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For a finitely generated module over a principal ideal domain, its invariant factors are nonzero nonunits $d_i$ in a decomposition $M\cong R^r\oplus {⨁}_{i}R/(d_i)$, ordered by $d_i\mid d_{i+1}$. They are unique up to multiplication by units and describe the torsion submodule. Over $F[t]$, making $t$ act as a linear operator recovers the invariant factors of a linear operator used in its rational canonical form.

Suppose the rational canonical form of $T$ is $C(f_1)\oplus \cdots \oplus C(f_r)$, with monic nonconstant invariant factors of a linear operator satisfying $f_1\mid \cdots \mid f_r$. Each block has minimal polynomial and characteristic polynomial $f_j$, because a cyclic operator has equal minimal and characteristic polynomials. The determinant of a block diagonal matrix is the product of block determinants, giving $c_T={\prod }_j{f}_j$. A polynomial kills the direct sum exactly when it kills each block, equivalently when every $f_j$ divides it. Thus $m_T$ is their least common multiple, which the divisibility chain reduces to $f_r$. On the zero vector space both polynomials are conventionally one.

A module isomorphism is a bijection of modules over the same ring preserving addition and scalar multiplication. Its inverse preserves these operations too. When a linear operator $T$ gives a vector space an $F[t]$-module structure by $t\cdot v=T(v)$, a module isomorphism is exactly an invertible linear map intertwining the two operators. This translates classification of modules into matrix similarity and rational canonical form.

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 iswith 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 $1,X,\dots ,X^{\mathrm{deg}{d}_i-1}$ 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,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 the minimal polynomial is read directly from that block.

Give $V$ a $k[t]$-module structure by $t\cdot v=\alpha v$. A finite vector space basis generates it as a module. It is a torsion module: the Cayley-Hamilton theorem gives a nonzero polynomial annihilating all of $V$. Since $k[t]$ is a Euclidean domain, cyclic decomposition giveswith monic nonconstant invariant factors of a linear operator $f_j$ and no free summand. On $k[t]/(f_j)$, use the basis $1,t,\dots ,t^{\mathrm{deg}{f}_j-1}$. Multiplication by $t$ then has the companion matrix $C(f_j)$, because the relation $f_j(t)=0$ expresses the last power in terms of earlier ones. Concatenating these bases yieldsthe rational canonical form of $\alpha$. In particular, its minimal polynomial is $f_s$ and its characteristic polynomial is ${\prod }_j{f}_j$.

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 $V$.

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,

The rational canonical form of $T$ is a block diagonal matrixsimilar to a matrix of $T$, where the $f_j\in F[t]$ are monic nonconstant polynomials and the sum of their degrees is $\dim V$. They are the invariant factors of a linear operator. The block $C(f_j)$ is the companion matrix for $f_j$, representing multiplication by $t$ on $F[t]/(f_j)$. The invariant factors are uniquely determined by $T$.

Every companion block has a cyclic vector and therefore has both minimal polynomial and characteristic polynomial equal to $f_j$, by part (a). The determinant of a block diagonal matrix is the product of its block determinants, givingA polynomial $h$ annihilates the whole matrix exactly when it annihilates every block. On block $j$, dividing $h$ by $f_j$ leaves a remainder of smaller degree; if that remainder annihilated the block without being zero, it would contradict minimality of $f_j$. Thus the annihilation condition is exactly $f_j\mid h$ for every $j$. The monic polynomial of least degree with this property is their least common multiple. The divisibility chain makes this $f_r$, soThese are the minimal and characteristic polynomials from invariant factors formulae. On the zero vector space the product is empty and both polynomials are defined to be one.

Let $J=\left(\begin{array}{cc}0& -2\\ 1& 0\end{array}\right)$, so the displayed rational canonical form of $T$ is $J\oplus J$. Since $J^2=-2I_2$, we have $T^2=-2I_4$.

If $a=0$, then $U=b{I}_4$ is a scalar matrix. Its invariant factors of a linear operator are four copies of $t-b$, so its rational canonical form isHere $m_U=t-b$ and $c_U=(t-b{)}^4$, including when $b=0$.

