TITLE: A square matrix has the same minimal polynomial over its base field as it has over an extension field
QUESTION [8 upvotes]: I think I have heard that the following is true before, but I don't know how to prove it:

Let $A$ be a matrix with real entries. Then the minimal polynomial of $A$ over $\mathbb{C}$ is the same as the minimal polynomial of $A$ over $\mathbb{R}$.

Is this true? Would anyone be willing to provide a proof?

Attempt at a proof:
Let $M(t)$ be the minimal polynomial over the reals, and $P(t)$ over the complex numbers. We can look at $M$ as a polynomial over $\Bbb C$, in which case it will fulfil $M(A)=0$, and therefore $P(t)$ divides it. In addition, we can look at $P(t)$ as the sum of two polynomials: $R(t)+iK(t)$. Plugging $A$ we get that $R(A)+iK(A)=P(A)=0$, but this forces both $R(A)=0$ and $K(A)=0$. Looking at both $K$ and $R$ as real polynomials, we get that $M(t)$ divides them both, and therefore divides $R+iK=P$. 
Now $M$ and $P$ are monic polynomials, and they divide each other, therefore $M=P$.
Does this look to be correct?

More generally, one might prove the following

Let $A$ be any square matrix with entries in a field$~K$, and let $F$ be an extension field of$~K$. Then the minimal polynomial of$~A$ over$~F$ is the same as the minimal polynomial of $A$ over$~K$.

REPLY [2 votes]: As Andrea explained, the statement in the question results immediately from the following one. 
Let $K$ be a subfield of a field $L$, let $A$ be an $m$ by $n$ matrix with coefficients in $K$, and assume that the equation $Ax=0$ has a nonzero solution in $L^n$. Then it has a nonzero solution in $K^n$. 
But this is obvious, because the algorithm giving such a solution (or its absence) depends only on the field generated by the coefficients of $A$.