref{pf.thm.block.puqv.ent1}).] \medskip
\end{vershort}

\begin{verlong}
[\textit{Proof of (\ref{pf.thm.block.puqv.ent1}):} Define the $1\times
1$-matrix $B=u\left(  \operatorname*{adj}\widetilde{A}\right)  p\in\left(
\mathbb{K}\left[  t\right]  \right)  ^{1\times1}$. Then, Lemma
\ref{lem.ent.map} (applied to $\mathbb{L}=\mathbb{K}\left[  t\right]  $ and
$\mathbb{M}=\mathbb{K}$ and $f=\varepsilon$) yields%
\begin{equation}
\operatorname*{ent}\left(  \varepsilon^{1\times1}\left(  B\right)  \right)
=\varepsilon\left(  \operatorname*{ent}B\right)  .
\label{pf.thm.block.puqv.ent1.pf.1}%
\end{equation}


However, applying the map $\varepsilon^{1\times1}$ to both sides of the
equality $B=u\left(  \operatorname*{adj}\widetilde{A}\right)  p=u\left(
\left(  \operatorname*{adj}\widetilde{A}\right)  p\right)  $, we obtain%
\begin{align*}
\varepsilon^{1\times1}\left(  B\right)   &  =\varepsilon^{1\times1}\left(
u\left(  \left(  \operatorname*{adj}\widetilde{A}\right)  p\right)  \right)
=\underbrace{\varepsilon^{1\times n}\left(  u\right)  }_{=u}\cdot
\underbrace{\varepsilon^{n\times1}\left(  \left(  \operatorname*{adj}%
\widetilde{A}\right)  p\right)  }_{\substack{=\varepsilon^{n\times n}\left(
\operatorname*{adj}\widetilde{A}\right)  \cdot\varepsilon^{n\times1}\left(
p\right)  \\\text{(by Theorem \ref{prop.functor.clas} \textbf{(c)}%
,}\\\text{applied to }\mathbb{K}\left[  t\right]  \text{, }\mathbb{K}\text{,
}\varepsilon\text{, }n\text{, }n\text{, }1\text{, }\operatorname*{adj}%
\widetilde{A}\text{ and }p\\\text{instead of }\mathbb{L}\text{, }%
\mathbb{M}\text{, }f\text{, }n\text{, }m\text{, }p\text{, }A\text{ and
}B\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Theorem \ref{prop.functor.clas} \textbf{(c)},}\\
\text{applied to }\mathbb{K}\left[  t\right]  \text{, }\mathbb{K}\text{,
}\varepsilon\text{, }1\text{, }n\text{, }1\text{, }u\text{ and }\left(
\operatorname*{adj}\widetilde{A}\right)  p\\
\text{instead of }\mathbb{L}\text{, }\mathbb{M}\text{, }f\text{, }n\text{,
}m\text{, }p\text{, }A\text{ and }B
\end{array}
\right) \\
&  =u\cdot\underbrace{\varepsilon^{n\times n}\left(  \operatorname*{adj}%
\widetilde{A}\right)  }_{=\operatorname*{adj}A}\cdot\underbrace{\varepsilon
^{n\times1}\left(  p\right)  }_{=p}=u\left(  \operatorname*{adj}A\right)  p.
\end{align*}
In view of this, we can rewrite (\ref{pf.thm.block.puqv.ent1.pf.1}) as
\[
\operatorname*{ent}\left(  u\left(  \operatorname*{adj}A\right)  p\right)
=\varepsilon\left(  \operatorname*{ent}B\right)  =\varepsilon\left(
\operatorname*{ent}\left(  u\left(  \operatorname*{adj}\widetilde{A}\right)
p\right)  \right)
\]
(since $B=u\left(  \operatorname*{adj}\widetilde{A}\right)  p$). This proves
(\ref{pf.thm.block.puqv.ent1}).] \medskip
\end{verlong}

Furthermore, we have $\varepsilon\left(  \lambda\right)  =\lambda$ for each
$\lambda\in\mathbb{K}$ (by the definition of $\varepsilon$). Applying this to
$\lambda=\det D$, we find%
\[
\varepsilon\left(  \det D\right)  =\det D.
\]