Now let $a\ne 0$ and putOn each two-dimensional invariant block, $A=aJ+b{I}_2$ satisfies $q(A)=0$. The vectors $e_1,A{e}_1$ form a cyclic vector basis, since their coordinate matrix is $P=\left(\begin{array}{cc}1& b\\ 0& a\end{array}\right)$ with determinant $a\ne 0$. Directly,Applying this basis change to both blocks givesThus $m_U=q$ and $c_U=q^2$. In particular there are two invariant factors $q,q$, rather than a single block for $q^2$. Also $q(t)>0$ for every real $t$, so it is an irreducible polynomial over $\mathbb{R}$. This is an instance of rational canonical form under an affine change.

Finally $T^2=-2I_4$ implies $T^{-1}=-T/2$. Substituting $a=-1/2$, $b=0$ in the preceding result gives $q(t)=t^2+1/2$, and henceIts minimal polynomial is $t^2+1/2$ and its characteristic polynomial is $(t^2+1/2{)}^2$. The normalized reciprocal of $t^2+2$ is also $t^2+1/2$, agreeing with rational canonical form of an inverse.

The nonconstant invariant factors of a linear operator satisfy $d_1\mid \cdots \mid d_k=m_A$. By part (a)(i), $m_A=X^2+1$ is an irreducible polynomial, so every $d_j$ must equal $X^2+1$. Their polynomial degrees sum to $n$, giving $n=2k$.

The companion matrix of $X^2+1$, in the basis $(1,X)$ of $\mathbb{Q}[X]/(X^2+1)$, isTherefore, for each fixed even $n$, every such $A$ has rational canonical formThere are no such matrices when $n$ is odd. This is an instance of the rational canonical form with irreducible minimal polynomial principle; equality of minimal polynomials alone does not determine rational canonical form in general.

Let an invertible matrix $T$ have monic invariant factors of a linear operator $f_j$ of degrees $d_j$. Their constant coefficients are nonzero, since their product is the characteristic polynomial and $\det T\ne 0$. On a cyclic block, $f(T)=0$ implies $T^{-d}f(T)=0$, so $f^{\vee }(T^{-1})=0$, where $f^{\vee }(t)=t^{d}f(1/t)/f(0)$ is monic of degree $d$. The vector $v$ remains cyclic for $T^{-1}$: multiplying $v,T^{-1}v,\dots ,T^{-(d-1)}v$ by the invertible map $T^{d-1}$ gives the reversed original cyclic basis. Thus this polynomial is the minimal polynomial of the inverse block.

Normalized reciprocation preserves divisibility because $(fg{)}^{\vee }=f^{\vee }{g}^{\vee }$ when both constant coefficients are nonzero. Consequently the rational canonical form of $T^{-1}$ is the direct sum of the companion matrices $C(f_j^{\vee })$. The normalization by $f_j(0)$ is essential to make each reciprocal polynomial monic.

Let $U=aT+bI$ on a nonzero finite-dimensional vector space over a field, with $a\ne 0$. An invariant block for $T$ is also invariant for $U$. If $v,Tv,\dots ,T^{d-1}v$ is its cyclic vector basis, then $v,Uv,\dots ,U^{d-1}v$ is another basis: the binomial theorem gives a triangular change-of-basis matrix with diagonal $1,a,\dots ,a^{d-1}$. The monic polynomial $g(t)=a^{d}f((t-b)/a)$ annihilates this block, and its degree is $d$, so it is the block's minimal polynomial. In the new cyclic basis the block is $C(g)$.

The substitution preserves divisibility of the monic invariant factors of a linear operator. Apply it to every block to obtain the rational canonical form of $U$. If $a=0$, then $U=bI$ is a scalar matrix instead: all invariant factors are $t-b$, one for each dimension. This exceptional case cannot be obtained by dividing by $a$.

Let $A$ act on a nonzero finite-dimensional vector space over a field $F$, with minimal polynomial an irreducible polynomial $f$ of polynomial degree $d$. The nonconstant invariant factors of a linear operator all divide $f$, so each equals $f$. Their degrees sum to the dimension of a vector space, giving $d\mid n$ and rational canonical form consisting of $n/d$ copies of the companion matrix $C_f$. Equivalently, make the vector space a module over the field $F[X]/(f)$ by letting $X$ act as $A$; a basis over this field gives the same block decomposition.