# How do I get the Rational Canonical Form from the minimal and characteristic polynomials?

Let's say I have the minimal polynomial and characteristic polynomial of a matrix and all its invariant factor compositions. How do I get a rational canonical form matrix from this?

For a matrix $A$ the invariant factors are determined by finding the Smith canonical form equivalent to $(xI - A)$. Let's say $$S(x) = \text{Dg}[1,\ldots,1,f_1(x),\ldots,f_k(x)]$$ is the Smith canonical form. As mentioned elsewhere $f_1(x),f_2(x),\ldots,f_k(x)$ are all monic and each lower index monic polynomial divides the following. These are the invariant factors. (Also note that $f_k(x)$ is the minimum polynomial)

Now, it is true that $A$ is similar to Dg$[C(f_1(x)),\ldots,C(f_k(x))]$ where $C(f_i(x))$ indicates the companion matrix associated with $f_i(x)$, but this is not the Rational Canonical Form, to the best of my knowledge, not according to my textbook anyway.

The rational canonical form is derived by finding the elementary divisors of $A$. Take any invariant factor $f_i(x)$, then we can write it as $p_1(x)^{ei1}p_2(x)^{ei2}\ldots p_t(x)^{eit}$ where the $p_i(x)$ are distinct, monic and irreducible. These are then the elementary divisors of $A$. The rational canonical form is constructed from these elementary divisors as $$\text{Dg}[\ldots,H(p_1(x)^{ei1}),H(p_2(x)^{ei2}),\ldots,H(p_t(x)^{eit}),\ldots]$$ where $$H(p(x)^e) = \begin{bmatrix} C(p(x)) & 0 & \cdots & 0 & 0 \\ N & C(p(x)) & \cdots & 0 & 0 \\ \vdots & & & & \vdots \\ 0 & 0 & \cdots & N & C(p(x)) \end{bmatrix}.$$ Here $N$ is a matrix that is all zero except for a 1 in the upper right hand corner, and the companion matrix $C(p(x))$ is repeated $e$ times. This is known as a hypercompanion matrix. Note that the rational canonical form reduces to Jordan canonical form when the field is algebraically closed.

Let's look at an example: suppose you have invariant factors $f_1(x) = (x^2 + 4)(x^2 - 3)$ and $f_2(x) = (x^2 + 4)^2(x^2 - 3)^2$, and suppose the field we are considering is $\mathbb{Q}$. Then the elementary divisors are $(x^2 + 4),(x^2 - 3),(x^2 + 4)^2$ and $(x^2 - 3)^2$ and so the rational canonical form is $$\text{Dg}\left [ \begin{bmatrix}0&-4\\1&0 \end{bmatrix}, \begin{bmatrix}0&3\\1&0 \end{bmatrix},\begin{bmatrix}0&-4&0&0\\1&0&0&0\\0&1&0&-4\\0&0&1&0 \end{bmatrix},\begin{bmatrix}0&3&0&0\\1&0&0&0\\0&1&0&3\\0&0&1&0 \end{bmatrix}\right ].$$

My source material for this theory: CG Cullen, Matrices and linear transformations (2nd edition), the chapter named: Similarity: Part II. There are more examples and in-depth discussion there...

I am not sure whether the terminology is uniform here. The invariant factors are certain monic, non-constant polynomials $f_{i}$ such that $$f_{1} \mid f_{2} \mid \dots \mid f_{k},$$ with $f_{k}$ the minimal polynomial, and $f_{1} f_{2} \dots f_{k}$ the characteristic polynomial.

Take companion matrices $A_{i}$ of each $f_{i}$. Your rational canonical form will be $$\begin{bmatrix} A_{1}\\ & A_{2}\\ & & A_{3} \\ & & & \ddots\\ & & & & A_{k}\\ \end{bmatrix},$$ where blanks denote zeros.

• So each individual invariant factor has its own companion matrix?? This is where I've been messing up then. I've been taking 1 large companion matrix of the characteristic polynomial and was unsure why. – JustCurious Apr 22 '16 at 18:59
• Yes. The relevant Wikipedia article gives some more details. – Andreas Caranti Apr 22 '16 at 19:05