Minimal polynomials for a matrice in a given field and in its extension For given matrices $X$ from the matrix ring $M_{n,n}
(F)$, where $F$ is a field,  the minimal polynomial for $X$ in $F[x]$ is the polynomial $P$ of the lowest degree with the following properties:
(*)$P\in F[x]$, $P$ is nonzero, monic (that is  its leading coefficient is one), $P(X)=0$.
Let $L$ be a field such that $F\subset L$. Then also $X\in M_{n,n}(L)$. 
My question is: are the minimal polynomials $P$, $Q$ of $X\in M_{n,n}(F)$ in $F[x]$ and in $L[x]$ the same?
Clearly, if $P$ satisfies (*) then also  $P\in L[x]$, $P$ is nonzero, monic, $P(X)=0$.
Thus $deg(Q) \leq deg(P)$ (in fact, by the properties of minimal polynomial, $Q|P$). 
 A: See the answer of Andrea in another question. Although that question is a special case of yours, Andrea answered the general case.
A: I think in most cases the two minimal polynomials are the same.
Suppose $K \subset L$ is a finite Galois extension, $A \in M_n(K)$, and let $P(T) \in K[T], Q(T) \in L[T]$ be the minimal polynomials of $A$.
Let $\sigma \in Gal_K(L)$. Then $0 = \sigma(Q(A)) = \sigma(Q)(\sigma(A)) = \sigma(Q)(A)$. Hence, $(\sigma(Q)-Q)(A) = 0$. Since $Q$ is monic, this is a new polynomial $R \in L[T]$ of degree less than that of $Q$, that annihilates $A$. Since $Q$ is the minimal polynomial over $L$, we must have $R = 0$, and we can conclude that $\sigma(Q) = Q$.
Therefore, $Q$ is left invariant by the Galois group, and must be a polynomial in $K[T]$. Since $Q$ divides $P$ and has degree at least that of $P$ (because $P$ is the minimal polyonmial over $K$), we must have $P=Q$.
This also works when an extension is finite and separable but non-Galois, by considering the Galois closure.

I am still thinking if there can be counter-examples when $K$ is not a perfect field.
