Suppose we have a linear operator $T : V \to V$, where $V$ is a vector space $V$ over a field $F$. Now if $F$ is not algebraically closed, we don't necessarily know that $T$ has a Jordan canonical form.
There are many problems in Dummit & Foote's abstract algebra that I am trying to solve which seem designed to be solved using the Jordan canonical form, yet the problems do not tell you to assume that the eigenvalues of $T$ are in the field. So my question is do we simply instead work over the splitting field $K$ of the characteristic polynomial of $T$? We know that $V \cong F^n$, so can we extend $T$ to a linear operator $T' : V' \to V'$, where $V' \cong K^n$, such that $T'$ restricted to $F^n \subset K^n$ equals $T$? Thanks.