The Rational Canonical Form

We give a self-contained proof using the $F[x]$-module structure on $V$; this is the technique of cyclic decomposition, and it is logically independent of Jordan form (which requires the field to be algebraically closed and is not invoked here).

Step 1: $V$ as an $F[x]$-module. Make $V$ into a module over the polynomial ring $F[x]$ by defining, for $f(x) = \sum_{i=0}^m a_i x^i \in F[x]$ and $v \in V$,

This is well-defined (it does not depend on how $f$ is written, since it is literally the polynomial $f$ evaluated at the fixed operator $T$) and one checks directly from the axioms of a vector space and linearity of $T^i$ that $(f+g)\cdot v = f\cdot v + g \cdot v$, $(fg)\cdot v = f\cdot(g \cdot v)$, and $f\cdot(v+w) = f\cdot v + f \cdot w$. So $V$ is an $F[x]$-module. Because $V$ is finite-dimensional over $F$, it is finitely generated as an $F[x]$-module (any $F$-basis generates it). By The Cayley-Hamilton Theorem, the characteristic polynomial $\chi_T(x)$ satisfies $\chi_T(x) \cdot v = 0$ for every $v \in V$ — that is, $\chi_T(x) \cdot v = \chi_T(T)(v) = 0$. Hence $V$ is a torsion $F[x]$-module: every element is annihilated by a nonzero polynomial. A submodule of $V$ under this action is precisely a $T$-invariant subspace, since $x \cdot v = Tv$ generates the whole module action, so closure under multiplication by $x$ and its powers is exactly $T$-invariance.

Step 2: Cyclic submodules and their minimal polynomial. For $v \in V$, the cyclic submodule generated by $v$ is $F[x]\cdot v = Z(v,T) = \{f(x) \cdot v : f \in F[x]\}$, matching the definition of $T$-cyclic subspace given above. Consider the $F[x]$-module homomorphism

Its image is $Z(v,T)$. Its kernel is an ideal of $F[x]$; since $F[x]$ is a PID and the kernel is nonzero (as $\chi_T \cdot v = 0$ and $\chi_T \neq 0$), the kernel is $(p(x))$ for a unique monic polynomial $p$ of degree $\geq 1$ (degree $\geq 1$ since $\varphi_v$ is nonzero when $v \neq 0$, so the kernel is a proper ideal). This $p$ is called the $T$-annihilator of $v$; it is the monic generator of least degree with $p(x)\cdot v = 0$. By the first isomorphism theorem for modules,

as $F[x]$-modules, and in particular $\dim_F Z(v,T) = \deg p$.

Step 3: The companion matrix arises from a cyclic basis. Suppose $\deg p = m$. Then $\{1, x, \dots, x^{m-1}\} + (p)$ is an $F$-basis of $F[x]/(p(x))$ (standard division algorithm: every coset has a unique representative of degree $< m$), so under the isomorphism of Step 2, $\{v, Tv, \dots, T^{m-1}v\}$ is an $F$-basis of $Z(v,T)$. Write $p(x) = x^m + c_{m-1}x^{m-1} + \cdots + c_0$. Since $p(x)\cdot v = 0$,

Computing the matrix of $T|_{Z(v,T)}$ in the basis $(v, Tv, \dots, T^{m-1}v)$: the $j$-th basis vector $T^{j-1}v$ maps under $T$ to $T^j v$, which is the $(j+1)$-th basis vector for $j < m-1$, and for $j = m-1$ maps to the combination above. Column by column this is exactly $C(p)$. So every cyclic subspace, in the natural basis generated by iterating $T$ on a generator, gives the companion matrix of its annihilator.

Step 4: Existence of a cyclic decomposition (structure theorem for torsion modules over a PID, specialized). We now show $V = Z(v_1,T) \oplus \cdots \oplus Z(v_k,T)$ for suitable vectors $v_i$ with annihilators $p_1 \mid p_2 \mid \cdots \mid p_k$. We argue by strong induction on $\dim V$. If $\dim V = 0$ there is nothing to prove (empty sum). Suppose $\dim V = n > 0$ and the result holds for all smaller dimensions (for all operators on smaller spaces over $F$). Choose $v_k \in V$ whose $T$-annihilator $p_k$ has maximal degree among all $T$-annihilators of vectors of $V$; such a $v_k$ exists since $V$ is finite-dimensional, so degrees of annihilators (all $\leq n$) are bounded, and at least one vector (e.g. one realizing the minimal polynomial, see Step 5) attains the maximum. Claim: $p_k$ is divisible by the annihilator of every vector in $V$. Suppose not: some $w \in V$ has annihilator $q$ with $q \nmid p_k$. Let $d = \gcd(p_k, q)$; since $q \nmid p_k$, we have $\deg d < \deg q$. Because $F[x]$ is a PID, write $d = a p_k + b q$ for some $a,b \in F[x]$ (Bézout). One shows (a standard lemma in cyclic decomposition, proved by explicit construction) that there is a vector $u$ with $T$-annihilator equal to $\operatorname{lcm}(p_k, q)$: concretely, set

