How to prove that the complexification of an operator T and T itself share the same minimal polynomial? The definition of the complexification of a vector space V is in this link.
And the complexification $T^c$of a linear transformation T is defined by $T^c(u+iv)=T(u)+iT(v)$, now consider the minimal polynomial $m_{T}$ of T. We have  $m_{T}(T^c)=(m_{T}(T))^c=0$, hence $m_{T^{c}}\left ( x \right )|m_{T}\left ( x \right )$, but how to prove the other direction?

Consider $B = \{v_1, \ldots, v_n\}$ a basis of $V$ and $B' = \{v_1 +i0, \ldots, v_n + i0\}$ a basis of $V^c$.
Then $[T^c]_{B'} = [T]_B \in M_n(\Bbb R) \subset M_n(\Bbb C)$.
Since $m_{T^c}$ is the minimal polynomial of this matrix, the question boils down to chekcing that the minimal polynomial of matrix of real numbers is the same as the minimal polynomial of the matrix viewed as one with complex entries.
This is always the case: fix $A$ a matrix with real entries and $p \in \mathbb{C}[X] \setminus \mathbb{R}[X]$ monic. If $p(A)$, then $0 = \overline{p(A)} = \overline{p}(A)$. By hypothesis we also know that $q = p-\overline{p}$ is nonzero, $\deg q < \deg p$ and $q(A) = 0$. In particular $p \neq m_A$.
Edit: the argument above is also considered here.
I think it can be generalized in the case of $F/k$ a Galois extension (perhaps this is stronger than needed) as follows: consider $A \in M_n(F)$ and $m_A \in F[X]$ its minimal polynomial. Fix $\sigma \in \mathbf{Gal}(F/k)$ and write $(-)^\sigma$ for the endomorpshism rings of $F[X], M_n(F)$ given by applying $\sigma$ coefficient/entry-wise.
If $A \in M_k(k)$, then $A^\sigma = A$. Hence $(m_A)^\sigma (A) = (m_A)^\sigma(A^\sigma) = (m_A(A))^\sigma = 0$. In particular, the polynomial $q = m_A - m_A^\sigma$ has degree less than $m_A$ and satisfies $q(A) = 0$. By degree considerations, it must be $q = 0$. But then each coefficient $a$ of $m_A$ satisfies $\sigma(a) = a$. In other words, the coefficients of $m_A$ lie in $F^{\mathbf{Gal}(F/k)} = k$, as claimed.
