Suppose V is a 3 dimensional vector space over $\displaystyle Z_2$ (the field of congruence classes mod 2) and that the characteristic polynomial of T is $\displaystyle (x-1)^3$. Then clearly the minimum polynomial is one of $\displaystyle (x-1)$, $\displaystyle (x-1)^2$, or $\displaystyle (x-1)^3$. I cannot see how we can use this to answer the final part of the question:

By considering the possible dimensions of eigenspaces, show that the matrix of T with respect to an arbitrary basis is exactly one of:

$\displaystyle \left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{ar ray}\right)$,$\displaystyle \left(\begin{array}{ccc}1&0&0\\0&1&1\\0&0&1\end{ar ray}\right)$, or $\displaystyle \left(\begin{array}{ccc}1&1&0\\0&1&1\\0&0&1\end{ar ray}\right)$