Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 3 / 13E / ii / Solution

Take a size-three Jordan block at one and a scalar block at three: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 $(x-1{)}^3(x-3)$.