Suppose $F$ is a field and $V$ is a finite-dimensional $F$-vector space. Let $T:V\to V$ be a fixed linear endomorphism. We can then consider $V$ as an $F[x]$-module, with the action of $x$ on $V$ given by the linear endomorphism $T$. In other words, for any vector $v\in V$ we have $x\cdot v=T(v)$. The general polynomial $p(x)={a}_{0}+{a}_{1}x+\cdots +{a}_{n}{x}^{n}\in F[x]$ then acts by

where ${T}^{k}=T\circ \cdots \circ T$ is repeated composition.

Since $V$ is finitely generated as an $F$-module, it is also finitely generated as an $F[x]$-module. The fundamental structure theorem for modules over a PID then provides a direct sum decomposition of $V$ as an $F[x]$-module. The free part of this decomposition must be trivial, since every nonzero free $F[x]$-module is infinite-dimensional as an $F$-vector space (since $F[x]$ is infinite-dimensional as an $F$-vector space). So, $V$ must be isomorphic to a direct sum of cyclic torsion $F[x]$-modules

where ${a}_{1}(x),\dots ,{a}_{m}(x)\in F[x]$ are nonunits (i.e., nonconstant polynomials) and ${a}_{1}(x)\mid {a}_{2}(x)\mid \cdots \mid {a}_{m}(x)$. These invariant factors are unique up to unit (which are the constant polynomials); if we require them to be monic, then they are unique.

Note that the annihilator of $V$ (as a torsion $F[x]$-module) is the ideal $({a}_{m}(x))\subseteq F[x]$. Note that this means ${a}_{m}(x)\cdot v={0}_{V}$ for every $v\in V$. By the definition of our action, this means that the endomorphism ${a}_{m}(T)$ is identically zero on the entire vector space $V$. Since we've currently assuming ${a}_{m}(x)$ is monic, this is exactly the minimal polynomial of $T$.

The minimal polynomial of $T$

The minimal polynomial ${m}_{T}(x)$ of the linear endomorphism $T$ is ${a}_{m}(x)$, the largest invariant factor in the above decomposition.

A warning about bases and generators

Suppose $\mathcal{B}=\{{v}_{1},\dots ,{v}_{k}\}$ is a basis for $V$ as an $F$-vector space. Then the set $\mathcal{B}$ generates $V$ as an $F$-vector space, and so every $v\in V$ can be written as an $R$-linear combination of the vectors in $\mathcal{B}$. It follows that every $v\in V$ can also be written as an $F[x]$-linear combination of the vectors in $\mathcal{B}$; i.e., the same set $\mathcal{B}$ generates $V$ as an $F[x]$-module.

However, the set $\mathcal{B}$ is not $F[x]$-linearly independent. In fact, as an $F[x]$-module the space $V$ is torsion and so there are no nonempty $F[x]$-linearly independent sets in $V$! Consequently, as an $F[x]$-module the space $V$ doesn't have a basis.

The rational canonical form

Consider one of the direct summands $F[x]/(a(x))$, where $a(x)={b}_{0}+{b}_{1}x+\cdots +{b}_{k-1}{x}^{k-1}+{x}^{k}$. As an $F$-vector space, a basis for $F[x]/(a(x))$ is the set $\{1,\stackrel{\u2015}{x},\dots ,{\stackrel{\u2015}{x}}^{k-1}\}$, where $\stackrel{\u2015}{x}=x+(a(x))$ is shorthand notation for the coset represented by $x$. Under this basis, the action of $x$ is simply:

This matrix is called the companion matrix of $a(x)$ and is denoted ${C}_{a(x)}$.

If we repeat this process for every direct summand in the invariant factor decomposition of $V$, we see that the matrix for $T$ with respect to that choice of basis (amalgamated from the bases for each summand, as described above) is

A matrix is said to be in rational canonical form if it is the direct sum of companion matrices for nonconstant monic polynomials ${a}_{1}(x),\dots ,{a}_{m}(x)$ with ${a}_{1}(x)\mid {a}_{2}(x)\mid \cdots \mid {a}_{m}(x)$. These polynomials are called the invariant factors of the matrix.

A rational canonical form for a linear endomorphism $T$ is a matrix representing $T$ that is in rational canonical form.

Example

Suppose $V$ is a finite-dimensional $\mathbf{\text{Q}}$-vector space and $T:V\to V$ is a $\mathbf{\text{Q}}$-linear transformation. Then $V$ can be given the structure of a $\mathbf{Q}[x]$-module by letting $x$ act via $T$. Suppose the invariant factors of $V$ as a $\mathbf{\text{Q}}[x]$-module are

This is an isomorphism of $\mathbf{\text{Q}}[x]$-modules, where the action of $x$ on the left is via $T$ and the action of $x$ in each summand on the right is by multiplication by $\stackrel{\u2015}{x}$, the coset represented by $x$ in the given quotient ring.

Let's look at each summand in turn. In the quotient $\mathbf{\text{Q}}[x]/(3+x)$ we have $3+\stackrel{\u2015}{x}=0$. So, if we use the $\mathbf{\text{Q}}$-basis ${\mathcal{B}}_{1}=\{1\}$ then the action of $x$ is given by $1\mapsto \stackrel{\u2015}{x}=-3$. The matrix for the action of $x$ on this summand is therefore the $1\times 1$ matrix

In the summand $\mathbf{\text{Q}}[x]/(6+5x+{x}^{2})$ we have $6+5\stackrel{\u2015}{x}+{\stackrel{\u2015}{x}}^{2}=0$. So, if we use the $\mathbf{\text{Q}}$-basis ${\mathcal{B}}_{2}=\{1,\stackrel{\u2015}{x}\}$ then the action of $x$ is given by $1\mapsto \stackrel{\u2015}{x}$ and $\stackrel{\u2015}{x}\mapsto {\stackrel{\u2015}{x}}^{2}=-6-5\stackrel{\u2015}{x}$. The matrix for the action of $x$ on this summand is therefore the $2\times 2$ matrix

In the summand $\mathbf{\text{Q}}[x]/(18+21x+8{x}^{2}+{x}^{3})$ we have $18+21\stackrel{\u2015}{x}+8{\stackrel{\u2015}{x}}^{2}+{\stackrel{\u2015}{x}}^{3}=0$. So, if we use the basis ${\mathcal{B}}_{3}=\{1,\stackrel{\u2015}{x},{\stackrel{\u2015}{x}}^{2}\}$ then the action of $x$ is given by

With all of this in mind, the isomorphism $\varphi $ corresponds to a basis $\mathcal{B}=\{{v}_{1},{v}_{2},{v}_{3},{v}_{4},{v}_{5},{v}_{6}\}$ for $V$ as an $F$-vector space such that the matrix for $T$ with respect to this basis $\mathcal{B}$ is

Also note that the $\mathbf{Q}$-vector space $V$ decomposes into a direct sum of three subspaces, namely ${V}_{1}=\mathrm{span}({v}_{1})$, ${V}_{2}=\mathrm{span}({v}_{2},{v}_{3})$, and ${V}_{3}=\mathrm{span}({v}_{4},{v}_{5},{v}_{6})$. These three spaces are invariant under the action of the linear transformation $T.$ Indeed, we have