How to complete proof about minimal polynomial

Question

I'm trying to prove that if $B=\begin{pmatrix} A & 0 \\ 0 & A\end{pmatrix} \in \Bbb C^{2n\times 2n}$ then $m_B(x)=m_A(x)$

My attempt:

$m_A(B)=\begin{pmatrix} m_A(A) & 0 \\ 0 & m_A(A)\end{pmatrix}=0 \implies m_B(x) |m_A(x)$

It remains to show that $m_A(x) |m_B(x)$. So, it suffices to prove that $m_B(A)=0$

We can see that $c_B(x)=(c_A(x))^2 \implies c_B(A)=0$ and $ c_A(x)|c_B(x)$

We know that $m_B(x)|c_B(x)$

So $\exists a(x),β(x)\in \Bbb F[x]: c_B(x)=a(x)c_A(x)+β(x)m_B(x)$

$\implies c_B(A)=a(A)c_A(A)+β(Α)m_B(A) \implies 0=0+β(Α)m_B(A)$

Is my proof so far correct? If so then, how do I show that $β(A)=0$ ?

Answer 1

A straightforward proof would be showing $m_B(A) = m_A(B) = 0$ which follows immediately from $B = A \oplus A$. This is because if we take a polynomial $P(x) = x^n + a_{n-1}x^{n-1}+ \dots + a_1x + a_0$ then $P(B) = P(A) \oplus P(A)$. So $P(A) = 0 \iff P(B) = 0$.