Usbrth

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 $i$.