## Deducing matrix from minimum polynomial

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