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.

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