Also, $v\left(  \operatorname*{adj}D\right)  q\in\mathbb{K}^{1\times1}$ and
thus $\operatorname*{ent}\left(  v\left(  \operatorname*{adj}D\right)
q\right)  \in\mathbb{K}$. Recall again that we have $\varepsilon\left(
\lambda\right)  =\lambda$ for each $\lambda\in\mathbb{K}$. Applying this to
$\lambda=\operatorname*{ent}\left(  v\left(  \operatorname*{adj}D\right)
q\right)  $, we obtain%
\begin{equation}
\varepsilon\left(  \operatorname*{ent}\left(  v\left(  \operatorname*{adj}%
D\right)  q\right)  \right)  =\operatorname*{ent}\left(  v\left(
\operatorname*{adj}D\right)  q\right)  \label{pf.thm.block.puqv.ent2}%
\end{equation}
(since $\operatorname*{ent}\left(  v\left(  \operatorname*{adj}D\right)
q\right)  \in\mathbb{K}$).

Now, Lemma \ref{lem.block.puqv-reg} (applied to $\mathbb{K}\left[  t\right]  $
and $\widetilde{A}$ instead of $\mathbb{K}$ and $A$) yields
\[
\det\left(
\begin{array}
[c]{cc}%
\widetilde{A} & pv\\
qu & D
\end{array}
\right)  =\det\widetilde{A}\cdot\det D-\operatorname*{ent}\left(  u\left(
\operatorname*{adj}\widetilde{A}\right)  p\right)  \cdot\operatorname*{ent}%
\left(  v\left(  \operatorname*{adj}D\right)  q\right)
\]
(since the element $\det\widetilde{A}$ of $\mathbb{K}\left[  t\right]  $ is
regular). Applying the map $\varepsilon$ to both sides of this equality, we
find%
\begin{align}
&  \varepsilon\left(  \det\left(
\begin{array}
[c]{cc}%
\widetilde{A} & pv\\
qu & D
\end{array}
\right)  \right) \nonumber\\
&  =\varepsilon\left(  \det\widetilde{A}\cdot\det D-\operatorname*{ent}\left(
u\left(  \operatorname*{adj}\widetilde{A}\right)  p\right)  \cdot
\operatorname*{ent}\left(  v\left(  \operatorname*{adj}D\right)  q\right)
\right) \nonumber\\
&  =\underbrace{\varepsilon\left(  \det\widetilde{A}\right)  }_{=\det A}%
\cdot\underbrace{\varepsilon\left(  \det D\right)  }_{=\det D}%
-\underbrace{\varepsilon\left(  \operatorname*{ent}\left(  u\left(
\operatorname*{adj}\widetilde{A}\right)  p\right)  \right)  }%
_{\substack{=\operatorname*{ent}\left(  u\left(  \operatorname*{adj}A\right)
p\right)  \\\text{(by (\ref{pf.thm.block.puqv.ent1}))}}}\cdot
\underbrace{\varepsilon\left(  \operatorname*{ent}\left(  v\left(
\operatorname*{adj}D\right)  q\right)  \right)  }%
_{\substack{=\operatorname*{ent}\left(  v\left(  \operatorname*{adj}D\right)
q\right)  \\\text{(by (\ref{pf.thm.block.puqv.ent2}))}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\varepsilon\text{ is a ring homomorphism}\right) \nonumber\\
&  =\det A\cdot\det D-\operatorname*{ent}\left(  u\left(  \operatorname*{adj}%
A\right)  p\right)  \cdot\operatorname*{ent}\left(  v\left(
\operatorname*{adj}D\right)  q\right)  . \label{pf.thm.block.puqv.5}%
\end{align}