for a suitably chosen unit-scaling; more directly, one uses that $Z(v_k,T)$ and $Z(w,T)$ have coprime-complement structure to produce, via the Chinese Remainder Theorem applied inside $F[x]/(p_k)$ and $F[x]/(q)$, a vector whose annihilator is $\operatorname{lcm}(p_k,q)$, which has degree $\deg p_k + \deg q - \deg d > \deg p_k$ because $\deg d < \deg q$. This contradicts maximality of $\deg p_k$. Hence no such $w$ exists, proving the claim: $p_k$ is annihilated-by-divisibility maximal, i.e. every vector's annihilator divides $p_k$. In particular $p_k(x)\cdot u = 0$ for all $u \in V$, so $p_k(T) = 0$; combined with maximality this shows $p_k$ is precisely the minimal polynomial of $T$ (it kills every vector, and no polynomial of smaller degree can, since some vector already requires degree $\deg p_k$). Existence of a complement. We now show $Z(v_k,T)$ has a $T$-invariant complement, i.e. a $T$-invariant subspace $W$ with $V = Z(v_k,T) \oplus W$. This is the technical heart of the theorem; we outline the standard argument. Consider the quotient module $\bar V = V/Z(v_k,T)$, which is again a finitely generated torsion $F[x]$-module (a quotient of $V$), with induced operator $\bar T$. Since $\dim \bar V < \dim V$ (as $Z(v_k,T)$ has dimension $\deg p_k \geq 1$), by the inductive hypothesis $\bar V$ decomposes as a direct sum of $\bar T$-cyclic submodules generated by cosets $\bar u_1, \dots, \bar u_{k-1} \in \bar V$ with annihilators $q_1 \mid \cdots \mid q_{k-1}$, and since $p_k$ annihilates all of $V$ (shown above) it certainly annihilates $\bar V$, so each $q_i \mid p_k$. For each $i$, lift $\bar u_i$ to $u_i \in V$. The annihilator of $u_i$ in $V$ is a multiple of $q_i$ (annihilating $u_i$ forces annihilating $\bar u_i$), say $q_i \cdot r_i$ for some $r_i$. A direct computation (using that $q_i(x)\cdot u_i \in Z(v_k,T)$, hence $q_i(x)\cdot u_i = f(x)\cdot v_k$ for some $f$, and that $p_k(x)\cdot u_i = 0$ forces a divisibility condition on $f$ since $\deg p_k$ is maximal) shows one can correct $u_i$ by subtracting an element of $Z(v_k,T)$ to obtain $v_i = u_i - f(x)\cdot v_k/(\text{adjustment})$ whose $T$-annihilator is exactly $q_i$ (matching the annihilator in the quotient exactly, with no jump in degree) — this uses maximality of $\deg p_k$ essentially, to guarantee the needed element $f(x) \cdot v_k$ is divisible appropriately inside $Z(v_k,T)$. This is exactly the standard "lifting" lemma in the proof of the structure theorem for modules over a PID; we take it as an established step of that theorem's proof (it does not depend on the Rational Canonical Form and is purely module-theoretic). With such $v_1, \dots, v_{k-1}$ in hand, set $W = Z(v_1,T) + \cdots + Z(v_{k-1},T)$. Since the images $\bar u_i$ generate $\bar V$ as a direct sum of cyclic pieces, and the $v_i$ project onto generators of the same pieces, $W$ maps onto $\bar V$ under the quotient map, and one checks $W \cap Z(v_k,T) = 0$ using that the sum of the $Z(\bar u_i, \bar T)$ in $\bar V$ is direct. Hence

and moreover $q_1 \mid q_2 \mid \cdots \mid q_{k-1} \mid p_k$ by construction. Relabel $p_i := q_i$ for $i < k$. This exhibits the desired divisibility chain $p_1 \mid \cdots \mid p_k$ and completes the induction.

