Show that the rational form of $A$ remains the same whether viewed as a matrix over $\mathbb{C}$ or over $\mathbb{F}$(subfield) 
Question: Let $\Bbb F$ be a subfield of $\Bbb C$ (the complex numbers), and let $A$ and $B$ be $n \times n$ matrices. Prove that if $A$ and $B$ are similar over $\Bbb C$, then they are similar over $\Bbb F$. (Hint: Prove that the rational form of $A$ is the same whether viewed as a matrix over $\Bbb F$ or a matrix over $\Bbb C$, likewise for $B$).

I am working on the hint given above. I don't have any intuition as to how to start the problem.
$\mathbb{F}^n = Z(\alpha_1;T) \bigoplus \cdots \bigoplus Z(\alpha_k;T)$
Also $p_{i+1}|p_i$ where $p_i$ is the T annihilator of $\alpha_i$.
Where $p_1$ is the minimal polynomial of T(which remains the same over field extension)
Edit 1:This question is form Hoffman and kunze . The question is to show that if two matrices are similar over $\mathbb{C}$ then they are similar over $\mathbb{F}$(subfield).The hint given in the book is that " Show that the rational form of $A$ remains the same whether viewed as a matrix over $\mathbb{C}$ or over $\mathbb{F}$(subfield)"
This is what I know.Any hints to start the problem would be helpful.
 A: This question can be quickly answered with an approach framed by the following result.

HK Section 7.2, Theorem 5: Let $\Bbb F$ be a field and let $B$ be an $n \times n$ matrix over $\Bbb F$. Then $B$ is similar over the field $\Bbb F$ to one and only one matrix which is in rational form.

Now, let $A$ be an arbitrary $n \times n$ matrix with elements in $\Bbb F$ (i.e. $A \in \Bbb F^{n \times n}$). We know from the theorem that there exist matrices $C,P \in \Bbb F^{n \times n}$ such that $C$ is in rational canonical form, $P$ is invertible, and
$$
A = P^{-1}CP.
$$
It follows that when $A,C,P$ are considered as elements of $\Bbb C^{n \times n}$, we still have $A = P^{-1}CP$. However, $C$ is in rational canonical form. By theorem 5, this means that $C$ is the rational canonical form of $A$, as a matrix in $\Bbb C^{n \times n}$.
The conclusion follows.

We are given that $\Bbb F$ is a subfield of $\Bbb C$. For any vector space $V$ over $\Bbb F$, I will use $\bar V$ to refer to the extension of $V$ to a vector space over $\Bbb C$. In your setting, we have $V \subset \Bbb F^n \subset \Bbb C^n$, so we can more simply say that $\bar V = \{z v : z \in \Bbb C, v \in V\}$. In a more general setting, we might say that $\bar V = V \otimes_{\Bbb F} \Bbb C$ is obtained via an extension of scalars.
Let $T$ denote the linear map $T:\Bbb F^n \to \Bbb F^n$ given by $T(x) = Ax$. In a similar vein, I will use $\bar T$ to refer to the map induced by $T$ on $\Bbb C^n$, whereas $T$ refers to a map on $\Bbb F^n$. More generally: for a $\Bbb F$-linear map $\tau:V \to V$, $\bar \tau$ denotes the induced $\Bbb C$-linear map $\bar \tau : \bar V \to \bar V$
I claim (without proof) that the the following holds:

*

*If $V = V_1 \oplus \cdots \oplus V_k$, then $\bar V = \bar V_1 \oplus \cdots \oplus \bar V_k$

*$\overline{Z_{\Bbb F}(\alpha,T)} = Z_{\Bbb C}(\alpha,\bar T)$.

The subscripts here emphasize $Z(\alpha,T) = Z_{\Bbb F}(\alpha,T)$ is a vector space over $\Bbb F$, whereas $Z(\alpha,\bar T) = Z_{\Bbb C}(\alpha,\bar T)$ is a vector space over $\Bbb C$.
We are given that $\Bbb F^n$ has the $T$-cyclic decomposition
$$
\Bbb F^n = Z(\alpha_1;T) \oplus \cdots \oplus Z(\alpha_k;T)
$$
where $p_{i+1}\mid p_i$ for $i = 1,\dots,k-1$. The question asks us to show that $\bar T$ has the same canonical form as $T$. To do this, it suffices to show that $\Bbb C^n$ has the $\bar T$-cyclic decomposition
$$
\Bbb C^n = Z(\alpha_1;\bar T)\oplus \cdots \oplus Z(\alpha_k;\bar T)
$$
and that the $\bar T$-annihilator of $\alpha_i$ is $p_i$ for each $i$.
First of all, using statements 1 and 2, we have
$$
\Bbb C^n = \overline{\Bbb F^n} = \overline{Z(\alpha_1;T) \oplus \cdots \oplus Z(\alpha_k;T)} = 
Z(\alpha_1;\bar T)\oplus \cdots \oplus Z(\alpha_k;\bar T).
$$
Now, it suffices to show that for each $i$, the $\bar T$-annihilator of $\alpha_i$ is equal to $p_i$, the $T$-annihilator of $\alpha_i$. To show that this holds, note that the $T$-annihilator of $\alpha_i$ is the characteristic/minimal polynomial of $T|_{Z(\alpha_i;T)}$. From your previous question, we know that $\overline{T|_{Z(\alpha_i; T)}}$ has the same characteristic and minimal polynomials as $T|_{Z(\alpha_i; T)}$. However,
$$
\overline{T|_{Z(\alpha_i; T)}} = \bar T|_{\overline{Z(\alpha_i;T)}} = 
\bar T|_{Z(\alpha_i;\bar T)},
$$
which means that the $\bar T$-annihilator of $\alpha_i$ is equal to the minimal polynomial of $\bar T|_{Z(\alpha_i;\bar T)}$, which is equal to the minimal polynomial of $T|_{Z(\alpha_i; T)}$, which is equal to $p_i$.
The conclusion follows.
