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$~K$.