However, $\left(
\begin{array}
[c]{cc}%
\widetilde{A} & pv\\
qu & D
\end{array}
\right)  $ is an $\left(  n+m\right)  \times\left(  n+m\right)  $-matrix in
$\left(  \mathbb{K}\left[  t\right]  \right)  ^{\left(  n+m\right)
\times\left(  n+m\right)  }$. Thus, Proposition \ref{prop.functor.det}
\textbf{(a)} (applied to $\mathbb{K}\left[  t\right]  $, $\mathbb{K}$,
$\varepsilon$, $n+m$ and $\left(
\begin{array}
[c]{cc}%
\widetilde{A} & pv\\
qu & D
\end{array}
\right)  $ instead of $\mathbb{L}$, $\mathbb{M}$, $f$, $n$ and $A$) yields
\begin{equation}
\varepsilon\left(  \det\left(
\begin{array}
[c]{cc}%
\widetilde{A} & pv\\
qu & D
\end{array}
\right)  \right)  =\det\left(  \varepsilon^{\left(  n+m\right)  \times\left(
n+m\right)  }\left(
\begin{array}
[c]{cc}%
\widetilde{A} & pv\\
qu & D
\end{array}
\right)  \right)  . \label{pf.thm.block.puqv.6}%
\end{equation}


On the other hand, Lemma \ref{lem.functor.block2x2} (applied to $\mathbb{K}%
\left[  t\right]  $, $\mathbb{K}$, $\varepsilon$, $m$, $n$, $m$,
$\widetilde{A}$, $pv$ and $qu$ instead of $\mathbb{L}$, $\mathbb{M}$, $f$,
$n^{\prime}$, $m$, $m^{\prime}$, $A$, $B$ and $C$) yields
\[
\varepsilon^{\left(  n+m\right)  \times\left(  n+m\right)  }\left(  \left(
\begin{array}
[c]{cc}%
\widetilde{A} & pv\\
qu & D
\end{array}
\right)  \right)  =\left(
\begin{array}
[c]{cc}%
\varepsilon^{n\times n}\left(  \widetilde{A}\right)  & \varepsilon^{n\times
m}\left(  pv\right) \\
\varepsilon^{m\times n}\left(  qu\right)  & \varepsilon^{m\times m}\left(
D\right)
\end{array}
\right)  =\left(
\begin{array}
[c]{cc}%
A & pv\\
qu & D
\end{array}
\right)
\]
(since $\varepsilon^{n\times n}\left(  \widetilde{A}\right)  =A$ and
$\varepsilon^{n\times m}\left(  pv\right)  =pv$ and $\varepsilon^{m\times
n}\left(  qu\right)  =qu$ and $\varepsilon^{m\times m}\left(  D\right)  =D$).
Taking determinants on both sides of this equality, we find%
\[
\det\left(  \varepsilon^{\left(  n+m\right)  \times\left(  n+m\right)
}\left(  \left(
\begin{array}
[c]{cc}%
\widetilde{A} & pv\\
qu & D
\end{array}
\right)  \right)  \right)  =\det\left(
\begin{array}
[c]{cc}%
A & pv\\
qu & D
\end{array}
\right)  .
\]
Hence,%
\begin{align*}
\det\left(
\begin{array}
[c]{cc}%
A & pv\\
qu & D
\end{array}
\right)   &  =\det\left(  \varepsilon^{\left(  n+m\right)  \times\left(
n+m\right)  }\left(  \left(
\begin{array}
[c]{cc}%
\widetilde{A} & pv\\
qu & D
\end{array}
\right)  \right)  \right) \\
&  =\varepsilon\left(  \det\left(
\begin{array}
[c]{cc}%
\widetilde{A} & pv\\
qu & D
\end{array}
\right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.block.puqv.6})}\right) \\
&  =\det A\cdot\det D-\operatorname*{ent}\left(  u\left(  \operatorname*{adj}%
A\right)  p\right)  \cdot\operatorname*{ent}\left(  v\left(
\operatorname*{adj}D\right)  q\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.thm.block.puqv.5})}\right)  .
\end{align*}
This completes the proof of Theorem \ref{thm.block.puqv}.
\end{proof}

