= 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:
$$
\boxed{m_T\in\{x,\ x^2,\ x-1,\ x(x-1),\ x^2(x-1)\}.}
$$
Here are corresponding <matrices>, all satisfying $T^3-T^2=0$:
$$
\begin{array}{c|c}
m_T&\text{matrix of }T\\\hline
x&0_{3\times3}\\
x^2&\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}\\
x-1&I_3\\
x(x-1)&\operatorname{diag}(0,0,1)\\
x^2(x-1)&\begin{pmatrix}0&1&0\\0&0&0\\0&0&1\end{pmatrix}
\end{array}
$$
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> $x\neq f$, providing the other requested example.
Back to article page
