# Rational Canonical Form: Example #2 (with Galois Theory)

Last week we saw an example of how to use the rational canonical form (RCF) to classify matrices of a given order in $\text{GL}_2(\mathbb{Q})$. Today we have a similar example* where now our matrices have entires in the finite field $\mathbb{F}_{13}$. The fact that our field is $\mathbb{F}_{13}$ instead of $\mathbb{Q}$ actually makes little difference in how to approach the solution, but I think this problem is particularly nice because part of it calls on some Galois Theory. So in the solutions below, I'll be assuming (without proof) a few facts both from both Galois theory and linear algebra.

## Example #2

### Part (a) (Galois Theory)

Recall that a linear transformation $T$ on an $F$-vector space $V$ is diagonalizable if and only if there exists a basis for $V$ consisting of eigenvectors of $T$. (The matrix representation of $T$ with repsect to this basis will have the eigenvalues of $T$ along its diagonal and zeros elsewhere.) Fortunately, we won't have to work so hard! One can also show (for instance, see Theorem 4.11 here) that $T$ can be diagonalized if and only if its minimal polynomial factors into distinct linear factors over $F$. Since our matrix $g\in \text{GL}_2(\mathbb{F}_{13})$ has order 7, we know its minimal polynomial must be a divisor of $x^7-1$ (and possibly $x^7-1$ itself).  So to show $g$ is diagonalizable over $\mathbb{F}_{169}$ but not over $\mathbb{F}_{13}$, it's enough to show $x^7-1$ splits completely over $\mathbb{F}_{169}$ but not over $\mathbb{F}_{13}$. This is where the Galois Theory comes in and makes our work easy.

Recall that the Galois group of any polynomial over a finite field is cyclic. In particular, let $p$ be a prime, let $k\geq 1$, set $q=p^k$ and let $\zeta_n$ denote a primitive $n$th root of unity. Then $\text{Gal}(\mathbb{F}_{q}(\zeta_n)/\mathbb{F}_{q})$ - the Galois group of $x^n-1$ over $\mathbb{F}_q$ - is cyclic of order $d$ where $d$ is the smallest positive integer such that $q^d\equiv 1(\text{mod }n)$. (This isn't too hard to see. Here's a sketch of the proof.)

Now if we can show that $\text{Gal}(\mathbb{F}_{169}(\zeta_7)/\mathbb{F}_{169})$ has order 1, we can conclude the degree of the splitting field $\mathbb{F}_{169}(\zeta_7)$ over $\mathbb{F}_{169}$ is also 1. This means $\mathbb{F}_{169}(\zeta_7)=\mathbb{F}_{169}$ which implies the polynomial $x^7-1$ will split completely into linear factors (all of which are distinct as $x^7-1$ is separable) over $\mathbb{F}_{169}$. Hence the minimal polynomial of $g$ will also split into linear factors, thus proving $g$ is diagonalizable.

So in the case when $p=13$ and $k=2$ with $q=169$, it is clear that $d=1$ is indeed the smallest integer such that $q^d\equiv 1(\text{mod }n)$ holds (since $168=7\times 24$). Thus $|\text{Gal}(\mathbb{F}_{169}(\zeta_7)/\mathbb{F}_{169})|=1$ and this allows us to conclude that $g$ can be diagnalized over $\mathbb{F}_{169}$.

That $g$ is not diagonalizable over $\mathbb{F}_{13}$ follows from the fact that the smallest $d$ such that $13^d\equiv 1(\text{mod }7)$ is not equal to 1. (In fact, it's 2.) In other words, the splitting field of $x^7-1$ over $\mathbb{F}_{13}$ is NOT $\mathbb{F}_{13}$. Thus, assuming $g$ is not the identity matrix, the minimal polynomial of $g$ (being one of the quadratic factors above, or some product thereof) does not split completely over $\mathbb{F}_{13}$, and so $g$ cannot be diagonalized.

### Part (b) (Rational Canonical Form)

This part is very similar to last week's example and so we'll use the same approach. Any two matrices which are similar (i.e. lie in the same conjugacy class) share the same rational canonical form (RCF). To find the possible RCFs of $g\in \text{GL}_2(\mathbb{F}_{13})$, we just need to find the possible invariant factors and corresponding companion matrices. To do so, we must keep in mind the following:

• The characteristic polynomial of $g$, $c_g(x)\in\mathbb{F}_{13}[x]$, must have degree 2 (since $g$ is a $2\times 2$ matrix)
• $c_g(x)$ is the product of all the invariant factors
• The minimal polynomial of $g$, $m_g(x)\in\mathbb{F}_{13}[x]$, must divide $x^7-1$
• If $a_1(x),\ldots,a_d(x)\in\mathbb{F}_{13}[x]$ are the invariant factors of $g$, then $a_d(x)=m_g(x)$ and $a_1(x)\mid\cdots\mid a_d(x)$
• $m_g(x)$ and $c_g(x)$ share the same roots and $m_g(x)\mid c_g(x)$
• The RCF of $g$ must be a $2\times 2$ matrix since $g$ itself is a $2\times 2$ matrix

#### Case 1: $m_g(x)=x-1$

In this case, we must have $c_g(x)=(x-1)^2=x^2+11x+1$ (remember we're working over $\mathbb{F}_{13}$) and so the RCF of $g$ is the companion matrix $$\mathscr{C}_{x^2+11x+1}= \begin{pmatrix} 0 & 1\\ 1 & 11 \end{pmatrix}.$$

#### Case 2: $m_g(x)=c_g(x)=x^2+3x+1$

Here the RCF is simply $$\mathscr{C}_{x^2+3x+1}= \begin{pmatrix} 0 & 1\\ 1 & 3 \end{pmatrix}.$$

#### Case 3: $m_g(x)=c_g(x)=x^2+5x+1$

Here we have $$\mathscr{C}_{x^2+5x+1}= \begin{pmatrix} 0 & 1\\ 1 & 5 \end{pmatrix}.$$

#### Case 4: $m_g(x)=c_g(x)=x^2+6x+1$

And similarly the RCF is $$\mathscr{C}_{x^2+6x+1}= \begin{pmatrix} 0 & 1\\ 1 & 6 \end{pmatrix}.$$

Thus there are four distinct conjugacy classes of matrices of order 7 in $\text{GL}_2(\mathbb{F}_{13})$ and they can be represented by the matrices

Footnotes:

* This is from CUNY's spring 2015 algebra qualifying exam.

Share
Related Posts