If $$T=\begin{pmatrix}A&0\\0&C\end{pmatrix}$$ is a block diagonal matrix, then (I assume) we have \begin{equation}\tag {1}m_T=\text{lcm}(m_A,m_C)\end{equation} for the minimal polynomials. If we have instead a block triangular matrix$$T=\begin{pmatrix}A & B \\ 0 & C\end{pmatrix}$$then $(1)$ does not have to hold. As a counterexample consider the matrix $$\begin{pmatrix} 0 & 1 \\0 & 0\end{pmatrix}$$ where $t^2\neq \text{lcm}(t,t)$. My question is: what if $A$ only has some eigenvalue $\lambda$ and $C$ only has some eigenvalue $\mu$ (with $\lambda \neq \mu$)? Do we then have a similar result in this case?
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In the block triangular case, $\text{lcm}(m_A,m_C)\mid m_T\mid m_A m_C$, so if $m_A$ and $m_C$ are relatively prime, (1) holds. This is true for example if $m_A=(x-\lambda)^p$ and $m_C=(x-\mu)^q$ with $\lambda\ne\mu$. – blargoner Jun 23 '23 at 03:01