3
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $i$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.