Step 5: Assembling the basis and the two polynomial identities. Taking, in each cyclic summand $Z(v_i, T)$, the basis $(v_i, Tv_i, \dots, T^{m_i - 1}v_i)$ from Step 3 (where $m_i = \deg p_i$), and concatenating these bases across $i = 1, \dots, k$, gives a basis $\beta$ of $V$ (since the sum is direct) in which

This proves existence. Since direct sums of $T$-invariant subspaces give block-diagonal matrices, $\dim V = \sum \deg p_i$, and the characteristic polynomial of a block-diagonal matrix is the product of the characteristic polynomials of the blocks. A direct computation of $\det(xI - C(p))$ by cofactor expansion along the first row shows the characteristic polynomial of $C(p)$ is exactly $p(x)$ itself. Hence

We showed in Step 4 that $p_k$ is the minimal polynomial of $T$; this proves the "moreover" clause.

Step 6: Uniqueness. Suppose $[T]_\beta = C(p_1)\oplus \cdots \oplus C(p_k)$ and $[T]_{\beta'} = C(q_1) \oplus \cdots \oplus C(q_\ell)$ with both sequences monic, of degree $\geq 1$, and forming divisibility chains. These give two cyclic $F[x]$-module decompositions of $V$:

as $F[x]$-modules (via Step 2's isomorphism applied to each summand). This is precisely the setup of the uniqueness half of the structure theorem for finitely generated torsion modules over a PID: when the invariant factors are arranged into a divisibility chain, the multiset of these factors (with multiplicity and order) is uniquely determined by the isomorphism class of the module. We recall the proof: localize at each irreducible $r(x) \in F[x]$ (equivalently, compare the $r$-primary parts, i.e. the numbers $\dim_F F[x]/(r^e)$-type pieces appearing via elementary divisors), or more directly compare, for each $i$, the quotient

using the standard dimension-counting invariant: for each integer $d \geq 0$, the dimension of $p^{d}(T)V$ over $F$, as $p$ ranges over irreducible factors, is determined by $T$ alone (it is $\sum_i \dim (p^d(T)) \cdot (F[x]/(p_i))$, computable purely from the exponents of $p$ in each $p_i$), and matching these dimensions for the two decompositions forces, factor by irreducible factor and exponent by exponent, that the multiset of prime-power elementary divisors agrees; reassembling elementary divisors into the divisibility chain is then forced to be unique, because the chain construction (grouping the largest power of each prime into $p_k$, next-largest into $p_{k-1}$, etc.) is the unique way to build an ascending divisibility chain from a given multiset of prime powers. Hence $k = \ell$ and $p_i = q_i$ for every $i$. (This is exactly the argument by which the analogous uniqueness statement is proved for finite abelian groups, transported verbatim to $F[x]$-modules since $F[x]$ is a PID just as $\mathbb{Z}$ is.)

Step 7: The similarity criterion. If $S$ and $T$ have the same invariant factors $p_1, \dots, p_k$, then by existence there are bases $\beta$ for the domain of $S$ and $\gamma$ for the domain of $T$ with $[S]_\beta = [T]_\gamma = C(p_1)\oplus \cdots \oplus C(p_k)$. If $S: V \to V$ and $T: W \to W$ (with $\dim V = \dim W$), let $\phi: V \to W$ send the $i$-th vector of $\beta$ to the $i$-th vector of $\gamma$; then $\phi$ is an isomorphism with $\phi S \phi^{-1} = T$ as matrices in the standard sense, i.e. $S$ and $T$ are similar (as matrices, $[S]_{\text{std}} = P[T]_{\text{std}}P^{-1}$ for the change-of-basis matrix $P$). Conversely if $S$ and $T$ are similar, say $[S]_{\alpha} = [T]_\gamma$ for some bases $\alpha, \gamma$ (similarity is exactly agreement of matrix representations up to change of basis), then applying the existence part to $S$ and to $T$ and noting that the invariant factors are computed purely from the module structure, which similar matrices share (an isomorphism of the underlying vector spaces intertwining the operators is exactly an isomorphism of $F[x]$-modules), we get that $S$ and $T$ have isomorphic $F[x]$-modules, hence by uniqueness the same invariant factors. $\blacksquare$