Let $F$ be a subfield of the field of complex numbers, and let $A$ and $B$ be $n\times n$ matrices over $F$. Prove that if $A$ and $B$ are similar over the field of complex numbers, then they are similar over $F$. (Hint: Prove that the rational form of $A$ is the same whether $A$ is viewed as a matrix over $F$ or a matrix over $\Bbb{C}$; likewise for $B$.)
My attempt: Let $A\in M_{n}(F)$ and $A\in M_{n}(\Bbb{C})$. By theorem 5 section 7.2, $\exists ! P\in M_n(F)$ such that $P$ is in rational form and $A$ is similar over $F$ to $P$, and $\exists ! Q\in M_n(\Bbb{C})$ such that $Q$ is in rational form and $A$ is similar over $\Bbb{C}$ to $Q$. Since $F\subseteq \Bbb{C}$, we have $P\in M_{n}(\Bbb{C})$ and $A$ is similar over $\Bbb{C}$ to $P$. By uniqueness, $P=Q$. Let $B\in M_{n}(F)$ and $B\in M_{n}(\Bbb{C})$. Then $\exists ! R\in M_n(F)$ such that $R$ is in rational form and $B$ is similar over $F$ to $R$, and $\exists ! S\in M_n(\Bbb{C})$ such that $S$ is in rational form and $B$ is similar over $\Bbb{C}$ to $S$. Similarly, $R=S$.
Since $A$ is similar over $\Bbb{C}$ to $B$ and $B$ is similar over $\Bbb{C}$ to $S$, we have $A$ is similar over $\Bbb{C}$ to $S$. By uniqueness, $Q=S$. Since $A$ is similar over $F$ to $P=Q=S=R$ and $R$ is similar over $F$ to $B$, we have $A$ is similar over $F$ to $B$. Is my proof correct?
