This is a question from linear algebra.
Suppose $A\in GL_{3}(\mathbb{F}_{q})$. Assume that the characteristic polynomial of $A$, is of the form $(x-\lambda)(x^2+ax+b)$, where $x^2+ax+b$, is a irreducible polynomial over $\mathbb{F}_{q}$. It is clear that $A$ is diagonalisable over $\mathbb{F}_{q^2}$. Now, we consider the matrix $A^2$. Now, if $\zeta$ is a root of $x^2+ax+b$ in $\mathbb{F}_{q^2}$, then the other root is $\sigma(\zeta)$, where $\sigma$ is the Frobenius automorphism. Now, we can say that the characteristic polynomial of $A^2$ , is $(x-\lambda^2)(x-\zeta^{2})(x-\sigma(\zeta)^2)$, over $\mathbb{F}_{q^2}$. Now, if we assume that $\zeta^{2}\in \mathbb{F}_{q}$, say $\zeta^2=\beta$, then $\sigma(\zeta)^2=\beta$, and hence the characteristic polynomial of $A^2$, over $\mathbb{F}_{q}$, becomes $(x-\lambda^2)(x-\beta)^2$.
My question is, what can we say about the minimal polynomial of $A^2$ over $\mathbb{F}_{q}$, or in other words, about the diagonalisability of $A^2$, over $\mathbb{F}_{q}$?
What I have been trying is as follows: Since $A$ is clearly diagonalisable over $\mathbb{F}_{q^2}$, then $\exists$ $P\in GL_{3}(\mathbb{F}_{q^2})$, such that $P^{-1}AP=diag(\lambda, \zeta, \sigma(\zeta))$, implies that $P^{-1}A^{2}P=diag(\lambda^{2}, \beta, \beta)$. But, now we have to find a $Q\in GL_{3}(\mathbb{F}_{q})$, such that $Q^{-1}A^2Q=diag(\lambda^{2}, \beta, \beta)$. But, I couldn't proceed further.
Thanks in advance for any kind of help.
If $\zeta^2\in\Bbb F_q$, so that $\zeta^2=\sigma(\zeta)^2$, then yes, $A^2$ is diagonalisable over $\Bbb F_q$. In this case $\sigma(\zeta)=-\zeta$ so that $$(A-\lambda I)(A-\zeta I)(A+\zeta I)=(A-\lambda I)(A-\zeta^2 I)=0$$ and then $$(A^2-\lambda^2 I)(A-\zeta^2 I)=(A+\lambda I)(A-\lambda I)(A-\zeta^2 I)=0.$$ As $A^2$ is annihilated by the square-free polynomial $(X-\lambda^2)(X-\zeta^2)$ with zeros in $\Bbb F_q$, then $A^2$ is diagonalisable over $\Bbb F_q$.