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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

  1. $\text{min}_A(x)=(x-1)(x-3)^3(x^3-1)$
  2. $\text{ch}_A(x)=(x-1)^2(x-3)^4(x^3-1)$

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

Wyatt Kuehster
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    I've not thought about the JCF so this isn't an answer. But you are astray on the RCF. You've got the last block corresponding to the minimal polynomial. So what's missing to get the correct characteristic polynomial? - answer $(x-1)(x-3)$. So we must have the $2\times 2$ block which is the companion of this quadratic. – ancient mathematician May 18 '21 at 06:51
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    But here's a thought for the JCF. Call the primitive cube roots of unity (which live as you have found in $\mathbb{F}_{25}$) $\omega, \omega^2$. Note that these are distinct else we'd have that $\omega=1$ and so $x^3-1=(x-1)^3$ which is false in $\mathbb{F}_5[x]$. So the "missing" block is just $J_1(\omega)\oplus J_1(\omega^2)$. – ancient mathematician May 18 '21 at 06:56
  • Aha! I forgot that $\text{ch}_A(x)$ had to be the product of the invariant factors, and also forgot that factor. Yes! I see that about the JCF too, thank you! So it's also completely determined, but up to $2\cdot 2\cdot 1 \cdot 1 \cdot 4!$ block permutations! – Wyatt Kuehster May 18 '21 at 18:44

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