Suppose V is a 3 dimensional vector space over Z_2 (the field of congruence classes mod 2) and that the characteristic polynomial of T is (x-1)^3. Then clearly the minimum polynomial is one of (x-1), (x-1)^2, or (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:

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