 Solution

Part (b) requires $m_T$ to be a monic divisor of $x^2(x-1)$. The constant divisor one cannot annihilate a nonzero vector space. Every remaining divisor can occur in dimension three:Here are corresponding matrices, all satisfying $T^3-T^2=0$:The last matrix contains a size-two nilpotent Jordan block, requiring the factor $x^2$, and an eigenvector with eigenvalue one, requiring $x-1$. Thus its minimal polynomial is exactly $f$. The zero matrix has minimal polynomial $x\ne f$, providing the other requested example.