Suppose is a field and is a finite-dimensional -vector space. Let be a fixed linear endomorphism. We can then consider as an -module, with the action of on given by the linear endomorphism . In other words, for any vector we have . The general polynomial then acts by
where is repeated composition.
Since is finitely generated as an -module, it is also finitely generated as an -module. The fundamental structure theorem for modules over a PID then provides a direct sum decomposition of as an -module. The free part of this decomposition must be trivial, since every nonzero free -module is infinite-dimensional as an -vector space (since is infinite-dimensional as an -vector space). So, must be isomorphic to a direct sum of cyclic torsion -modules
where are nonunits (i.e., nonconstant polynomials) and . 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 (as a torsion -module) is the ideal . Note that this means for every . By the definition of our action, this means that the endomorphism is identically zero on the entire vector space . Since we've currently assuming is monic, this is exactly the minimal polynomial of .
The minimal polynomial of
The minimal polynomial of the linear endomorphism is , the largest invariant factor in the above decomposition.
A warning about bases and generators
Suppose is a basis for as an -vector space. Then the set generates as an -vector space, and so every can be written as an -linear combination of the vectors in . It follows that every can also be written as an -linear combination of the vectors in ; i.e., the same set generates as an -module.
However, the set is not -linearly independent. In fact, as an -module the space is torsion and so there are no nonempty -linearly independent sets in ! Consequently, as an -module the space doesn't have a basis.
The rational canonical form
Consider one of the direct summands , where . As an -vector space, a basis for is the set , where is shorthand notation for the coset represented by . Under this basis, the action of is simply:
So, with respect to our chosen basis, the matrix for the action of is
This matrix is called the companion matrix of and is denoted .
If we repeat this process for every direct summand in the invariant factor decomposition of , we see that the matrix for with respect to that choice of basis (amalgamated from the bases for each summand, as described above) is
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 with . These polynomials are called the invariant factors of the matrix.
A rational canonical form for a linear endomorphism is a matrix representing that is in rational canonical form.
Example
Suppose is a finite-dimensional -vector space and is a -linear transformation. Then can be given the structure of a -module by letting act via . Suppose the invariant factors of as a -module are
In other words, there is a -module isomorphism
This is an isomorphism of -modules, where the action of on the left is via and the action of in each summand on the right is by multiplication by , the coset represented by in the given quotient ring.
Let's look at each summand in turn. In the quotient we have . So, if we use the -basis then the action of is given by . The matrix for the action of on this summand is therefore the matrix
In the summand we have . So, if we use the -basis then the action of is given by and . The matrix for the action of on this summand is therefore the matrix
In the summand we have . So, if we use the basis then the action of is given by
The matrix for the action of on this summand is therefore the matrix
With all of this in mind, the isomorphism corresponds to a basis for as an -vector space such that the matrix for with respect to this basis is
Also note that the -vector space decomposes into a direct sum of three subspaces, namely , , and . These three spaces are invariant under the action of the linear transformation Indeed, we have