Suppose that $F$ is a subfield of a field $E$ and, for  

$n\times n$ matrices $A\_1,\dots,A\_m, B\_1,\dots,B\_m$
over $F$, there exists a matrix $T\in{\rm GL}\_n(E)$
such that $T^{-1}A\_iT=B\_i$ for all $i$.



> 
> Does this imply that such a matrix $T$ can be chosen from ${\rm GL}\_n(F)$?
> 
> 
> 


It is easy to see that the answer is


* **yes** if $m=1$;
* and **yes** if the field $F$ is infinite.



Is simultaneous similarity of matrices independent from the base field?
This question is [answered in comments](https://mathoverflow.net/questions/369154/is-simultaneous-similarity-of-matrices-independent-from-the-base-field?__=1674402206#comment931814_369154):



> 
> "As everyone is saying, this follows from Noether-Deuring.
> See [mathoverflow.net/questions/28469/hilbert-90-for-algebras](https://mathoverflow.net/questions/28469/hilbert-90-for-algebras) for a quick proof.
> I also asked this question a while back [math.stackexchange.com/questions/305696](https://math.stackexchange.com/questions/305696)."
> 
> 
> 


        –
[David E Speyer](https://mathoverflow.net/users/297/david-e-speyer)


Thanks to all!


