= Solution
The <structure theorem for finitely generated modules over a principal ideal domain> applies because every <Euclidean domain> is a <principal ideal domain>. It says that a finitely generated $R$-module is isomorphic to
$$
R^s\oplus R/(d_1)\oplus\cdots\oplus R/(d_k),
\qquad d_1\mid d_2\mid\cdots\mid d_k,
$$
where the nonzero nonunits $d_i$ are unique up to multiplication by units. They are the <invariant factors of a linear operator>[invariant factors].
For $V_\alpha$, multiplication by $X$ is the <linear map> $\alpha$. Since $V$ is finite-dimensional over $F$, $V_\alpha$ is a finitely generated <torsion module> over the <polynomial ring> $F[X]$, so it has no free summand. Choosing each invariant factor $a_i$ to be monic gives
$$
V_\alpha\cong\bigoplus_{i=1}^k F[X]/(a_i),
\qquad a_1\mid\cdots\mid a_k.
$$
The basis $1,X,\ldots,X^{\deg a_i-1}$ of each cyclic summand makes multiplication by $X$ a <companion matrix>. Concatenating these bases therefore puts $\alpha$ in <rational canonical form>.
On $F[X]/(a_i)$, a polynomial annihilates multiplication by $X$ exactly when it is divisible by $a_i$. It follows that the <minimal polynomial> and <characteristic polynomial> are
$$
\boxed{m_\alpha(X)=a_k(X)},
\qquad
\boxed{\chi_\alpha(X)=\prod_{i=1}^k a_i(X)}.
$$
The second identity follows blockwise from the characteristic polynomial of a companion matrix. Since every $a_i$ divides $a_k$, the product $\chi_\alpha$ annihilates every cyclic summand. Thus $\chi_\alpha(\alpha)=0$, which is the <Cayley-Hamilton theorem>.
For the displayed matrix, the given generators of $\ker\theta$ are the columns of
$$
XI-A=
\begin{pmatrix}
X&-1&0\\
4&X-4&0\\
2&-1&X-2
\end{pmatrix}.
$$
Elementary row and column operations over $\mathbb R[X]$ give the <Smith normal form>
$$
\operatorname{diag}\bigl(1,X-2,(X-2)^2\bigr).
$$
Consequently the nonunit invariant factors are
$$
\boxed{X-2\quad\text{and}\quad(X-2)^2},
$$
and, as a check,
$$
m_\alpha(X)=(X-2)^2,
\qquad
\chi_\alpha(X)=(X-2)^3.
$$
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