Solution
= Solution
Take a size-three <Jordan block> at one and a scalar block at three:
$$
\boxed{A=\begin{pmatrix}1&1&0&0\\0&1&1&0\\0&0&1&0\\0&0&0&3\end{pmatrix}.}
$$
Triangularity gives $\chi_A(x)=(x-1)^3(x-3)$. On the first block, $(A-I)^3$ is zero but $(A-I)^2$ is not; on the last component the <eigenvalue> is three. Thus both factors with exactly those powers are needed, and their product annihilates $A$. Its <minimal polynomial> is therefore also $(x-1)^3(x-3)$.