Sorry for the long question and attempt, but I want to really make sure I understand and I'm not doing anything stupid, and that I've done everything I can.
I am asked to find all the the possible Jordan normal forms and rational canonical forms for a matrix $A$ over $\mathbb{F}_5$ which satisfies
- $\text{min}_A(x)=(x-1)(x-3)^3(x^3-1)$
- $\text{ch}_A(x)=(x-1)^2(x-3)^4(x^3-1)$
So far in my life I have only computed JCF over $\mathbb{R}$ and $\mathbb{C}$ given a minimal and characteristic polynomial. How do I approach this over $\mathbb{F}_{p^n}$?
My first question is if there's a reason it's written like this, since we can still factor out an $(x-1)$ from $(x^3-1)$ even over $\mathbb{F}_5$. I know that the rational canonical form exists even over $\mathbb{F}_5$ and that it's the diagonal direct sum of the companion matrices for the invariant factors of $A$.
Since $\min_A(x)=(x-1)^2(x-3)^3(x^2+x+1)$, the last block of the RCF must be 7x7, and looks like
$$
C(\text{min}_A(x)):=\begin{pmatrix}0&0&0&0&0&0&27\\ 1&0&0&0&0&0&-54\\ 0&1&0&0&0&0&36\\ 0&0&1&0&0&0&-37\\ 0&0&0&1&0&0&55\\ 0&0&0&0&1&0&-36\\ 0&0&0&0&0&1&10\end{pmatrix}
$$
For the rest of the RCF we have three options for invariant factors of degree 2:
$(x^2+x+1)$,$(x-1)^2$, or $(x-3)^2$. So the other 2x2 block is one of
$$ C(x^2+x+1):= \begin{pmatrix}0&-1\\ 1&-1\end{pmatrix}$$
$$ C_2(1):= \begin{pmatrix}0&-1\\ 1&2\end{pmatrix}$$
$$ C_2(3):= \begin{pmatrix}0&-9\\ 1&6\end{pmatrix} $$
I think I got that so far (?)
NOW here's the problem. I know that the JCF doesn't exist over $\mathbb{F}_5$ because the characteristic polynomial doesn't split over $\mathbb{F}_5$. And in order to know how many (elementary) Jordan blocks and their sizes, don't I need to know the roots of $x^2+x+1$ in its splitting field over $\mathbb{F}_5$? Or at least whether they are distinct?
I know the JCF has one 3x3 block for the eigenvalue 1 since the multiplicity in the characteristic polynomial is 3, and it contains one 2x2 block $J_2(1)$ since the multiplicity in the minimal polynomial is 2. There's one 4x4 block for the eigenvalue 3, and the block contains a 3x3 block $J_3(3)$. Then there's a 2x2 block for the other two 3rd roots of unity, $\alpha,\beta$. I don't know what this block looks like, but so far, the possibilities are
$$(J_2(1)\oplus J_1(1))\oplus (J_3(3)\oplus J_1(3))\oplus ?$$ $$(J_1(1)\oplus J_2(1))\oplus (J_3(3)\oplus J_1(3))\oplus ?$$ $$(J_1(1)\oplus J_2(1))\oplus (J_1(3)\oplus J_3(3))\oplus ?$$ $$(J_2(1)\oplus J_1(1))\oplus (J_1(3)\oplus J_3(3))\oplus ?$$
Up to permutations of the blocks for each eigenvalue.
How do I determine the $?$ Jordan block? I know that we're looking for a finite field with $5^n$ elements which contains the 3rd roots of unity. By this answer https://math.stackexchange.com/a/1457559/594721 I know that we want $5^n\equiv 1\pmod 3$. Clearly n=2 is the smallest such integer, so then $(x^3-1)$ should split into linear factors over $\mathbb{F}_{25}\cong\frac{\mathbb{F}_p}{\langle x^2+x+1\rangle}$. How do I proceed?
Thanks so much for your help