T is an operator on $R ^3$ with M(T) $\left( \begin {matrix} 1&1&0 \\ 0&1&0\\0&0&3 \end{matrix} \right)$ with respect to the standard basis.
This is the answer from the book : the subspace of generalised eigenvectors for the eigenvalue 3 is span{(0,0,1)}
I have no idea how they worked this out.
I always thought since 3 only appears once then it can have at most 1 generalized eigenvector but this is obviously not right.