Note that Theorem \ref{thm.block.puqv} generalizes the following known fact
(e.g., \cite[Exercise 6.60 \textbf{(a)}]{detnotes}):

\begin{corollary}
\label{cor.block.bordered}Let $n\in\mathbb{N}$. Let $u\in\mathbb{K}^{n\times
1}$ be a column vector with $n$ entries, and let $v\in\mathbb{K}^{1\times n}$
be a row vector with $n$ entries. (Thus, $uv$ is an $n\times n$-matrix,
whereas $vu$ is a $1\times1$-matrix.) Let $h\in\mathbb{K}$. Let $H$ be the
$1\times1$-matrix $\left(
\begin{array}
[c]{c}%
h
\end{array}
\right)  \in\mathbb{K}^{1\times1}$. Let $A\in\mathbb{K}^{n\times n}$ be an
$n\times n$-matrix. Then,%
\[
\det\left(
\begin{array}
[c]{cc}%
A & u\\
v & H
\end{array}
\right)  =h\det A-\operatorname*{ent}\left(  v\left(  \operatorname*{adj}%
A\right)  u\right)  .
\]

\end{corollary}

\begin{proof}
[Proof of Corollary \ref{cor.block.bordered} (sketched).]Consider the
$1\times1$ identity matrix $I_{1}=\left(
\begin{array}
[c]{c}%
1
\end{array}
\right)  $. Then, $I_{1}$ is both a row vector and a column vector, and we
have $u=uI_{1}$ and $v=I_{1}v$. Moreover, the adjugate of any $1\times
1$-matrix is $\left(
\begin{array}
[c]{c}%
1
\end{array}
\right)  $ (since the determinant of a $0\times0$-matrix is defined to be
$1$). Thus, in particular, $\operatorname*{adj}H=\left(
\begin{array}
[c]{c}%
1
\end{array}
\right)  =I_{1}$. Furthermore, from $H=\left(
\begin{array}
[c]{c}%
h
\end{array}
\right)  $, we obtain $\det H=h$. Now, from $u=uI_{1}$ and $v=I_{1}v$, we
obtain
\begin{align*}
\det\left(
\begin{array}
[c]{cc}%
A & u\\
v & H
\end{array}
\right)   &  =\det\left(
\begin{array}
[c]{cc}%
A & uI_{1}\\
I_{1}v & H
\end{array}
\right) \\
&  =\det A\cdot\underbrace{\det H}_{=h}-\operatorname*{ent}\left(  v\left(
\operatorname*{adj}A\right)  u\right)  \cdot\operatorname*{ent}\left(
I_{1}\underbrace{\left(  \operatorname*{adj}H\right)  }_{=I_{1}}I_{1}\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Theorem \ref{thm.block.puqv}, applied to }1\text{, }H\text{,
}u\text{, }I_{1}\text{, }I_{1}\text{ and }v\\
\text{instead of }m\text{, }D\text{, }p\text{, }q\text{, }v\text{ and }u
\end{array}
\right) \\
&  =\underbrace{\det A\cdot h}_{=h\det A}-\operatorname*{ent}\left(  v\left(
\operatorname*{adj}A\right)  u\right)  \cdot\operatorname*{ent}%
\underbrace{\left(  I_{1}\cdot I_{1}\cdot I_{1}\right)  }_{=I_{1}}\\
&  =h\det A-\operatorname*{ent}\left(  v\left(  \operatorname*{adj}A\right)
u\right)  \cdot\underbrace{\operatorname*{ent}\left(  I_{1}\right)  }_{=1}\\
&  =h\det A-\operatorname*{ent}\left(  v\left(  \operatorname*{adj}A\right)
u\right)  .
\end{align*}
This proves Corollary \ref{cor.block.bordered}.
\end{proof}

Another particular case of Theorem \ref{thm.block.puqv} is the following:

\begin{corollary}
\label{cor.block.01}Let $n$ and $m$ be two positive integers. Let
$A\in\mathbb{K}^{n\times n}$ and $D\in\mathbb{K}^{m\times m}$ be two square
matrices. Let $B$ be the $n\times m$-matrix whose $\left(  n,1\right)  $-th
entry is $1$ and whose all other entries are $0$. Let $C$ be the $m\times
n$-matrix whose $\left(  1,n\right)  $-th entry is $1$ and whose all other
entries are $0$. Then,%
\[
\det\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  =\det A\cdot\det D-\det\left(  A_{\sim n,\sim n}\right)  \cdot
\det\left(  D_{\sim1,\sim1}\right)  .
\]
(Recall that we are using the notations from Definition
\ref{def.submatrix.minor}.)
\end{corollary}

