= Solution
An $R$-<module isomorphism> $T:V_\alpha\to V_\beta$ is a $K$-linear isomorphism satisfying $T\alpha=\beta T$, by applying it to multiplication by $X$. Conversely that identity implies $Tf(\alpha)=f(\beta)T$ for every <polynomial>, so a similarity is an $R$-<module isomorphism>.
The cyclic decomposition theorem over the <principal ideal domain> $R=K[X]$ states that any <finitely generated module> is
$$
R^r\oplus R/(d_1)\oplus\cdots\oplus R/(d_s),\qquad d_1\mid d_2\mid\cdots\mid d_s,
$$
with nonunit monic invariant factors, uniquely determined, and free rank $r$ uniquely determined. A finite-dimensional $K$-space has $r=0$. In each cyclic summand the basis $1,X,\ldots,X^{\deg d_i-1}$ makes multiplication by $X$ the <companion matrix> of $d_i$. Their block diagonal sum is the <rational canonical form>.
Here the two <polynomials> are already in divisibility order. With the companion convention having ones below the diagonal,
$$
\boxed{\operatorname{RCF}(\alpha)=\operatorname{diag}\left(\begin{pmatrix}0&-1\\1&2\end{pmatrix},\begin{pmatrix}0&0&1\\1&0&-3\\0&1&3\end{pmatrix}\right).}
$$
Integer coefficients are interpreted in $K$, so this remains valid in characteristics two and three.
In the cyclic case there is one companion block $C(d)$. Its defining <polynomial> is both characteristic and minimal: the vector represented by $1$ is cyclic and a <polynomial> annihilates it exactly when divisible by $d$. Thus \b[the <minimal polynomial> is read directly from that block].
Back to article page
