why we look at A-I and not A
?
for generalized eigenvectors, you pick the first eigenvector v1 like you normally do, and continue until you run out (in this case, one is all there is).
ker(A - I) has basis {(3,1,1)} now we want to pick our next generalized eigenvector so that:
$\displaystyle \begin{bmatrix}0&-3&3\\-2&-7&13\\-1&-4&7 \end{bmatrix} \begin{bmatrix}x\\y\\z \end{bmatrix} = \begin{bmatrix}3\\1\\1 \end{bmatrix}$
solve this for (x,y,z), which should give you (0,-2,-1), which is in ker((A-I)^2), but not in ker(A-I).
do it again, and you should get e1 as the 3rd generalized eigenbasis vector.