because all of A's eigenvalues are 1?
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:
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.