## The minimal polynomial of a direct sum is the least common multiple of minimal polynomials

Let $M = A \oplus B = \begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix}$ be a direct sum of square matrices $A$ and $B$. Prove that the minimal polynomial of $M$ is the least common multiple of the minimal polynomials of $A$ and $B$.

Given a linear transformation $T$ on $V$, we let $\mathsf{Ann}_T(V)$ denote the annihilator in $F[x]$ of $V$ under the action induced by $x \cdot v = T(v)$.

Let $p(x) \in \mathsf{Ann}_M(V)$. If $p(x) = \sum r_ix^i$, we have $\sum a_iM^i = 0$ as a linear transformation. Note that $M^k = \begin{bmatrix} A^k & 0 \\ 0 & B^k \end{bmatrix}$. So we have $\begin{bmatrix} \sum r_i A^i & 0 \\ 0 & \sum r_i B^i \end{bmatrix} = 0$, and thus $\sum r_i A^i = 0$ and $\sum r_i B^i = 0$. So $p(x) \in \mathsf{Ann}_A(W_1) \cap \mathsf{Ann}_B(W_2)$, where $V = W_1 \oplus W_2$. Conversely, if $p(x) \in \mathsf{Ann}_A(W_1) \cap \mathsf{Ann}_B(W_2)$, then $p(A) = 0$ and $p(B) = 0$ as linear transformations. Then $p(M) = \sum r_i M^i$ $= \begin{bmatrix} \sum r_iA^i & 0 \\ 0 & \sum r_iB^i \end{bmatrix}$ $= 0$, so that $p(x) \in \mathsf{Ann}_M(V)$. So we have $\mathsf{Ann}_M(V) = \mathsf{Ann}_A(W_1) \cap \mathsf{Ann}_B(W_2)$.

That is, $(m_M) = (m_A) \cap (m_B)$, where $m_T$ is the minimal polynomial of $T$.

Lemma: In a principal ideal domain $R$, if $(a) \cap (b) = (c)$, then $c$ is the least common multiple of $a$ and $b$. Proof: Certainly $c \in (a)$ and $c \in (b)$, so that $c$ is a multiple of both $a$ and $b$. If $d$ is a multiple of $a$ and of $b$, then $d \in (a) \cap (b) = (c)$, so that latex d\$ is a multiple of $c$. $\square$

The result then follows.