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\begin{document}
\section*{Characteristic Values}
\begin{defn}
Let $V$ be a vector space over the field $F$ and let $T$ be a linear operator on $V$. A characteristic value of $T$ is a scalar $c$ in $F$ such that there is a non-zero vector $\alpha \in V$ with $T\alpha = c\alpha$. If $c$ is a characteristic value of $T$, then
\begin{enumerate}[(a)]
 \item any $\alpha$ such that $T\alpha = c\alpha$ is called a characteristic vector of $T$ associated with the characteristic value $c$;
 \item the collection of all $\alpha$ such that $T\alpha = c\alpha$ is called the characteristic space associated with $c$.
\end{enumerate}
\end{defn}

\begin{thm}
Let $T$ be a linear operator on a finite-dimensional space $V$ and let $c$ be a scalar. The following are equivalent:
\begin{enumerate}[(a)]
 \item $c$ is a characteristic value of $T$.
 \item The operator $(T-cI)$ is singular (and invertible).
 \item $\det(T-cI) = 0$.
\end{enumerate}
\end{thm}

\begin{defn}
If $A$ is a $n \times n$ matrix over the field $F$, a characteristic value of $A$ in $F$ is a scalar $c$ in $F$ such that the matrix $(A-cI)$ is singular (not invertible). The monic polynomial of degree $n$, $f = \det(xI - A)$, is called the characteristic polynomial of $A$.
\end{defn}

\begin{lemma} Similar matrices have the same characteristic polynomial.
\end{lemma}

\begin{defn}Let $T$ be a linear operator on the finite dimensional space $V$. We say that $T$ is diagonalizable if there is a basis for $V$ each vector of which is a characteristic vector of $T$.
\end{defn}

\begin{lemma}
Suppose that $T\alpha = c\alpha$. If $f$ is any polynomial, then $f(T)\alpha = f(c)\alpha$.
\end{lemma}

\begin{lemma}
Let $T$ be a linear operator on the finite-dimensional space $V$. Let $c_1, \ldots, c_k$ be the distinct characteristic values of $T$ and let $W_i$ be the space of characteristic vectors associated with the characteristic value $c_i$. If $W = W_1 + \cdots + W_k$, then
	$$ \dim W = \dim W_1 + \cdots + \dim W_k. $$
In fact, if $\B_i$ is an ordered basis for $W_i$, then $\B = (\B_1, \ldots, \B_k)$ is an ordered basis for $W$.
\end{lemma}

\begin{thm}Let $T$ be a linear operator on a finite-dimensional space $V$. Let $c_1, \ldots, c_k$ be the distinct characteristic values of $T$ and let $W_i$ be the null spaces of $(T-c_iI)$. The following are equivalent:
 \begin{enumerate}[(i)]
  \item $T$ is diagonalizable.
  \item The characteristic polynomial for $T$ is 
  $$ f = (x-c_1)^{d_1} \cdots (x-c_k)^{d_k}$$
  and $\dim W_i = d_i$, $i=1, \ldots, k.$
  \item $\dim W_1 + \cdots + \dim W_k = \dim V.$
 \end{enumerate}
\end{thm}

\section*{Annihilating Polynomials}

\begin{defn}
 Let $T$ be a linear operator on a finite-dimensional vector space $V$ over $F$. The minimal polynomial for $T$ is the (unique) monic generator of the ideals of polynomials over $F$ which annihilate $T$.
\end{defn}

\begin{thm}
Let $T$ be a linear operator on a $n$-dimensional vector space $V$ [or, let $A$ be an $n \times n$ matrix]. The characteristic and minimal polynomials for $T$ [for $A$] have the same roots, except for multiplicities.
\end{thm}

\end{document}