Is the minimal polynomial also minimal over the closure of the base field? [closed]

Let $V$ be a finite dimensional vector space over the field $F$ and $T: V\to V$ be linear. Then the minimal polynomial of $T$ is the least degree monic polynomial in $F[x]$ that annihilates $T$. Is it true that it must also be the least degree monic annihilator of T in $\overline F[x]$?

I know this is true when $F=\mathbb{R},\overline F =\mathbb{C}$, but only by seeing $\mathbb{C}$ as a finite dimensional vector space over $\mathbb{R}$, but I wouldn't know how to extend that to arbitrary fields. I also tried applying the division algorithm on the minimal polynomial $m_A \in F[x]$ and a hypothetical minimal polynomial $\mu_A \in \overline F[x]$ of lesser degree to reach a contradiction.

Yes, it is true. Take a standard basis $e_1,...,e_n$ and let $e=e_i$. Consider the sequence of vectors $e, Te, T^2e,...,T^m e,...$, all these vectors have coordinates in $F$.

There exists the minimal $m=m(e)$ such that the vector $T^me$ is a linear combination $\lambda_0e+\lambda_2Te+...+\lambda_{m-1}T^{m-1}e$ (over $\bar F$). Note that since $T^k e$ have coordinates in $F$, all $\lambda_i$ are in $F$. Let $p_e$ be the polynomial $t^m-\lambda_{m-1}t^{m-1}-...-\lambda_0$. Note that it is in $F[t]$. Then $p_e(T)e=0$.

If $e_1,...,e_n$ is the standard basis then the minimal polynomial $p(t)$ of the matrix $T$ is the lcm of all $p_{e_i}(t)$, and so it belongs to $F[t]$ too.