Rational Canonical Form I - Definition

Suppose F is a field and V is a finite-dimensional F-vector space. Let T:VV 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 vV we have xv=T(v). The general polynomial p(x)=a0+a1x++anxnF[x] then acts by

p(x)v=a0v+a1T(v)++anTn(v),

where Tk=TT 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

VF[x]/(a1(x))F[x]/(am(x)),

where a1(x),,am(x)F[x] are nonunits (i.e., nonconstant polynomials) and a1(x)a2(x)am(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 (am(x))F[x]. Note that this means am(x)v=0V for every vV. By the definition of our action, this means that the endomorphism am(T) is identically zero on the entire vector space V. Since we've currently assuming am(x) is monic, this is exactly the minimal polynomial of T.

The minimal polynomial of T

The minimal polynomial mT(x) of the linear endomorphism T is am(x), the largest invariant factor in the above decomposition.

A warning about bases and generators

Suppose B={v1,,vk} is a basis for V as an F-vector space. Then the set B generates V as an F-vector space, and so every vV can be written as an R-linear combination of the vectors in B. It follows that every vV can also be written as an F[x]-linear combination of the vectors in B; i.e., the same set B generates V as an F[x]-module.

However, the set 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)=b0+b1x++bk1xk1+xk. As an F-vector space, a basis for F[x]/(a(x)) is the set {1,x,,xk1}, where x=x+(a(x)) is shorthand notation for the coset represented by x. Under this basis, the action of x is simply:

1xxx2x2x3xk2xk1xk1xk=b0b1xbk1xk1.

So, with respect to our chosen basis, the matrix for the action of x is

[00b010b101b2001bk1]

This matrix is called the companion matrix of a(x) and is denoted Ca(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

[Ca1(x)Ca2(x)Cam(x)]
Definition of rational canonical form

A matrix is said to be in rational canonical form if it is the direct sum of companion matrices for nonconstant monic polynomials a1(x),,am(x) with a1(x)a2(x)am(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 Q-vector space and T:VV is a Q-linear transformation. Then V can be given the structure of a Q[x]-module by letting x act via T. Suppose the invariant factors of V as a Q[x]-module are

a1(x)=3+x,a2(x)=(3+x)(2+x)=6+5x+x2,a3(x)=(3+x)2(2+x)=18+21x+8x2+x3.

In other words, there is a Q[x]-module isomorphism

ϕ:V(Q[x]/(3+x))(Q[x]/(6+5x+x2))(Q[x]/(18+21x+8x2+x3)).

This is an isomorphism of 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 x, the coset represented by x in the given quotient ring.

Let's look at each summand in turn. In the quotient Q[x]/(3+x) we have 3+x=0. So, if we use the Q-basis B1={1} then the action of x is given by 1x=3. The matrix for the action of x on this summand is therefore the 1×1 matrix

Ca1(x)=[3].

In the summand Q[x]/(6+5x+x2) we have 6+5x+x2=0. So, if we use the Q-basis B2={1,x} then the action of x is given by 1x and xx2=65x. The matrix for the action of x on this summand is therefore the 2×2 matrix

Ca2(x)=[0615].

In the summand Q[x]/(18+21x+8x2+x3) we have 18+21x+8x2+x3=0. So, if we use the basis B3={1,x,x2} then the action of x is given by

1xxx2x2x3=1821x8x2.

The matrix for the action of x on this summand is therefore the 3×3 matrix

Ca3(x)=[00181021018].

With all of this in mind, the isomorphism ϕ corresponds to a basis B={v1,v2,v3,v4,v5,v6} for V as an F-vector space such that the matrix for T with respect to this basis B is

M(T;B)=[30000000600001500000000180001021000018].

Also note that the Q-vector space V decomposes into a direct sum of three subspaces, namely V1=span(v1), V2=span(v2,v3), and V3=span(v4,v5,v6). These three spaces are invariant under the action of the linear transformation T. Indeed, we have

T(v1)=3v1,T(v2)=v3andT(v3)=6v25v3T(v4)=v5andT(v5)=v6andT(v6)=18v421v58v6.