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 $i$.

  • $\begingroup$ I am very sorry, I made a mistake writing the question. A and B should be matrices over K, and not over L. Your point 3 solution is still working?I think it does, but there is one problem: How do you know that there exists a finite basis of F over K? $\endgroup$ – Omer Apr 8 at 14:33
  • $\begingroup$ Nevermind, I think it's because there is finite amount of entries of P? $\endgroup$ – Omer Apr 8 at 14:38
  • $\begingroup$ Just finished/fixed my full answer. $\endgroup$ – Andreas Caranti Apr 8 at 14:38

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