Intuition on Rational Canonical Form I was studying the theory of rational canonical from Dummit & Foote's abstract algebra. 
From what I understand, a vector space $V$ over field $F$ can be 'transformed' into a $F[X]$ module by letting the multiplication of the indeterminate $X$ in $F[X]$ onto any vector element $v$ be defined as the action of some linear transformation $T$ (acting from $V$ to itself) onto $v$. i.e. $xv=T(v)$.
Moreover we have the following $F[X]$-isomorphism:
$$V \cong F[X]/(a_1(x))\oplus F[X]/(a_2(x))\oplus\cdots\oplus F[X]/(a_m(x))$$
where $a_1(x),\ldots,a_m(x)$ are unique monic polynomials (invariant factors) in $F[X]$. 
The text mentioned that by choosing a basis in the form of $\left\{ \overline1, \overline{x},\overline{x^2},\ldots \right\}$ for each $F[X]/(a_i(x))$ and we can obtain the companion matrix ${C_{a_i}}$ for each $a_i(x)$. Moreover, the Rational canonical form of $T$ is the matrix direct sum of all $C_{a_i}$.
My question is: Why can we assert that the rational canonical form of $T$ (which acts on the LHS) must be equivalent to the matrix direct sum of all companion matrices (which is produced at RHS)? To me it is not intuitively obvious that this must be true. I know that both sides are related by a module isomorphism but to me it is not intuitively obvious that this must be true.
Or am I missing some links in between? Would appreciate if anyone can help enlighten!
 A: Let's go through a concrete example. Suppose $F = \mathbb C$ and $V=\mathbb C^3$. Suppose we have the isomorphism of $\mathbb C[x]$-modules:
$$ \phi \ : \  \mathbb C[x]/(x-2) \oplus \mathbb C[x]/(x - 2)^2  \overset{\cong}{\longrightarrow} V.$$
[I switched the left- and right-hand sides for convenience.]
Before we can proceed, we need to spell out how elements of $\mathbb C[x]$ act on either side. On the left-hand side, every polynomial in $\mathbb C [x]$ simply acts by multiplication. On the right-hand side, complex scalars again act by multiplication, but as you said, $x$ acts as the linear transformation $T$.
Since $\phi$ is a  $\mathbb C[x]$-module isomorphism, $\phi$ is bijective, and it obeys three key properties:
$$ \phi(f + g) = \phi(f) +\phi(g)$$
$$ \phi(\alpha f) = \alpha \phi(f), \ \ \ \ ( \alpha \in \mathbb C),$$
$$ \phi(xf) = T( \phi(f)).$$
Note that, since both sides are $\mathbb C[x]$-modules, they are certainly also $\mathbb C$-vector spaces, where the  $\mathbb C$ that I am referring to is the set of constant polynomials within $\mathbb C[x]$. Furthermore, $\phi$ is certainly a $\mathbb C$-vector space isomorphism between the left- and right-hand sides.
Clearly, $\{\bar 1 \}$ forms a $\mathbb C$-basis for $\mathbb C[x]/(x-2)$, and clearly, $\{\bar 1, \bar x\}$ forms a $\mathbb C$-basis for $\mathbb C[x]/(x-2)^2$.
Therefore,
$$\{ (\bar 1, \bar 0), (\bar 0, \bar 1), (\bar 0, \bar x) \}$$ forms a $\mathbb C$-basis for $\mathbb C[x]/(x-2) \oplus \mathbb C[x]/(x - 2)^2 $.
[I am using commas to separate the element of the $\mathbb C[x]/(x-2)$ summand from the element of the $\mathbb C[x]/(x-2)^2$ summand.]
Therefore, the vectors
$$ v_1 = \phi(\bar 1, \bar 0), \ \ \ v_2 = \phi(\bar 0, \bar 1), \ \ \ v_3 = \phi (\bar 0, \bar x)$$
form a $\mathbb C$-basis for $V$.
Now, let us work out the action of the linear transformation $T$ on the basis vectors $v_1, v_2 \in V$. We will do this using the properties of the $\mathbb C[x]$-module isomorphism $\phi$.
\begin{align*} T(v_1) & = T(\phi(\bar 1, \bar 0)) = \phi(x \times (\bar 1, \bar 0)) = \phi(\bar x, \bar 0) = \phi(2(\bar 1, \bar 0)) = 2 \phi(\bar 1, 0) = 2v_1, \\
 T(v_2) & = T(\phi(\bar 0, \bar 1)) = \phi(x \times (\bar 0, \bar 1)) = \phi(\bar 0, \bar x) = v_3, \\
 T(v_3) & = T(\phi(\bar 0, \bar x)) = \phi(x \times (\bar 0, \bar x)) = \phi(\bar 0, \bar x^2) = \phi(-4 (\bar 0, \bar 1) +4 (\bar 0,\bar x )) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ =  - 4\phi(\bar 0, \bar 1) + 4\phi(\bar 0, \bar x) = -4v_2 + 4v_3.\end{align*}
[I used the fact that $x \equiv 2$ modulo $x-2$ in the first line, and $x^2 \equiv -4 + 4x$ modulo $(x-2)^2$ in the third line. I also used the three properties of the $\mathbb C[x]$-module isomorphism $\phi$ mentioned above.]
Therefore, the matrix representation of $T$ relative to the basis $\{ v_1 , v_2, v_3 \}$ is
$$ \left( \begin{array}{ccc}2 & 0 & 0 \\ 0 & 0 & -4 \\ 0 & 1 & 4\end{array}\right).$$
This is precisely the direct sum of the matrices $C_{(x-2)}$ and $C_{(x-2)^2}$ defined in Dummit and Foote.
