Let $F$ be a sub-field of the field of complex numbers, and let $A$ be an $n \times n$ matrix over $F$. Let $p$ be the minimal polynomial for $A$. If we regard $A$ as a matrix over $C$, then $A$ has a minimal polynomial $f$ as an $n \times n$ matrix over $C$. Use a theorem on linear equations to prove $p = f$. Can you also see how this follows from the cyclic decomposition theorem?
I am not able to do the problem. Help Needed.
A method by using theorem for linear system is as follows:
Let $p(t)=a_0 + a_1 t + \cdots a_m t^m$. Then, a linear system in $F$ given by
$$ x_0 I + x_1 A + \cdots + x_m A^m = 0, $$
gives a one-dimensional solution set $\{c(a_0, a_1, \ldots , a_m)| c\in F\}$ since $p$ is the minimal polynomial for $A$. Note that the way of solving linear system would not change after extension of fields. The solution set would change to $\{c(a_0, a_1, \ldots, a_m)| c\in \mathbb{C}\}$. Thus, the minimal polynomial considered over $\mathbb{C}$ does not change.
Another method by using cyclic decomposition theorem is as follows:
This is because similarity is not changed under field extension. The $F[x]$-module $M^A$ in which $x$ acts as a left $A$ multiplication, is decomposed as $$ M^A \simeq F[x]/(f_1) \oplus \cdots \oplus F[x]/(f_r) $$ where $f_1|f_2 | \cdots | f_r=p$. Then by applying a tensor product $\mathbb{C}\otimes_F -$ both sides, $M^A$ transforms to $$ \mathbb{C}[x]/(f_1) \oplus \cdots \oplus \mathbb{C}[x]/(f_r).$$ This structure shows also that the minimal polynomial considered over $\mathbb{C}$ does not change.