# Matrices similarity in a bigger field $K$ Implies matrices similarity in the smaller field $F$.

I have a Linear Algebra exercise and I have trouble solving a part of it.
The follwing question shows us that if $$K \subseteq L$$ is a field extension such that both $$L,K$$ are infinite ($$L,K$$ are fields) and $$A,B \in M_n(K)$$ such that $$A,B$$ are similar in the field $$L$$ (or: there exists an invertible matrix $$P \in M_n(L)$$ such that $$PA=BP$$) then $$A,B$$ are already similar in the field $$K$$ (or: there exists an invertible matrix $$P \in M_n(K)$$ such that $$PA=BP$$). I need to prove that this way:

(1). Show that every non-zero polynomial $$f \in K[x_1,...,x_n]$$ there exists $$\lambda_1,...,\lambda_n \in K$$ such that $$f(\lambda_1,...,\lambda_n)\neq 0$$. Do that using induction and show that this is necessary that $$K$$ is infinite (find a counter example for finite $$K$$)

(2). Suppose $$f \in K[x_1,...,x_n]$$ is a polynomial such that there are $$\lambda_1,...,\lambda_n \in L$$ such that $$f(\lambda_1,...,\lambda_n) \neq 0$$.
Show that there are $$\mu_1,...,\mu_n \in K$$ such that $$f(\mu_1,...,\mu_k) \neq 0$$.

(3) Assume that there exists invertible $$P \in M_n(L)$$ such that $$PA=BP$$.
Show that there exists scalars $$a_1,...,a_r \in L$$ and matrices $$P_1,...,P_r \in M_n(K)$$ such that the set $$\{a_1,...,a_r\}$$ is $$K$$-linearly independent and also $$a_1P_1+...+a_rP_r = P$$. Show that $$P_iA = BP_i$$ for all $$i$$.

(4) Show that there exists $$b_1,...,b_r \in K$$ such that $$b_1P_1+...+b_rP_r$$ is invertible (Hint: use (2) with $$f(x_1,...,x_r) = det(x_1P_1+...+x_rP_r)$$

(5) Conclude from (4) and (3) that there exists an invertible matrix $$Q \in M_n(K)$$ such that $$QA=BQ$$.

I was able to solve everything but part (3). I tried searching that in google and all I found was this: Similar matrices and field extensions
And there he just uses part (3) as guaranteed. How do I prove that?

Consider the finite-dimensional $$K$$-vector subspace $$V$$ of $$L$$ spanned by all the entries of the matrix $$P$$. Let $$a_{1}, \dots, a_{r}$$ be a basis of $$V$$ over $$K$$. Then the $$(j, k)$$ entry of $$P$$ can be written as (I use exponents for the entries of a matrix, since indices are taken for another role) $$P^{jk} = a_{1} P^{jk}_{1} + \dots + a_{r} P^{jk}_{r}$$ for suitable $$P^{jk}_{i} \in K$$. Now $$P_{i}$$ is the matrix whose $$(j, k)$$ component is $$P_{i}^{j k}$$.
We have $$a_{1} (P_{1} A) + \dots + a_{r} (P_{r} A) = P A = B P = a_{1} (B P_{1}) + \dots + a_{r} (B P_{r}).$$ Now consider each component of this matrix identity: $$a_{1} (P_{1} A)^{jk} + \dots + a_{r} (P_{r} A)^{jk} = a_{1} (B P_{1})^{jk} + \dots + a_{r} (B P_{r})^{jk}.$$ Since the $$a_{i}$$ are independent over $$K$$, and the $$(P_{i} A)^{jk}, (B P_{i})^{jk}$$ are in $$K$$, this shows that $$(P_{i} A)^{jk} = (B P_{i})^{jk}$$ for each $$i, j, k$$, so that $$P_{i} A = B P_{i}$$ for all $$i$$.