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Suppose that $F$ is a subfield of a field $G$ and, for
$n\times n$ matrices $A_1,\dots,A_m, B_1,\dots,B_m$ over $F$, there exists a matrix $T\in{\rm GL}_n(G)$ 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.

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.

Suppose that $F$ is a subfield of a field $G$ and, for
$n\times n$ matrices $A_1,\dots,A_m, B_1,\dots,B_m$ over $F$, there exists a matrix $T\in{\rm GL}_n(G)$ 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 tha the answer is

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

Suppose that $F$ is a subfield of a field $G$ and, for
$n\times n$ matrices $A_1,\dots,A_m, B_1,\dots,B_m$ over $F$, there exists a matrix $T\in{\rm GL}_n(G)$ 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.