Given a field F and a subfield K of F. Let A, B be n×n matrices such that all the entries of A and B are in K. Is it true that if A is similar to B in Fn×n then they are similar in Kn×n?

Any help … thanks!

**Answer**

If the fields are infinite, there is an easy proof.

Let F⊆K be a field extension with F infinite. Let A,B∈Matn(F) be two square matrices that are similar over K. So there is a matrix M∈GLn(K) such that AM=MB. We can write:

M=M1e1+⋯+Mrer,

with Mi∈Mn(F) and {e1,…,er} is a F-linearly independent subset of K. So we have AMi=MiB for every i=1,…,r. Consider the polynomial

P(t1,…,tr)=det(t1M1+⋯+trMr)∈F[t1,…,tr].

Since detM≠0, P(e1,…,er)≠0, hence P is not the zero polynomial. Since F is infinite, there exist λ1,…,λr∈F such that P(λ1,…,λr)≠0. Picking N=λ1M1+⋯+λrMr, we have N \in \mathrm{GL}_n(F) and A N = N B.

**Attribution***Source : Link , Question Author : Melesia , Answer Author : Andrea*