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Since finite fields are not algebraically closed, this suggests to me that there may be matrices over finite fields whose characteristic polynomials don't split over that finite field, and thus do not have Jordan normal forms.

As a simple example, is it the case that there exists a $3 \times 3$ matrix with entries from $\mathbb{F}_3$ that does not have a Jordan normal form? I'm wondering if I can just find a polynomial $f \in \mathbb{F}_3[x]$ that doesn't split in $\mathbb{F}_3[x]$ and then try to build a matrix whose characteristic polynomial is $f$, but I'm not sure that this is a feasible approach to constructing such a matrix.

Rodrigo de Azevedo
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Booker09
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1 Answers1

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Consider the companion matrix

$$A=\begin{pmatrix}0&0&-1\\1&0&1\\0&1&0\end{pmatrix}\in M_3({\Bbb F}_3)$$

Its characteristic polynomial $$p(\lambda)=\lambda^3-\lambda+1$$ is irreducible over ${\Bbb F}_3$. The idea is to extend ${\Bbb F}_3$ into ${\Bbb F}_{27}$, the field with $27$ elements that contains the three roots of $p$. We put ${\Bbb F}_{27} = \mathbb{Z}_3[x]/(p(x))$ (the quotient with the ideal generated by $p$), that is ${\Bbb F}_{27} = \left\{ a + bx + cx^2 ; a,b,c \in \mathbb{Z}_3 \right\}$ with the condition $x^3=x-1$. It's easy to see that the required rots of $p$ are $x,x+1,x+2$. Then $A$ is similar over ${\Bbb F}_{27}$ to $\mbox{diag}(x,x+1,x+2)$.

Rodrigo de Azevedo
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