How does extending a field affect matrix similitude? Suppose we have fields $\mathbb{K}_1, \mathbb{K}_2$ such that $$\mathbb{K}_1 \subset \mathbb{K}_2.$$ 

Question Let $A, B$ be square $n\times n$ matrices over $\mathbb{K}_1$. If there exists $P_2 \in \mathrm{GL}(n, \mathbb{K}_2)$ s.t. 
$$ P_2^{-1}AP_2=B, $$
then there exists $P_1 \in \mathrm{GL}(n, \mathbb{K}_1)$ s.t. 
  $$ P_1^{-1}AP_1=B. $$
  Is it true?

 A: Yes. It is true. $A$ and $B$ are similar iff $\lambda I-A$ and $\lambda I-B$ are equivalent iff they have the same invariant factors or the same determinate factors. These are only depend on matrix itself.
see Similar matrix
A: Here is a somewhat more elementary proof that works if the field $\mathbb{K}_1$ is sufficiently big (for example it works with infinite fields). Assume $A$ and $B$ are similar over $\mathbb{K}_2$, so you have $P \in \textrm{GL}_n(\mathbb{K}_2)$ such that
$$P A = B P$$
Now write $P = (p_{i,j})$, and pick a basis $(e_1, \ldots, e_r)$ of $\textrm{Vect}_{\mathbb{K}_1}(p_{i,j})$ as a $\mathbb{K}_1$-vector space. We can thus write
$$P = \sum_{i = 1}^r e_i P_i$$
with $P_i \in M_n(\mathbb{K}_1)$ for $i \in [\!|1,r|\!]$ (note that the $P_i$ need not be invertible in general). And because $(e_1, \ldots, e_r)$ is free, we get $P_i A = B P_i$ for all $i \in [\!|1,r|\!]$. Now consider the polynomial
$$f(X_1, \ldots, X_r) = \det \left(\sum_{i = 1}^r X_i P_i \right) \in \mathbb{K}_1[X_1, \ldots, X_r]$$
Since $P$ is invertible, we have $f(e_1, \ldots, e_r) \neq 0$, so $f$ is non zero. And if $|\mathbb{K}_1| > n$, then there exist $(\lambda_1, \ldots, \lambda_r) \in \mathbb{K}_1^r$ such that $f(\lambda_1, \ldots, \lambda_r) \neq 0$ (see lemma below for an explanation). So the matrix $P' = \sum_{i = 1}^r \lambda_i P_i$ is in $\textrm{GL}_n(\mathbb{K}_1)$ and satisfies $A = P'^{-1} B P'$.
Lemma :
Let $f \in \mathbb{K}[X_1, \ldots, X_r]$ be a non zero polynomial. Assume 
$$|\mathbb{K}| > \deg(f) = \max_{1 \le i \le r} (\deg_{i}(f))$$
Then there exist a point $(\lambda_1, \ldots, \lambda_r) \in \mathbb{K}^r$ such that $f(\lambda_1, \ldots, \lambda_r) \neq 0$.
Proof :
We will prove this by induction on the number $r \ge 1$ of variables. The case $r = 1$ follows from the fact that a polynomial of degree $d$ has at most $d$ roots. If $r \ge 2$, write $d = \deg_{r}(f)$ and
$$f = \sum_{k = 0}^{d} a_k X_r^k$$
where the $a_i$ are polynomials in $r-1$ variables with $\deg(a_i) \le \deg(f)$. Now $a_d$ is non zero, so by induction, there is a point $(\lambda_1, \ldots, \lambda_{r-1}) \in \mathbb{K}_1^{r-1}$ such that $a_{d}(\lambda_1, \ldots, \lambda_{r-1}) \neq 0$. Finally, since $g(X) = f(\lambda_1, \ldots, \lambda_{r-1}, X)$ is a non zero polynomial with $\deg(g) \le \deg(f)$, you can find $\lambda_n$ such that $g(\lambda_n) \neq 0$. Which concludes.
