A square matrix has the same minimal polynomial over its base field as it has over an extension field

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$~K$.

Written before/while the OP was adding his/her own proof, which is essentially the same as what follows.

Let $\mu_{\mathbb{R}}(x)$ be the minimal polynomial of $A$ over $\mathbb{R}$, and let $\mu_{\mathbb{C}}(x)$ be the minimal polynomial of $A$ over $\mathbb{C}$.

Since $\mu_{\mathbb{R}}(x)\in\mathbb{C}[x]$ and $\mu_{\mathbb{R}}(A) = \mathbf{0}$, then it follows by the definition of minimal polynomial that $\mu_{\mathbb{C}}(x)$ divides $\mu_{\mathbb{R}}(x)$.

I claim that $\mu_{\mathbb{C}}[x]$ has real coefficients. Indeed, write $$\mu_{\mathbb{C}}(x) = x^m + (a_{m-1}+ib_{m-1})x^{m-1}+\cdots + (a_0+ib_0),$$ with $a_j,b_j\in\mathbb{R}$. Since $A$ is a real matrix, all entries of $A^j$ are real, so $$\mu_{\mathbb{C}}(A) = (A^m + a_{m-1}A^{m-1}+\cdots + a_0I) + i(b_{m-1}A^{m-1}+\cdots + b_0I).$$ In particular, $$b_{m-1}A^{m-1}+\cdots + b_0I = \mathbf{0}.$$ But since $\mu_{\mathbb{C}}(x)$ is the minimal polynomial of $A$ over $\mathbb{C}$, no polynomial of smaller digree can annihilate $A$, so $b_{m-1}=\cdots=b_0 = 0$. Thus, all coefficients of $\mu_{\mathbb{C}}(x)$ are real numbers.

Thus, $\mu_{\mathbb{C}}(x)\in\mathbb{R}[x]$, so by the definition of minimal polynomial, it follows that $\mu_{\mathbb{R}}(x)$ divides $\mu_{\mathbb{C}}(x)$ in $\mathbb{R}[x]$, and hence in $\mathbb{C}[x]$. Since both polynomials are monic and they are associates, they are equal. QED

Another way of proving this fact may be observing that ''you do not go out the field while using Gaussian elimination''. More precisely:

Proposition. Let $K \subseteq F$ be a field extension let $v_1, \dots, v_r \in K^n$. If $v_1, \dots, v_r$ are linearly dependent over $F$, then they are linearly dependent over $K$.

Proof. We'll prove the contrapositive of the statement. Suppose that the $v_i$'s are linearly independent over $K$. Let $\lambda_i \in F$ such that $\sum_i \lambda_i v_i = 0$. We can find $e_j \in F$ linearly independent over $K$ such that $\lambda_i = \sum_j \alpha_{ij} e_j$, with $\alpha_{ij} \in K$. Now from $\sum_{i,j} e_j \alpha_{ij} v_i = 0$ we deduce that $\sum_i \alpha_{ij} v_i = 0$, for every $j$. From the independence of $v_i$'s over $K$, we have $\alpha_{ij} = 0$, so $\lambda_i = 0$. $\square$

Now consider a field extension $K \subseteq F$ and a matrix $A \in M_n(K)$. Let $\mu_K$ and $\mu_F$ the minimal polynomials of $A$ over $K$ and $F$, respectively. Considering $I, A, A^2, \dots, A^r$ in the vector space $M_n(K)$, from the proposition you have $\deg \mu_K \leq \deg \mu_F$. On the other hand it is clear that $\mu_F$ divides $\mu_K$. So $\mu_F = \mu_K$.

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 $A$.