\begin{example}
Let us see what Corollary \ref{cor.block.01} says in the case when $n=2$ and
$m=3$. Indeed, let $n=2$ and $m=3$ and%
\[
A=\left(
\begin{array}
[c]{cc}%
a_{1,1} & a_{1,2}\\
a_{2,1} & a_{2,2}%
\end{array}
\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ D=\left(
\begin{array}
[c]{ccc}%
d_{1,1} & d_{1,2} & d_{1,3}\\
d_{2,1} & d_{2,2} & d_{2,3}\\
d_{3,1} & d_{3,2} & d_{3,3}%
\end{array}
\right)  .
\]
Then, the matrices $B$ and $C$ defined in Corollary \ref{cor.block.01} are%
\[
B=\left(
\begin{array}
[c]{ccc}%
0 & 0 & 0\\
1 & 0 & 0
\end{array}
\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \ C=\left(
\begin{array}
[c]{cc}%
0 & 1\\
0 & 0\\
0 & 0
\end{array}
\right)  .
\]
Hence,%
\[
\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  =\left(
\begin{array}
[c]{ccccc}%
a_{1,1} & a_{1,2} & 0 & 0 & 0\\
a_{2,1} & a_{2,2} & 1 & 0 & 0\\
0 & 1 & d_{1,1} & d_{1,2} & d_{1,3}\\
0 & 0 & d_{2,1} & d_{2,2} & d_{2,3}\\
0 & 0 & d_{3,1} & d_{3,2} & d_{3,3}%
\end{array}
\right)  .
\]
Therefore, the claim of Corollary \ref{cor.block.01} rewrites as follows in
our case:
\begin{align*}
&  \det\left(
\begin{array}
[c]{ccccc}%
a_{1,1} & a_{1,2} & 0 & 0 & 0\\
a_{2,1} & a_{2,2} & 1 & 0 & 0\\
0 & 1 & d_{1,1} & d_{1,2} & d_{1,3}\\
0 & 0 & d_{2,1} & d_{2,2} & d_{2,3}\\
0 & 0 & d_{3,1} & d_{3,2} & d_{3,3}%
\end{array}
\right) \\
&  =\det\left(
\begin{array}
[c]{cc}%
a_{1,1} & a_{1,2}\\
a_{2,1} & a_{2,2}%
\end{array}
\right)  \cdot\det\left(
\begin{array}
[c]{ccc}%
d_{1,1} & d_{1,2} & d_{1,3}\\
d_{2,1} & d_{2,2} & d_{2,3}\\
d_{3,1} & d_{3,2} & d_{3,3}%
\end{array}
\right)  -\det\left(
\begin{array}
[c]{c}%
a_{1,1}%
\end{array}
\right)  \cdot\det\left(
\begin{array}
[c]{cc}%
d_{2,2} & d_{2,3}\\
d_{3,2} & d_{3,3}%
\end{array}
\right)  .
\end{align*}

\end{example}

\begin{verlong}
Our proof of Corollary \ref{cor.block.01} will rely on two lemmas:

\begin{lemma}
\label{lem.entvBu}Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $u=\left(
\begin{array}
[c]{c}%
u_{1}\\
u_{2}\\
\vdots\\
u_{m}%
\end{array}
\right)  \in\mathbb{K}^{m\times1}$ be a column vector, and let $v=\left(
\begin{array}
[c]{cccc}%
v_{1} & v_{2} & \cdots & v_{n}%
\end{array}
\right)  \in\mathbb{K}^{1\times n}$ be a row vector. Let $B=\left(
b_{i,j}\right)  _{1\leq i\leq n,\ 1\leq j\leq m}\in\mathbb{K}^{n\times m}$ be
an $n\times m$-matrix. Then,%
\[
\operatorname*{ent}\left(  vBu\right)  =\sum_{i=1}^{n}\ \ \sum_{j=1}^{m}%
u_{j}v_{i}b_{i,j}.
\]

