Let $F$ be a field. We say a matrix $A\in M_n(F)$ is regular over $F$ if the minimal polynomial of $A$ over $F$ equals the characteristic polynomial of $A$.
Suppose $L$ is a field extension of $F$. Show that for any $A\in M_n(F)$, $A$ is regular over $F$ is and only if $A$ is regular over $L$.
Assume that the minimal polynomial of $A$ (over $K$ or over $L$, depends on which implication you want to prove) equals the characteristic polynomial of $A$ (denoted by $f_A$). Note that $f_A\in K[X]$. Then $XI_n-A\in M_n(K[X])$, the characteristic matrix of $A$, has only one invariant factor (over $K$ or over $L$), namely $f_A$, so $A$ is similar with the companion matrix of $f_A$. Now use the result of this topic.
The minimal polynomial of $A$ is the largest invariant factor of the $F[X]$-module defined by$~A$, and the characteristic polynomial is the product of the invariant factors. As such both are invariant under field extensions, as is the condition that they are equal (i.e., that there are no other invariant factors).
The invariant factors can be found by computing the Smith normal form over $F[X]$ of $XI_n-A$; the operations involved are purely rational (no factoring of polynomials) so the algorithm and the resulting invariant factors do not change under extensions of the field$~F$.
The invariance of characteristic and minimal polynomial under field extensions can also be seen directly: for the characteristic polynomial it is obvious (it is just a determinant), and for the minimal polynomial it follows from the fact that it can be found by solving a sequence of linear problems (finding the first linear dependence between powers of $A$), and whether a system of equations has any nonzero solutions does not change under field extensions.
