Show that the minimal polynomial of linear transformation $T:\mathbb{K}^n \mapsto \mathbb{K}^n$ remains the same over field extension using the cyclic decomposition theorem.
Let $\mathbb{C}$ be a field and let $\mathbb{K}$ be a subfield of $\mathbb{C}$.
Then we see that the minimal polynomial of $T$ over $\mathbb{K}$ (say $p_K(x)$) divides the minimal polynomial over $\mathbb{C}$(say $p_C(x)$)
Now assume that $p_C(x)= (x-c_1)p_K(x)$
I don't understand where do I use the cyclic decomposition theorem.Some hints rather than the complete answer would be helpful
Edit 1:
Here's my attempt to the outline given in the answer:
We try to show that $p_{\mathbb{C}}(x) | p_{\mathbb{K}}(x)$.
Now $p_{\mathbb{K}}(T)(e_i) = 0, 1 \le i \le n$ . [ As $e_i \in \mathbb{K}^n , \forall i s.t. 1 \le i \le n]$
Now $v \in \mathbb{C}^n$ then $v = d_1.e_1 + \cdots + d_n.e_n$ where $d_i \in \mathbb{C}$ then $p_{\mathbb{K}}(T)(d_1.e_1 + \cdots d_n.e_n) = p_{\mathbb{K}}(T)(d_1.e_1) + \cdots + p_{\mathbb{K}}(T)(d_n.e_n) $ then $p_{\mathbb{K}}(T)(v) = 0$ for all $v \in \mathbb{C}$.
Then by the definition of minimal polynomial we see that $p_{\mathbb{C}}(x) | p_{\mathbb{K}}(x)$.
Now to show the converse we know that by the cyclic decomposition theorem we can write :
$\mathbb{K}^n = Z(\alpha_1;T) \bigoplus \cdots \bigoplus Z(\alpha_k;T)$
Then $T_{Z(\alpha_i;T)} = T|_{Z(\alpha_i;T)}$ then as $Z(\alpha_i;T)$ is a cyclic subspace then we know that the minimal polynomial of $T_{Z(\alpha_i;T)}$ is the same as the c.p. of $T_{Z(\alpha_i;T)}$.
Now the c.p.($T$) = m.p.($T_{Z(\alpha_1;T)}$).mp.($T_{Z(\alpha_i;T)}$)…. m.p.($T_{Z(\alpha_j;T)}$
Also the minimal polynomial of $T$ over $\mathbb{K}$ is the l.c.m.( m.p.($T_{Z(\alpha_1;T)}$),mp.($T_{Z(\alpha_i;T)}$),…., m.p.($T_{Z(\alpha_j;T)})$
Thus $p_{\mathbb{K}}(x) = $ l.c.m.( m.p.($T_{Z(\alpha_1;T)}$),mp.($T_{Z(\alpha_i;T)}$),…., m.p.($T_{Z(\alpha_j;T)})$
Now by the definition of the characteristic polynomial of $T$ it remains the same over any field.
Now $p_{\mathbb{C}}(x) | $ c.p.$(x)$ then by the property of the l.c.m. we can claim that $p_{\mathbb{K}}(x) | p_{\mathbb{C}}(x)$
Hence $p_{\mathbb{K}}(x) = p_{\mathbb{C}}(x)$.
Best Answer
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.