\end{lemma}

Lemma \ref{lem.entvBu} is precisely \cite[Proposition 7.263]{detnotes} (except
that the column vector $\left(
\begin{array}
[c]{c}%
u_{1}\\
u_{2}\\
\vdots\\
u_{m}%
\end{array}
\right)  $ is written as $\left(  u_{1},u_{2},\ldots,u_{m}\right)  ^{T}$, and
that the row vector $\left(
\begin{array}
[c]{cccc}%
v_{1} & v_{2} & \cdots & v_{n}%
\end{array}
\right)  $ is written as $\left(  v_{1},v_{2},\ldots,v_{n}\right)  $ in
\cite[Proposition 7.263]{detnotes}). Thus, we will not prove it here.

\begin{lemma}
\label{lem.uv}Let $n\in\mathbb{N}$ and $m\in\mathbb{N}$. Let $u=\left(
\begin{array}
[c]{c}%
u_{1}\\
u_{2}\\
\vdots\\
u_{m}%
\end{array}
\right)  \in\mathbb{K}^{m\times1}$ be a column vector, and let $v=\left(
\begin{array}
[c]{cccc}%
v_{1} & v_{2} & \cdots & v_{n}%
\end{array}
\right)  \in\mathbb{K}^{1\times n}$ be a row vector. Then,%
\[
uv=\left(  u_{i}v_{j}\right)  _{1\leq i\leq m,\ 1\leq j\leq n}.
\]

\end{lemma}

\begin{proof}
[Proof of Lemma \ref{lem.uv}.]We have $u=\left(
\begin{array}
[c]{c}%
u_{1}\\
u_{2}\\
\vdots\\
u_{m}%
\end{array}
\right)  =\left(  u_{i}\right)  _{1\leq i\leq m,\ 1\leq j\leq1}$ and
$v=\left(
\begin{array}
[c]{cccc}%
v_{1} & v_{2} & \cdots & v_{n}%
\end{array}
\right)  =\left(  v_{j}\right)  _{1\leq i\leq1,\ 1\leq j\leq n}$. Hence, the
definition of the product of two matrices yields%
\[
uv=\left(  \underbrace{\sum_{k=1}^{1}u_{i}v_{j}}_{=u_{i}v_{j}}\right)  _{1\leq
i\leq m,\ 1\leq j\leq n}=\left(  u_{i}v_{j}\right)  _{1\leq i\leq m,\ 1\leq
j\leq n}.
\]
This proves Lemma \ref{lem.uv}.
\end{proof}
\end{verlong}

\begin{proof}
[Proof of Corollary \ref{cor.block.01} (sketched).]What follows is by far not
the easiest proof of Corollary \ref{cor.block.01}, but it puts the corollary
in the context of Theorem \ref{thm.block.puqv}.

We let $p\in\mathbb{K}^{n\times1}$ be the column vector whose $n$-th entry is
$1$ and whose all other entries are $0$.

We let $q\in\mathbb{K}^{m\times1}$ be the column vector whose $1$-st entry is
$1$ and whose all other entries are $0$.

We let $v\in\mathbb{K}^{1\times m}$ be the row vector whose $1$-st entry is
$1$ and whose all other entries are $0$.

We let $u\in\mathbb{K}^{1\times n}$ be the row vector whose $n$-th entry is
$1$ and whose all other entries are $0$.

Thus,%
\begin{align*}
p &  =\left(
\begin{array}
[c]{c}%
0\\
0\\
\vdots\\
0\\
1
\end{array}
\right)  \in\mathbb{K}^{n\times1},\ \ \ \ \ \ \ \ \ \ q=\left(
\begin{array}
[c]{c}%
1\\
0\\
0\\
\vdots\\
0
\end{array}
\right)  \in\mathbb{K}^{m\times1},\\
v &  =\left(
\begin{array}
[c]{ccccc}%
1 & 0 & 0 & \cdots & 0
\end{array}
\right)  \in\mathbb{K}^{1\times m},\ \ \ \ \ \ \ \ \ \ u=\left(
\begin{array}
[c]{ccccc}%
0 & 0 & \cdots & 0 & 1
\end{array}
\right)  \in\mathbb{K}^{1\times n}.
\end{align*}


\begin{vershort}
Now, it is easy to see (using just the definition of a product of two
matrices) that the following four claims hold:
\end{vershort}

