# Rational Canonical Form

This post proves the existence of rational canonical form, using the structure theoerm of finite modules over PID.

Another common canonical form is the rational canonical form. While Jordan form is more like a substitute for non-diagonalizable matrix and thus does not exist when the eigenvalues do not lie in the ground field, the rational canonical form exploits the properties of the characteristic/minimal polynomial.

Theorem 1 (Existence of rational canonical form, invariant factor version)Let be an matrix with coefficients in a field . Then it is similar to a direct sum of companion matrices such that .

Recall that a **companion matrix** is a matrix of the form

with the property that both the characteristic polynomial and the minimal polynomial are exactly .

**Proof**

Regard as the matrix of a linear map , where is a finite dimensional vector space over . Regard as an -module and apply the structure theorem for finitely generated modules over PID. Then we may write

such that are the **invariant** **factors**, up to units.

Finite dimensionality of implies that there is no free part in the decomposition. For each , it is clear that with respect to the basis ,

acting on takes the form of a companion matrix . Putting all these companion matrices together, we have shown the desired result.

Using rational canonical form, it is then clear that

- The minimal polynomial of is , the characteristic polynomial of the largest cyclic block.
- The characteristic polynomial of is .

Similarly, we also have the elementary divisor version by using the “primary decomposition” in the structure theorem of finitely generated modules over PID.

Theorem 1′ (Existence of rational canonical form, primary decomposition version)Let be an matrix with coefficients in a field . Then it is similar to a direct sum of companion matrices such that each is a prime power.

So how can we compute the rational canonical form? The first question is how should the rational canonical form look like, and the second question is how to find one such basis.

To answer the first question, it suffices to find the invariant factors provided by the structure theorem. This is obtained by

- finding a presentation for the finitely generated module. In our case, we need to find a presentation for as an -module if we regard as the matrix of the linear map .
- Find the relations matrix, and apply Smith normal form to it. The resulting invariants are the invariant factors we want.

1. **A presentation for **

Let has an -basis for which is represented by the matrix . The most natural presentation is to consider the map defined by

We then need to find the relations matrix, i.e. the kernel of .

Claim 1The kernel of is the -submodule generated by .

**Proof**

It is obvious that lies in .

For the other inclusion, let , i.e. . For each involved, we rewrite it as , and take the first term out. This way the degree of the polynomials in will decrease by 1. Repeat this process until all the s are eliminated. We are left with a linear expression in that equals 0. Therefore every coefficient in this last relation is 0.

Therefore equals the parts taken away. Since each part taken away is a multiple of , we have shown the other inclusion.

2. We now have the relations matrix , where is just an independent variable. Working out its Smith Normal Form give us the invariants needed.

**Primary decomposition**

If we factorize the invariants factors and put the prime powers separately (using Chinese remainder theorem), we would get the required primary decomposition.