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A cyclic vector gives a companion matrix, whose characteristic relation annihilates its generator and hence every iterate. If there is no cyclic vector spanning the whole space, a nonzero proper cyclic subspace is invariant. Induction annihilates the restriction and the quotient; the quotient polynomial first sends the whole space into the invariant subspace, and the restriction polynomial then kills it. The block-triangular characteristic polynomial is the product of these two polynomials.

The $n$ spanning vectors form a basis. WriteIn this cyclic basis the matrix is the companion matrixExpanding $\det (tI-C)$ along its last column givesFor example, the term containing $a_j$ has minor $t^j$ and the signs from the subdiagonal cancel the cofactor sign; the diagonal term supplies $t^n$. The defining relation says ${\chi }_T(T)x=0$. Every polynomial in $T$ commutes with $T$, soIt therefore vanishes on a basis, provingThis direct argument for a cyclic vector does not assume the Cayley-Hamilton theorem.

The structure theorem for finitely generated modules over a principal ideal domain applies in particular to a Euclidean domain $R$: every finitely generated module has an isomorphismwith each $d_i$ nonzero and a nonunit. The rank of a free module $r$ and the invariant factors of a finitely generated module $d_i$ are unique up to multiplication by units; empty sums are allowed. This is the classification theorem being used, without a proof of that theorem.

The rational canonical form theorem states that for a linear operator $T$ on a finite-dimensional vector space over any field $F$, there are unique monic nonconstant polynomials $p_1\mid \cdots \mid p_k$ such that some basis givesHere, for $p(t)=t^d+a_{d-1}{t}^{d-1}+\cdots +a_0$, the companion matrix has ones on its subdiagonal and last column $(-a_0,\dots ,-a_{d-1}{)}^T$. The empty matrix covers a zero-dimensional space. Two linear operators have matrix similarity exactly when they have the same invariant factors of a linear operator.

For the proof, make $V$ an $F[t]$-module by $t\cdot v=T(v)$. It is a finitely generated module, since a vector-space basis also generates it as a module. It is a torsion module: the powers of $T$ have linear dependence in the finite-dimensional space ${\mathrm{End}}_F(V)$, so some nonzero polynomial annihilates every vector. Because $F[t]$ is a Euclidean domain, the classification theorem givesThere is no free summand, since a nonzero free module over $F[t]$ is not a torsion module. Choose monic generators of the ideals. On $F[t]/(p_i)$, the classes of $1,t,\dots ,t^{d_i-1}$ form an $F$-basis by polynomial division. Multiplication by $t$ is exactly $C(p_i)$ in that basis, proving existence. Uniqueness in the classification theorem proves uniqueness of the rational canonical form. Finally, an $F[t]$-module isomorphism is precisely an invertible $F$-linear map intertwining the operators, establishing the assertion about matrix similarity. In particular the characteristic polynomial is ${\prod }_i{p}_i$ and the minimal polynomial is $p_k$ for nonzero $V$.

Make $F^n$ an $F[X]$-module by $Xv=Av$. It is finitely generated and torsion, so the structure theorem gives cyclic summands $F[X]/(d_i)$ with $d_1\mid \cdots \mid d_k$. In the power basis of each summand, multiplication by $X$ has the companion matrix of $d_i$. Their block diagonal sum is the rational canonical form, unique because the invariant factors are unique.

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 towhere the nonzero nonunits $d_i$ are unique up to multiplication by units. They are the 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 givesThe basis $1,X,\dots ,X^{\mathrm{deg}{a}_i-1}$ 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 areThe 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 ofElementary row and column operations over $\mathbb{R}[X]$ give the Smith normal formConsequently the nonunit invariant factors areand, as a check,

For the standard basis vector $e_j$, matrix multiplication selects the $j$th column, soConsequentlyand the columns of $PM$ are $P{c}_1,\dots ,P{c}_n$.

Suppose first that $v,Av,\dots ,A^{n-1}v$ are linearly independent, and defineThen $S$ is invertible. The first $n-1$ columns of $AS$ are the last $n-1$ columns of $S$. By the Cayley-Hamilton theorem,This is exactly the last-column rule for the companion matrix $C$, so

Conversely, suppose $S^{-1}AS=C$, or $AS=SC$, and write the columns of $S$ as $s_1,\dots ,s_n$. Comparing the first $n-1$ columns givesHenceSince $S$ is invertible, its columns are linearly independent. Taking $v=s_1$ therefore makeslinearly independent. Thus $A$ is similar to $C$ exactly when it has the stated cyclic vector.

If $V=V_1\oplus V_2$ were a nontrivial invariant decomposition, each restricted minimal polynomial would be a power of $x-\alpha$ of degree at most $\dim V_i<n$. Their least common multiple is the minimal polynomial of $\varphi$, contradicting its degree $n$.

Since $m_{\varphi }$ has degree $n$, $\varphi$ has a cyclic vector $b_1$. Put $b_k={\varphi }^{k-1}{b}_1$. Then $\varphi (b_k)=b_{k+1}$ for $k<n$. In this basis the matrix is the companion matrix with ones immediately below the diagonal and last columnobtained by expanding $(x-\alpha {)}^n$.

The prime ideals of $\mathbb{R}[x]$ are $(0)$ and the maximal ideals generated by the monic irreduciblesFor a nonzero prime $I=(p)$, use the residue classes of $1,x,\dots ,x^{N-1}$, where $N=n\mathrm{deg}p$. Multiplication by $x$ on $\mathbb{R}[x]/(p^n)$ is represented by the companion matrix of the monic polynomial $p^n$. In particular, for $p=x-a$ this may equivalently be written as one size-$n$ Jordan block with eigenvalue $a$. The ideal $(0)$ does not give a finite-dimensional quotient.