Show that the minimal polynomial of $T:\mathbb{K}^n \mapsto \mathbb{K}^n$ remains the same over field extension

Solution 1:

Outline of solution: One approach is to show that $p_{\Bbb C} = p_{\Bbb K}$ by showing that $p_{\Bbb C} \mid p_{\Bbb K}$ and $p_{\Bbb K} \mid p_{\Bbb C}$, where $p \mid q$ means "$p$ divides $q$".

The easy direction of the proof is showing that $p_{\Bbb C} \mid p_{\Bbb K}$. Argue that $p_{\Bbb K}(T) v = 0$ for all $v \in \Bbb C^n$, then apply the definition of the characteristic polynomial.

One way to show that $p_{\Bbb K} \mid p_{\Bbb C}$ is to cyclic decomposition theorem (CDT). First, note that by the CDT, it suffices to consider a transformation $T$ for which $\Bbb K^n$ is a cyclic subspace. Then, note that for such a transformation, the minimal and characteristic polynomials of $T$ (over a given field) are the same. Finally, using the definition of the characteristic polynomial, argue that $T$ and its extension to a map over $\Bbb C^n$ have the same characteristic polynomials.