Are the minimal polynomials of $A$ and $A^T$ related? I'm wondering if the minimal polynomials of $A, A^T\in\mathbb{M}_{n\times n}(\mathbb{F})$ are somehow related? Are they the same?
If $m_A(t)$ is the minimal polynomial of $A$, then $m_A(A) = 0$. So, $(m_A(A))^T = 0$. If two matrices $P,Q$ commute then $(PQ)^T = P^TQ^T$. Using this, we can see that $m_A(A^T) =0$.
Is this enough to conclude that $m_A(t)$ is also the minimal polynomial for $A^T$?
P.S.
Does this depend on whether or not the minimal polynomial splits over $\mathbb{F}$?
 A: You don't need to consider any factorisation. Let $p(x)=\sum_{i=1}^ka_ix^i$ be any monic polynomial. If $p(A)=0$, then
$$
p(A^T)=\sum_{i=1}^ka_i(A^T)^i=\left(\sum_{i=1}^ka_iA^i\right)^T=p(A)^T=0.
$$
Since $A=(A^T)^T$, it is also true that $p(A^T)=0\Rightarrow p(A)=0$. Therefore $A$ and $A^T$ share the same set of annihilating polynomials. The monic polynomial of the lowest degree in this set is the common minimal polynomial of $A$ and $A^T$.
A: $m_A$ and $m_{A^T}$ are the minimal polynomials for $A$ and $A^T$ respectively.
So we have $m_A(A) = 0$ and $m_{A^T}(A^T) = 0$ by definition. Let $m_A(t) = p_1(t)p_2(t)...p_k(t)$ where $p_i(t)\forall 1\leq i\leq k$ are the irreducible factors of $m_A(t)$.
$m_A(A) = 0 \implies (m_A(A))^T = 0$, which is just $(p_1(A)p_2(A)...p_k(A))^T = 0$. Since polynomials in $A$ commute, this is just $(p_1(A))^T(p_2(A))^T...(p_k(A))^T = 0 \implies p_1(A^T)p_2(A^T)...p_k(A^T) = 0$. So $m_A(A^T) = 0$, which means that $m_{A^T}\vert m_A$.
Starting with $m_{A^T}(A^T) = 0$, we can similarly prove that $m_A\vert m_{A^T}$.
So $m_A(t) = m_{A^T}(t)$