\begin{verlong}
Now, it is easy to see that the following four claims hold\footnote{Indeed,
Claim 1 and Claim 2 follow easily from Lemma \ref{lem.uv}, whereas Claim 3 and
Claim 4 follow easily from Lemma \ref{lem.entvBu}.}:
\end{verlong}

\begin{statement}
\textit{Claim 1:} We have $qu=C$.
\end{statement}

\begin{statement}
\textit{Claim 2:} We have $pv=B$.
\end{statement}

\begin{statement}
\textit{Claim 3:} Every $m\times m$-matrix $Y$ satisfies%
\begin{equation}
\operatorname*{ent}\left(  vYq\right)  =\left(  \text{the }\left(  1,1\right)
\text{-th entry of }Y\right)  .\label{pf.cor.block.01.entvYq=}%
\end{equation}

\end{statement}

\begin{statement}
\textit{Claim 4:} Every $n\times n$-matrix $X$ satisfies%
\begin{equation}
\operatorname*{ent}\left(  uXp\right)  =\left(  \text{the }\left(  n,n\right)
\text{-th entry of }X\right)  .\label{pf.cor.block.01.entuXp=}%
\end{equation}

\end{statement}

Now, Theorem \ref{thm.block.puqv} yields%
\[
\det\left(
\begin{array}
[c]{cc}%
A & pv\\
qu & D
\end{array}
\right)  =\det A\cdot\det D-\operatorname*{ent}\left(  u\left(
\operatorname*{adj}A\right)  p\right)  \cdot\operatorname*{ent}\left(
v\left(  \operatorname*{adj}D\right)  q\right)  .
\]
In view of $pv=B$ and $qu=C$ and%
\begin{align*}
\operatorname*{ent}\left(  u\left(  \operatorname*{adj}A\right)  p\right)   &
=\left(  \text{the }\left(  n,n\right)  \text{-th entry of }%
\operatorname*{adj}A\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.cor.block.01.entuXp=}), applied to }X=\operatorname*{adj}A\right)  \\
&  =\underbrace{\left(  -1\right)  ^{n+n}}_{\substack{=\left(  -1\right)
^{2n}=1\\\text{(since }2n\text{ is even)}}}\det\left(  A_{\sim n,\sim
n}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of an
adjugate}\right)  \\
&  =\det\left(  A_{\sim n,\sim n}\right)
\end{align*}
and%
\begin{align*}
\operatorname*{ent}\left(  v\left(  \operatorname*{adj}D\right)  q\right)   &
=\left(  \text{the }\left(  1,1\right)  \text{-th entry of }%
\operatorname*{adj}D\right)  \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.cor.block.01.entvYq=}), applied to }Y=\operatorname*{adj}D\right)  \\
&  =\underbrace{\left(  -1\right)  ^{1+1}}_{=1}\det\left(  D_{\sim1,\sim
1}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of an
adjugate}\right)  \\
&  =\det\left(  D_{\sim1,\sim1}\right)  ,
\end{align*}
we can rewrite this as%
\[
\det\left(
\begin{array}
[c]{cc}%
A & B\\
C & D
\end{array}
\right)  =\det A\cdot\det D-\det\left(  A_{\sim n,\sim n}\right)  \cdot
\det\left(  D_{\sim1,\sim1}\right)  .
\]
This proves Corollary \ref{cor.block.01}.
\end{proof}

\begin{noncompile}
[Additional material that could find its way into this note:

-- The $\operatorname*{Tr}\left(  A^{p}\right)  \equiv\left(
\operatorname*{Tr}A\right)  ^{p}$ congruence.

-- Arnold's congruences (require $\psi$-structure on $\mathbb{K}$).

-- Another formula for derivations on determinants: $f\left(  \det\left(
a_{i,j}\right)  \right)  =\sum_{k=1}^{n}\det\left(  f^{\delta_{k,i}}%
a_{i,j}\right)  $.

-- If $A\in\mathbb{K}^{n\times n}$ and $A^{k}=0$, then $\left(
\operatorname*{Tr}A\right)  ^{n\left(  k-1\right)  +1}=0$ (Almkvist's
theorem). Provided I find a nice proof...]

See also todo1.
\end{noncompile}

\begin{thebibliography}{99999999999}                                                                                      %


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\end{document}